Bauer Solucionario Tomo2

Chapter 21: Electrostatics

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Chapter 21: Electrostatics InClass Exercises

21.1. d 21.2. a 21.3. e 21.4. e 21.5. c 21.6. b 21.7. a 21.8. a 21.9. c 21.10. e Multiple Choice

21.1. b 21.2. b 21.3. b 21.4. d 21.5. b 21.6. b 21.7. a 21.8. a 21.9. c 21.10. b Questions

21.11. The given quantities are the charge of the two particles, 1Q Q and 2 .Q Q They are separated by a

distance .d The Coulomb force between the changed particles is 2

1 22 2

Q Q QF k k

d d . If the change on

each particle is doubled so that 1 22Q Q Q and the separation distance is 2d d the then the

Coulomb Force is given by: 2 2

2 2

4

4

Q QF k k

d d so the force is the same as it was in the initial situation.

21.12. The gravitational force between the Sun and the Earth is S Eg 2

M MF G

r where G is the gravitational

constant and is equal to 11 2 26.67 10 N m / kg , SM is the mass of the Sun (301.989 10 kg ) and EM is

the mass of the Earth ( 245.974 10 kg ). The Coulomb force is given by the equation 1 2C 2

Q QF k

r

where k is Coulomb’s constant (k = 9 2 28.99 10 N m / C ). In this question 1 2Q Q Q and is the charge given to the Earth and Sun to cancel out the gravitational force.

2S E

C g S E2 2

GM MkQ GF F Q M M

kr r

Therefore,

30 2411 2 217

9 2 2

(1.989 10 kg)(5.974 10 kg)6.67 10 N m / kg2.97 10 C.

8.99 10 N m / CQ

I can get the number of elementary charges, ,n by dividing Q by 191.602 10 C (the charge of one

electron): 17

36

19

2.97 10 C1.85 10 .

1.602 10 Cn

To estimate the number of elementary change of either sign for the Earth I can assume the mass of the Earth is due to the mass of the protons, neutrons and electrons of which it is primarily composed. If I assume that the Earth’s mass is due to the proton and neutron masses primarily (became an electrons mass is much smaller than a protons) and I assume that there are an equal number of protons and neutrons than I can get the number of protons by dividing the Earth’s mass by two times the mass of a proton. The mass of a proton is 27

P 1.6726 10 kg,m so you can estimate

the number of elementary charges on the Earth, En by: 24

51EE 27

P

5.97 10 kg3.57 10 .

1.67 10 kg

mn

m

So the

percentage of the Earth’s changes that would be required to cancel out the gravitational force is 14

E/ 100% 5.18 10 %,n n a very small percentage.

21.13. One reason that it took such a long time to understand the electrostatic force may have been because it was not observed as frequently as the gravitational force. All massive objects are acted on

Bauer/Westfall: University Physics, 1E

820

by the gravitational force; however, only objects with a net charge will experience an electrostatic force.

21.14. The accumulation of static charge gives the individual hairs a charge. Since like charges repel and because the electrostatic force is inversely proportional to the charges separation distance squared, the hairs arrange themselves in a manner in which they are as far away from each other as possible. In this case that configuration is when the hairs are standing on end.

21.15. The given quantities are the charge which is 1 2Q Q Q and the separation distance of 2 .d The

third charge is 3 0.2Q Q and it is positioned at .d Charge 3Q is then displaced a

distance x perpendicular to the line connecting the positive charges. The displacement .x d The question asks for the force, ,F on charge 3 .Q For x d the question also asks for the approximate motion of the negative charge.

13 23 ,F F F

where 13F

is the force 3Q feels due to 1Q and 23F

is the force 3Q feels due to charge

2 .Q Because 1Q and 2Q have the same sign and are of equal charge there is no net force in the x ‐

direction. The forces in the y ‐direction are given by:

1 313 2

1

sinQ Q

F kr

and 2 323 2

2

sin ,Q Q

F kr

where 2 21r d x and 2 2

2r d x and the negative signs denote that there will be an attraction

between the positive and negative charges. To simplify we can substitute 1 1sin /x r and

2 2sin /x r into force equations. So we can write the force equation as:

1 3 2 3

2 2 2 22 2

313 23 12 2 22 2 2

3/

kQ Q kQ Q kxQx xF F F Q Q

d x d xd x d x d x

,

Substituting 1 2Q Q Q and 3 0.2Q Q gives:

2 2

3/2 3/22 2

3/22 2 2 2

2 0.20.2 0.4k Q xkx Q kQ xF Q Q

d x d x d x

Since ,x d it is reasonable to use the approximation 2 2 3/2 2 3/2 3) (( .) dx dd Hence, 2

3

0.4kQ xF

d .

This solution is similar in form to Hooke’s law which describes the restoring force due to the compression or expansion of a spring, springF kx

where k is the spring constant. The motion of

the negative charge can therefore be approximated using simple harmonic motion.

21.16. As the garment is dried it acquires a charge from tumbling in the dryer and rubbing against other clothing. When I put the charged garment on it causes a redistribution of the charge on my skin and this causes the attractive electric force between the garment and my skin.


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21.17. The initial separation of the spheres is 1x . The magnitude of the force on each of the spheres at

separation 1x is1 2

1 21

Q QF k

x . The force after the distance change is 1 2

2 22

,Q Q

F kx

where the new

distance is 2x . Because the charge is conserved I can equate the forces 1F and 2 .F 1 2

1 21

Q QF k

x and

1 22 2

2

,Q Q

F kx

so 2 21 2 1 1 2 2 ,kQ Q Fx F x or 2 2

2 1 2 1/ .x F F x Substituting 2 19F F into the equation gives:

2 2 212 1 2 1 1

1

1 1 .

9 9 3

Fx x x x x

F Therefore the distance would have to decrease to a factor of a

third of its original value to achieve nine times the original force.

21.18. An electrically neutral atom can exert electrostatic force on another electrically neutral atom if they do not have symmetric charge distribution. In the case of two atoms where one atoms electron or electrons were closer to the proton of the other atom. This type of situation can occur when atoms undergo polar bonding to form a molecule.

21.19. The scientist could convince themselves that the electrostatic force was not a variant of the gravitational force in various ways. One distinction is that gravitating objects attract but in the electric force like charged objects repel. For Earth bound experiments the scientists may observe that massive objects are pulled towards the ground by the gravitational force at a constant acceleration. If they performed careful experiments with objects of the same charge they would observe that the gravitational force downward on one of the charged objects could be diminished or balanced by the electrostatic force that object felt due to the second like charged object that was placed underneath it.

21.20. The electrostatic force is an inverse square force, of the same form as the Newtonian gravitational force. As long as the bodies are not moving too rapidly (i.e., not at speeds near the speed of light), the problem of determining their motion is the same as the Kepler problem. The motion of the two particles decomposes into a center of mass motion with constant velocity, and a relative motion which traces out a trajectory which can be either a portion of a straight line (for zero angular momentum, i.e., head on collisions) or a Keplerian ellipse (including a circle), parabola, or hyperbola, in the case of opposite charges. For charges of the same sign, for which the force is repulsive, the relative motion must be either a straight line or a hyperbola, an open orbit.

21.21. The wall does not have to be positively charged. The negatively charged balloon induces charges on the wall. The repulsive force between electrons in the balloon and those in the wall cause the electrons in the wall to redistribute. This leaves the portion of the wall that is closest to the balloon with a positive charge. The negatively charged balloon will be attached to the positively charged region of the wall even though the net charge of the wall is neutral.

21.22.


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The electric lines flow from the positive charge to the negative charge as is shown in the sketch below.

There is nowhere on the line between the charged particles that I could place a test charge without it moving. This is due to the eclectic charges on the line having opposite charge, so a test charge (of either sign) that is placed between these two charges would be attracted by one and repelled by the other.

21.23.

In order for the test charge to feel no net force it would have to be at a location where the force it

felt due to the charge 2 4 C.Q For convenience I can say that the charge 1 2 CQ is located

at 1 0x , and charge 2 4 CQ is located at 2x L and charge 3Q is located at a position, 3x which is

between 0 and L . I can equate the expressions for the electric force on 3Q due to 1Q and the electric

force on 3Q due to 2Q to solve for 3x as these forces would have to balance for the charge 3Q to feel no net force.

13 23

1 3 2 32 2

3 3

2 21 3 2 3

2 2 21 3 3 2 3

2 21 2 3 1 3 1

( )

( )

( 2 ) 0

( ) 2 0

F F

kQ Q kQ Q

x L x

Q L x Q x

Q x x L L Q x

Q Q x Q x L Q L

Note that in the second step of the calculation above, it is shown that the sign and magnitude of 3Q will not impact the answer. I can solve using the quadratic equation:

2 2 2 2 2 2 21 1 1 2 1

31 2

2 4 4( )( ) 2(2 C) 4(2 C) 4(4C )0.414 , 2.414

2( ) 4C

Q L Q L Q Q Q L L L Lx L L

Q Q

The correct answer is 3 0.414x L because this point is between 1Q and 2 .Q

21.24. When a positively charged rod is brought near to an isolated neutral conductor without touching it the rod will experience an attractive force. The electric charge on the rod induces a redistribution of charge in the conductor. The net effect of this distribution is that electrons move to the side of the conductor nearest to the rod. The positively charged rod is attracted to this region.

21.25. Using a metal key to touch a metal surface before exiting the car, which will discharge any charge I

carry. When I begin to fuel a car, I can touch the gas pump and the car before pumping the gas, discharging myself. If I get back into the car, I can re‐charge myself, and when I again get out of the


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car and touch the fuel nozzle without grounding myself first, I can get a spark, which might ignite the gasoline.

Problems

21.26. The charge of each electron is 191.602 10 C. The total number n of electrons required to give a total charge of 1.00 C is obtained by dividing the total charge by the charge per electron:

18

19

1.00 C6.18 10 electrons.

1.602 10 C/electron

Qn

e

21.27. The number of atoms or molecules in one mole of a substance is given by Avogadro’s number, 23 1

A 6.022 10 mol .N The faraday unit is A ,F N e where e is the elementary charge of an electron

or proton and is equal to 191.602 10 C. To calculate the number of coulombs in 1.000 faraday you can multiply AN by the elementary charge:

23 19A1.000 F (6.022 10 atoms/mol)(1.602 10 C) 96470 C.N e

21.28. 2 51 dyne 1 g cm / s 1 10 N and it is a unit of force. An electrostatic unit or esu is defined as follows: Two point charges, each of 1 esu and separated by one centimeter exert a force of exactly one dyne on each other. Coulomb’s law gives the magnitude of the force on one charge due to another, which is 2

1 2 /F k q q r (where 9 2 28.99 10 N m / C ,k 1q and 2q are electric charges and r is the

separation distance between charges.) (a) By substituting the values given in the question into Coulomb’s law, the relationship between the esu and the Coulomb can be determined:

2 2 55 10

2 9 2 2

(1 esu) (0.01 m) (1 10 N)1 10 N 1 esu 3.34 10 C

(0.01 m) 8.99 10 N m / C

k

(b) The result of part (a) shows that 101 esu 3.34 10 C. The elementary charge on an electron or

proton is 191.602 10 C.e To get the relationship between the esu and elementary charge, can divide 1 esu by the charge per electron (or proton).

109

19

3.34 10 C1 esu 2.08 10

1.602 10 C/e

e

21.29. The given quantities are the current, 35.00 10 AI and the exposure time, 10.0 st . One coulomb is equal to1 A s. To calculate the number of electrons that flow through your skin at this current and during this time, multiply I by t and then divide by the elementary charge per electron which is

191.602 10 .C

3

17

19

5.00 10 A 10.0 s 0.0500 A s 0.0500 C;

0.0500 C3.12 10 electrons.

1.602 10 C /

I t

e

21.30. THINK: Consider a mass, 1.00 kgm of water. To calculate how many electrons are in this mass, a relationship must be found between mass, the number of water atoms presents and their charge. Let denote the number of electrons. SKETCH:


824

RESEARCH: The molecular mass of water ( 2H O ), W 18.015 g/mol.m The number of moles of water can be found by dividing the mass of water by its molecular mass. The number of electrons present in the water can be found from the atomic numbers, ,Z for hydrogen and oxygen ( 1Z and

8Z respectively). The total number of water molecules can be found by multiplying the number

of moles of water present by Avogadro’s number, 23 1A 6.022 10 mol .N

SIMPLIFY: AW 2

10 electrons

H O atom

mN

m

CALCULATE: 3

23 1 261.00 10 g6.022 10 mol 10 electrons 3.34277 10 electrons

18.015 g/mol

ROUND: The values in the question were provided to 3 significant figures, so the answer is 263.34 10 electrons.

DOUBLECHECK: Considering that there are approximately 55 moles of 2H O per kilogram of water

and there are 10 electrons per 2H O atom, it makes sense that the answer is approximately 550 times greater than Avogadro’s number.

21.31. THINK: Protons are incident on the Earth from all directions at a rate of 21245.0 protons / m s .n

Assuming that the depth of the atmosphere is 120 km 120,000 md and that the radius of the Earth is 6378 km 6,378,000 m,r I want to determine the total charge incident upon the Earth’s atmosphere in 5.00 minutes. SKETCH:

RESEARCH: Modeling the Earth like a sphere, the surface area A can be approximated as 24 .A r The total number of protons incident on the Earth in the time t can be found by multiplying the rate,n by the surface area of the Earth and the time, .t The total chargeQ can be found by multiplying the total number of protons, P by the charge per proton. The elementary charge of a proton is 191.602 10 C.

SIMPLIFY: 24 ,P nAT n r t

191.602 10 C /Q P P

CALCULATE: 2 2 201245.0 protons / (m s) 4 (6378 km+120 km) (300. s) 1.981800 10 protons,P 20 191.981800 10 protons 1.602 10 C / protons 31.74844 CQ

ROUND: 31.7 C


825

DOUBLECHECK: The calculated answer has the correct units of charge. The value seems reasonable considering the values that were provided in the question.

21.32. The charges obtained by the student performing the experiment are listed here: 193.26 10 C, 196.39 10 C,

195.09 10 C, 194.66 10 C,

191.53 10 C. Dividing the above values by the smallest measured value will give the number of electrons, en found in each measurement.

The number of

electrons en must be rounded to their closest integer value because charge is quantized. Dividing the observed charge by the integer number of electrons gives the charge per electron. Taking the average of the observed charge/integer value data the average charge on an electron is calculated to be 19(1.60 0.03) 10 C. The error in a repeated measurement of the same quantity is:

error

standard deviation

number of measurements

N

.

21.33. THINK: An intrinsic silicon sample is doped with phosphorous. The level of doping is 1 phosphorous atom per one million silicon atoms. The density of silicon is 3

S 2.33 g/cm and its

atomic mass is S 28.09 g/mol.m The phosphorous atoms act as electron donors. The density of

copper is 3C 8.96 g/cm and its atomic mass is C 63.54 g/mol.m

SKETCH:

RESEARCH: Avogadro’s number is 23 1A 6.022 10 mol .N It gives the number of atoms or

molecules per mole of a substance. Density, / ,m V where massm and volumeV . SIMPLIFY:

Observed charge en Integer

value

Observed charge (integer value)

193.26 10 C 2.13 2 191.63 10 C

196.39 10 C 4.17 4 191.60 10 C

195.09 10 C 3.32 3 191.69 10 C

194.66 10 C 3.04 3 191.55 10 C

191.53 10 C 1 1 191.53 10 C


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(a) There will be 1 conduction electron per 61.00 10 silicon atoms. The number of silicon atoms

per 3cm is S S S A/n m N . The number of conduction electrons per 3cm is 6e S / (1.00 10 )n n .

(b) The number of copper atoms is C C C A/n m N . The number of conduction electrons in the

copper is Cn . The ratio of conduction electrons in silicon to conduction electrons in copper is e C/n n . CALCULATE:

(a) 3

23 1 22 3C

2.33 g/cm6.022 10 mol 4.995 10 /cm

28.09 g/moln

2216 3

e 6

4.995 104.995 10 conduction electrons / cm

1.00 10n

(b) 3

23 1 22 3C

8.96 g/cm6.022 10 mol 8.4918 10 /cm

63.54 g/moln

167e

22C

4.995 105.88215 10

8.4918 10

n

n

ROUND: There were three significant figures provided in the question so the answers should be: (a) 16 3

e 5.00 10 conduction electrons / cmn

(b) There are 75.88 10 conduction electrons in the doped silicon sample for every conduction electron in the copper sample.

DOUBLECHECK: It is reasonable that there are approximately 75 10 less conduction electrons in the doped silicon sample compared to the copper sample.

21.34. The force between the two charged spheres is F1 kqaqb

d12 initially. After the spheres are moved the

force is 2 22

.a bq qF k

d Taking the ratio of the force after to the force before gives:

F2 / F1 kqaqb

d22

/ k

qaqb

d12

d1

2 / d22 4 . The new distance is then d2 d1

2 / 4 d1 / 2 4 cm .

21.35. The charge on each particle is q . When the separation distance is 1.00 m,d the electrostatic force

is 1.00 N.F The charge q is found from 2 2 2

1 2 / / .F kq q d kq d Then, 2 2

5

9 2 2

(1.00 N)(1.00 m)1.05 10 C.

8.99 10 N m / C

Fdq

k

The sign does not matter, so long as each particle has a charge of the same sign, so that they repel.

21.36. In order for two electrons to experience an electrical force between them equal to the weight of one of the electrons, the distance d separating them must be such that.

2 2g Coulomb / .eF F m g ke d Then,

29 2 2 192

31 2

8.99 10 N m / C 1.602 10 C5.08 m

(9.109 10 kg)(9.81 m/s )e

ked

m g

21.37. In solid sodium chloride, chloride ions have a charge 19Cl 1.602 10 C,q e while sodium ions

have a charge 19Na 1.602 10 C.q e These ions are separated by about 0.28 nm.d The Coulomb

force between the ions is

9 2 2 19 2

9 9Cl Na2 9 2

8.99 10 N m / C (1.602 10 C)2.94285 10 N 2.9 10 N.

(0.28 10 m)

kq qF

d


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The negative sign indicates that the force is attractive.

21.38. In gaseous sodium chloride, chloride ions have a charge 19Cl 1.602 10 C,q e while sodium ions

have a charge 19Na 1.602 10 C.q e These ions are separated by about 0.24 nm.d Another

electron is located 0.48 nmy above the midpoint of the sodium chloride molecule. Find the magnitude and the direction of the Coulomb force it experiences.

The x‐component of the force is

- -Cl, e Na, e

2

22 2 2 2

2 2 2 2 3/222 2 2 2 2

9 2 2 19 2 9

39 2

2

2

29

/ 22

/ 4cos cos 2 cos

4 4 4 4 4

8.99 10 N m / C (1.602 10 C) (0.24 10 m)

(0.24 10 m)(0.48 10 m)

4

4.5717

x x xF F F

dke

d yke ke ke ke d

d d d d dy y y y y

10 1010 N -4.6 10 N

By symmetry, the ycomponents cancel; that is - -Cl, e Na, e.

y yF F The magnitude is therefore

104.6 10 NF ; The electron is pulled in the x direction (in this coordinate system).

21.39. The two up quarks have identical charge 19(2 / 3) (2 / 3) 1.602 10 C .q e They are

150.900 10 md apart. The magnitude of the electrostatic force between them is

2

9 2 2 192

2 15 2

28.99 10 N m / C (1.602 10 C)

3127 N.

(0.900 10 m)

kqF

d

This is large, however the proton does not ‘break apart’ because of the strength of the strong nuclear force which binds the quarts together to form the proton. A proton is made of 2 up quarks, each with charge (2 / 3) ,e and one down quark with charge (1/ 3)e . The net charge of the proton is e .

21.40. Coulomb’s Law can be used to find the force on 1 2.0 μCq due to 2 4.0 μC,q where 2q is

0.200 mr to the right of 1.q


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9 2 21 2 1 2212 1 2 2 2

2.0 μC 4.0 μC8.99 10 N m /C 1.8 N

0.200 m

q q q qF k r k x x x

r r

The 4.0 μC charge pulls the 2.0 μC charge to the right.

21.41. THINK: The two identical spheres are initially uncharged. They are connected by an insulating spring of equilibrium length 0 1.00 mL and spring constant 25.0 N/mk . Charges q and q are then placed on metal spheres 1 and 2, respectively. Because the spring is insulating, the charges cannot neutralize across the spring. The spring contracts to new length 0.635 m,L due to the attractive force between the charges spheres. Determine the charge .q If someone coats the spring with metal to make it conducting, find the new length of the spring. SKETCH:

RESEARCH: The magnitude of the spring force is S SF k x . The magnitude of the electrostatic force

is 21 2 /F kq q r . For this isolated system, the two forces must be in balance, that is SF F . From this

balance, the chargeq can be determined. The spring constant is denoted by Sk to avoid confusion with the Coulomb constant, k.

SIMPLIFY:

22

S 01 2S S S 02 2

( ) ( )

k L L Lkq q kqF F k x k L L q

kr L

CALCULATE:

2

5

9 2 2

25.0 N/m 0.635 m (1.00 m 0.635 m)2.02307 10 C

8.99 10 N m / Cq

If someone were to coat the spring such that it conducted electricity, the charge on the two spheres would distribute themselves evenly about the system. If the charges are equal in magnitude and opposite in sign, as they are in this case, the net charge in the system would be zero. Then the electrostatic force between the two spheres would be zero, and the spring would return to its equilibrium length of 1.00 m. ROUND: To three significant figures, 52.02 10 C.q

DOUBLECHECK: Dimensional analysis confirms that the answer is in coulombs, the appropriate unit for charge.

21.42. THINK: A point‐like charge of 1 3q q is located at 1 0,x and a point‐like charge of 2q q is

located on the x‐axis at 2 ,x D where 0.500 m.D Find the location on the x‐axis 3x where will a

third charge 3 0q q experiences no net force from the other two charges. SKETCH:


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RESEARCH: The magnitude of the electrostatic force is 21 2 /F kq q r . The net force on the third

charge 3q is zero when the sum of the forces from the other two charges is zero:

net,3 13 23 13 230 .F F F F F The two forces 13F and 23F must be equal in magnitude, but

opposite in direction. Consider the following three possible locations for the charge 3q . Note that

this analysis is independent of the charge of 3q . In the case 3 1 0,x x the two forces 13F and 23F will

be opposite in direction but they cannot be equal in magnitude: the charge 1q at 1x is greater in

magnitude than the charge 2q at 2x and 3x would be closer to 1x . (Remember that the electrostatic force increases as the distance between the charges decreases.) This makes the magnitude of 13F greater than that of 23F . In the case 30 m 0.500 mx , the two forces are in the same direction

and therefore cannot balance. In the case 3 2x x D , the two forces are opposite in direction, and

in direct opposition to the first situation, the force 13F and 23F can now be balanced. The solution will

have a positive x position, or more accurately, the third charge 3q must be placed near the smaller

fixed charge, 2q , without being between the two fixed charges 1q and 2q SIMPLIFY: Since 3 2x x , consider only the magnitudes of the forces. Since only the magnitudes of the forces are compared, only the magnitudes of the charges need be considered.

2 22 21 3 2 3

13 23 1 3 2 2 3 3 32 23 3 2

3kq q kq q

F F q x x q x q x D qxx x x

2 2 2 23 3 3 33 0 2 6 3 0x D x x x D D

Solving for 3x :2 2

3

6 36 4(2)(3 )

4

D D Dx

CALCULATE: 2 2

3

6(0.500 m) 36(0.500 m) 24(0.500 m)1.1830 m, 0.3170 m

4x

ROUND: Since 3 2x x , 3 1.18 m.x

DOUBLECHECK: The solution fits the expected location that was determined above (where 3 2x x ).

21.43. THINK: Identical point charges 632 10Q C are placed at each of the four corners of a rectangle of dimensions 2.0 mL by 3.0 m.W Find the magnitude of the electrostatic force on any one of the charges. Note that by symmetry the magnitude of the net force on each charge is equal. Choose to compute the net electrostatic force on 4 .Q SKETCH:


830

RESEARCH: The magnitude of the force between two charges is 2

2112 1 2 21/ .F kq q r r

The total

force on a charge is the sum of all the forces acting on that charge. The magnitude of the force is

found from 1/22 2 ,x yF F F where the components xF and yF

can be considered one at a time.

SIMPLIFY:

2 22

14, 24, 34, 2 2 2 2 3/22 2

1-component: cos 0x x x x

kQ kQ Wx F F F F kQ

W W L W W L

2 2

214, 24, 34, 2 2 2 3/2 2

2 2

2 2net

1-component: 0 siny y y y

x y

kQ kQ Wy F F F F kQ

W L L LW L

F F F

CALCULATE:

9 2 2 6 2

2 3/22 2

1 3.0 m8.99 10 N m / (32 10 C) 1.612 N

3.0 m 3.0 m 2.0 mxF C

9 2 2 6 2

3/2 22 2

2 2

net

2.0 m 18.99 10 N m / (32 10 C) 2.694 N

2.0 m3.0 m 2.0 m

1.612 N 2.694 N 3.1397 N

yF C

F

ROUND: Since each given value has 2 significant figures, net 3.1 NF

DOUBLECHECK: Since L is less than ,W the y‐component of netF should be greater than the x‐component.

21.44. THINK: Charge 81 1.4 10q C is at 1 (0,0).r Charge 8

2 1.8 10q C is at 2 (0.18 m,0 m),r and

charge 83 2.1 10q C is at 3 (0 m,0.24 m).r Determine the net force (magnitude and direction) 3F

on charge 3q . SKETCH:


831


31212 1 2 12 1 2 12 12/ / .F kq q r r kq q r r

The total force on charge 3q is the sum of all the forces acting on it. The magnitude of 3F is found

from 1/22 2

3 1 2 ,F F F and the direction is found from 1tan / .y xF F

SIMPLIFY:

net, 3 13 23

1 3 13 2 3 233 3

13 23

1 3 3 1 3 1 2 3 3 2 3 2

3/2 3/22 2 2 23 1 3 1 3 2 3 2

1 3 2 33 2 33 2 2 3/2

3 2 3

ˆ ˆ ˆ ˆ( ) ( ) ( ) ( )

( ) ( ) ( ) ( )

ˆ ˆ ˆ( )

F F F

kq q r kq q r

r r

kq q x x x y y y kq q x x x y y y

x x y y x x y y

kq q kq qy y x x y y

y x y

CALCULATE:

9 2 2 8 8

net, 3 3

9 2 2 8 8

3/22 2

5 5 5

8.99 10 N m / C (1.4 10 C)(2.1 10 C) 0.24 mˆ

0.24 m

ˆ ˆ8.99 10 N m / C ( 1.8 10 C)(2.1 10 C) 0.18 m 0.24 m

0.18 m 0.24 m

ˆ ˆ(4.5886 10 N) (2.265 10 N) (3.0206 10 N)

F y

x y

y x

5 5

ˆ

ˆ ˆ2.265 10 N 1.568 10 N

y

x y

2 2 5 2 5 2 5net, 3

51 1

5

(2.265 10 N) (1.568 10 N) 2.755 10 N

1.568 10 Ntan tan 34.69 above the horizontal

2.265 10 N

x y

y

x

F F F

F

F


832

ROUND: With 2 significant figures in each given value, the final answers should be rounded to

5 5 5net, 3 ˆ ˆ2.265 10 N 1.568 10 N 2.8 10 NF x y

and 35 .

DOUBLECHECK: Due to the attraction between 2q and 3q and that 1q is directly underneath 3q ,

the x component of net, 3F

has to be positive.

21.45. THINK: A positive charge Q is on the y‐axis at a distance a from the origin and another positive charge q is on the x‐axis at a distance b from the origin. (a) Find the value(s) ofb for which the x‐

component of the force onq is a minimum. (b) Find the value(s) of b for which the x‐component of the force on q is a maximum. SKETCH:

RESEARCH: The electrostatic force is2

/ .F kqQr r The x‐component of this force

is 2( / )cos .xF kqQ r The values of b for which xF is a minimum can be determined by inspection;

the values of b for which xF is a maximum can be found by calculating the extrema of xF , that is,

taking the derivative of xF with respect tob , setting it to zero, and solving forb .

SIMPLIFY: 2 3 3/2

2 2cosx

kqQ kqQb kqQbF

r r a b

a) Minima: By inspection, the least possible value of xF is zero, and this is attained only when 0.b

b) Maxima: 0xdF

db

2 2 25/2

2 2

3/2 5/22 2 2 2

2 2 2

33 2 0 0

2

3 0 2

kqQ a b kqQbkqQkqQ a b b

a b a b

aa b b b

CALCULATE: Reject the negative solution, since distances have to be positive: .2

ab


833

ROUND: Not applicable DOUBLECHECK: It makes sense that the possible values of b should be symmetrically distributed

about the origin (above which lies the chargeQ ).

21.46. THINK: Two protons are placed near one electron as shown in the figure provided. Determine the electrostatic force on the electron. The charge of the electron is eq e and the charge of each

proton is pq e , where 191.602 10 C.e

SKETCH:

RESEARCH: By symmetry the forces in the vertical direction cancel. The force is therefore due solely to the horizontal contribution cosF in the x direction: the Coulomb force is 2

21 1 2 21/ .F kq q r

SIMPLIFY: By symmetry, and with the two protons,

2 2

pe 2 3/22 2

ˆ ˆ ˆ2 cos 2 2 .ke x ke x

F F x x xrr x d

CALCULATE:

29 2 2 19

26

3/22 2

8.99 10 N m /C 1.602 10 C 0.0700 mˆ ˆ2 ( 5.0742 10 N)

0.0700 m 0.0500 mF x x

ROUND:

26 ˆ5.07 10 NF x

DOUBLECHECK: This is a reasonable force as the charges are as small as they can possibly be and the separation is large.

21.47. THINK: The positions of the three fixed charges are 1 1.00 mCq at 1 (0,0),r 2 2.00 mCq at

2 (17.0 mm, 5.00 mm),r and 3 3.00 mCq at 3 ( 2.00 mm,11.0 mm).r Find the net force on the

charge 2 .q SKETCH:

RESEARCH: The magnitude force is 2

31212 1 2 12 1 2 12 12/ / .F kq q r r kq q r r

The net force on 2q is the sum

of all the forces acting on 2 .q

SIMPLIFY: 1 2 1 2 1 3 2 3 2 3net, 2 12 32 2 3/2 3/22 2 2 2

2 1 2 1 2 3 2 3

ˆ ˆ ˆ ˆ( ) ( ) ( ) ( )

( ) ( ) ( ) ( )

q x x x y y y q x x x y y yF F F kq

x x y y x x y y

CALCULATE: Without units,


834

9net, 2 3/2 3/22 2 2 2

8 7

ˆ ˆ(1.00) 17.0 5.00 ˆ ˆ(3.00)(19.0 16.0 )8.99 10 ( 2.00)

17.0 5.00 19.0 16.0

ˆ ˆ1.2181 10 7.2469 10 .

x y x yF

x y

Then, the units of net, 2F

are:

2 2net, 2 3/2 3/22 2 2 2

(mC)(mm mm) (mC)(mm mm)N m / C (mC) N

mm mm mm mmF

Altogether , 8 7net, 2 ˆ ˆ1.2181 10 N 7.2469 10 N .F x y

The magnitude of the force is

2 22 2 8 7 8net, 2 1.2181 10 N 7.2469 10 N 1.4174 10 Nx yF F F

ROUND: 8 7 8net, 2 net, 2ˆ ˆ1.22 10 N 7.25 10 N and 1.42 10 N.F x y F

DOUBLECHECK: The charges are large and the separation distance are small, so net, 2F should be

very strong.

21.48. THINK: the masses of the beads are 510.0 mg 1.00 10 kgm and they have identical charge. They

are a distance 0.0200 md apart. The coefficient of static friction between the beads and the surface is 0.200. Find the minimum chargeq needed for the beads to start moving. SKETCH:

RESEARCH: Assume the surface is parallel to the surface of the Earth. The frictional force is f N ,

where .N mg The electrostatic force is 2 2/ .F kq d The beads will start to move as soon as F is

greater than ,f enabling one bead to move away from the other. Then the minimum chargeq can be

found by equating f and .F

SIMPLIFY: 2

2

2 /

kqF f mg q d mg k

d

CALCULATE:

2 5 2

10

9 2 2

0.0200 m (0.200)(1.00 10 kg)(9.81 m / s )9.3433 10 C

8.99 10 N m / Cq

ROUND: All of the given values have three significant figures, so

109.34 10 C.q DOUBLECHECK: The units of the solution are those of charge. This is a reasonable charge required

to overcome the frictional force.

21.49. THINK: The ball’s mass is 1 0.0300 kg;m its charge is 1 0.200 μ .q C The ball is suspended a

distance of 0.0500 md above an insulating floor. The second small ball has mass

2 0.0500 kgm and a charge 2 0.400 μC.q Determine if the second ball leaves the floor. Find the

tension T in the string when the second ball is directly beneath the first ball. SKETCH:


835

RESEARCH: The electrostatic force between two charges is 21 2 / .F kq q r The force of gravity is

g .F mg The ball will leave the floor if the electrostatic force between the two balls is greater that

the force of gravity, that is if gF F , and if the charges are opposite. The tension in the rope can be

found by considering all of the vertical forces acting on the first ball. SIMPLIFY: The electrostatic force is: 2

1 2 / .F kq q d The gravitational force is: g 2F m g . The

forces acting on 1m in the y‐direction sum to: coulomb 10 .T F m g So the tension is coulomb 1 .T F m g

CALCULATE: 9 2 2 6 6 28.99 10 N m / C ( 0.200 10 C)(0.400 10 C)/(0.0500 m) 0.28768 N,F

2g (0.0500 kg)( 9.81 m/s ) 0.4905 N,F

20.28768 N (0.0300 kg)( 9.81 m/s ) 0.58198 N.T

Since g ,F F the second ball does not leave the ground.

ROUND: With all given values containing three significant figures, round the tension to 0.582 N.T

DOUBLECHECK: The balls are not quite close enough to overcome the force of gravity, but the magnitude of coulombF is comparable to gF , despite the small charges (on the order of 710 C ).

21.50. THINK: A 1 3.00 mCq charge and a 2 4.00 mCq charge are fixed in position and separated by

5.00 m.d Take the position of 1q to be at 1 0,x and position of 2q to be at 2 5.00 m.x (a) Find

the location, 3 ,x of a 3 7.00 mCq charge so that the net force on it is zero. (b) Find the

location, 3 ,x of a 3 7.00 mCq charge so that the net force on it is zero. SKETCH:

RESEARCH: The electrostatic force between two charges is 21 2 /F kq q r . The net force on a third

charge is zero: net,3 13 23 13 230 .F F F F F The two forces must be equal in magnitude, but

opposite in direction. Consider the following three possible locations for the charge 3q . Note that

this analysis is independent of the charge of 3q : At 3 5.00 m,x the two forces 13F and 23F will be

opposite in direction but they cannot be equal in magnitude: the charge 2q at 2 5.00 mx is greater

in magnitude than the charge 1q at 1 0x and 3x would be closer to 2x . (Remember that the electrostatic force increases as the distance between the charges decreases.) This makes the magnitude of 23F greater than that of 13F . Next, consider values of 3x satisfying: 30 m 5.00 m.x

The two forces are in the same direction and therefore cannot balance. At 3 0 m,x the two forces

are opposite in direction, and in direct opposition to the first situation, the force 13F and 23F can

now be balanced. The solution will have a negative position, or more accurately, the third charge 3q


836

must be placed near the smaller fixed charge, 1,q without being between the two fixed charges 1q and 2 .q This answer is independent of the charge of 3.q

SIMPLIFY: With 3 0,x and 13F opposite in direction to

23 ,F the force are balanced when

2 2 2 21 3 2 3

13 23 1 2 3 2 3 1 2 3 1 2 3 1 22 2

2 33

2 0.kq q kq q

F F q x x q x q q x q x x q xx xx

Solving for 3x :

2 2 21 2 1 2 1 2 1 2

31 2

2 4 4( ).

2( )

q x q x q q q xx

q q

CALCULATE: 2 2 2

3

2(3.00 mC)(5.00 m) mC m 4(3.00) (5.00) 4(3.00 4.00)(3.00)(5.00)32.321 m, 2.305 m

2(3.00 mC 4.00 mC)x

By the convention established in this solution, 3x is negative. (The second solution places 3q a

between 1q and 2q , a possibility which has been ruled out.)

ROUND: All given values have three significant figures, so 3 32.3 m.x

DOUBLECHECK: Inserting the calculated value of 3x back into the expressions for the Coulomb

force:

1 3

13 2 23

3.00 mC 7.00 mC181 N

32.3 m

kkq qF

x

and

2 323 2 2

2 3

4.00 mC 7.00 mC181 N.

5.00 m 32.3 m

kkq qF

x x

21.51. THINK: Four point charges, each with charge q , are fixed to the four corners of a square with a

sides of length 10.0 μm.d An electron is suspended above a point at which its weight is balanced

by the electrostatic force due to the four electrons: 15 nmz above the center of the square. The

mass of an electron is 319.109 10 kgem , and the charge is 191.602 10 Ceq e . Find the value of q of the fixed charges, in Coulombs and as a multiple of the electron charge. SKETCH:

RESEARCH: The electrostatic force between two charges is 2

1 2 / .F kq q r By symmetry, the net force in the horizontal direction is zero, and the problem reduces to a balance of the forces in the vertical direction, with one fixed charge having a quarter of the charge of the electron. The vertical component of the electrostatic force is sinF . The weight of the electron is eW m g .


837

SIMPLIFY: Balancing the forces in the vertical (z) direction yields

coulomb 2

1 1 sin .

4 4e

e

kqqF W m g

r

Solving for q:

3/222

2 3 2 2 3/2 2( )1.

4 sin 4 4 4

e

e e e

e e e

dm g z

m gr m gr m g L zq

kq kq z kq z kez

CALCULATE:

3/2231 2 2

9 2 2 19

29 10

(10.0 μm)9.109 10 kg (9.81 m / s ) (15 nm)

2

4 8.99 10 N m / C (1.602 10 C)(15 nm)

3.6562 10 C, or - 2.282 10

q

e

ROUND: With 2 significant figures in ,z 29 103.7 10 C 2.3 10 .q e

DOUBLECHECK: The gravitational force on an electron is extremely small, on the order of 3010 N. The force charges q need only an extremely small charge to balance the gravitational force on the electron.

21.52. THINK: A uniformly charged thin rod of length L has a total charge .Q Find an expression for the electrostatic force strength acting on an electron, whose magnitude of charge is e, is positioned on the axis of the rod at distance d from the center. SKETCH:

RESEARCH: The electrostatic force between two charges is 2/F kqQ r . The net electrostatic force acting on a charge q is the sum of all the electrostatic forces acting on q . In the event of a continuous and linear distribution of charge of length L and total chargeQ , the force due to an

infinitesimal amount of chargedp from the distribution acting on the charge q is: 2/ ,dF kq dq x

where ( / ) .dp Q L dx dx ( is the linear charge density.) In this case, the total force on the electron is then

/2

2/2,

d L

d L

keF dx

x

where the integration runs over the length of the rod, starting from the point closest to the electron / 2d L

and ending with the point farthest from the electron / 2 .d L

SIMPLIFY:

/2

2 2 2 2 2 2/2

/2 /2

/2 /2

1 1 1 4 42

2 2 4 4

dd L d

dL d

LL

Ld L

ke ke L keQF dx ke dx ke x ke

d L d Lx x d dL L

CALCULATE: Not applicable ROUND: Not applicable

DOUBLECHECK: The answer is in the correct units of force:

2

2

2

N mC C

CN.

mF

21.53. THINK: A negative charge q is located and fixed at (0, 0) . A positive charge q is initially at

( , 0).x The positive charge will accelerate towards the negative charge. Use the binomial expansion


838

to show that when the positive charge moves a distance x closer to the negative charge, the

force on it increases by 2 32 / .F kq x SKETCH:

RESEARCH: The Coulomb force is 22121 1 2 21/ ,F kq q r r

where 21r is the unit vector that points from

charge 2 to charge 1. To first order, the binomial expansion is (1 ) 1nx nx for 1.x

SIMPLIFY: The initial force on q (when it was at ( , 0)x was 2

2.

kqF x

x

After moving closer to q

by 1 the new force on q is

22 2 2

2 2 2

2

1 .

1

kq kq kqF x x x

xxxx

x

Using the binomial

expansion, 2 2

2 21 ( 2) ... 1 2

kq kqF x x

x xx x

(to first order in ). Then,

2

3

2kqF F F x

x

and

2

3

2,

kqF

x

as desired.

CALCULATE: Not applicable. ROUND: Not applicable.

DOUBLECHECK: The charge in force has the correct units for force:

2

2

2

N m CC m

mC N.m

F

21.54. THINK: Two charges, both q , are located and fixed at coordinates ( ,0)d and ( ,0)d in the xy plane.

A positive charge of the same magnitude q and of massm is placed at coordinate (0,0) . The positive

charge is then moved a distance d along the +y direction and then released. It will oscillate between co‐ordinates (0, ) and (0, ) . Find the net force netF acting on the positive charge when it is

moved to (0, ) and use the binomial expansion to find an expression for the frequency of the resulting oscillation. SKETCH:

RESEARCH: The Coulomb force is 22121 1 2 21/ ,F kq q r r

where 21F

is the force on the charge 1 by

charge 2, and 21r points from charge 2 to charge 1. To first order, the binomial expansion is, in


839

general, (1 ) 1nx nx for 1.x The restoring force of a simple harmonic oscillator obeys Hooke’s

Law, 2 ,F mx where is the characteristic angular frequency, and / (2 ).f

SIMPLIFY: 2 2 2

3121 1 31 2net 3 3 2 2 3/2 2 2 3/2 2 2 3/2

21 31

3/22 2 2

3/2 3 223

2

2ˆ ˆ ˆ ˆ ˆ( ) ( )( ) ( ) ( )

2 2ˆ ˆ1

1

kq q rkq q r kq kq kqF dx y dx y y

r r d d d

kq kqy y

d dd

d

Note the binomial expansion of 3

222 2 21 / 1 3 / 2 / .d d

Neglecting the term 2 2/ d

(keeping only terms linear in ), the net force is 2 3net ˆ2 / .F kq d y

Then from 2 ,F mx

/F mx with ,x the angular frequency

is 2 3 2 32 / 2 / / 2 /kq md kq md q d k md and the frequency

is / 2 2 / / / 2 .f q d k md q d k md

CALCULATE: Not applicable ROUND: Not applicable

DOUBLECHECK: The frequency of oscillation should depend directly on the magnitude of the charges and inverse on the distance separating the charges. This lends support to the formulas found above.

21.55. The gravitational force between the Earth and Moon is given by 2g Earth Moon EM/ .F GM m r The static

electrical force between the Earth and the Moon is 2 2EM/ ,F kQ r where Q is the magnitude of the

charge on each the Earth and the Moon. If the static electrical force is 1.00% that of the force of gravity, then the charge Q would be:

2Earth Moon Earth Moon

g 2 2EM EM

0.0100 0.01000.01 .

GM m GM mkQF F Q

kr r

This gives

11 2 24 2212

9 2 2

0.0100(6.67 10 N m / kg)(5.97 10 kg)(7.36 10 kg)5.71 10 C.

8.99 10 N m / CQ

21.56. The gravitational force between the Earth and Moon is given by 2g Earth moon EM/ .F GM m r If this is due

solely to static electrical force between the Earth and Moon, the magnitude of Q would be: 2

Earth Moon Earth Moong 2 2

EM EM

.M m GM mQ

F G k Qkr r

So,

24 2211 2 13

9 2 2

(5.97 10 kg)(7.36 10 kg)(6.67 10 N m / kg) 5.71 10 C.

8.99 10 N m / CQ

This is a large amount of charge, on the order of 3110 electrons worth of charge. This is equivalent to about 60 million moles of electrons.

21.57. THINK: The radii of the electron orbits are 2n Br n a , where n is an integer (not 0) and

11B 5.29 10 m.a Calculate the electrostatic force between the electron (charge e and mass

31e 9.109 10 kgm ) and the proton (charge e and mass

27p 1.673 10 kgm ) for the first 4 orbits

and compare them to the gravitational interaction between the two. Note 191.602 10 .e C


840

SKETCH:

RESEARCH: The Coulomb force is 21 2 / ,F k q q r or 2 2

n n/F ke r in this case. The gravitational

force is 2g 1 2 / ,F Gm m r or 2

g, n e p n/ .F Gm m r

SIMPLIFY: 2 2

e p e p

1 g, 1 2 g, 22 2 2 2B B B B

1: ; , 2 : ; (4 ) (4 )

Gm m Gm mke ken F F n F F

a a a a

2 2e p e p

3 g, 3 4 g, 42 2 2 2B B B B

3 : ; , 4 : ; (9 ) (9 ) (16 ) (16 )

Gm m Gm mke ken F F n F F

a a a a

CALCULATE: Note that:

9 2 2 19 228

2 211

B

8.99 10 N m /C (1.602 10 C)8.2465 10 N

5.29 10 m

ke

a

and

11 2 31 27

e p 47

2 211B

(6.67 10 N m / kg)(9.109 10 kg)(1.673 10 kg)3.632 10 N.

5.29 10 m

Gm m

a

2e p8 47

1 g, 12 2B B

2e p9 48

2 g, 22 2

B B

2e p9 49

3 g, 32 2

B B

2

4 2

B

Then for 1: 8.2465 10 N; 3.6342 10 N

2 : 5.1515 10 N; 2.2712 10 N4 4

3: 1.1081 10 N; 4.4863 10 N9 9

4 : 16

Gm mken F F

a a

Gm mken F F

a a

Gm mken F F

a a

ken F

a

e p10 49g, 4 2

B

3.2213 10 N; 1.4195 10 N16

Gm mF

a

ROUND: Since Ba has three significant figures, 8

1 8.25 10 N,F

47g, 1 3.63 10 N,F

92 5.15 10 N,F

48g, 2 2.27 10 N,F

93 1.12 10 N,F

49g, 3 4.49 10 N,F

104 3.22 10 N,F

49g, 4and 1.42 10 N.F In

every case the gravitational force between the proton and the electron is almost forty orders of magnitude smaller than the electrostatic force between them.

DOUBLECHECK: Asn increases, the distance between the proton and the electron increases. Since each force follows an inverse‐square law with respect to the distance, the forces decrease asn increases


841

21.58. THINK: The net force on the orbiting electron is the centripetal force, CF . This is due to the electrostatic force between the electron and the proton, F . The radius of the hydrogen atom is

115.29 10 mr . The charge of an electron is 19e 1.602 10 Cq e , and the charge of a proton is

19p 1.602 10 Cq e . Find the velocity v and the kinetic energy K of the electron orbital. The mass

of an electron is 31e 9.109 10 kg.m

SKETCH:

RESEARCH: The centripetal force is 2C e / .F m v r The electrostatic force is 2

1 2 / .F k q q r The

kinetic energy is 2 / 2.K mv SIMPLIFY: Solve for v :

1/22 2 2 2 2C e 1 2 e / / / /F F m v r k q q r ke r v ke rm

Solve for K : 2e / 2.K m v

CALCULATE: 9 2 2 19 2

6

11 31

8.99 10 N m / C (1.602 10 C)2.18816 10 m/s

(5.29 10 m)(9.109 10 kg)v

231 6

18(9.109 10 kg) 1.5915 10 m/s

1.14106 10 J 7.1219 eV2

K

ROUND: 62.19 10 m/s, and 7.12 eV.v K DOUBLECHECK: Because the electron has very little mass, it is capable of approaching speeds on

the order of 0.01c or 0.1c (where c is the speed of light). For the same reason, its kinetic energy is small (on the order of a few electron volts, in the case of the hydrogen atom).

21.59. For the atom described in the previous question, the ratio of the gravitational force between the electron and proton to the electrostatic force is:

2 2g e p 1 2

2e p

11 3 2 31 27

9 2 2 19 2

40

/ / /

/

6.6742 10 m / (kg s ) (9.109 10 kg)(1.673 10 kg)

8.99 10 N m / C (1.602 10 C)

4.41 10

F F Gm m r k q q r

Gm m ke

This value is independent of the radius; if this radius is doubled, the ratio does not change.

21.60. THINK: The Earth and the Moon each have a charge 61.00 10 C.q Their masses are 24

E 5.97 10 kgm and 2M

27.36 10 kgm , respectively. The distance between them is

384.403 km,r center‐to‐center. (a) Compare their electrostatic repulsion, F , with their gravitational attraction, gF . (b) Discuss the effects of the electrostatic force on the size, shape

and stability of the Moon’s orbit around the Earth.

SKETCH:


842

RESEARCH: Treat each object as a point particle. The electrostatic force is 2

1 2 /F k q q r , and the

gravitational force is 2g / .F GMm r

SIMPLIFY: (a) 2 2

E M2

g/ ; /F kq r F GM m r

(b) Not applicable. CALCULATE:

(a)

9 2 2 6 2

28

8.99 10 N m / C ( 1.00 10 C)60839.6 N

3.84403 10 mF

11 3 2 24 22

20g 2

8

6.6742 10 m / kg s 5.9742 10 kg 7.36 10 kg1.986 10 N

3.84403 10 mF

(b) The force of gravity is about 16 orders of magnitude greater than the electrostatic repulsion. The electrostatic force is an inverse‐square central force. It therefore has no effect on the shape or stability of the Moon’s orbit. It could only affect the size of the orbit, but given the orders of magnitude in difference between this and gF , the effect is probably undetectable.

ROUND: (a) 46.08 10 NF and

Fg 1.99 1020 N

DOUBLECHECK: gF should greater than ,F otherwise the Moon would not remain in the Earth’s

orbit.

21.61. Eight 1.00-μC charges are aligned on the yaxis with a distance 2.00 cmy between each closest pair:

The force on the charge at 4.00 cm,y 3 ,q is:

8

tot, 3 , 3 13 23 43 53 63 73 83 13 23 43 53 63 73 831, 3

( )n

n n

F F F F F F F F F F F F F F F F y

All terms have in common the factor 3k q . Then,

1 2 4 5 6 7 8tot, 3 3 2 2 2 2 2 2 2

1 3 2 3 4 3 5 3 6 3 7 3 8 3

q q q q q q qF k q

y y y y y y y y y y y y y y

Since 1 2 8...q q q q ,


843

2tot, 3 2 2 2 2 2 2 2

2

tot, 3 2 2 2 2 2 2

9 2 2 6

2

2

1 1 1 1 1 1 1

(2 ) ( ) ( ) (2 ) (3 ) (4 ) (5 )

1 1 1 1 11 1

( ) 2 2 3 4 5

8.99 10 N m / C 1.00 10 C 769

36000.0200 m

ˆ4.80 N

F kqy y y y y y y

kqF y

y

y

y

21.62. The distance between the electron (charge eq e ) and the proton (charge pq e ) is 115.2 10 m.r

The net force on the electron is the centripetal force, 2c e c e /F m a m v r . This is due to the Coulomb

force,

21 2 / .F k q q r That is, 2 2

c e 1 2 / / .F F m v r k q q r The speed of the electron is:

9 2 2 19 22 22 6 6

e 31 11e

8.99 10 N m / C (1.602 10 C) 2.207 10 m/s 2.2 10 m/s.

(9.109 10 kg)(5.2 10 m)

ke kem v v

r m r

21.63. The radius of the nucleus of 14C is 0 1.505 fm.r The nucleus has charge 0 6 .q e

(a) A proton (charge q e ) is placed 3.01 fmd from the surface of the nucleus. Treating the nucleus as a point charge, the distance between the proton and the charge of the nucleus is

0 .r d r The force is repulsive due to the like charges. The magnitude of this force is

9 2 2 19 220

2 2 215 150

8.99 10 N m / C 6(1.602 10 C)6 67.908 N 67.9 N

3.01 10 m 1.505 10 m

k q q k eF

r d r

(b) The proton’s acceleration is: 2

28ec e c 2

p

2

7

67.908 N4.06 10 m/s

1.673 10 kg

m v FF m a a

r m

21.64. The original force is 21 2 / 0.10 N.F k q q r Now 1q becomes 1(1/ 2) ,q while r becomes 2 .r The

new force is:

1 21 2

2 2

1

2 1 1 1 = (0.10 N) 0.013 N

8 8 82

k q qk q q

F Frr

21.65. The charge and position of three point charges on the x‐axis are:

11

22

33

19.0 μC; 10.0 cm

57.0 μC; 20.0 cm

3.80 μC; 0

q x

q x

q x

The magnitude of the total electrostatic force on 3q is:

3 1 3 2 1 213 23tot, 3 13 23 13 23 32 2 2 2

1 21 3 2 3

9 2 2

2 2

19.0 μ 57.0 μ8.99 10 N m / C 3.80 μ 114 N

0.100 m 0.200 m

k q q k q q q qF F F F F F F k q

x xx x x x

C CC

21.66. The charge and position of three point charges on the x‐axis are:


844

11

22

33

64.0 μC; 0.00 cm

80.0 μC; 25.0 cm

160.0 μC; 50.0 cm

q x

q x

q x

The magnitude of the total electrostatic force on 1q is:

2 1 3 1 3 221 31tot, 1 21 31 12 2 2 2

3 22 1 3 1

9 2 2

2 2

160.0 μC 80.0 μC8.99 10 N m / C 64.0 μC 368 N.

0.500 m 0.250 m

k q q k q q q qF F F F F k q

x xx x x x

21.67. The charge of the Earth is 56.8 10 C.Q The mass of the object is 1.0 g.m For this object to be

levitated near the Earth’s surface ( E 6378 kmr ), the Coulomb force and the force of gravity must be the same. The charge q of the object can be found from balancing these forces:

2E

g Coulomb 2E

2 6 25

9 2 2 5

0.0010 kg (9.81 m / s )(6.378 10 m)6.5278 10 C 65 μC.

8.99 10 N m /C 6.8 10 C

k Qq mgrF F mg q

k Qr

q

Since Q is negative, and the object is levitated by the repulsion of like charges, it must be that 65 μCq .

21.68. The mass of the cat is 7.00 kg. The distance between the cat and the metal plate is 2.00 m. The cat is suspended due to attractive electric force between the cat and the metal plate.

The attractive force between the cat and the metal plate is 2/ .F kQQ d Since the cat is suspended

in the air, this means that .F mg Therefore 2 2/ .mg kQ d Solving for Q gives 2 / / .Q mgd k d mg k Substituting 7.00 kgm , 29.81 m / sg , 9 2 28.99 10 N m /k C and

2.00 md yields 2

4

9 2 2

7.00 kg 9.81 m / s2.00 m 1.748 10 .

8.99 10 N m / CQ e

The number of electrons that must be extracted is 4

15

19e

1.748 10 C1.09 10 electrons.

1.602 10 C

QN

q


845

21.69. THINK: A 10.0 g mass is suspended 5.00 cm above a non‐conducting flat plate. The mass and the plate have the same charge .q The gravitational force on the mass is balanced by the electrostatic force. SKETCH:

RESEARCH: The electrostatic force on the mass m is 2 2

E / .F kq d This force is balanced by the

gravitational force gF mg . Therefore, E gF F or 2 2/ .kq d mg

SIMPLIFY: The charge on the mass m that satisfies the balanced condition is / .q d mg k

CALCULATE: Putting in the numerical values gives:

3 2

7

9 2 2

10.0 10 kg 9.81 m / s0.0500 m 1.6517 10 .

8.99 10 N m / Cq e

The number of electrons on the mass m is:

712

19

1.6517 101.0310 10 electrons.

1.602 10

q eN

e e

The additional mass of electrons is 12 31 191.0310 10 9.11 10 kg 9.39263 10 kg.m

ROUND: Rounding to three significant figures gives 71.65 10 ,q e and 199.39 10 kg.m DOUBLECHECK: It is expected that m is negligible since the mass of electron is very small.

21.70. THINK: This problem involves superposition of forces. Since there are three forces on 4 ,Q the net force is the vector sum of three forces. SKETCH:


846

RESEARCH: The magnitude of the forces between two charges, 1q and 2 ,q is 21 2 / .F kq q r The

forces on 4Q are

3 41 4 2 41 2 32 2 2

14 24 34

ˆ ˆ ˆ ˆ ˆsin cos , sin cos , and .Q QQ Q Q Q

F k x y F k x y F k yr r r

SIMPLIFY: By symmetry, the horizontal components of 1F and 2F cancel, and 3F has no horizontal component. The net force is

31 21 2 3 4 2 2 2

14 24 34

ˆcos .QQ Q

F F F F kQ yr r r

Since 1 2Q Q and 14 24r r , the above equation simplifies to

314 2 2

14 34

2 cosˆ .

QQF kQ y

r r

CALCULATE: The distance 14r and 34r are 2 2

14 343 cm 4 cm 5 cm; 4 cm.r r Therefore

cos 4 / 5. Substituting the numerical values yields:

3 39 2 2 3

2 22 2

2 1 10 C 4 1.024 10 C8.99 10 N m / C 2 10 C 0 N.

55 10 m 4 10 mF

ROUND: Not needed DOUBLECHECK: It is clear from the symmetry of the problem that this is a reasonable outcome.

21.71. THINK: Three 5.00‐g Styrofoam balls of radius 2.00 cm are tied to 1.00 m long threads and suspended freely from a common point. The charge of each ball is q and the balls form an equilateral triangle with sides of 25.0 cm. SKETCH:


847

RESEARCH: The magnitude of the force between two charges, 1q and 2 ,q is 2

12 1 2 / .F kq q r The

magnitude of F in the above figure is2 2/ .F kq r Using Newton’s Second Law, it is found that

sinyT T mg and cos 2 cos .xT T F

SIMPLIFY: Eliminating T in the above equations yields tan / 2 cos .mg F Rearranging gives,

2 2/ 2tan cos / .F mg kq r Therefore, the charge q is

2

.2 tan cos

mgrq

k

From the sketch, it is clear that the distance of the ball to the center of the triangle is / 2cos .d r

Therefore 2 2tan / .L d d

CALCULATE: Substituting the numerical values, 0.250 m,r 35.00 10 kg,m 29.81 m/s ,g 1.00 mL and 30 (exact) gives

0.250 m0.1443 m

2cos(30 )d

2 21.00 m 0.1443 m

tan 6.8560.1443 m

23 2

7

9 2 2

5.00 10 kg 9.81 m / s 0.250 m1.69463 10 C

2 8.99 10 N m / C 6.856cos(30 )q

ROUND: 0.169 μCq DOUBLECHECK: This charge is approximately 11 orders of magnitude larger than the elementary

charge e. The charge required to deflect 5.00 g balls by a distance of 25.0 cm would need to be fairly large.

21.72. THINK: Two point charges lie on the x‐axis. A third point charge needs to be placed on the x‐axis such that it is in equilibrium. This means that the net force on the third charge due to the other charges is zero. SKETCH:


848

RESEARCH: In order for the third charge to be in equilibrium, the force on it due to 1,q 1,F

must be

equal in magnitude and opposite in direction to 2F the force due to 2 .q Note that the sign of the

third charge is irrelevant, so I can arbitrarily assume it is positive. Since 1 2 ,q q the third charge

must be closer to 2q than to 1q . Also, since 1q and 2q are oppositely charged, the forces on a particle between them will be in the same direction and hence cannot cancel. The third charge must

be in the region 20.0 cm.x The net force on 3q is 1 3 2 3

net 2 23 3 2

.k q q k q q

Fx x x

SIMPLIFY: Solving net 0F for 3x yields 222 3 1 3 2 2 3 2or .q x q x x q x x Therefore the

position of 3q is 1 2

3

1 2

.q x

xq q

CALCULATE: Putting in the numerical values yields

3

6.0 μC 20.0 cm47.32 cm.

6.0 μC 2.0 μCx

ROUND: Using only two significant digits, the position 3x is 3 47 cmx

DOUBLECHECK: This is correct since 3 2x x .

21.73. THINK: In this problem, a gravitational force on an object is balanced by an electrostatic force on the object. SKETCH:

RESEARCH: The electric force on 2q is given by

21 2 / .EF kq q d The gravitational force on 2m is

g 2F m g .

SIMPLIFY: 2 21 2 2 2 1 2/ / .kq q d m g m kq q gd

CALCULATE: Substituting the numerical values, 1 2 2.67 μe, 0.360 mq q d produces

29 2 2 6

2 22

8.99 10 N m / C 2.67 10 C0.05041 kg.

9.81 m / s 0.360 mm

ROUND: Keeping only three significant digits gives 2 50.4 g.m

DOUBLECHECK: This makes sense since EF is small.

21.74. THINK: Because this is a two‐dimensional problem, the directions of forces are important for determining a net force. SKETCH:


849


1 2 / .F k q q r The net force on 1q

is 1 2 1 312 13net 2 2

1 2

ˆ ˆ .k q q k q q

F F F x yr r

The direction of the net force is tan1Fy

Fx

.

SIMPLIFY: Not needed

CALCULATE:

9 2 2 9 2 2

net 2 2

9 9

8.99 10 N m / C (2.0 C)(5.0 C) 8.99 10 N m / C (2.0 C)(3.0 C)ˆ ˆ

3 m 4 m

ˆ ˆ9.99 10 N 3.37 10 N

F x y

x y

The magnitude of netF

is Fnet 9.992 3.372 109 N 10.5109 N. The direction of netF

is

tan1 3.37 109 N

9.99109 N

161.36with respect to the positive x‐axis, or 18.64 above the negative x‐

axis (the net force points up and to the left, in the II quadrant).

ROUND: Keeping only two significant digits yields 8 9net

ˆ ˆ1.0 10 N 3.4 10 NF x y and

Fnet 11109 N at 19 above the negative x‐axis.

21.75. THINK: To solve this problem, the force due to the charges and the tension in the string must balance the gravitational force on the spheres. SKETCH:

RESEARCH: The force due to electrostatic repulsion of the two spheres is 2 2 2

E 1 2 / / .F kq q d kq d

Applying Newton’s Second Law yields (I) EsinxT T F and (II) cos .yT T mg 30.45 m, 2.33 10 kg, 10.0 .L m


850

SIMPLIFY: Dividing (I) by (II) gives 2 2Etan / / .F mg kq d mg After simple manipulation, it is

found that the charge on each sphere is q d 2mg tan / k 2Lsin mg tan / k using 2 sin .d L

CALCULATE: Substituting the numerical values, it is found that

q 2 0.45 m sin10.0

2.33103 kg 9.81 m / s2 tan 10.0 8.99109 N m2 / C2 1.0464 107 C.

ROUND: Keeping only two significant digits gives q 0.10 C DOUBLECHECK: This is reasonable. The relatively small spheres and small distance will mean the

charge is small.

21.76. THINK: I want to find the magnitude and direction of the net force on a point charge 1q due to

point charges 2q and 3 .q The charges 1,q 2 ,q and 3q are located at (0,0), (2.0,0.0),

and (0, 2.00), respectively. SKETCH:

RESEARCH: The magnitude of the force between two charges is 21 2 / .F k q q r The net force on 1q

is 1 2 1 3net 12 13 2 2

1 2

ˆ ˆ .k q q k q q

F F F x yr r

SIMPLIFY: Not needed CALCULATE: Putting in the numerical values yields


851

9 2 2 9 9 9 2 2 9 9

net 2 2

5 5

8.99 10 N m / C (100. 10 C)(80.0 10 C) 8.99 10 N m / C (100. 10 C)(60.0 10 C)ˆ ˆ

2.00 m 2.00 m

ˆ ˆ1.798 10 N 1.348 10 N

F x y

x y

The magnitude of netF

is 2 2 5 5

net 1.798 1.348 10 N 2.247 10 N.F

The direction of netF

is

1 1.348tan 36.860 .

1.798

ROUND: Rounding to three significant digits, it is found that 5net 2.25 10 NF

and 36.9 below

the horizontal. DOUBLECHECK: Since both forces acting on 1q are attractive, it is expected that the direction of the net force would be between the two contributing force vectors.

21.77. THINK: If it is assumed that the third charge is positive, then the third charge experiences a repulsive force with 1q and an attractive force with 2 .q SKETCH:

RESEARCH: Because 1 2q q and the force between 1q and 3q is attractive, the possible region

where 3q can experience zero net force is in the region 0x . The net force on 3q is

1 3 2 3

net 2 2

3 2 3

.0

k q q k q qF

x x x

SIMPLIFY: Solving net 0F for 3x yields 223 2 1 2 3 implies: x q q x x

3 2 1 2 3 3 2 1 2 3(I) or (II) x q q x x x q q x x

Equation (I) gives 3 0x and equation (II) gives 3 0.x Therefore the correct solution is the solution

of Equation (II). Solving (II) yields 1 2

3

2 1

.q x

xq q

CALCULATE: Substituting 1 1.00 μC,q 2 2.00 μCq and 2 10.0 cmx into above equation gives

3

1.00 μC 10.0 cm24.142 cm

2.00 μC 1.00 μCx

ROUND: 3 24.1 cmx

DOUBLECHECK: The negative value of x indicates that 3q is located in the region 0x , as expected.

21.78. THINK: The electrostatic force on a bead is balanced by its gravitational weight. SKETCH:


852

RESEARCH: The repulsive force between two charged beads is 1 2E 2

.q q

F kd

Using Newton’s Second

Law, 1 2E 22

sinq q

F k m gd

.

SIMPLIFY: Therefore the distance d is 1 2

2

.sin

kq qd

m g

CALCULATE: Substituting the numerical values into the above equation gives

9 2 2 6 6

3 2

8.99 10 N m / C (1.27 10 C)(6.79 10 C)1.638 m.

3.77 10 kg 9.81 m/s sin 51.3d

ROUND: Keeping only three significant digits gives 1.64 m.d DOUBLECHECK: The beads are very light, so a small charge is sufficient to cause a relatively large

separation.

21.79. THINK: Since this is a two dimensional problem, electrostatic forces are added as vectors. It is assumed that AQ is a positive charge. SKETCH:


853

RESEARCH: To balance the forces 1F and 2F , the charge on 0Q must be positive. The electrostatic

forces on AQ are A A 0 A

1 2 02 2 21 2 0

, , and .k q Q k q Q kQ Q

F F Fr r r

Applying Newton’s Second Law, it is

found that 0 1 2x xF F F or 20 A 0 1 2/ cos cos .kQ Q r F F Using 1 2r r this becomes

A0 A2 2

0 1

2cos .k q QkQ Q

r r

SIMPLIFY: Solving the above equation for 0Q gives the charge 0 ,Q 2

0 0 1/ 2cos .Q r r q From the

above figure, it is noted that 2 2

0 2 2 2 2,r a a a 2 21 2 5,r a a a and

2 2 2 3 2 3cos cos(45 ) cos45 cos sin 45 sin cos 10.

2 2 2 5 105 5

a a

a a

Therefore the magnitude of charge 0Q is2

0 2

8 3 482 10 10 .

10 505

aQ q q

a

CALCULATE: Substituting 1.00 nCq yields 0

4810 1.00 nC 3.036 nC.

50Q

ROUND: Rounding to three significant figures gives 0 3.04 nC.Q

DOUBLECHECK: Since 0r is larger than 1,r it is expected that 0Q is larger than 2 2 nC.q


854

21.80. THINK: The two balls both have a mass of 0.681 kg.m The electrostatic force between two balls

is 2 2E / .F kq d The angle 20.0 . The charge on each ball is the same, 18.0 μC.q Find L.

SKETCH:

RESEARCH: Decompose the tension T into horizontal and vertical components. Newton’s Second Law on the left ball yields:

(I)2

E 2sinx

qT T F k

d and (II) cos .yT T mg

Use the two equations to eliminate T. From the sketch, 2 sin .d L Substitute for d, and solve for L. SIMPLIFY: Dividing the left hand side of (I) by the left hand side of (II) and equating the result to the quotient obtained by dividing the right hand side of (I) by the right hand side of (II) gives:

2

2tan .

kq

mgd Using 2 sin ,d L it is found that

2

2 2tan .

4 sin

kq

mg L

After simple rearrangement

the length L is 2

2.

4 sin tan

kqL

mg

CALCULATE:

29 2 2 6

2 2

8.99 10 N m / C 18.0 10 C1.6000 m

4 0.681 kg 9.81 m/s sin 20.0 tan 20.0L

ROUND: 1.60 mL

DOUBLECHECK: 1.60 meters is a realistic length for a string in this situation. The units of meters are appropriate for a measurement.

21.81. THINK: The net force on a point charge is a sum of two repulsive forces due to interaction with 1q and 2 ,q are positive, this means that the location of zero net force is located in between 1q and 2 .q The values given in the question are:

1 3.94 μC,q 1 4.7 m,x 2 6.14 μC,q 2 12.2 m,x and 3 0.300 μC.q SKETCH:

RESEARCH: The net force on 3q is given by

1 3 2 3net 13 23 2 2

3 1 2 3

.kq q kq q

F F Fx x x x

SIMPLIFY: Solving net 0F for 3x gives 2 2

3 1 2 1 2 3 .x x q q x x There are two solutions of this

equation:


855

(I) 3 1 2 1 2 3x x q q x x and (II) 3 1 2 1 2 3 .x x q q x x

The possible solutions for 3x are

(I) 1 2 2 13

1 2

q x q xx

q q

and (II) 1 2 2 1

3

1 2

.q x q x

xq q

The correct solution is the first solution since the two original charges have the same sign, and therefore the point where the forces balance is between 1x and 2 .x

1 2 2 13

1 2

q x q xx

q q

CALCULATE: Substituting the numerical values into the above equation gives

3

3.94 μC 12.2 m 6.14 μC 4.7 m2.817 m

3.94 μC 6.14 μCx

ROUND:

3 2.8 mx

DOUBLECHECK: The total distance b 1x and 2x is 16.9 m, and the point 3x is between 1x and 2x but closer to the weaker charge.

Chapter 22: Electric Fields and Gauss’s Law

Chapter 22: Electric Fields and Gauss’s Law InClass Ex ciseer s

22.1. c 22.2. e 22.3. (a) True (b) False (c) False (d) True (e) True 22.4. c 22.5. e 22.6. a 22.7. a 22.8. c 22.9. d

Multiple Choice

22.1. e 22.2. d 22.3. a 22.4. a 22.5. d 22.6. c 22.7. c 22.8. c 22.9. a 22.10. a & d

Questions

22.11. The metal frame and sheet metal of the car form a Faraday cage, excluding the electric fields induced by the lightning. The current in the lightning strike flows around the outside of the car to ground. The passengers inside the car can be in contact with the inside of the car with no ill effects, but should not stick their hands out an open window.

22.12. Since lightning can strike the tree and have the current flow through the wet tree, the current would jump to any object near the tree. To avoid lightning, go inside the house or a car. If I were outside, I would go to a low place and avoid trees or tall buildings. I should not lie down on the ground since the current can flow along the surface of the Earth.

22.13. If electric field lines crossed, there would be a charge at the crossing point. It is known that the electric field lines extend away from a positive change and the lines terminate at a negative charge. If in the vicinity of the crossing point there is no charge, then the lines cannot cross. Moreover, if we put a test charge on the crossing point, there would be two directions of the force; this is not possible; therefore the lines cannot cross.

22.14. The net flux through a closed surface is proportional to the net flux penetrating the surface, that is, the flux leaving the volume minus the flux entering the volume. This means that if there is a charge within a surface, the flux due to the charge will only exit through the surface creating a net flux no matter where the charge is located within the surface. If a charge moves just outside the surface, then the net flux crossing the surface would be zero since the flux entering the volume must be equal to the flux leaving the volume as shown in the figure:

22.15. Because of the spherical symmetry of this problem, Gauss’s Law can be used to determine electric fields. The image below shows a cross‐section of the nested spheres:

855


856

plied

Gauss’s Law is ap on four surfaces, G , G2 , G and G4 as shown in the figure.

ng re. 1 3

(a) In the region 1 < ,r r the electric field is zero because it is inside the conducti sphe

(b) Applying Gauss’s Law on the surface G gives the electric field in the region i.e., 2 1 ,r r r

204 3 /E r Q / 40r

2 .

ince it is inside a c

or r

E 3Q

2

(c) In the region r2 3 , the electric field is zero s onductor.

(d) In the region using Gauss’s Law yields

r

r r3 , 204 3 /E r Q . Therefore, the electric field is

E

22.16.

3Q / 40r2 .

(a) If you are very, very close to the rod, then the field is approximately constant since the surface of the rod appears large. Thus, the electric field is constant. (b) If you are a few centimeters away from it; then the electric field can be approximated by the field produced by a very long rod, that is E is proportional to 1/ .r (c) If you are very far away from it, then the electric field behaves as if it is from a point charge. herefore, the field is proportional to T

22.17.

21/ .r

The total electric flux through a closed surface is equal to the net charge, encq , divided by the

constant 0 or enc 0net / .q This is known as Gauss’s Law. The strength of a dipole is p qd.

Because the dipole is completely enclosed by the spherical surface, the enclosed charge will be Thus the net flux through the closed surface will be zero.

q

22.18

enc q q 0.

.


Consider two small elements dx at x and x as shown in the above figure. Due to the symmetry of the problem, it is found that the component of 1E in the y‐direction, 1y ,E is equal in magnitude, but

in the opposite direction, to the ycomponent of 2 .E Therefore, only the x‐components of electric

fields contribute to the net field. Integrating over the length of wire yields a

20 4E x

Using

0

2sin ˆ .dq

r

,dq dx it simplifies to

E d x

q a

200

.2 r

Substituting

sin 2 2r x y and sin /x r

yields

a

3/2.

xdx0 2 2

0

ˆ

2

xE

x y

2z x Using the substitution yields:

2 22 1/2 1/2a 2 2

200 00 0 0

2 2

ˆ ˆ ˆ12 / 2 /

2 2 2 2

ˆ

a ax dz x xE z y z y

z y

0

1/ 1/ .2

x y a y

22.19. Since the conductor has a negative charge, this means that the electric field lines are toward the conductor. Electrons inside the conductor can move freely and redistribute themselves such that the repulsion forces between electrons are minimized. As a consequence of this, the electrons are distributed on the surface of the conductor.

22.20. St. Elmo’s Fire is a form of corona discharge; the same phenomenon whereby lightening rods bleed off accumulated ground charge to prevent lightening strokes. Lightning rods are not supposed to conduct a lightning strike to ground except as a last resort. In stormy weather, a ship or aircraft can become electrically charged by air friction. The charge will collect at the sharp edges or points on the structure of the ship or plane because the electric field is concentrated in areas of high curvature. Sufficiently large fields ionize the air at these areas, as the molecules of nitrogen and oxygen de‐ionize they give off energy in the form of visible light. The ghostly glow known since the

857

days of “wooden ships and iron men” is St. Elmo’s Fire.

22.21. Consider the surface layer of charge to be divided into two component; a ‘tile’ in the vicinity of some point, and the ‘rest’ of the charge on the surface. Seen from close enough to the given point on the surface, the ‘tile’ appears as a flat plane of charge. Gauss’s Law applied to the cylindrical surface pierced symmetrically by such a plane, implies that the ‘tile’ produces an electric field with the


858

component 0/ 2 perpendicularly outward from the surface on the outside, inward on the inside. But Gauss’s Law applied to a short cylinder (‘pillbox’) partially embedded in the conductor, implies that the entire charge layer produces an electric field with component / perpendicularly outward outside the surface, and zero inside. To yield this result, the ‘rest’ must produce electric field 0/ 2 , outward, in the vicinity of the ‘tile’ inside and out. It is this electric field which exerts force on the ‘tile’, carries charge per unit area . Hence, every portion of the charge layer

experiences outward force per unit area stress of magnitude 2 / 2 .0 Note that the outward direction of the stress is independent of sign of .

22.22. The net force on the dipole is zero, so there will be no translational motion of dipole. The net orque; however, is not zero, so the dipole will rotate. With the force on the positive charge to the ight and the force on the negative charge to the left, the dipole will rotate counter‐clockwise. tr

Problems

22.23. The electric field produced by the charge is:

9 2 2 9

4

2 2

8.99 10 N m /C 4.00 10 C57536 N/C 5.75 10 N/C.

0.0250 m

kqE

r

22.24.

The electric field vector will be i

2 2 21 2 1

i

/ / /E E kq r x kq r y k r q x q y 2 .

The magnitude

of the vector is:

9 2 2 2 22 2 2 2 9 9

2 2

8.99 10 N m /C1.6 10 C 2.410 C 25.931 N/C 26 N/C.

1.0 mx y x y

kE E E q q

r

22.25.

The electric field at the origin is 2 2i 1 1 2 2

i

ˆ/ /E E k q r x k q r y .

The direction is

ta n / .y xE E


22 22 2 1 21 1 1 1

0 21 1

tan tan tan/ .0xE k q r

The elect

2 22 1

/ 4.000 m 24.00 nCtan 12.53 .

6.000 m 48 0 nC

yE k q r r q

r q

ric field lies in the 3rd quadrant so 0180.00 12.53 192.53

22.26. THINK: The electric field is the sum of the fields generated by the two charges of the corner

triangle. The first charge is 51 1.0 10 Cq and is located at 1 0.10 m .r y

The second charge is

2 Cq located at

50

SKETCH:

1.5 1 0.20 m .r x2

RESEARCH The electric field is given : by the equation 2/ .E kq r r

SIMPLIFY: The magnitude of the field is 21 1 2 2/ /E kq r y kq r x

2 .

2 2

2 2 1 22 2

1 2

,x y

q qE E E k

r r

and has a direction 1 1 2 2 1 21 1 2 2 2 1 1 2tan / tan / / / tan /y xE E q r q r r q r q 2 where is in the

second quadrant.

CALCULATE:

2 25 5

9 2 2 6

2 2

1.0 10 C 1.5 10 C8.99 10 N m /C 9.6013 10 N/C

0.100 m 0.200 mE

2 5

1

2 569.444

0.100 m 1.5 10 C

or 110.56 .

0.200 m 1.0 10 Ctan

ROUND: The least precise value given in the question has two significant figures, so the answer should also be re at the corner is

aported to two significant figures. The electric field produced t 110 from the 69.6 10 N/CE x ‐axis.

DOUBL

859

ECHECK: Dimensional analysis confirms the answer is in the correct units.

22.27. THINK: We want to find out where the combined electric field from two point charges can be zero. Since the electric field falls off as the inverse second power of the distance to the charge, and since both charges are located on the x‐axis, only points on the same line have any possible chance of canceling the electric field from these two charges, resulting in a net zero electric field. The first charge, 1 5.0 Cq , is at the origin. The second charge, 2 3.0 Cq , is at 1.0 m.x Let’s think about in which region of the x‐axis it can be possible to have zero electric field. On the sketch we have marked three regions (I, II, and III). If we place a positive charge anywhere in region 2, then the 5 C will repel it and the ‐3 C will attract it, resulting in the positive charging moving towards the right. If we place a negative charge into the same reason, it will always move to the left. So we know that the electric field cannot be zero anywhere in region II. Region I is closer to the 5 C charge. Since this is also the charge with the larger magnitude, its electric field will dominate region I, and thus we cannot ha e any place in region I where the electric field is 0. This leaves only region III, where the two elect fields from the point charges can cancel.

vric

SKETCH:


860

RESEARCH: The electric field due to the charge at the origin is The other charge

produces a

E0 kq0 /x 2 .

field of E1 kq1 / x x1 2 .

SIMPLIFY: The combined electric field is E kq0 /x 2 kq1 / x x1 2 . Setting the electric field to zero,

olve for :s x kq

0

x 2 1

x x1 2kq

0 0

x 2

kq 1

x x1 2kq

x x1 2 q0 x 2q1 x x1 2 q0 x 2 q1

We could now solve the resulting quadratic equation blindly and would obtain two solutions, each of which we would have to evaluate for validity. Instead, we can make use of the thinking we have done above. In the last step we used the fact that the charge at the origin is positive and the other is negative, replacing them with their absolute values. Now we can take the square root on both sides, leaving us with

x x1 q0 x q1 x x1 q0 / ( q0 q1 )

CALCULATE: x 1.0 m 5.0 C / ( 5.0 C 3.0 C ) 4.43649 m ROUND: The positions are reported to two significant figures. The electric field is zero at x 4.4 m. DOUBLECHECK: This is a case where we can simply insert our result and verify that it does what it is supposed to: E(x=4.4 m) k(5 C)/(4.4 m)2 k(3 C)/(4.4 m 1 m)2 0 .

22.28. THINK: Let’s fix the coordinate notation first. The charges are located at points (0,d), (0,0), and (0,‐d) on the y‐axis, and the point P is P = (x,0). In order to specify the electric field at a point in space, we need to specify the magnitude and the direction. Lets first think about the direction. The distribution of the charges is symmetric with respect to the x‐axis. Thus if we flip the charge distribution upside down, we see the same picture. This means also that we can do this for the electric f ld generated by these charges. Right away this means that the electric field anywhere on the x‐axi cannot have a y‐component and can only have an x‐component.

ies

SKETCH:

RESEARCH: The electric field strength is given by , and the electric fields from different charges add as vectors. We need to add the x-components of the electric fields from all charges. They are (from top to bottom along the y-axis):

E kQ / r2


E1 kq

d 2 x 2

E1,x kq

d 2 x 2

x

d2 x 2

kqx

d 2 x 2 3/2

E2,x 2kq

x 2

E3 E1 kq

d 2 x 2

E3,x E1,x kqx

d 2 x 2 3/2

SIMPLIFY: All we have to do is add the individual x-components to find our expression for the x-component of the electric field along the x-axis:

Ex (x ,0) E1,x E2,x E3,x 2kq

x 2

2kqx

d2 x 2 3/2 2kq1

x 2

x

d 2 x 2 3/2

(This is the expression for x>0; for x<0 it has the opposite sign so that it always points away from the origin.)

CALCULATE: Not applicable. ROUND: Not applicable. DOUBLE-CHECK: For x 0 we see that the first term diverges as we get very close to the positive charge at the origin, which is as expected. For large distances, we expect at most a very weak electric field because the net

charge of our configuration is 0. We can factor out the 1

x , d / x 0,

/x2 term to get

Ex (x ,0) 2kq

x 21

x 3

d2 x 2 3/2

2kq

x 21

1

d

x

2

1

3/2

2kq

x 21

d

x

2

1

3/2

.

For

the binomial expansion gives us

d2 /x 2 = 1,

d 2

x 21

3/2

13

2

d 2

x 2.

The electric field then simplifies to

Ex (x ? d ,0)

2kq

x 21 1

3

2

d 2

x 2

2kq

x 2

3

2

d 2

x 2

3kqd2

x 4.

Thus the electric field strength of this configuration, called a “quadrupole”, falls with the inverse fourth ower of the distance to the origin for large distances. (… as compared to the electric field from a dipole, hich falls with the inverse third power).

pw

22.29. The dipole is just two charges fixed together of opposite sign. The electric field at a point is the sum of the fields produced by each charge. The figure indicates that the electric field produced is created by the component of the field perpendicular to line x.

861


862

2 22 2iy 2y 1 2

2 2d x

+ sin sin / / 2 sin / / 2 sin

2 sin

/ 2

E E E E E kq d x kq d x

kq

Note that

2 2sin / 2 / / 2 .d d x This means the field is:

3/2 3/22 2 22 2 22 / 2 / / 2E

d x d d x

If

3/2

2.

2

kqd kqd kp

x

x d then . The field along

the axis of the dipole is 2E indicating that the field strength falls off more rapidly perpend

3/E kp x

3/ ,kp x

icular to the dipole axis.

22.30. THINK: The field due to a dipole moment at a point h along the x‐axis is 3( ) 2 /E h k qd h . I want to find the p int perpendicular to the x‐axis as measured from the origin (i.e., along the y‐axis), where the electr field has this same value.

oic

SKETCH:

RESEARCH: From the previous problem, the electric field along the y‐axis is 2

3/22

( ) ./ 4

kqdE l

d l

Set and solve for l . ( ) ( )E l E h

SIMPLIFY:

2

2 23/2

2 343 3/2 3 3/2 32

4

2 2 1 2

22d

d

k qd qd h dl h l

l

2 224

dh hl

CALCULA ble.

. k

TE: Not applica

ROUND: Not applicable.

DOUBLECHECK: According to this expression, l will always be less than h. This is consistent with the previous result that the electric field strength along a line perpendicular to the dipole axis falls off more an the fie rapidly th ld strength along the dipole axis.

22.31. THINK: As the 4.0 gm ball falls the force of gravity acting on it will cause it to accelerate downwards. At the same time, the force due to the electric field acts on the ball causing it to accelerate towards the east. The forces act perpendicular to each other. The problem is solved by finding ea h component of the velocity. In order to find the velocity due to the electric field, the time requ ed for the ball to travel 30.0 cm downwards is needed.

cir

SKETCH:


RESEARCH: The velocity in the downward direction is found using 2 20 2 .y yv v gdy The time it takes

to reach this velocity t The acceleration eastward is calculated using / .yv g .F ma qE The

velocity is then .x xv a t

SIMPLIFY: The y‐component of the velocity is 2yv gdy because the ball starts from rest. The

time it takes for the ball to fall 30.0 is

cm 2 /t gdy .g The acceleration eastward is The

velocity eastw is

/ .a qE m

ard / 2 / / 2 / .dy g

CALCULATE: v

x E m gdy g qE m v q

y 2 9.81 m/s2 0.300 m 2.4261 m/s downward

3

2

12 N/C 0.300 m10 C 2 3.7096 m/s

0.0040 kg 9.81 m/s5.0xv

eastward

ROUND: The ve city of locity is report to two significant figures. The ball reaches a velo ˆ ˆ3.7 m/s /s .2.4 mx y

DOUBLECHECK: This is a reasonable answer considering the size of the values given in the question

.

22.32. THINK: A line of charge along the y‐axis has linear charge density from 0 to y y a , and from 0 to .y y a I want to find an expression for the electric field at any point x along the x‐axis. It is oted that the charge configuration is similar in structure to a dipole. By symmetry, the xcompone s of the field cancel out, and the net field is in the ydirection.

nnt

SKETCH:

RESEARCH: The electric field resulting from a charge distribution is the integral over the differential charge: 2/ .dE kdq r The y‐component of the field is sin / ,ydE kdq r where 2 is the angle

between the electric field produced by and the y‐axis. Also, dq 2 2 ,r x y si / .n y r From

0 to , a d ,q dy and from 0 to , a dq .dy

SIMPLIFY: , 2 2 2 2 2 2 2

siny

kdq k dy y k ydydE dE

r x y x y x y

3/2

863


864

, 2 2 2 3/2

2 2siny

kdq k dy y k ydydE dE

y

2 2r x y x y x

The field due to the positive charge distribution is:

+ 3/2 3/20 02 2 2 2.

a ak ydy ydyE k

x y x y

Similarly,

the field due to the negative charge distribution is: 3/2 3/20 02 2 2 2

.a ak ydy ydy

E kx y x y

CALCULATE: Let then 2u x y 2 2du ydy then:

1/2

2 23/2 1/2 1/2 1/20 2 200 0

2 1,

2 2

aa aa kk du k k

E kx y xu u u x a

1

and

1/22 23/2 1/2 1/2 1/20 220 0

2 1.

2 2

aaa kk du k k

E kx y xu u u x a

1

The total

electric field at x is: + - 1/22 2

1 1+ 2E E E k .

xx a

ROUND: Not applicable. DOUBLE CHECK: The electric field decreases inversely proportionally to the distance from the wire, as e

xpected.

22.33. THINK: A semicircular rod carries a uniform charge of +Q along its upper half, and –Q along it’s lower half. I want to determine the magnitude and direction of the electric field at the center of the semicircle. The rod has a length of .L R The charge density of the upper half of the rod is

. / / 1/ 2 2 /Q L Q R Q R Similarly, the lower half of the rod is 2 / .Q R

SKETCH:

RESEARCH: From the symmetry of the semi‐circle, the x‐components of the field cancel, and the resulting electric field only has a y‐component. The y‐component of the ele tric field for the upper segment of the rod is given by

c

2 2cos / cos / cdE kdq R k dx R+y osdE ,

where .dx Rd Therefore, 2, / cos cos /ydE k Rd R k d R .


SIMPLIFY: Integrating both sides gives:

/2 /2 2+y 00

/ cos / sin | / 1 0 / 2 / .dE k R d k R k R k R k Q R

The lower half of the semicircle also contributes the same y‐component. The tot l electric field at the origin is

a

2 20 04 / 4 / 4 /k Q R y Q R y Q R y

2 .

+ -y y y yE

TE: Not

+ 2y y

E E

CALCULA

E

applicable. ROUND: Not applicable.

DOUBLE CHECK: The resulting field points in the direction from the positive charge to the negative charge, as required.

22.34. THINK: Two semicircular rods, with uniformly distributed charges of 1.00 μC and 1.00 μC, respectively, form a circle of radius 10.0 cm.r I want to determine the magnitude and direction on the electric field at the center of the circle. SKETCH:

RESEARCH: The charge densities of the positively charged and negatively charged rods are

d - / ,Q R/ anQ R respectively. The differential element of the electric field is given by 2/ ,dE kdq R where the differential element of charge along the line is dq .dx Rd It is also

necessary t consider the x‐ and y‐components of the differential elemo ents.

SIMPLIFY: , 2 2 3 2R R

sin sin sin sinx

dx k d kQRd kQ d

R

kdE

R

. Similarly, ,ydE2

cos;

kQ

R

d

, ,2

sin cos; .x y

kQ d kQ ddE dE

R 2R

Integrating both sides of each expression gives:

, 2 2 00

, 2 2 00

2 2

, 2 2

2 2

2

2

2

2sin cos

cos sin 0

2sin cos

sin 0

x

y

x

kQ kQ kQE d

R R RkQ kQ

E dR RkQ kQ kQ

E dR R R

kQ kQ

R

, 2cosyE d

R

The total electric field at the center is given by:

, , , , 2 2

2 2 40 0x y x y

kQ kQ kQE E E E E

R R 2

.R

865


866

CALCULATE:

9 2 2 6

6

21.1446 10 N/

0.100 mE

ROUND: The elec all of the y‐ components

4 8.99 10 N m /C 1.00 10 CC

tric field is reported to three significant figures: 61.14 10 N/C.E Because are zero, the resultant field is in the positive x‐direction.

DOUBLE iCHECK: Given the symmetry of the charge configuration, this s a reasonable result.

22.35. THINK: The charge Q is uniformly distributed along the rod of length L. The rod has linear charge density / .Q L The electric field at a position x = d can be calculated by integrating over the differential electric field due to the differential charge on the rod. The electric field differential

2/ ,dE kdq r where the differential is along the y‐axis, and 2 2 .R d y The x‐ and y‐components

of the field must be considered individually. The x‐component of the field differential is given by cos ,xdE dE and the y‐component is given by sin .ydE dE

SKETCH:

SIMPLIFY: 2 2 2 2

2 2cos cos cos cosx

k dy kQdy kQdykdQdE

R R LR L d y

2 3/22 2 2 22 2 2 2

3/22 2 2 2

cos

x

y

kQdy dkQdyd d ddE

R d y d yL d y L d y

kQdy y y ykQdydE

R d y L d y

22 2sin ; sinydE

L d y

Integrate both expressions.

3/2 3/2 2 2 20 02 2 2 20

2 23/2 3/20 02 2 2 20

1

1

LL L

x

LL L

y

ydkQdy dkQ kQdE dy

L L d d yL d y d y

ykQdy ykQ kQE dy

L L d yL d y d y

ˆ ˆ( ) x yE d E x E y

CALCULATE: 2 2 2 2 2 2 2 2

0

0

L

x

ykQd kQd L kQE

L Ld d y d d L d d L


2 2 2 2 2 20

1 1 1L

y

kQ kQ kQ kQE

L L d dLd y d L L d L

2 2x

dLd L

ROUND: Not appl

2 2L d L icable.

DOUBLE le

ˆ( )

kQ kQ kQE d y

d

CHECK: The magnitude of the e ctric field decreases as d increases, as expected.

22.36. THINK: A wire bent into an arc of radius R and carrying a uniformly distributed charge Q will have a linear charge density of / 2 .Q R By the symmetry of the charge distribution, the y‐components cancel, and only the x‐component of the charge contributes to the electric field. SKETCH:

RESEARCH: An electric field produced by an infinitesimal segment of the arc is

2 2 2/ / / / .dE kdq R k dx R k Rd R k d R The total electric field can be calculated by integrating over the differential elements of the field. Since the y‐component of the field is zero,

cos .x

k dE E

R

SIMPLIFY:

2

2

2 2

2

cos cos cos sin sin sin2 2 2

sinsin sin

2

k k kQ kQ kQE d d d

R R R R R

kQ kQ

R R

CALCULATE:

DOUBLE CHECK: As 0ROUND: Not applicable.

the field is the same as that of a point charge and becomes zero as the

point is enclose by a shell of charge. This is because 0 2 2 0 2

sin sinlim lim .

kQ kQ kQ

R R R

22.37. THINK: The washer will create an electric field that should be not to different from the electric field of the thin ring of charge we encountered in Solved Problem 22.1. The washer has a total charge

7.00Q with inner and outer radius of the washer are ri nC, 2.00 cm and The electric

field at 30.0 cmz away from the center of the washers is desired.

ro 5.0 cm.

o SKETCH:

867


868

RESEARCH: The surface density is /Q A where the area is 2 2

o i .A r r

2/ .dE kdq R

The field will point in

along the z‐axis due to symmetry. The field due to a segment is The distance from the

segment of charges is 2 2ox z R and 2 2

o ocos / .z x z

SIMPLIFY:

o o o

i i i

2 2 oo o2

cos .ER

Evaluating the single integral gives:

3 3/23 2 2 2 2o o 2 2o i o i o

2r r r

r r r

kQz x dxd x dxk dAz kQzkdq

R R r r r r x z

o

i

o o o2 2 1/2 1/22 2 2 2 2 2 2 2 2 22 2 2 2

oo i o i o i i o o oo o i o

12 2 21 1 1 1.

r

r

kQz kQz kQzE

x zr r r r r r r z r zr z r z

CALCULATE:

E 2 8.99 109 N m2 /C2 7.00 109 C 0.300 m

0.0500 m 2 0.0200 m 21

0.0200 2 0.300 2

1

0.0500 2 0.300 2

1

m

682.715 N / CROUND: The values are given to three significant figures. The electric field is pointing towards

E 6.83102 N / Cthe positive z‐axis.

DOUBLECHECK: In Solved Problem 22.1 we found for the thin ring: E kQ . Using

the average of our outer and inner radius we then find from this formula:

zo / r2 z02 3/2

E 8.99109 N m2 /C2 7.00109 C 0.300 m

( 0.0350 m 2 0.300 m 2 )3/2 685.2 N / C

ince this is fairly close to our result for a ring with finite thickness, we have added confidence in ur result. So

22.38. The force on the particle is .F qE The charge is 2q e so the force is

19 3 152 1.60 10 C 10.0 10 N / C 3.20 10 N.E 2F qE e

22.39. The torque due to the field is

15 3 3

153.46

sin sin 5.00 10 C 0.400 10 m 2.00 10 N / C sin60.0p E pE qdE

10 N m.

22.40. The maximum torque occurs when the dipole is perpendicular to the field. The electric field is

30 28sin 1.05 D 3.34 10 C m/D 160.0 N/C sin90 5.61 N m.p E pE

22.41. The force acting on the electron is .

10

F ma qE The acceleration is then / .a qE m Assuming the electron is moving in the same direction as the electric field, the acceleration will oppose the

motion. The velocity is given by / .v v Solving this equation for 2 20 0ax v 2 2

02 2 / 2qEx m v eEx m


x: 20 or 22 E/e m x v v 2 2

0 / 2 E.x m v v e

The distance traveled is

26 kg 27.5 10 m / s

11400 N / C

31 20 m.

192 1.602 10 Cx

189 m before i

p q

9.109 10

0.189

The electron travels 0. t stops.

22.42. The dipole moment is .d ed The torque experienced by the dipole is

19 9 3 25E Esin Esin 1.602 10 C 0.68 10 m 4.4 10 N / C sin 45 3.4 10 N m.p p ed

22.43. THINK: The net force on falling object in an electric field is the sum of the force due to gravity and the force due to the electric field. If the falling object carries a positive charge, then the force on the object due to the electric field acts in the direction opposite to the force of gravity.

SKETCH:

RESEARCH: The net force acting on the g e . object is F Ma F F Mg QE

This corresponds to an acceleration of / .a g QE M Recall that the velocity of an object in free fall

is given by 2 .v ah SIMPLIFY:

(a) 2v

a g 2 2 /2

QEv ah v h g QE M

h M

(b) If the value /g QE M isal an

CALCULA

less than zero, then the argument of the square root is negative. This means the d the body does not move. value is non‐re

TE: Not applicable. ROUND: Not applicable. DOUBLE CHECK: Dimensional analysis confirms that the units of the expression reduce to m/s, the correct units for velocity.

22.44. THINK: The force in between the charge and the dipole moment is equal to the force acting on each pole of the dipole. The dipole moment is 30 and is 1.00 cmr6.20 10 C mp from the charge

1.00 μC.Q SKETCH:

RESEARCH: The force due to an electric field is ,F qE r

where the electric field is

2/ .E r kQ r r

869


870

SIMPLIFY: The total force is .F qE r r qE

r From the fundamental theorem of calculus,

2 3

22ˆ ˆ( )kpQd d kQ

3ˆF q r E r p r pkQ r r

dr dr r r r

CALCULATE:

9 2 30 6

19

30.0100 m

F 2 8.99 10 N m / C 6.2 10 C m 1.00 10 C

ˆ 1.11476 10 Nr

ROUND: The force is reported to 3 significant figures. (a) The force between the dipole and the charge is 191.11 10 N. (b) The molecule of the sign of the charge of oppo ser to the cha

is attracted to the charge regardless the charge. This occurs because site sign on the dipole will move clo rge creating an attractive force.

DOUBLECHECK: The mass of a water molecule is 263.01 10 kg, meaning the force is relatively large. To view a dipole attracted to a charge, place a charged rod or comb near running water from a faucet.

22.45. THINK: Assuming that the wire is made of a conducting material, the charges will be uniformly distributed over its length. The wire will produce an electric field. This field in turn produces a force on a roton, causing the proton to accelerate. The wire has a length of 1.33 mL and a total

charge of 63.05 10 .e The proton is 0.401 mx

p

Q away from the center of the wire. SKETCH:

RESEARCH: The linear density of the wire is / .Q L Due to the symmetry around the center of the wire the field produced is only along the x‐axis. The electric field due to a segment of charge is

2/ cos .dE kdq R The distance from the charge to the segment of the wire is 2 2 .R x y The

force on the proton is .F ma q

SIMPLIFY: The electric field is:

E r

/2/2 /2

2 2 3/2 2 2/2 /2 2 2/2

1/2 1/2 1/2 1/22 2 2 2 2 2 2 2

cos

/ 2 / 2

2

/ 4 / 4 / 4

LL L

L L

/ 4

L

yq k dy dyxx k x

RR R x y xx y

k k L k QL L

x x L x x x x L

k dE k

x L L

The acceleration of the proton is

1/22 2

E.

/ 4

q k q Qa

m mx x L

CALCULATE:

9 2 6 19

1/22 2

8.99 10 N m / C 3.05 10 1.602 10 C0.0141062 N/C

0.401 m 0.401 m 1.33 / 4E

19

2

27

1.602 10 C0.0141062 N/C 1351561 m/s

1.672 10 kga


RO NU nt fig rD: The values are reported to 3 u significa es. (a) The electric field produced by the wire at 0.401 m from its center is 0.0141 N/C.

(b) The acceleration of the proton is 6 21.35 10 m/s . (c) The force is a ton is positively charged. The force points t

ttractive since the wire is negatively charged and the proowards the wire.

DOUBLECHECK: These are reasonable answers with appropriate units.

22.46. The flux through a Gaussian surface is the sum of the total charges within the surface divided by the permittivity of free space 0 . i 0 03 2 7 / 3 / .Q q q q q q

22.47. The sum of the flux through each surface is equal to the charge enclose divided by 0 . 0/ .ii

Q

The charge is then

12 2 2 20 8.85 10 C / N m 70.0 300.0 300.0 300.0 400.0 500.0 N mi

i

Q 81.00 10 C.

22.48. The Gaussian surface of the sphere gives 2 2enc 0 04 / 4 /EdA E R Q R

or 0/ ,E which

is the same result from the Gaussian sphere. The electric field is 0 /kQ R the same as a point charge at the center of the sphere.

2 20/ /E Q 4 R

9 2 6

6

20.15

8.99 10 N m / C 6.1 10 C2.4 10 N/C

m

22.49. The cube does not contain any charges, th must be zero.

E

us the total flux 0,A B C D E F i

i

A E E E E E E and therefore,

15.0 N 25.0 N/C 20.0 N/C

60 N/C.

F A B C D EE E E E E E

/C 20.0 N/C 10.0 N/C

he field on the face F is 60.0 into the face of the cube.

T

22.50

N/C

.

The charge inside the sphere induces a charge of 3e on the inside surface of the sphere. The 3e charge must come from somewhere. In this case the 3e charge is removed from the outer surface charge. The outer surface charge is then 2 .e The total charge within the material of the sphere is

5 .e

22.51. Gauss’s Law states that enc 0/ .EdA q

The integral over the sphere gives 2enc 0/ 4 .q E R The

electric field outside a uniform distribution of charge is identical to the field created by a point charge of the same magnitude, located at the center of the distribution. Since the radius of the balloon

871

never reaches ,R the charge enclosed is constant and the electric field does not change.

22.52. THINK: The charges on the surface of the shell may be found using Gauss’s Law. The inner and outer radius of the shell is i 8.00 cmr and o 10.0 cmr respectively. The electric field at the surface o he inner radius is 80.0 N/C and points towards the center of the sphere. The electric field at th surface of the outer radius is 80.0 N/C pointing away from the center of the sphere.

f te

SKETCH:


872

RESEARCH that: Gauss’s Law states enc 0/e q .EdA

SIMPLIFY: The charge equals 2enc 0 0 4 .q EdA E R

A Gaussian surface inside the sphere

gives the charge at the center of the sphere. This is also equal to the total surface charge at the inner

radius of the conductor. 2i o i i4q E r or 2

i o i i4 .q E r The Gaussian surface around the

whole sphere contains the charge in the sphere and the charge of the shell. 20 sheq q

The charge of the shell equals the sum of the inside and outside charge of the sphere or

ll o i o4 .E r

i 0 shellq q q

0 shell i .q q q

0 shell i shell 0 0 04 .q q q q q E r

CALCULATE: 212 2 2 11i 8.854 10 C / N m 80.0 N/C 4 0.0800 m 5.6966 10 C

212 2 2 11o 8.854 10 C / N m 80.0 N/C 4 0.100 m 8.9010 10 Cq

ROUND: Roundin

q

g to three significant figures, the inside and outside total charges over the surface of the sphere are 115.70 10 C and 118.90 10 C, respectively. DOUBLECHECK: These are reasonable answers with appropriate units. As you would expect, the ratio of the charges is the ratio of the square of the radii: 2 28 5.70 :81 .: 0 .90

22.53. THINK: The electric field at various points can be found using Gauss’s Law. This law can also be used to find the charge on the outside surface of the conductor. There is a 6.00 nCq charge at

the center of the sphere. The shell has inner and outer radii of i 2.00 mr and o 4.00 mr respectively. The shell has a total charge of 7.00 nC.Q SKETCH:

RESEARCH: LGauss’s aw states that enc 0/ .EdA q

SIMPLIFY: The electric field of charges with spherical symmetry are given by 2

enc 04EdA E r q /

or 2enc 0/ 4 .E q r The electric field at 1 ir r is

1 04 .E r The electric field inside any conductor is always zero: 2enc 0 1/ 4 /q r q 2

1r 2 0E r

where i 2 o The electric field outside of the conductor is 3 o .r r

2 23 enc 0 3 0 3/ 4 / 4 .E r q r Q q r Because the field inside the conductor must be zero, Gauss’s Law

indicates that the charge at the center of the shell is the mirror to the charge on the inside of the shell: E r

.r r r

2 enc0 / 4q 2 oq0 3r q 2 0 2/ 4 r iq or .q The charge on the sphere is equal to the sum


of charges on the inner and outer surfaces of the shell i o .q q Q Thus, the outer surface charge is

i o/ 4Q q r Q 2 2o o o/ 4 / 4 .q r q r2

1 1.00 mr

CALCULATE: The electric field at

is

9

1 212 2 2

6.00 10 C53

.00 m .951 N/C

4 8.85 10 C / N m 1E

. The electric field at is

2 0 N/C.E The electric field at 3 5.00 mr

2 3.00 mr

is

9 9

2

C

5.00 m

12 2

7.00 10 C

.85 10 C 2

6.00 100.3597 N/C.

/ N m

4E

The surface charge on the outside part of the shell is

8

9 9

12 2

24.974 10

4.00 m

U ave an accuracy of o e s gnificant figure. .

7.00 10 C 6.00 10 C C/m .

4

i is

RO ND: All the values h n(a) The electric field at N/1 1.00 mr 54.0 C

m(b) The elect ric field at 0 is

4.9

2

(c)

3.0 0 Nr /C.

The electric field at 5.00 mr is 0.360 N/C.

(d) The surface d 3

ensity on the outside surface is e r

a

12 210 C/m .7 DOUBLECHECK: These ar easonable results.

22.54. Inside the sphere of radius , the charge density is 3tot tot/ / 4 / 3Q V Q a and is zero anywhere

else. Gauss’s Law states enc 0/ .EdA q

The area of the Gaussian surface is always taken to be 24A r and points radically. Thus, enc 0/EdA q

gives

2

3 3r Qr a

2enc / .r kq r r The enclosed charge is then

3 3enc / 4 4 / 3 /q V Q a if .r a

encE q 0 enc/ /A r q 04 r

Otherwise, the surface encloses the whole charge

.Q The electric field is then 3kQr a r if r a3 3E kQr a r 2/ / and 2/ r r if .r a E kQ

22.55. Using Gauss’s Law 2 . Solving for the charge gives Earth 0 Earth/ 4EA q E r

22 12 2 2 3Earth 0 Earth4 8.85 10 C / N m 150. 10 m 6.7711 10 C

6

r N/C 4 6371

5.77 1

5

0 C.

22.56. Let 10.0 cm

q E

be the radius of the solid sphere, the distance between the solid sphere and the inner part of the hollow sphere, and the thickness of the hollow sphere. Let 15.0 cmPr be the

distance from the center to the point P, and let 35.0 cmQr be the distance from the center to the

point Q.

873


874

(a) The Gaussian surface at pr encloses the charge on the inner sphere. 0enc / .2

1 4 pE r q The charge

on the inner sphere is

22 12 2 2 80 14 8.85 10 C / N m 4 10000 N/C 0.150 m 2.50 10 C 25.0 nC.pq E r

(b) For the electric field inside the shell to be zero, the charge on the inner surface of the shell must be equal to the negative of the charge on the inner sphere. 0/ 4 / 4 0E q 2 2

enc 0 ir q q r or

ner surface oi .q q The charge on the in f the shell is then i 25.0 nC 25.0 nC.q q

(c) The Gaussian surface at mQ 35.0 cr from the center encloses the inner charge and the charge

on the shell: 22 enc 0 shell 04 / /QE r q q q or 2

shell 0 24 .Qq q E r The charge on the shell is the

sum of the charge on the inner and outer surfaces of the shell: i o .qshellq q The charge on the outer surface of the shell is

2 2o shell i 0 2 i 0 2

212 2 742

4 4

8.85 10 C / N m 4 1.00 0 N/C 0.350 m 1.36 10 μC.

Q Qq q q E r q q E r

1

22.57.

The field due to either of the two sheets is 0/ 2 . For the positively charged plate the field points normally away from it. The negatively charged plate has field lines pointing towards it. Adding these fields together gives zero on the outside of the two plates, and 0 02 2 / 2 /E E 0 within the two plates, directed towards the negative plate. The field is

6 2 2

5

12 2 28.85 10 C / N mE

the negative

1.00 10 C / m1.13 10 N/C,

and points from the positive plate to plate.

22.58. The magnitude of an electric field is 31.23 10 N/C at a distance 50.0 cm perpendicular to the wire. The direction of the electric field is pointing toward the wire.


Applying Gauss’s Law on the surface shown above gives: EdA enc 0 enc 0 enc 0/ / 2 / .q E dA q E rL q

Therefore, the charge density of the

wire is

12 2 2 3/ 2 2 0.500 m 8.85 10 C / N m 1.23 10 N/Cq L r E enc 0

83.42 10 C / m. The number of electrons per meter is

8

11

19

112.14

3.42 10 C / m2.135 10 electrons/m

1.602 10 C

10 electrons per meter.N

22.59. THINK: A f radius R has a non‐uniform charge density 2 .

N

solid sphere o Ar Integrate the sphere. SKETCH: Not required. RESEARCH: The total charge is given by

Sphere

.Q dV

SIMPLIFY: Integrating in the spherical polar coordinate yields:

2 22 2 2

0 00 0 0 0 0

5 5 5

sin sin 4

/ 5 4 / 5 4 / 5 .

RR R

r R

r o

Q r r d d dr d d Ar r dr A r dr

r A R AR

4 1A CALCUL

4

ATE: NotROUND: Not requ

required. ired.

DOUBLECHECK: This makes sense.

22.60. THINK: This is a superposition of two electric fields. SKETCH:

RESEARCH: The magnitude of the electric field of a charged wire at a distance r from the wire is

0/ 2 ,E r where is the linear charge density of the wire. The net electric field at P is given

by 0 0

ˆ ˆ ˆsin os sin cos2 2

E x y xr r

net yc

x

SIMPLIFY:

net 0ˆ/ 2 sin .E r

875


876

CALCULATE: 1.00 μC/m, 2 23.00 40.0 cm 40.11 cm,r 3.00 cm

sin 0.0747940.11 cm

and

60 10 C / m 0

net

1.0 .07479ˆ ˆ6707 N/C .

2E x x

12 2 28.85 10 C / N m 0.4011 m

R eping

OUND: Ke three significant figures yields net ˆ6.71 kN/C .E x

DOUBLECHECK: Since the vertical components cancel out, it makes sense that the answer is in the x‐direction.

22.61. THINK: Since this problem has a spherical symmetry, it is possible to apply Gauss’s Law. SKETCH:

1r is the radius of a sphere with a charge density 3120 nC/cm . 2r is the inner radius of a

conducting s ll. The shell has a net harhell. 3r is the outer radius of the conducting she c ge s .q

RESEARCH: For this problem, four Gaussian surfaces, the sphere),G 2G (between the

sphere and shel are used. By applying Gauss’s Law on eac can

1 (within

the shell), 3 (within the l),G and 4 (outside the shell)G

h surface, the electric field be determined. SIMPLIFY: For the Gaussian surface applying Gauss’s Law gives 1,G

enc enc enc

0

EdA

0 0

.q q V

E dA

Using 3en 4 / 3 ,aV r

c the electric field is 2 31 4 4 / 3a aE r r 0/ , 0/ 3 .aE r For the Gaussian

surface applying Gauss’s Law yields: 2 ,G

3 2 3enc 0 1 0 1 0/ 4 / 3 / 4 4 / 3 .bEdA q r E r r

Thus, 3 2

1 0/ 3 .bE r r

For the Gaussian surface 3 ,G the electric field is zero since the surface is in a conductor. For the

Gaussian surface Applying Gauss’s Law gives 4 .G

2 3enc s1

0 0

4 4 / 3d

q qEdA E r r .

Thus, E 3 2s 03 / 4 / 3 .dq

CALCULATE: Substituting the numerical values, 3 3m 0.12 C/m , 1 0.12r

1r r

120 nC/c m,

3 0.500 0.100 0.2r 0.300 m, m,r m,ar 200 mbr and 0.800 mdr yields the electric fields:

(a)

3

8

12 2 20

0.12 C/m 0.100 m4.52 10 N/C

3 3 8.85 10 C / NmarE


(b)

3338

2

0.12 C/m 0.12 m1.953 10

m

r

1

2 12 2 203 3 8.85 10 C / Nm 0.200br

0 since it is in the conducting shell.

N/CE

(c) E

(d)

33 331 s 7

212 2 2

0.12 C/m 0.12 m 3 2 10 C / 43 / 41.59 10

0 C / Nm 0.800 m

r qE

2

03 3 8.85 1dr

UND: Keep 2 significant figures.

N/C

RO E(a) 5 84. 10 N/C

82.0 10 N/C 0E since it i

71 N

(b)(

Ec) s in the conducting shell.

(d) 1.6 0 /CE DOUBLECHECK: The values of electric fields have the correct units and are of reasonable orders of magnitude.

22.62. THINK: U ing the symmetry of a cylinder, Gauss’s Law can be applied. sSKETCH:

Note that the Gaussian surfaces 1G and 2G are cylindrical surfaces with radii 1r and 2r and a length

RESEARCH det w on the Gaussian surfaces and

.L: The electric field can be ermined by applying Gauss’s La

SIMPLIFY: For the Gaussian surface applying Gauss’s Law produces 1G 2 .G

1,G

enc1 1 1

0 0 0

2E dA E r L

Similarly for the Gaussian surface using Gauss’s Law gives

10 1

/ 2 / 2ˆ ˆ .

4

L Lr E r

r

q

2 ,G

enc2 2 2

0 0

2E dA E r L 2

0 0 2

/ 2 2 / 2 2 4ˆ .4

L RL L RL Rr E

r

q

re, the electric fields are:

Therefo expressions of the

(a) For the electric field is ,r R0

ˆ.4

E rr

(b) For r ld is , the electric fieR0 2

4 ˆ.4

RE r

r

CALCUL red. ATE: Not requiROUND: Not required.

877


878

DOUBLECHECK: Since the metal cylinder is a conductor, all its charge resides on its outer surface. This means that the field inside the cylinder is not affected by the charge on the cylinder. Therefore, for ,r R the electric field is only due to the wire. For ,r R the charge on the cylinder produces an electric field as if all its charge was concentrated in the center of the cylinder. Therefore, the electric field can be found by replacing / 2 with / 2 2 R as th w linear density of a wire. e ne

22.63. THINK: Use the values from the question: 21 3.00 μC/m , and 2

2 5.00 μC/m . (a) The total field can be determined by superposition of the fields from both plates. The field contributions from the two charged sheets are opposing each other at point P, to the left of the first sheet. (b) The s ation is similar to a) except that the fields due to both charged sheets point in the same direction point

itu at

SKETCH: .P

RESEARCH:

(a) At point the field due to sheet #1 is given by

,P 1 1 0/ 2 ,E x and the field due to sheet #2 is

given by 2 2 0/ 2 .E x Note that total 1 2 .E E E

(b) At point the field due to sheet #1 is given by ,P 1 1 0/ 2 ,E x

and the field due to sheet #2 is

given by Again, 2 0/ 2 .E x 2

total 1 2 .E E E

SIM

(a)

PLIFY:

1 21 2

0 0 02 2 2E x x

(b) 1 21 2

0 0 02 2 2E x x

CAL

(a)

CULATE:

Etotal 3.005.00 106 C/m2

2 8.85 1012 C2 / N m2 x 1.130 105 N/C x

(b) E

total 3.00 5.00 106 N/C

2 8.851012 C2 / N m2 x 4.520 105 N/C x

ROU i

ND: (a) n the positive x direction 5

total 1.13 10 N/CE

(b) 5total 4.52 10 N/CE in the positive x‐direction

DOUBLECHECK: The results are reasonable because the answer in (b) is four times larger than that found in (a) since in (a) the fields are opposing each other and in (b) the fields are in same direction.

THINK: the field due to (a) The field due to a charged sphere outside the radius of the sphere is equivalent to

center of the sphere. erpendicular to the surface of the sphere.

22.64.

a point charge of equal magnitude at the(b) The electric field radiates outward, p(c) The field inside a conductor is zero.


SKETCH:

RESEARCH:

0.271nC and 2 (a) The field is given by: 2 23.1 cmq 2 2

1.1 cm 0 cm .r

(b) The angle is given by 23.1 cm or 1tan 1.1 cm / 23.1 cm . tan 1.1 cm /

(c) The field a conductor. i ero inside SIMPLIFY: N required.

s zot

CULATE: CAL

(a) E0.271 nC 109 C/nC

4 8.851012 C2 / N m2 0.231 2 0.011 2 N/C 45.56 N/C

(b)

1 1.1 cmtan

23.1

0 N

/

2. 37 cm

(c)

C

/C UND: RO

N(a) 45.6(b) 2.7

E

(c) 0 N/C DOUBLECHECK: (a) Not required.

ponent is much less than the x‐component I expected the angle to be small, which (b) Since the y‐comit is. (c) Not r

879

equired.

22.65. THINK: The spherical symmetry of the charged object allows the use of Gauss’s Law to calculate the electric field. To do this, separate Gaussian surfaces must be considered for r a and r a . SKET(a)

CH:

(b)


880

RESEARCH:

(a) The total charge inside the Gaussian surface is given by 2 10 4 .dr The charge density is

0

0 s

rq r

phere/Q V , and the volume is 3sphere 4/3 .V a

(b) The total charge is simply the charge of the non‐conducting layer and the gold layer: 2Q Q

A

Total Charge .q Q

Gauss’s Law states enc 0/ .E d q

Since the Gaussian surface in this case is a sphere, Gauss’s Law

simplifies t Eo 204 / .r q

SIM

(a)

PLIFY:

2 1 2 1 3 3 30 0 00 04 4 4/3 / 4/3 2/3

r rq r dr r dr r Q a

.r Substituting

30 / 4/3 ,Q a 3 3/ .q Qr a 3

0 0/ / /4q Qr a E Qr a 2 34E r 3 r0 ,

The direction is

y from the center of the sphere.

< .r a

awa

(b) ,q Q

0 0 /4 /4 .E Q r r a E Q r r2 2

0 04 / /E r q Q The direction is

rds the center of the sphere.

2

towa(c)

The discontin ue to the surface charge density of the gold. The charge on the gold layer cau the total charge resulting in a discontinuity in the electric fields.

uity at r a is dke in

CALCUL ble. ses a sudden spiATE: Not applica

ROUND: Not applicable. DOUBLECHECK: (a) The electric field increases r gets larger since the charge inside the Gaussian surface increases as a function of 3r while the area increases as a function of 2 .r Since the increase of the area

2 increases the field by it is reasonable that decreases the field by a function of r and the charge.r rce is a

3rthe field increases, as a function of (b) The sphere acts like a point sou s expected. (c) There is a discontinuity in the graph due to the presence of a surface charge density on the gold

v. E r layer, which is expected.

22.66. THINK: constructing Gaussian surfaces in both regions r RBy and ,r R the electric field can be calculated sing Gauss’s Law. uSKETCH:


RESEARCH: The total charge inside the Gaussian surface is given by 2004 .

rq r dr The charge

density is given by sin / 2 .r/r r For the Gaussian surface outside the sphere

r R the total charge is given by 2004 .

rq r dr

R

, The electric field can be calculate using Gauss’s

Law: enc 0/ ,A qE d w

hich for a spherical Gaussian surface is 204 / .E r q

SIMPLIFY: For the case r R, 2

0 0

00

sin 4 4 sin2 2

2 24 cos cos

2 2

r r

rr

r rq r dr r

r R R

Rr r R rdr

R R

dr

Integration by parts: 2

0

2

2 24 cos sin

2 2

2 24 cos sin .

2 2

r

Rr r R rq

R R

Rr r R r

R R

For the case is given by

,r R q

2

2

0

0

2 2 2 2

2 2

2 24 4 cos sin

2 2

2 4 2 4 164 0 sin 4 0 1

2

R

R R R R R Rq r r dr

R R

R R R R

2

.

The electric field is given by

2

20 0

4 4

qE r E

r

.

q

For the case r R,

2

20

2

20

2 2 14 cos sin

2 2 4

2 2cos sin 1

2 2

Rr r R rE

R R r

Rr r R r

R Rr

,r R

For 2 2

2 2 20

16 1 4.

4

Rq

r r

0

(2)

For

,r R

2 2 2

20

2 2 20 0

2 2 2 4 41 cos sin 0 1

2 2

R R R RR R R

R RR R

2

2 20 0

4 42

1 2

R

R

881


882

The expr n essions are equal whe .r R CALCULATE: Not applicable. ROUND: Not applicable. DOUBLECHECK: The two expressions are equal at ,r R which should be the case since there are no surface charge densities present cause discontinuities for the objects act like a point source,

,r Rwhich is expected from a charged sphere.

22.67. THINK: The principle of superposition can be used to find the electric field at the specified point. The electric field at the point 2.00,1.00 is modeled as the sum of a positively charged cylindrical

rod with no hole and a negatively charged cylindrical rod whose size and location are identical to those of the cavity. Let’s first think about the case of the positively charged cylindrical rod without a hole. Since the point of interest is inside the rod, the entire charge distribution of the rod cannot contribute. Instead we draw our Gaussian surface as a cylinder with our point of interest on its rim (see sket h below, where the dashed circle in the cross‐sectional view represents the Gaussian cylinder)

c.

SKETCH:

RESEARCH: In section 22.9 of the textbook it was shown that for cylindrical symmetry of the charge distribution the electric field outside the charge distribution can be written as E 2k / r , where r is the distance to the central axis of the charge distribution and is the charge per unit length. In the problem here the charge was initially uniformly distributed over the entire cross‐sectional area, which means that the value of for the Gaussian surface and for the hole are proportional to their cross‐sectional area: Gauss rod (r / R)2 , and hole rod (rhole / R)2 . Now we have the tools to calculate the magnitudes of the individual electric fields of the rod and of e h a v ed y he

th ole. What is left is to add the two, which is ector addition. So we have to determine th x an components of the fields individually and t combine them. If 1E is the field from the dashed cylinder and is that of the cavity then from considering the

geometry the relation are given by: E

2E

22 11x E12 / 2 1/2, E2x E2 2 / 22 12 1/2

, 1/22 2

1 1 / 2 1yE E

and 1/22 2

2 20.5 / 2 0.5 .yE E

The net electric field is given by the following relations 1 2x xE E E x and 1 2 .y yE E E y

SIMPLIFY:

1 auss / rod ( rod

E2 2k le / r2 2krod (rhole / R)2 / r2

where is the distance between ou nt of interest and the center of th ole.

E 2kG r 2k r / R)2 / r 2k r / R2

ho

r2 r poi e h

1 21/2 1/22 2 2 2

2 2,

2 1 2 0.5xE E E

and

1 2 1/22 2 2 2

1 0.5.

2 1 2 0.5yE E E

CALCULATE: r .0

0.01 m 2 0.0200 m 2 1/2

0 2236 m

1/22 2 m 0.0200 m 0.02062 m 2 0.00500r


E1 2(8.99 109 Nm2 /C2 )(6.00 107 C/m) 0.02236 m

0.0300 m 2 2.680 105 N/C

E21

2(8.99 109 Nm2 /C2 )(6.00 107 C/m)

0.02062 m 0.0100 m

0.0300 m

2

0.581105 N/C

Ex 1.833105 N/C, y

kN/C, Ey 134 kN/C

DOUBLECHECK: We can calculate the magnitude and direction of the combined electric field and

find: E Ex

2 Ey2

E 1.339 105 N/C

ROUND: Ex 183

227 kN/C, and tan1(Ey / Ex ) 36.1 . If the hole would not have been drilled,

the magnitude would have been he magnitude we calculated above for E1 , E1 268 kN/C , and it

would have pointed along the rr vector with an angle of 26.6°. This means that our result states that

the magnitude of the electric field is weakened due to the presence of the hole, and that it does not point radial outward any more, but further away from the x‐axis. Both of these results are in accordance with expectations and add confidence to our result: the hole modifies the electric field somewhat, but does not do so radically.

22.68. THINK: U e the principle of superposition and model the problem as a positive infinite plane and a negative c cular disc.

sir

SKETCH:

RESEARCH: The electric field contributed by the plane is given by: plane 0/ 2 .E One can find the

electric field of a disc by adding up the contributions from each small area. From the symmetry one many conclude that the field points vertically. The contribution of each small area to the field in the y‐direction is given by:

20/ 4 cos / ,dE dA r cos / ,h r 2 2 2 ,r h disc .E dE total plane discE E E .

2 C/m .0.200 m 0.050 m, 1.3h R ,

SIMPLIFY:

3/22 220 0

cos 4 4

hd dedAdE

hr

2 1/22 2

disc 3/2 1/20 0 2 2 2 200 0 0

1 12

4 4 2

RRh h hE d d h

hh h

R

total disc plane 1/2 1/22 2 2 20 0 0

1 1

2 2 2

h hE E E

h h R h R

883


884

CALCULATE:

10total 1/212 2 2

1.3 0.200N/C 7.125 10 N/C

2 8.85 10 0.200 0.050E

ROUND: 10total 7.1 10 N/CE

OUBLECHECK: The plot shows that for large the result is the same as that of an infinite plane ithout a hole as one would expect.

Dw

h

Additional Problems

22.69. Regardless of what orientation the cube is in, we can always enclose it in a Gaussian surface that just covers the cube. Gauss’s Law states that enc 0/ .EdA q

Now consider the flux through one particular face given by EA1.

There exists a flux through the

opposite face given by 2EA

with the relation 1 2EA EA

since 1A

and 2A

point the opposite way.

The sum of the flux contributed between the two opposite sides is 1 2 0.EA EA

If this calculation ust be 0 by Gauis done for each side then the total flux is 0 and hence the total charge m ss’s Law.

22.70. The electric field is required to measure the acceleration of the proton. ,F qE

.F ma

The electric

field of an infinite wire is given by 2 / ,E k r

where r is the distance to the wire. 0.620 m,

124.81 10 C / m, 9 2 28.99 10 Nm / C ,k 271.67 10 kgm

r

p .22 kqF k

.qE q

m r ra

m m m

roton charge 19 and 7 21 p

22.71

1.6q 0217 10 C 6 28 10 m/s 1.3413.3a 0 m/s .

.


Consider a cylindrical Gaussian surface with a radius of 4.00 cm . By Gauss’s Law, enc 0/ .EdA q

The charge inside the cylinder is so the field is given by 2 ,q r l

8 3/m 0.02 6.40 10 C 400 mr l r

2

12 3 1 4 20 0

2 1.45 10 N / C2 2 8.854 10 m kg s A

E rl E

T i a i of i

p

away from the y‐axis. he nformation concerning the r d us the cylinder s irrelevant.

22.72. The dipole moment is given by qd where d is the distance between the charges.

30 10 20/ 8.0 10 C m / 1.2 10 m 10 C.q p d The maximum torque is when the field is

perpendicular to the dipole moment.

6.7

The torque is then 30 278.0 10 C m 500.0 N/C 4.0 10 N m.qEd pE

22.73.

(a) Construct a Gaussian surface (spherical) with radius between 20.0 cm and 24.0 cm. Gauss’s Law states that the total flux is equal to 0/ ,q since the electric field inside the last metallic shell is zero, the flux must be zero and hence the total charge must be zero. Since the total charge to be zero:

inside wall inside wall10.00 μC 5.00 μC 0 5.00 μC.q q (b) Constructing a Gaussian sphere that contains all the shells, it can be determined that since the electric field is zero, outside the largest shell that the flux is also zero and hence the total charge must be zero.

outside wallq , which then inside wall outside wall10.00 μC 5.00 μC 0 5.00 μC 5.00 μC 10.00 μC 0q q implies outside wall 0.q

22.74. The electric force and the gravitational force must balance.

qE

(a)

0 / ,mg E mg q

31electron 9.109 10 kg,m

29.81 m/sg

1.602 10q 19 C,

31 2

11

19

9.109 10 kg 9.81 m/s5.58 10 N/C

1.602 10 CE

885


886

27 19

with the field directed down.

(b) proton 1.672 10m kg, 1.602 10 C,q

27 2

7

19

1.672 10 kg 9.81 m/s1.02 10 N/C

1.602 10 CE with

the field directed up.

22.75. The sum of the forces on the electron is given by .total gravity coulomb+F F F mg q E

191.602 10 C, 3110 kg.m Thus,

150. N/C,E

17 3110 N / 9.11 10 kg 2.64 10 13 2a m/s .2.405q 9.11

22.76.

The fields from both plates are always perpendicular to each other. The field E1 from plane 1

always points away from plane 1. The field E2 from plane 2 always points toward plane 2. The combined field E points in different directions depending on where you measure it, but the magnitude of the field is the same everywhere.

2 22 21 2 1 0 2 0

2 22 22 2

1 2

12

/ 2 / 2

30.0 pC/m 40.0 pC/m2.82 N/C

m

E

E

3 1 4 202 2 8.85 10 kg s A

22.77. This problem can be solved using Gauss’s Law. total 0Flux / .q

E E

The approximation can be made that

the flux leaving the ends of the rod are negligible, so 0 0ltotalFlux / /q where l is the length of the rod.

12 6 2

0 58.85 10 1.46 10 N m /CFlux

4.31 10 C/m0.300l m

22.78. This problem can be solved using Gauss’s Law. 21

total 0Flux / 0 N m / C.nq E da E da

Since E ,nda E da 2 12 11total 0

2 2 210 N m /C 8.85 10 C 10 N m /C 9 10/ N .m Cq

22.79. THINK: ant to find the charge, 1q needed to balance out the force of gravity. After finding ,q I can determine the number of electrons based on the charge of a single electron.

I w

SKETCH:


RESEARCH: The net force on the object must equal zero in order for the object to remain motionless. total gravityF F coulomb+ 0F , gravity g,F m coulomb ,F Eq 0/2E for an infinite plane. The

number of electrons is

F Felectron/ .q q

SIMPLIFY: 0,g Eqtotal gravity coulomb total+ 0 F F m

q m0 0 / 2 / 2 .mg q gEq mg Number of electrons 0 electron2 /mg q .

2s , 5 23.5 10 C/m , 1.0 g.m9.81 m/g CALCULATE:

Number of electrons

3 2 12 3 1 4 2

10

5 2 19

2 1.0 10 kg 9.81 m/s 8.85 10 m kg s A3.097 10 electrons

3.5 10 C/m 1.602 10 C

.

ROUND: 3.1 10 electrons DOUBLECHECK: This number, though large, is reasonable since the amount of charge on each electron

10

is tiny.

22.80. THINK: I first wire and the secon firSKETCH: N

st need to find the relationship between the d wire. ot required.

RESEARCH: The field due to the first wire is given by: 1 2 / 2.73 N / CE k r The field due to the

second wire is given by 2 2 0.81 / 6.5 .E k r

SIMPLIFY:

2

CALCUL TE:

1

.81 20.81 0.816.5 6.5 6.5

k k

r r

2 0E E

A 2 1

0.81 0.812.73 N/C 0.3402 N/C

6.5 6.5ROUND: 0.3 DOUBLECHECK: The answer is comparable to the electric field of the original wire which makes it reasona

E E

4 N/C

ble.

22.81. THINK: T e net electric field can be determined by integrating the differential elements of the field due to each piece of the wire.

h

SKETCH:

887


888

RESEARCH: Based on the symmetry, it can be seen that the net field on A is in the y‐direction. The contribution of the infinitesimal piece dl in the y‐direction is given by 01/ 4 sin .dldE The

total field is 00/ 4 sin ,E r d

6.0800 10 C / m, 8. 0 cm,l / .r l

SIMPLIFY:

0 20

0 0 0 0

sin cos | 24 4 4 2 2

r r rE d

0

r l

CALCULATE: 125 10 2 8 28.00 10 0.0600 m / 2 8.8 7.477 N/CE

ROUND: Reported to 3 significant figures gives 27.5 N/CE in the y‐direction. DOUBLECHECK: It is reasonable that the field in the y‐direction is zero since the field in the x‐

direction is given by

000 0

cos ,sin4 4

r rd which evaluates to zero.

22.82. THINK: (a) The necessary electric field strength can be determined by finding the acceleration required to

peed achieve the desired deflection. The final s of the proton can be found through the relation between the proton’s initial velocity and its angle of deflection. (b) The electric field strength required to give the protons a specific acceleration will impart a different celeration to the kaons due to difference in mass. acSKETCH:

RESEARCH: (a,b) Initially the velocity in the y‐direction, ,yv is zero. The only part of the velocity affected by the

electric field is ,yv xv is the same before and after the deflection. ,atyv t is the time the proton

spends in between the plates. 0 ,F m a Eq tan / ,y xv v 2 2 2x yv v v , where v is the new speed.

/ ,xt l is the distance the proton has to traverse between the plates.

.

v l 31.50 10 rad, 15.0 cm,l 15.0 km/sxv

(c) The mass of a proton 1.67 1027 kg. The mass of a kaon is 8.811028 kg. The speed of the kaon is given by setting the momentum of a kaon equal to the momentum of a proton:

m v

LIFY: kaon

SIMP(a,b)

kaon proton proton .m v

vy vx tan at vy / t a

Eqma

mvy

t

mvx tant

mvx tan

l / vx

mvx

2 tanl

Emvx

2 tanlq

v2 vx

2 vy2 vx

2 vx tan 2 vx2 1 tan2

(c) Take the result from part (a) to find .

Emvx

2 tanlq

qEl

mv2 tanx

tan1 qEl

mvx2

CULATE: CAL

(a)

E1.67 1027 kg 15.0 103 m / s 2 tan 1.50 103 rad

0.150 m 1.6021019 C 0.023455 N/C


(b) v 15 15.0 103 m / s 1 tan2 1.50103 rad

1/2

.000017 km/s

(c) vkaon 8.811028 kg

1.67 1027 kg15.0 103 m/s 28434 m/s

With the results from part (a), the electric field is Emvx2 tan / lq .

31.50 10 rad,

227 3 3

19

1.67 10 kg 15.0 10 m/s tan 1.50 10 rad0.02345507 N/C

0.150 m 1.602 10 CE

mv

ND:

19

1 1 4

2 228

1.602 10 C 0.02345507 N/C 0.150 mEtan tan 7.91295 10 rad

8.81 10 kg 28434 m/sx

q l

ROU(a)

v

0.0235 N/CE

(b) s 41.50 10 km/

(c) 47.91 10 rad DOUBLECHECK: The change in speed is small compared to the magnitude of the speed, which is expected since the deflection was also small. The deflection of the kaon is less that the deflection of a proton with the same momentum because the kaon has a higher speed.

22.83. THINK: U e density, Gauss’s Law can be used to find the electric field as a function of the radiu

sing the chargs.

SKETCH: N t iro requ ed. RESEARCH: The charge inside a spherical Gaussian surface is given by sphere .q V

3sphere 4 / 3 ,V r and 63.57 10 C / m 3 0.530 m.r Gauss’s Law gives the field

2EdA E r 04 / .q

SIMPLIFY: 32

2 2 20 0 0 0

4 / 31 1 14

34 4 4

rq q V rE r E

r r r

0

CALCULATE:

6

4

12

3.57 10 0.530N/C 7.127 10 N/C

3 8.85 10

ROUND:

E

410 N/C DOUBL

7.13E

889

ECHECK: The result was independent of the actual radius of the sphere as it should be.

22.84. THINK: uss’s Law can be used to determine the electric field as a function of radius for the three cases r R 2R r R and 2 .r R

Ga,

SKETCH:

RESEARCH: The electric field through the surface of a sphere of radius r is given by Gauss’s Law:


890

204 /A E r q .E d

For the enclosed charge is given by:

,r R

21 10

4 ,r

q r dr

where

1 4 /

For the enclosed charge is given by:

3.

3

Q

R

2 ,R r R

22 2 4 ,

r

Rq Q r dr

where

2 3 3.

4 / 3 2

Q

R R

e enclosed charge is

For ,2r R th 3 0.q Q Q SIMPLIFY: For :r R

3

2 21 3 3 3

3

0 0

3 3 34

34

r rQ Q Q r Qq r dr r dr r

R R R R

3

2 313 3

0 0 0

4 4r R r R

q Q QE r r E

R R

r

For 2 ,R r R

2 2 22 2 3 3

3 34 4

28 7 7

r r r

R R R

Q Q Qq Q r dr Q r dr Q r dr Q r R

R R R

3 3

3

3 32 2 3 2

0

14 7 4r R Q r R

r R r For r rge is zero, by Gauss’s Law

3 2 30

31 1 8

7 287R

Q Q r Q rE

R r R

0

l cha

2 :R Since the tota 2 0.r RE CALCULATE: Not applicable. ROUND: Not applicable. DOUBLECHECK: It is expected that the expression for r R and 2R r R are equal at r R and the expressions for and 2r R 2 ,R r R are equal at 2 .r R For :r R

3 20 04 4r R

Qr QE

R R

2 2 3 2 3 20 0 0

8 8

28 28 4R r R r R

Q r Q R QE E

r R R R R

For 2 :r R

2 2 3 2 30 0

8 8 2

28 28 4 >22 2

2 20

28 0R r R

Q r Q RE

r R R R

The exp

r R

QE

R R

ressions are equal, so the solution is reasonable.

22.85. THINK: The electric field due to the charge induces a charge distribution on the floor below it. As a result, the charge experiences a force directed toward the floor. Since the charge and its ‘mirror image” describe a dipole, the electric field lines are perpendicular to the floor. I want to determine the force ting on the charge, the electric field just above the floor, the surface charge density and the total surface charge induced on the floor.

ac

SKETCH:


A Gaussian pill box may be drawn along an infinitesimally small area as follows:

RESEARCH: The electric field due to the charge is given by 2/ ,E kq r where q is the magnitude of the charge and r is the distance from the charge to the floor. The force experienced by the charge is

given by Coulomb’s law; 20 1 21/ 4 / .F q q r Since the electric field points in the negative y‐

direction, only the y‐contribution from each charge nee found. The y‐contribution is given by d be

20

cos ,4

qE

r

1

cos

r a , and 1/2

2 2 .r a

Since the y‐component from both charges is the same (i.e. since the charges are equal in magnitude),

the total electric field is then: total 20

2cos .

4E

r

q

Using Gauss’s Law on the pillbox,

0/ .EdA dq dq dA The total charge is given by infiniteplane

.q d A

SIMPLIFY:

(b)

1 22

04 2

q qF

a

(c) total 2 3 2 2

0 0 0

2 2 1cos

4 4 2

Q Qa QE

r r a

3/2

a

(d) 0 0

,dq dA

EdA

0

E

and 0 3/22 2

1

2

adqE

a

(e)

1/22 2

3/2 3/20 02 2 2 2 02 2

2 2 2

Qa Qa aQq dA dp d a

a a

Q

891


892

CALCULATE:

(b)

F4 8.85 10

1.00 106 C 1.00106 C 12 C2 / N m21.00 m 2

8.9918 103 N

(c) Not applicable. (d) Not applicable.

Not applicable. (e) ROUND: (b) 38.99 10 N downwardF

C is sym

DOUBLECHE K: (a) The sketch metric as it should be. (b) The force is downward as it should be since the positive charge is attracted to the negative charge. (c) The field gets weaker as gets larger as expected since the source is farther away with increasing . (d) The surface charge density gets smaller as gets larger since the source is farther away with increasing . (e) Since all the field lines coming from the charge go onto the top of the slab it is not unreasonable that the total charge induced is equal to the charge in magnitude.

Chapter 23: Electric Potential

893


In-Class Exercises

23.1. e 23.2. e 23.3. a 23.4. a 23.5. e 23.6. d 23.7. e 23.8. b 23.9. a Multiple Choice

23.1. a 23.2 c 23.3. c 23.4. c 23.5. a 23.6. d 23.7. a 23.8. d 23.9. a and c 23.10. c

Questions

23.11. Birds are safe on a power line because there is no current flowing through the birds. A potential difference is needed in order for a current to flow. Since the potential in the bird is the same as the high voltage wire, the potential difference is zero. Therefore, the birds are safe resting on the wire.

23.12. It is unsafe to stand under a tree during an electrical storm because lightning is more likely to strike trees. This is due to the trees being made of materials making the tree a better conductor then air, which provide an easy path of least resistance for the electricity to the ground. After a strike trees have a high electric field in their vicinity, which helps initiate and guide lightning to the ground. The electricity can travel through the ground to inflict damage or even strike directly from the trees outer surface.

23.13. An equipotential line is defined as a line connecting points of the same potential. This means that if two equipotential lines were to cross, at the cross point, the potential would have two values at the same point. The equipotential lines are also always perpendicular to the electric field. If they were to cross, then there would have to be two different electric fields acting at the same point. If a point charge were put at this point where the electric fields crossed, there would be two separate forces acting from the two different electric fields. Both of these situations are not possible. Therefore two equipotential lines cannot cross.

23.14. In the vicinity of a pointy protrusion, the electric field can be very high. This can lead to a spark inside an electronic device which can make the device to stop functioning.

23.15.

Applying Gauss’s law on a spherical Gaussian surfaces as shown above gives:

0

.qdAE =

Since the spherical symmetry of the Gaussian surface, the above equation simplifies to:

( )2

0 0

4 .q qE dA E rπ= =


894

Thus, the electric field is π

= 204

.qEr

This is exactly the field of a point change with the same charge, q,

located at the center of the spherical uniform charge. Using the relation between potential and field, f

fi

i ,V V E ds− = − yields: 2

0

0 .4i

r r

qdr drr

EVπ

∞∞

− = − = − Thus,

0 0 0 0

1 1 1 14 4 4 4i

r

q qVr

q qr r rπ π π π

∞ − − − − = = ∞ = =

.

This potential is identical to the potential of a point change. Substituting r = R gives the potential at R:

0

.4

qVRπ

=

Changing the charge distribution to non-uniform but spherical (radial) symmetry yields the same result as above. That is, its potential is the same as the potential produced by a point change in the center of spherical charge.

23.16. Since the distances of all points in the ring to the center of the ring are the same, the potential is

04qV

Rπ=

. The electric field is zero since electric field lines cannot cross where they would converge at

the center of the ring.

23.17.

The potential due to a small element dA is given by: 04

dAr

dV σπ

=

, where .dA ad daθ= Integrating over the

area of a half disk gives: 0 0

0

.4

R

rV ad da

π σ θπ

= Substituting 2 2r aH= + yields:

0 0 02 2 2 200

.44

R Ra adaV d da

aHaH

π σ σθπ +

=+

= Now, for integration by substitution, let 2 2 ,z H a= +

which makes 2 .dz a da= Then the previous integral becomes

( )2 2 2 2

0 0 00 0 0 0

1 .4 2 4 4 4

a Ra Ra R

a a a

dz z H a H R Hz

σ σ σ σ===

= = == = + = + −

23.18. As an electron moves away from a proton, it encounters a decreasing potential. Since the electron has a negative charge and the potential energy is defined as U = qV, then the potential energy increases as the electron moves away.


895

23.19.

The sphere is divided into many spherical shells. Consider two shells with radii 1r and 2 .r The potential

energy of the two shells is: 1 212

0 24U QQ

rπ=

, where 1Q and 2Q are the total changes of shells of radii 1r and

2 .r The total charges are ( )211 1 14Q r r drπ ρ= and ( )2

2 2 2 24Q r r drπ ρ= . Therefore,

( ) ( )ρ ρ= 212 11 2

02 12

1 .U r r drr drr

The total potential energy is obtained by integrating over an interval (0, ∞ ) for the variable 2r and an interval (0, 2r ) for 1.r Thus,

( ) ( )ρ ρ∞ ∞

= = 2 2

2 12

12 2 2 20 0 1 100

0 1 .1r rU dr dr r r dr rr drU

Problems

23.20. The work required to increase the distance between these ions is:

π π π

− = − = Δ = − =

−

1 2 1 2f i 1 2

0 0 i 0 f i

1 1 1 1 14 4 4

.f

q qW U U q

r r r rq q

U q

Substituting 191 1.602 0 C,1eq −= − ⋅= 1

291.602 0 C,1eq −= = ⋅ 2

f 1.0 m10r −⋅= and 9i 0.24 m10r −⋅= yields:

( ) ( )( )9 2 2 19 192 9

19

1 110 10 101.0 10

8.99 N 0.24 10

10

m /C 1.602 C 1.602 C m m

9.6 J.

W − −− −

−

−⋅ ⋅

− = ⋅ ⋅ − ⋅ ⋅ ⋅

= ⋅

23.21. THINK: As the positively charged ball approaches the positively charged plane, its potential energy will increase and its kinetic energy will decrease. At the point when the ball stops, all its initial kinetic energy will have been converted into potential energy. Work must be done on the ball to accomplish this change. The force necessary to do this work is supplied by the electric field created by the charged plane. SKETCH:

RESEARCH: Since work is force times distance, the stopping distance can be calculated by exploiting the relationship between the work done on the ball, and the force exerted on the ball by the electric field. The electric field due to the charged plane is given by 0/ .E σ= The net force acting on the ball is given by


896

0/ .qF qE σ= = The work done on the ball is equal to the change in kinetic energy, Δ = −f i .K K K The

work-energy relation states •= − .W F d f i0

.q dW K K K F d qE d σε

−= Δ = − = − ⋅ = − ⋅ =

SIMPLIFY: ( )0 f if i

0

.K Kq dK K dq

εσε σ

−−− = = − Since f 0,K = i 0 .q

Kd

σ=

CALCULATE: ( )( )

( )( )8 12 2 2

3 2

6.00 J C /N m0.2655 m;

5

10 8.85 10

10 4.00.00 C/mCd

−

−

⋅

⋅

⋅= = Therefore, the final distance from the

plane is = − =1 m 0.2655 m 0.7345 m.x ROUND: Keeping three significant figures gives d = 0.266 m. DOUBLE-CHECK: This is a reasonable distance and less than the initial distance of 1 m.

23.22.

The change in potential energy, Δ = Δ .U Vq From the conservation of energy, Δ = −Δ .U K So

( ) Δ = −Δ = − − = − −

2 2f i f i

1 1 .2 2

Vq K K K mv mv Since i 0v = ,

( )( )−

−

−− ΔΔ = − =− ⋅

⋅=⋅

=19

731

2f f

370 V 1.602 1010

10

2 C1 2 1.1 m/s.2 9.11 kg

VqVq mv vm

23.23. The work done by the electric field on a proton is given by ( )f i .W U q V q V V= −Δ = − Δ = − − Substituting

f 60.0 V,V = − i 180. VV = + and 191.602 10q −⋅= C yields:

( )19 1710 60.0 V 180. V( 1.602 C 3.85 10) J.W − −− − == ⋅− ⋅

23.24. The work done by an electric field is given by .W K q V= Δ = − Δ This means that the potential difference is

/ .V W qΔ = − Putting in ( )192 1.602 C10q −= ⋅ and ( )3 1910 1.200 k 602eV 200 J10W −⋅= = ⋅ yields:

( )( )

3 193

19

200 J100 V or 100 kV.

2 1.602

10 1.602 1010

C10V

−

−Δ = − = − ⋅ −

⋅

⋅ ⋅

23.25. Using the work-energy relation and ,W q V= − Δ it is found that:

2 2f i

1 1 .2 2

W K q V mv mv q V= Δ = − Δ − = − Δ Since the proton is initially at rest, i 0v = :

2f f

1 2 .2

q Vmv q V vm

− Δ= − Δ = 19101.602 C,q −= ⋅ 500. VVΔ = − , and 27167 kg01.m −= ⋅ , and this

means that: ( )( )

( )1

f 75

9

2

2 1.602 10 C 500. V3.10 m/s.

1.67 10 g10

kv

−

−⋅

− ⋅ −= =

⋅

23.26. The kinetic energy is calculated using the work-energy relation and ,W q V= − Δ that is:

f i .W K q V K K q V= Δ = − Δ − = − Δ Since the initial kinetic energy is zero, the find kinetic energy is f .K q V= − Δ


897

(a) Inserting 191.602 C10q −= − ⋅ and 10 VVΔ = yields the final kinetic energy:

( )19 18f (1.602 C) 10.0 V 1.610 10 J.0K − −= ⋅ = ⋅

(b) The final velocity of the electron is:

( )186f

f f 312f

2 1.602 J21 1.88 m/s.2 9.11 kg

1010

10K

mv K vm

−

−= = = ⋅⋅

=⋅

23.27. THINK: The force due to the electric field opposes the motion of the proton. Work must be done on the proton to move it toward the plate of higher potential. This work must come from the initial kinetic energy of the proton. If the proton has sufficient kinetic energy to provide the necessary work, the proton will reach plate B. SKETCH:

RESEARCH: The work done is equal to the change in kinetic energy. The work is also equal to the scalar product of the force and the distance over which the force acts. So, W F dK= Δ = •− . The force on the proton due to the electric field is = ,F qE and electric field due to the two plates is = Δ / .E V L So

Δ Δ Δ= = = −− − Δ

.K K KLdF qE q V

SIMPLIFY:

(a) ( )( )

f i

B

;A

K K Ld

q V V−

= −−

Since =f 0K and = 0,AV the distance is: = =2

f

B B

.2

K L mv LdqV qV

(b) If d is less than 5.0 cm, then the proton will turn around at x = 5.0 cm + d. If d is greater than 5.0 cm, the proton will reach plate B.

(c) The proton will reach the plate A with speed determined from the work-energy relation: ( )B A2

f2

fi i1 1

2 .

2 2 2q V V LLW K K F mv mv

L

−− = − =

= Thus, 2 B A

f i .V Vv v qm− = +

CALCULATE:

(a) This distance, d, is: ( )( ) ( )

( )( )

227 3

19

kg 150.0 m/s0.029319

1.602 C 400.0 V

1.67 10 10 0.10

0

0

V

m m.

2 10d

−

−

⋅

⋅=

⋅=

− Therefore the

proton will not reach the plate B. (b) The proton will turn around at a distance x = (10.0/2) cm + 2.9319 cm = 7.9319 cm from the plate A. (c) The speed of the proton when it reaches the plate A is:

( )23 19f 27

400.0 10 101.67

V 0 V150.0 m/s 1.6010

2 C 246721 m/s.kg

v −−⋅

⋅ −= ⋅ + =


898

ROUND: (a) 0.0293 m (b) 7.93 cm (c) f 247 km/sv =

DOUBLE-CHECK: The work required to move the proton all the way to plate B is = = ⋅ 173.2 10 J.W qEd

The initial kinetic energy of the proton is −= = ⋅2 170.5 1.9 10 J.K mv This is not enough energy to provide the work required. The proton experiences constant acceleration in the electric field. The proton moves toward plate B, stops, and then moves back toward plate A, passing through its original position where it has the same speed as it had initially. The proton continues to accelerate until it strikes plate A. Thus it makes sense that the magnitude of its final velocity is greater than the initial velocity.

23.28. THINK: 32S ions are accelerated from rest using a total voltage of 91.00 10 V⋅ . 32S has 16 protons and 16 neutrons. The accelerator produces a beam of 126.61 ion .10 s/s⋅ Note that the total power is the total energy absorbed per second. SKETCH:

RESEARCH: The kinetic energy of each ion is determined from the work-energy relation, that is,

.W K q V= Δ = − Δ Since i 0,K = the final kinetic energy of the ion is f .K q V= Δ SIMPLIFY: The total power is equal to f ,NKP Nq V= = Δ where N is the number of ions per second. CALCULATE: Substituting the numerical values yields:

( )( )( )( )12 19 910 16 16.61 ions/s 1.602 C V 16.94 k0 1.00 10 W.P −= =⋅⋅ ⋅

ROUND: 16.9 kWP = (keeping three significant digits). DOUBLE-CHECK: Watts are an appropriate unit of power.

23.29.

(a) The potential at the point A is given by:

( )6 6

9 2 2 41 1A

1

2 2

12 2

1.0 C 3.0 C8.99 N m /C 1.8 V 18 kV.0.25

10 1010 1m 0.5 m

0kq kq q q

V kr r r r

− − −= + = + = + = =

⋅ ⋅⋅ ⋅

(b) The potential difference between points A and B is:


899

( )

( )( )

1 11

1 1

9

2 22

2 2

2 2 6 6

1 2

4

1 1

1 18.99 N 10 1m /C 1.0 C 3.0 C 7.2 V 72 kV.0.25 m 0.50

1m

0 0 10

BAB Aq q q q

V V V k k k q qr r r r r r

− −

= − = + − + = − −

= − − − = − = − ⋅

⋅

⋅ ⋅

23.30. Each charge, q = 1.61 nC, is the same distance from the center of the square, ( ) ( )2 2/ 2 / 2 .r a b= + Since there are four point charges, the total electric potential is sum of four individual electric potentials:

( )( )( )

( ) ( )

−

=

⋅= = = =

+ +

⋅

9 2 94

1

2

2 2 2 2

8 8.9875 N m /C 1.61 10 C4 19.9 V.1/ 2 3.00 m 5.00 m

10

i

kq kqVr a b

23.31. For the van de Graff generator, all the excess charge is on the surface, so the electric potential is: 51.00 10 V,kQV

r= = ⋅ where r = d/2 is the radius.

The total charge, Q, is: = = .2

Vr VdQk k

The total charge is a result of the number of electrons, ,Q n e=

where n is given by:

( )( )( )( )

512

9 2 2 19

1.00 10 V 0.200 m6.95 electrons.

2 2 8.9875 N m10

/c10 101.602 CVdnk e −

⋅⋅ ⋅

⋅= = =

23.32. The electric potential of a charged uniform sphere is = / ,V kq r where V = 100 V and r = 1.0 m, so the total charge is:

( )( )9 2 2

100. V 1.00 m11.1 nC.

8.987 15 N m /C0kVrq = = =

⋅

23.33. All the charge, Q = 5.60 μC , is equidistant from the center, R = 4.50 cm, so the electric potential at the centre is:

( )( )( )

−⋅=

⋅=

69 2 28.9875 N m /C 5.60 10 C= 11.2 MV.

0.041

50 0

mkQVr

23.34. When considering a charged conducting sphere, the sphere can be considered to be a point charge Q = 8.0 nC for any distance r > R, where R is radius of sphere and R = 5.0 cm. Since all the charge is spread out evenly across the surface of the sphere, every point inside the sphere has the same electric potential.

(a) ( )( )−⋅

= = ⋅⋅ 9 2 2

1

92

8.9875 N m /C 8.0 10 C= 9.0 10 V

0.080

0

m

1

kQVr

(b) ( )( )−⋅

= = =⋅ 9

2

9 2 28.9875 N m /C 8.0 10 C= 1400 V

0.0

10

50 mkQ kQVr R

(c) = =3

= 1400 VkQ kQVr R


900

23.35.

Since the wire is a half circle of radius, = 8.00 cm,R the length of the wire is θ= .L R The total charge of the wire is λ= ,q L where 83.00 10 C/m.λ −= ⋅ The charge for a small length of the wire is λ= .dq dL For constant R, .dq R dλ θ= The electric potential is then:

( )( )0

9 2 2 8

08.9 10875 N m /C 3.00 1 C/m 847 V.0dq RV k k d

r Rk d k V

π πλ θ λ θ λπ π−= = = ⋅ ⋅= = =

23.36. THINK: To find the electric potential, consider the dipole as a system of two point charges, +q and –q. The two charges are a distance d away from each other. The potential as a function of θ and x can be found by summing the potentials due to each charge. SKETCH:

RESEARCH: Using the law of cosines, the two distances can be determined: 2

2 21 os2 c

24d dr x x = + −

θ and 2

2

22 2

2cos(180 ).

4d dr x x = + − ° −

θ

The electric potential is: =

=2

1.i

i i

Vrqk

SIMPLIFY: 21 2

1

1

2 1 2 1 2

1 1 .kq q r rq q

r rkV k k

r r r r −

− =

=

= + Since ( )cos(180 ) cos ,° − = −θ θ the electric

potential is 2 2

2 2

2 22 2

cos cos4 4

cos cos4 4

.

d dx xd x xdV q

d dx xd x xk

d

+ − + −=

+ −

+

+ +

θ θ

θ θ

CALCULATE: Not applicable. ROUND: Not applicable. DOUBLE-CHECK: In the case when x = 0 it is seen that V = 0, which is expected for the point between two opposite charges. Then consider next the limit .x d>> For this limit, the denominator simplifies to:


901

2 2 2cos coscos co 1 1s .d dx xd x xd x x xx x

− + = − + ≈

θ θθ θ

The numerator is simplified to (using series expansion): 1/2 1/2

2 2 cos coscos cos

cos cos c

1 1

1 11 1 s2

o2

.

d dx xd x xd x xx x

d dx x dx x

θ θθ θ

θ θ θ

+ − − = + − − ≈ + − − =

The potential simplifies to: 2 2

coscos ,kpdkqx x

V = =

θθ where p = qd is the electric dipole moment of two

point charges. These two cases confirm that our answer is correct.

23.37. THINK: The water droplet can be thought of as a solid insulating sphere of diameter 50. μm0 d = and a total charge of 20.0 pC.q = The potential is then found by integrating the electric field it produces from infinity to the centre. The electric fields inside and outside the sphere are different. SKETCH:

RESEARCH: The electric potential is found by: ( ) ( ) ( ) .

rV r V E r dr

∞− ∞ = − Since the water droplet is a

non-conducting sphere, the electric field outside the sphere is = 21 / ,kq rE while inside the sphere is

= 32 / ,kqr RE where / 2R d= is the radius of the sphere.

SIMPLIFY:

(a) The potential on its surface, ,r R= is: ∞ ∞

∞

− = − = − −

= = − 2 2

1( ) 0 .R

R R drdkq kqV R kr q kqr Rr r

(b) The potential inside the sphere at center, 0,r = must be broken into 2 parts. 0 0 0

0

2 3 3

3

21 2 2

2

1(0) 02

1 3 30 (0) ( ).2 2 2 2

R

R R

R R R

k k k k k kE dr E dr dr dr rdq qr r

k k k k k

q q q qV rR Rr R R R

q q q q qR V V RR R R RR

∞ ∞

− = = − − = − − = + =

− − = − − =

=

CALCULATE:

(a) ( )( )9 2 28.9875 N m /C 20.0 pC

( ) 7190 V50

10μm / 2.0

V R⋅

= =


902

(b) ( )= =3(0) 7190 V 10785 V2

V

ROUND: (a) ( ) 7.19 kVV R = (b) (0) 10.8 kVV = DOUBLE-CHECK: Though these values seem large, the droplet has a charge density of 3300 C/m , which is quite large for an object. Therefore, the values seem reasonable.

23.38. THINK: Both a proton and electron of a Hydrogen atom have a charge of .q e= ± If the electron orbits the

proton at a distance of 10100.529 m,a −= ⋅ then the electric force is the same as the centripetal force. The escape speed of an object is the speed needed for its kinetic energy to equal its potential energy. The kinetic energy the electron needs to escape minus the potential energy it has in orbit is then the energy needed to remove the electron from orbit. SKETCH:

RESEARCH: The electric force the electron feels is = 2 2e / .F ke a The centripetal force to keep electron in

orbit is = 2c e / .F m v a The potential energy of electron in orbit is = 2 / .U ke a The kinetic energy it has for

escape speed is = 22 ee / 2.K m v The kinetic energy the electron has in orbit is = 2

e1 / 2.K m v . SIMPLIFY: (a) Since the electric force is the only force acting on the electron:

= = =22 2

ece 2

e

.m vke keF F v

a m aa

(b) If electron escapes its orbit, it needs enough kinetic energy to counter its potential energy:

= = = =2 2

22 e e e

e

1 2 2 .2

ke keK U m v v va m a

(c) The additional energy the electron needs to escape is equal in the change in kinetic energy:

( ) = Δ = − = − = − = − =

2 2 22 2 2 2

2 1 e e e e e ee e

1 1 1 1 2 1 .2 2 2 2 2

ke ke keE K K K m v m v m v v mm a m a a

CALCULATE:

(a) ( )( )

( )( )

29 2 2 196

31 10

8.9875 N m /C 1.602 C2.188 m/s

9.109 kg 0.529

10

m

1010

10 10v

−

− −

⋅= ⋅

⋅ ⋅

⋅=

(b) ( )⋅ ⋅= =6 6e 2 2.188 m/10 10s 3.094 m/sv

(c) ( )( )

( )

29 2 2 1918

10

8.9875 N m /C 1.602 C2.18 J 13.6 eV

2 0.5

·10 1010

29 m10E

−−

−=

⋅⋅

⋅= =

ROUND: (a) ⋅= 612.1 s09 m/v (b) ⋅= 6

e 13.0 s09 m/v


903

(c) 13.6 eVE = DOUBLE-CHECK: Both velocities are less than speed of light, so they make sense. Also, 13.6 eV is the experimentally found energy of an electron in a ground state of a hydrogen atom, so it makes sense too.

23.39. THINK: Each charge, three at q = 1.5 nC and one at -q, are placed the corners of a square of sides l = 2a = 5.4 cm. Since the point P in space is located above the very center of the square, each charge is the exact same distance from P. The point P is a distance c = 4.1 cm above the center of the square. The electric potential can be determined as the sum of the four individual point charges. SKETCH:

RESEARCH: The distance from each charge to point P is: 2 2 2

2 2 .2 2 2l l lr c c = + + = +

The electric

potential at this point is =

=4

1.i

ii

kqVr

SIMPLIFY: − = + + + = =

+2

2

2

2 2q q q q q qVr r r r r l

kkc

k

CALCULATE: ( )( )

( ) ( )

9

2

92 2

2

2 8.9875 N m /c 1.5 10 nC481.2 V

0.054 cm0.041 cm

10

2

V−⋅ ⋅

= =

+

ROUND: 480 VV = DOUBLE-CHECK: Given the charges and distances involved, this value seems reasonable.

23.40. THINK: The electric potential at a point P, a distance y above the end of a rod, can be derived by simply integrating the charge over the length of the rod, L. The distance to the point P, from a point on the rod is found by using the Pythagorean theorem. The charge distribution of the road is .cxλ = SKETCH:


904

RESEARCH: The total charge of the rod is q Lλ= , so a small element of length dx has a charge

.dq dxλ= At any given point along the rod, the distance from it to P is 2 2 .r x y= + The electric

potential at point P is = 0/ .

LqV d rk

SIMPLIFY:

( )λ + += = = = += −

+ 2 2 2 2

0 0 0 02 2 2 02 =

L L L LLkdq dx cxdx xV y y yr r y

dxk k kc kc x kcy

Lx x

CALCULATE: Not applicable. ROUND: Not applicable. DOUBLE-CHECK: From the expression, if y L>> (far away point) then V = 0, which would be expected. Likewise, if ,L y>> the potential becomes e ,V K cL= which is constant. This is expected for an infinite distance, so it makes sense.

23.41. THINK: The electric field, 0 ˆ,xE E xe x−=

has a maximum when its derivative with respect to x is zero. The electric potential is found by integrating the electric field between the two points 0 and max .x SKETCH:

RESEARCH: Electric field is at maximum when / 0.dE dx = The potential difference between 0 and maxx

is max

0.

xdxV E= −

SIMPLIFY:

(a) ( ) ( ) ( ) ( )0 0 0 .x

x x x xd ed xdE d E xe E e x E e xe

dx dx dx dx

−− − − −

= = + = −

If max0 : 1x xdE e xe xdx

− −= = =

(b) ( ) ( ) ( )1 1 1 10 0 0 0 0

1

0 00 01 1 2 1x x x xV E xe E xe E x e E x e Ed edx x− − − − − = − = − = − − + = + = −

CALCULATE: There is no need to calculate. ROUND: There is no need to round. DOUBLE-CHECK: The answer is reasonable.

23.42. THINK: The electric potential at a point a distance x from the center of a disk with inner radius 1R and outer radius 2R is found by integrating the charge over the radius of the disk and considering a ring of charge for a given radius. The distance to the point of interest and any point along a ring of given radius is found using the Pythagorean theorem.


905

SKETCH:

RESEARCH: Assuming the disk has a uniform charge distribution, the total charge is ,q Aσ= where A is the area of the disk and σ is area charge density. The area of a thin ring along disk is 2 .dA rdrπ= The

distance from a point along the disk to a point x along the central axis of the disk is 2 2 .l r x= + A small element of charge along the disk is written as 2 .dq dA rdrσ πσ= = The potential then at a point along the

x-axis is = / .k dq lV

SIMPLIFY:

( )πσ πσ πσ πσ = = = + = + + +−

+

22 2

2 2 1

2 2 2 2 2 222 2 2 2 1 2 2 2 2

RR R

R R RV k k k xrdr rdr r R

rxRx

rk x

x

CALCULATE: Not applicable. ROUND: Not applicable. DOUBLE-CHECK: To determine if this value is reasonable, determine the electric field it produces:

2 2 21

20 2

.2x

dV X XEdx xR R x

σ = − = − + +

If 1R → ∞ and 2 0R → , the disk is an infinite plane and the electric field is 0/ 2 ,xE σ= so it makes sense.

23.43. The electric field is related to the potential difference by = −Δ Δ/ .E V x So, when 20 ,V V x= where

20 270 V/m ,V = the x-component of the electric field at x = 13 cm is then:

( )( )20 13 cm

13 cm

2 2 270 V/m 0.13 m 70. V/m.x xx

dVE V xdx =

=

= − = − = − =

23.44.

(a) The left plate has a potential of = +1 200.0 VV and the right plate has a potential of = −2 100.0 V,V so the potential difference across the plates is Δ = − =1 2 300.0 V.V V V The electric field from plate to plate is:

,dV VEdx x

Δ= − ≈Δ

where 1.00 cm.xΔ = Therefore, = = ⋅ 4300.0 V 3.00 10 V/m.0.0100 m

E

(b) If the electron only travels / 2,d x= Δ the change in electric potential is ′Δ = = 150.0 V.V Ed Since all its initial potential energy becomes kinetic energy:

( )( )− −′= = Δ = ⋅ = ⋅19 17i 1.602 10 C 150.0 V/m 2.40 10 JK U e V or K = 150. eV.


906

23.45. The electric field from an electric potential, ( ) 2 31 2 ,V x V x V x= − where 2

1 2.00 V/mV = and 3

2 3.00 V/mV = is found by:

( )2 3 21 2 2 13 2 .dV dE V x V x V x V x

dx dx= − = − − = −

This field produces a force on a charge, 1.00 μC,q = of .F qE= The acceleration of the charge is / / ,a F m qE m= = where 2.50 mg.m = Therefore,

( ) ( ) ( )( ) ( )( )( )

26 3 222 1 2

6

1.00 10 C 3 3.00 V/m 2.00 m 2 2.00 V/m 2.00 m3 211.2 m/s .

2.50 10 kg

q V x V xa

m

−

−

⋅ −− = = =⋅

23.46. In three dimensions, the electric field is:

( ) ˆ ˆ ˆ, , .V V VE x y z x y zx y z

∂ ∂ ∂= − + + ∂ ∂ ∂

Therefore, if ( ) 2 2, , ,V x y z x xy yz= + + 22 ,V x yx

∂ = +∂

2V xy zy

∂ = +∂

and :V yz

∂ =∂

( ) ( ) ( )( ) ( ) ( )( ) ( )( ) ( )( ) ( )

2

2

ˆ ˆ ˆ, , 2 2

ˆ ˆ ˆ ˆ ˆ ˆ3, 4, 5 2 3 4 2 3 4 5 4 22 29 4

E x y z x y x xy z y yz

E x y z x y z

= − + − + −

= − + − + − = − − −

23.47. THINK: The electric potential, ( )2 23000 5 / m V,V x= − is a function of x and thus acts only in one

dimension. The electric field is found by differentiating the electric potential. The acceleration of a proton (q = +e, x = 4 m and 27

p 1.673 10 kg)m −= ⋅ is then found by relating the electric field to the force on the proton. Since the electric field is not constant, kinematics cannot be used to determine the final speed. Conservation of energy must then be used to relate its final kinetic energy to the initial potential energy it has. SKETCH:

RESEARCH: The electric field is determined by ( ) ( ) / .E x dV x dx= − The force on the proton is given by

( ) ( ),F x qE x= and this force is also related to acceleration by ( ) ( )p .F x m a x= The change in electric

potential from 1x and 2x is ( ) ( )2 1 ,V V x V xΔ = − so the change in potential energy is .U q VΔ = Δ From conservation of energy: .U KΔ = −Δ SIMPLIFY:

(a) ( )2

22

53000 V 10 V/mm

d xE x xdx

= − − =

(b) ( ) ( ) ( ) ( ) ( ) 2

pp p p

10 V/m F x qE x qxF x m a x a xm m m

= = = =

(c) ( ) ( ) ( )2 22 1 2 12

5 Vm

V V x V x x xΔ = − = − −


907

Therefore, ( )2 22 12

V5 .m

U q x xΔ = − − Use the equation: ( )2 2 2p f i p f

1 12 2

K m v v m vΔ = − = , when

( ) ( )2 2 2 2 2 2 2p f 2 1 f 2 1

p

1 10 5 V/m V/m .2

qK U m v q x x v x xm

Δ = −Δ = − = −

CALCULATE: (a) Not applicable.

(b) ( ) ( )( )19 29 2

27

10 1.602 10 C 4.00 m V/m4.00 m 3.8302 10 m/s

1.673 10 kga

−

−

⋅= = ⋅

⋅

(c) ( ) ( ) ( )

192 2 2 5

f 27

10 1.602 10 C10.0 m 4.00 m V/m 2.836 10 m/s

1.673 10 kgv

−⋅ = − = ⋅ ⋅

ROUND: (a) ( ) 210 V/mE x x=

(b) 9 23.83 10 m/sa = ⋅ (c) 5

f 2.84 10 m/sv = ⋅ DOUBLE-CHECK: The units for the ( )E x expression are valid. The final velocity is lower than the speed of light, so it is reasonable. The acceleration is high; however, the purpose of the device is to accelerate particles to large speeds over short distances.

23.48. THINK: All points in space will be influenced by the infinite plane of charge with surface charge density, 24.00 nC/m ,σ = and a point charge, q = 11.0 nC, located 0 2.00 mx = in a perpendicular direction from

the plane. The plane produces a constant electric field. The overall potential at any point between the two will be the sum of the two individual potentials. The minimum is found by differentiating the potential in one dimension and setting it to zero. The derivative of the potential with respect to position is also the electric field. Therefore, when the potential is a minimum, the electric field is zero. SKETCH:

RESEARCH: The electric field produced by the plane of charge is p 0/ 2 .E σ ε= The electric potential

from a constant electric field is Ex = V. The electric potential from a point charge is = / ,V kq r where

0 .r x x= − The electric potential is at a minimum when / 0.dV dx E= = SIMPLIFY: (a) The electric potential from the plane along the x-axis is σ ε π σ= = =1 p 0/ 2 2 .V E x x k x The electric

potential from the charge, q, is ( )= −2 0/ .V kq x x The total electric potential is:


908

πσ

= + = + − tot 1 2

0

2 .qV V V k xx x

(b) ( )( )

π σ π σ−= + − = − =−

10 2

0

2 2 0dV dx d kqk kq x x kdx dx dx x x

( )2 2 2 2 20 0 0 0 02 2 0

2 2q qx x x x x x x x x xπσ πσ

− = − + = − + − =

( )( )

2 20 0 0

0 0 0

2 4 4 / 2 / 2 / 2 .

2

x x x qx x q x q x x

πσπσ πσ

± − − = = ± = − <

(c) E is zero at the same position of minimum in V. CALCULATE: (a) Not applicable.

(b) ( )2

11.0 nC2.00 m 1.338 m2 4.00 nC/m

xπ

= − =

(c) 1.338 mx = ROUND: (a) Not applicable. (b) 1.34 mx = (c) 1.34 mx = DOUBLE-CHECK: The minimum is located closer to the point charge than it is to the plane of charge, as it should be.

23.49. THINK: The position, r, must be defined for three-dimensional space so that each derivative has a non-zero answer. While the potential is a scalar, each derivative is actually a vector that points in that direction, i.e. xE points in the x-direction. SKETCH: Not applicable.

RESEARCH: The position in three-dimensional space is given by 2 2 2 .r x y z= + + The electric field in

direction α is δ α δα= −

i ˆ / ,E V where , , .x y zα =

SIMPLIFY: ( ) ( ) ( )δ δδ δ

−= − = − + + = + + =

3/22 2 2 2 2 23

ˆ ˆ ˆ ˆ2 .2x

V kq kqE x kq x y z x x y z x x xxx x r

Likewise,

=

3ˆy

kqE yyr

and =

3ˆ.z

kqE zzr

Therefore, ( ) ( )= + +

3ˆˆ ˆ .kqE r xx yy zz

r

CALCULATE: Not applicable. ROUND: Not applicable. DOUBLE-CHECK: In vector notation, ˆˆ ˆxx yy zz+ + can be written as ˆ.r rr= This is evident if you let

2 .r r r= Therefore, the expression for electric field for a point charge can be written as:

( ) = =

3 2ˆ ˆ.kq kqE r rr r

r r

Since the potential was for a point charge, this makes sense.

23.50. THINK: Given the electric potential, ( ) 2 ,V x Ax= the potential energy can be determined. The force a particle feels is related to the derivative of the potential energy of a particle. If the particle is to behave like a harmonic oscillator, then the force needs to be related to a force resulting from a spring. This will yield a spring constant, k, which is then related to the period of the motion. The units of A are 2V/m . To avoid confusing the spring constant with the Coulomb constant, the spring constant will be denoted K here.


909

SKETCH:

RESEARCH: Given an electric potential, ( ),V x the potential for an electron is ( ).eV x The force such a

potential causes is ( ) ( ) / .F x dU x dx= − If this force causes simple harmonic motion, it should resemble

the force of a spring, = − .F Kx The period of an oscillating spring is given by π= 2 / .T m K SIMPLIFY: The force of this potential is ( ) 2 / 2 .F x dAex dx Aex= − = − Relating this force to = − :F Kx

= − = − =2 2 .F Aex Kx K Ae

The period of this oscillation is then: e2 .2m

TAe

π=

CALCULATE: Not applicable. ROUND: Not applicable. DOUBLE-CHECK: Checking the units in the expression for the period:

( )2 2 2

kg kg kg kg 1 s.N/C C/m N/mV/m C kg m/s /m 1/s

T = = = = = =

Units of time are necessary for the period.

23.51. THINK: The electric potential is given as ( ) 2 2/0 .r aV r V e −= The electric field and charge density are

related to the first and second derivative of the electric potential. The total charge is the charge density integrated over all space. Rather than work in Cartesian coordinates, remain in r-space. SKETCH:

RESEARCH: Given an electric potential ( ),V r the electric field it produces is ( ) / .rE dV r dr= − The

charge distribution is given by an electric field, ( ),E r as ( ) ( )0 / .p r dE r drε= The total charge is then:

( )0

.Q p r dr∞

=

SIMPLIFY:

(a) The electric field is: ( ) ( ) ( )2 2 2 2 2 2/ / /00 0 2 2

22 .r a r a r adV r V rd rE r V e V e edr dr a a

− − − = − = − = − − =

(b) The charge distribution is then: ( ) ( ) 2 2 2 22

/ /0 0 00 0 2 2 2

2 21 2 .r a r adE r V r Vd rp r e e

dr dr a a aεε ε − − = = = −


910

(c) The total charge is: 2 2

2/0 0

2 20

21 2 .r aV rQ e dr

a aε ∞ −

= −

Let 0 02

2,

VA

aε

= r drx dxa a

= = and the

above equation becomes ( )2 22

02 .x xQ Aa e x e dx

∞ − −= − Referring to a table of definite integrals:

2

0 2xe dx π∞ − = and ( )22

0

3 / 22 2 .

2 2xx e dx π∞ − Γ

= = Therefore, ( )/ 2 / 2 0.Q Aa π π= − = The total

net charge is zero, i.e. there is equal negative charge and is positive charge. A plot of ( )p r vs. r is below.

Notice that the area above the r-axis (positive charge) is equal to the area below the r-axis (negative charge). CALCULATE: Not applicable. ROUND: Not applicable. DOUBLE-CHECK: Given that the electric potential is a Gaussian distribution, which shows symmetry, then symmetry in charge is expected. Such symmetry in charge would mean equal negative and positive charge, resulting in a zero net charge.

23.52. THINK: For this problem, assume the directions of the electric fields across the plates are appropriate to cause the necessary deflection, so only the magnitudes must be considered. The electron,

31e 9.109 10 kgm −= ⋅ and ,q e= is deflected from ( )0,0 to ( )0,8.00 cm . Since the deflection is only in

the y-direction, the second pair of plates ( )= =5 cm, 4 cmd D that cause horizontal deflection must have no potential across them, so only the first set of plates cause deflection. The voltage across the plates causes an electric field and then in turn a force that causes the electron to accelerate vertically in this area. Once out of this field, the electron is moving with constant velocity. Kinematics can then be used to determine what voltage is needed to cause the proper deflection. L = 40.0 cm and 7

i 2.00 10 m/s.v = ⋅ SKETCH:

RESEARCH: The electric potential across the second set of plates is V 0,V = while across the first set the potential is H H .V E D= The force that the horizontal plates cause is H e .yF qE m a= = During the whole

trajectory, the horizontal velocity, i ,v is constant. The time it take to cross the first set of plates is

1 i/ ,t d v= while its vertical displacement is 21 1 / 2.yy a tΔ = After the first plate, its vertical velocity remains


911

constant as 0 12 .yv a y= Δ The time after the first plate is ( )2 i/ .t d L v= + Then the vertical displacement

is 2 0 2 .y v tΔ = The total y-displacement is then 1 2 .y y yΔ = Δ + Δ

SIMPLIFY: The vertical acceleration is: HH

e e

.y

e VqEa

m m D= = The total y-displacement is then:

( ) ( )

2 22 2 2

1 2 1 0 2 12 2i i1 1

22 2H

12 2 2 2i ei i i i

1 22 2 2

2 2.

2 2 2 2

y yy y y

y y

a d a dd L d Ly y y a t v t a y a tv vv v

e Vd d L d d d Ld d L da t av m Dv v v v

+ +Δ = Δ + Δ = + = + Δ = +

+ + + += + = + =

The potential across the plates is then: ( ) 12

eH 2

i

2.

2d d d Lm D y

Ve v

− + +Δ

=

CALCULATE:

( )( )( ) ( ) ( ) ( ) ( )( )( )

1231

H 19 27

9.109 10 kg 4.00 cm 8.00 cm 5.00 cm 2 5.00 cm 5.00 cm 40.0 cm1.602 10 C 2 2.00 10 m/s

306.45 V

V

−−

−

⋅ + + = ⋅ ⋅

=

ROUND: H 306 VV = DOUBLE-CHECK: This is the same principle that a TV works by and 300 V is within the range of electric potentials that a TV can produce.

23.53. If the proton comes to a complete stop at 151.00 10 m,r −= ⋅ then all of its initial kinetic energy is converted to potential energy:

( )( )22 9 2 2 1913

15

8.9875 10 N m /C 1.602 10 C2.31 10 J or 1.44 MeV.

1.00 10 mk e

Ur

−−

−

⋅ ⋅= = = ⋅

⋅

23.54. The barium nucleus has a charge of 1 56q e= and the krypton nucleus has a charge of 2 36 .q e= Their combined kinetic energy is =f 200. MeV,K which is equal to their initial potential energy, =i 1 2 / .U kq q r r is the separation of the two atoms, assumed to be the average size of the uranium atom, so:

( )( ) ( )( )( )( ) ( )( )

−−

−

⋅ ⋅= = = = = ⋅

⋅

22 9 2 2 19141 2

i f 19f

8.9875 10 N m /C 2016 1.602 10 C36 56 1.45 10 m.

1.602 10 J/ 1 eV200. MeV

k ekq qU K rr K

23.55. Assuming the first ion is brought in from an infinite distance, then the work needed to bring it a distance of −= 1410 mr to the other ion is the potential energy of the two ions:

( )( )22 9 2 2 19

14 19

8.9875 10 N m /C 1.602 10 C 1 eV 140,000 eV.10 m 1.602 10 J

k eU

r

−

− −

⋅ ⋅ = = = ⋅

23.56. THINK: If each charge initially starts at an infinite distance, then the work done to move each charge to its final position is simply the potential energy of each charge in that position (the potential energy at infinity is zero). The charges are 1 1.0 pC,q = 2 2.0 pCq = and 3 3.0 pC.q = Since they are on the corners of an equilateral triangle, each charge is the same distance, l = 1.2 m, from the others.


912

SKETCH: (a) (b)

(d)

RESEARCH: In general, the potential for point charges is: = i j

i,j ij

,kq q

Ur

where ijr l= for all i and j.

SIMPLIFY: The work done to bring in each charge is i j

i,j ij

.kq q

Wr

=

(a) Since there is no charge for 1q to interact with, 1 1 0 J.U W= =

(b) Charge 1q is present as 2q is moved to its corner, so 1 22 2 .

kq qU Wl

= =

(c) Charges 1q and 2q are present as 3q is moved to its corner, so

( )1 3 2 33 3 1 3 2 3 .

kq q kq q kU W q q q ql l l

= = + = +

(d) The total energy is tot 1 2 3 .U U U U= + + CALCULATE: (a) 1 0 JW =

(b) ( )( )( )

( )9 2 2 12 12

142

8.99 10 N m /C 1.0 10 C 2.0 10 C1.498 10 J

1.2 mW

− −−

⋅ ⋅ ⋅= = ⋅

(c) ( ) ( )( ) ( )( )

9 2 212 12 12 12

3

14

8.99 10 N m /C1.0 10 C 3.0 10 C 2.0 10 C 3.0 10 C

1.2 m6.743 10 J

W − − − −

−

⋅ = ⋅ ⋅ + ⋅ ⋅

= ⋅

(d) ( ) ( ) ( )14 14 14tot 0 J 1.498 10 J 6.743 10 J 8.241 10 JU − − −= + ⋅ + ⋅ = ⋅

ROUND: (a) 1 0 JW =

(b) 142 1.5 10 JW −= ⋅

(c) 143 6.7 10 JW −= ⋅

(d) 14tot 8.2 10 JU −= ⋅

DOUBLE-CHECK: These small energy values are reasonable for such small amounts of charge.


913

23.57. THINK: Two balls have masses, 1 5.00 gm = and 2 8.00 g,m = and charges, 1 5.00 nCq = and

2 8.00 nC.q = Their center separation is l = 8.00 mm, and although the balls are not point charges, use the center separation to determine the potential energy stored in the two. Conservation of momentum and energy will allow the velocities of each to be determined. Since they are like charges, they repel and so the velocities will be in different directions. SKETCH:

RESEARCH: The balls have no initial momentum, so by the conservation of momentum: 1 1 2 2 .m v m v= The initial potential energy of the two balls is given by =i 1 2 / .U kq q l The final kinetic energy of the balls is

given by ( ) ( )2 2f 1 1 2 2/ 2 / 2 .K m v m v= +

SIMPLIFY: From the conservation of momentum: 2 1 1 2/ .v m v m= Conservation of energy then requires: = = + = +

22 2 21 2 1 1

i f 1 1 2 2 1 1 22

1 1 1 .2 2 2

kq q m vU K m v m v m v ml m

Therefore,

= + = +

221 2 1 1 2 2

1 1 1 22 1 2 1

2 2 .

kq q m kq q mm v vl m l m m m

CALCULATE:

( )( )( )( ) ( )

9 2 2

1 2

2 8.9875 10 N m /C 5.00 nC 8.00 nC 0.00800 kg0.008.00 m 0.00500 kg 0.00800 kg 0.00500 kg

0.1052 m/s

v ⋅ = +

=

( )2

5.00 g 0.1052 m/s0.06575 m/s

8.00 gv = =

ROUND: 1 0.105 m/sv = and 2 0.0658 m/sv = DOUBLE-CHECK: The charges are small and the masses relatively large, so the velocities obtained for the masses should be small.

Additional Problems

23.58. Conservation of energy can be considered to relate the change in kinetic energy to the change in potential energy by: f i i f f i f .K U K K U U K U UΔ = −Δ = − = − = − Each proton has the same mass,

27p 1.673 10 kg,m −= ⋅ and thus has the same kinetic energy, so the total kinetic energy is 2

f p .K m v= Therefore,


914

( )( )( )

22 22 f i f i

f i f pi f i f p i f

29 2 2 19

27

1 1

8.9875 10 N m /C 1.602 10 C 10.00 mm 1.00 mm 11.1 m/s.10.00 mm 1.00 mm1.673 10 kg

k er r r rK U U m v k e k e v

r r r r m r r

−

−

− −= − = − = =

⋅ ⋅ −= = ⋅

23.59. The battery places an electric potential of 12 V on the entire conducting surface of the hollow metal sphere. Inside the conducting shell, the electric field is zero and the electric potential remains at 12 V.

23.60.

Since the object is a solid conducting sphere, the electric potential is distributed evenly through the sphere, so it is the same at 1 0 mr = and 2 3 mr = by = / ,V kq R where 4 mCq = and R = 3 m. Outside the sphere, it acts as a point charge, so the electric potential is = 3/ .V kq r

(a) ( )( )9

32 2 38.9875 10 N m / C 4.00 10 C

1.20 10 kV3.00 m

kqVR

−⋅ ⋅= = = ⋅

(b) ( )( )9

32 2 38.9875 10 N m / C 4.00 10 C

1.20 10 kV3.00 m

kqVR

−⋅ ⋅= = = ⋅

(c) ( )( )9

32 2 38.9875 10 N m / C 4.00 10 C

7.19 10 kV5.00 m

kqVR

−⋅ ⋅= = = ⋅

23.61. The infinite plate of surface charge density, 6 23.5 10 C/m ,σ −= ⋅ produces a constant electric field, 0/ 2 .E σ ε= In going from point A to B, any movement perpendicular to the electric field results in no

change in electric potential. Therefore, the only displacement of importance is 1.0 m.yΔ = − The change in potential is independent of the charge Q, and since the electric field is constant, it is the product of the electric field times the displacement:

( )( )( ) ( )σ

ε

−

−

⋅Δ = − Δ = − Δ = − − = ⋅

⋅

6 25

12 2 20

3.5 10 C/m1.0 m 2.0 10 V.

2 2 8.854 10 C / N mV E y y

23.62. Conservation of energy means the change in kinetic energy is equal to the magnitude of change in potential energy, .K UΔ = Δ The change in potential is ,U q VΔ = Δ where q = +e and 21.9 kV.VΔ =

The initial velocity is zero, so the change in kinetic energy is 2e f / 2,K m vΔ = where 31

e 9.11 10 kg.m −= ⋅

Therefore, ( )( )−

−

⋅ΔΔ = Δ = = Δ = = = ⋅⋅

192 4

e f f 31e

2 1.602 10 C 21.9 kV1 2 8.78 10 km/s.2 9.11 10 kg

e VK U m v e V vm


915

23.63.

Since the object is a solid conducting sphere, the electric potential is distributed evenly through the sphere, so it is the same at points B and C and is given by = / ,V kq R where 66.1 10 Cq −= ⋅ and R = 18 cm.

Therefore, ( )( )9 2 2 6

5B C

8.9875 10 N m / C 6.1 10 C3.0 10 V.

0.18 mV V

−⋅ ⋅= = = ⋅ Outside the sphere, at A 24 cm,r =

the electric potential is that of a point charge, so:

( )( )−⋅ ⋅= = = ⋅

9 2 2 65

AA

8.9875 10 N m / C 6.1 10 C2.3 10 V.

0.24 mkqVr

23.64. The electric field of a spherical conductor is the same as that of a point charge at the center of the sphere with a charge equal to that of the spherical conductor. The potential outside the sphere is therefore also the same as a point charge:

( )πε

−

−

⋅ = = ⋅ = ⋅ ⋅

69 2 2 4

surface 10

1 1.00 10 C8.99 10 N m / C 8.99 10 V.4 1.00 10 m

qVr

23.65. First, determine the relationship between the electric field and the potential. The electric field is given by 2/ .E kq r= The potential is given by / .V kq r= Therefore, the maximum voltage is

( )( )6 5max max 2.00 10 V/m 0.250 m 5.00 10 V.V E r= = ⋅ = ⋅

The maximum charge that it can hold is

( ) ( )( )

2 625max

max 9 2 2

0.250 m 2.00 10 V/m1.39 10 C.

8.99 10 N m / Cr E

qk

−⋅

= = = ⋅⋅

23.66. Consider the conservation of energy to solve the problem. The potential energy is given by .U qV= 2

kq kqV Ur r

= =

The moving proton will stop a distance r from the stationary proton, where the electric potential energy is equal to the initial kinetic energy:

( )( )( )( )

22

p

29 2 2 1929

2 227 4p

1 2

2 8.99 10 Nm / C 1.602 10 C2 1.82 10 m.1.673 10 kg 1.23 10 m/s

kqK U m vr

kqrm v

−−

−

= =

⋅ ⋅= = = ⋅

⋅ ⋅

23.67. (a) First an expression must be determined for the surface charge on each sphere. The surface area of a sphere is 24 .rπ The surface charge density is given by:


916

11 2

14q

rσ

π= for the first sphere, and 2

2 224

qr

σπ

= for the second sphere.

2 21 1 2

2 2 1

20.0 cm 4 :110.0 cm

q rq r

σσ

= = =

(b) The charge flow stops when the potential is equal. If 1q and 2q are the final charge distributions after the potential of the two spheres are equal, then the following equations describe the potentials:

11

0 1

1 ,4

qV

rπε

=

22

0 2

1 .4

qV

rπε

=

1 2 11 2 1 2 2 2

1 2 2

10.0 cm 1 .20.0 cm 2

q q rV V q q q q

r r r = = = = =

Also, 1 2 200. μC.q q+ = Solving the two equations yields 1 200. / 3 μC 66.7 μCq = = and

2 400. / 3 μC 133.3 μC.q = = The amount of charge that flows through the wire is then

1 2 / 2 66.7 μC 133.3 μC / 2 33.3 μC.q q− = − =

23.68. The potential of a sphere is given by ( ) 0/ 4 ,V r q rπε= where q is the total charge of the sphere. The total

charge is given by 2s4 ,q rπ σ= where sr is the radius of the sphere. The potential difference between the

surface of the sphere and the point, P, is then given by:

( ) ( ) ( )

( )( )

12

s 0s p 2

0 s p 0 s p s ps

11210 2

2 3

41 1 1 1 1 112.566 V 12.566 V4 4

8.85 10 F/m 1 1 12.566 V 9.27 10 C/m .0.200 m 0.500 m0.200 m m

rqV r V rr r r r r rr

π εσ σπε πε

−

−−−

− = − = − = = −

⋅ = − = ⋅

23.69. Consider the conservation of energy to determine the final kinetic energy:

1 2 1 2 1 2final initial initial final final

0 initial 0 final 0 initial final

1 1 1 1 .4 4 4

q q q q q qK U K K U U K

r r r rπε πε πε

Δ = −Δ − = − = − = −

Thus, ( )( )( )9 2 2final

1 18.99 10 N m / C 5.0 μC 9.0 μC 2.02 J 2.0 J.0.10 m 0.20 m

K = ⋅ − = ≈

23.70. The potential of a spherical object with a uniform charge distribution is the same as that of a point charge

at the center of the sphere: ( )9 2 2 60

0

2.00 μC1/ 4 8.99 10 N m /C 8.99 10 V.4 2.00 mm

qV q rr

πεπε

= = = ⋅ = ⋅

The

potential difference has no angular dependence. If the potential in defined in terms of a charge distribution that depends on θ ,

( )0

1 ,4

V dVr

ρ θπε

=

the potential difference will have an angular dependence. Note that dV in the integral stands for differential volume.

23.71. THINK: First determine the total charge in each sphere based on the field. The charge from one sphere will flow into the other after they are connected until the potential of the two spheres are equal.


917

SKETCH:

RESEARCH: The electric fields of the two spheres are given by:

11 2

1

kqE

r= and 2

2 22

.kq

Er

=

The potentials are given by: 1

11

kqV

r′

= and 22

2

,kq

Vr

′=

where 1q ′ and 2q ′ are the charges of the first and second sphere after they reach the same potential (when

1 2 ).V V= Conservation of charges requires that, 1 2 1 2 .q q q q′ ′+ = + The final field strengths are given by:

11 2

1

kqE

r′

′ = and 22 2

2

.kq

Er

′′ =

The given values are 1 10 cm,r = =2 5 cmr and 1 2 3600. V/m.E E E= ≡ =

SIMPLIFY: The charge on each sphere before the two are connected is 21 1 /q Er k= and 2

2 2 / .q Er k= Once the spheres are connected, their potentials are equal:

1 2 21 2 2 1

1 2 1

.q q r

V V q qr r r′ ′

′ ′= = =

( ) ( )( )

( )( )

2 21 22 22

1 2 1 2 1 1 2 11 2 1

2 2 2 21 2 1 2

1 2 22 1 1 1 1 2

1 1 /

1 1 /

E r rr Eq q q q q r r qr k k r r

E r r r rE k Ek r r r r r r

+ ′ ′ ′ ′+ = + + = + = +

+ + ′ = = + +

Using 22 1

1

rq qr

′ ′= and 22 2 2/E kq r′ ′= gives

2 1 1 1 12 12 2

1 2 22 1

.r kq r kq rE Er r rr r

′ ′′ ′= = =

CALCULATE: ( ) ( )( ) ( )( )

( )2 2

1 2

10.0 cm 5.00 cm3600. V/m 3000. V/m

10.0 cm 10.0 cm 5.00 cmE

+′ = =

+


918

( )( ) ( )2

10.0 cm3000. V/m 6000. V/m

5.00 cmE′ = =

ROUND: 31 3.00 10 V/mE′ = ⋅ and 3

2 6.00 10 V/mE′ = ⋅ DOUBLE-CHECK: Since 1 2r r> it is expected that 2 1E E′ ′> for the electric fields at the surfaces of the spheres.

23.72. THINK: First determine the potential for each infinitesimal part of the ring and then sum over the whole ring. Using the relationship between the potential field and the electric field, E can be determined. SKETCH:

RESEARCH: The potential of each small dQ is given by 0/ 4 .dV dQ rπε= The total potential is then

0/ 4 ,V Q rπε= where 2 2 .r x R= + From the symmetry, it can be inferred that E

is pointing in the x-direction. The relation xE and x is given by / .xE V x= −∂ ∂

SIMPLIFY: 2 2

0 0

1 14 4

Q QVr x Rπε πε

= = +

( ) ( )3/2 3/22 2 2 20 0

1 24 2 4x

V Q x Q xEx x R x Rπε πε

∂ = − = − − = ∂ + +

CALCULATE: Not applicable. ROUND: Not applicable. DOUBLE-CHECK: Check if E is truly zero in the y- and z-directions:

0yVEy

∂= − =∂

and 0,zVEz

∂= − =∂

which confirms the result.

23.73. THINK: (a) First determine an expression for the total potential from both charges. After finding the expression, the potential can be determined. (b) The derivative of the expression determined in part (a) can be used to determine the minimum point. SKETCH: A sketch is not necessary. RESEARCH: (a) Let 1 0.681 nCq = and 2 0.167 nCq = be the two charges with positions 1 0r = and 2 10.9 cm,r = respectively. The total potential is given by:

1 2tot

0 1 0 2

1 1 .4 4

q qV

r r r rπε πε= +

− −

There are three cases, depending on the value of r:


919

1 2tot

0 1 2

14

q qV

r r r rπε

= + − − for 1 2, ,r r r> 1 2

tot0 1 2

14

q qV

r r r rπε

= − − − for 1 2r r r< < and

1 2tot

0 1 2

14

q qV

r r r rπε

= − − − − for 1 2, .r r r<

(b) The minima occur each time the derivative is equal to zero: tot / 0.V r∂ ∂ = SIMPLIFY: (a) There is nothing to simplify. (b) Take the derivative for all three cases.

1 2, :r r r> ( ) ( )

tot 1 22 2

0 1 1 2

1 .4

V q qr r r r rπε

∂ = − −

∂ − − The expression is equal to zero at infinity.

1 2 :r r r< < ( ) ( )

tot 1 22 2

0 1 2

1 .4

V q qr r r r rπε

∂ = − +

∂ − − The expression is zero when:

( ) ( ) ( )( )1 2 1 2

12 2 2 21 2 2

0 cmq q q q

rrr r r r r r

= = =− − −

( )2 22 2 2 2 2 2 2 2 2

21 1 1 1 2 1

1 1 1 ./ 1

r r q r q r q r q rr

q r q r q r qr q q−

= − = − = − = = +

CALCULATE:

(a) ( )9 2 2tot

0.681 nC 0.167 nC8.99 10 N m / C 46.78 V20.1 cm 0 20.1 cm 10.9 cm

V = ⋅ − = − −

(b) ( )

10.9 cm 7.28997 cm0.167 nC / 0.681 nC 1

r = =+

ROUND: (a) 46.8 V (b) 7.29 cm DOUBLE-CHECK: (a) The potential is positive and the potential from both charges is the sign that one would expect. This makes sense, since if a test charge was placed at 20.1 cm, it would move away from either one of the charges. (b) An equilibrium point will exist between the two charges, where the force from one is balanced by the other. Note that 0 7.29 cm 10.9 cm.< <

23.74. THINK: (a) The total potential of the origin can be determined using superposition. (b) The expression for the potential determined in part (a) can be used to find the point where the potential is zero. SKETCH: A sketch is not necessary. RESEARCH:

(a) 1 2tot

0 1 2

1 ,4

q qV

r rπε

= +

( ) ( )2 22 2 21 1 1 2.5 m 3.2 m ,r x y= + = + ( ) ( )2 22 2 2

2 2 2 2.1 m 1.0 mr x y= + = − +

1 22.0 μC and 3.1 μC.q q= = −


920

(b) Think of this as a one-dimensional problem with 1q at the origin. The distance between 1q and 2q is

given by ( ) ( )2 21 2 1 2 1 2 .d r r x x y y= − = − + − If r is the distance from 1q to the point where tot 0,V =

then:

Vtot = k q1

r+ q2

d − r

= 0 for .r d>

To determine the new point, simply switch to Cartesian coordinates:

2 1zero

2 1

,r r

r rr r

−= −

xzero = x1 + x2 − x1

dr and yzero = y1 + y2 − y1

dr.

SIMPLIFY: (a) Nothing to simplify.

(b)

Vtot = k q1

r+ q2

d − r

= 0 q1

r+ q2

d − r= 0 q1

r= − q2

d − r d − r

r= − q2

q1

dr

−1 = − q2

q1

r = d1− q2 / q1

CALCULATE:

(a) ( )( ) ( ) ( ) ( )

9 2 2 3tot 2 2 2 2

2.0 μC 3.1 μC8.99 10 N m / C 7.554 10 V2.5 m 3.2 m 2.1 m 1.0 m

V − = ⋅ + = − ⋅ + − +

(b) ( ) ( )2 22.5 m 2.1 m 3.2 m 1.0 m 5.099 m,d = + + − = ( )5.099 m 2.000 m

1 3.1 μC/2.0 μCr = =

− −

xzero = 2.5 m + −2.1 m- 2.5 m

5.099 m

2.000 m( )= 0.6957 m, yzero = 3.2 m + 1.0 m − 3.2 m5.099 m

2.000 m( )= 2.337 m

ROUND: (a) 37.6 10 V− ⋅ (b) ( )0.70 m,2.3 m DOUBLE-CHECK: (a) The total voltage has appropriate units: volts. (b) The point is between the two points, as one would expect because when going from a negative potential to a positive potential, the zero point is expected to be between the negative and positive charges.

23.75. THINK: (a) Since the electric field of a conducting sphere is the same as that of a point charge its center, the expression for the potential is the same. (b) The charge flow will stop when the potential of the two surfaces is equal. SKETCH: A sketch is not necessary. RESEARCH:

(a) sphere0

1 ,4

QVRπε

=

64.2 10 C,Q −= ⋅ R = 0.40 m.

(b) 1 21 2

0 1 0 2

1 1 ,4 4

Q QV V

R Rπε πε

= = =

61 2 4.2 10 C,Q Q Q −+ = = ⋅ 1 0.40 m,R = 2 0.10 mR =

SIMPLIFY: (a) Nothing to simplify.

(b) 1 2 11 2

1 2 2

Q Q R

Q QR R R

= =


921

Substitute this expression into 1 2Q Q Q+ = to get:

12 2 2

2 1 2

,1 /

R QQ Q Q QR R R

+ = =+

1 2 .Q Q Q= −

Charge flow is 2 .Q 11 2

0 1

14Q

ERπε

=

and 2

2 20 2

1 .4Q

ERπε

=

CALCULATE:

(a) ( )6

9 2 2 4sphere

4.2 10 C8.99 10 N m / C 9.44 10 V0.40 m

V− ⋅= ⋅ = ⋅

(b) 6

62

4.2 10 C 0.84 10 C,1 0.40 m / 0.10 m

Q−

−⋅= = ⋅+

6 6 61 4.2 10 C 0.84 10 C 3.36 10 CQ − − −= ⋅ − ⋅ = ⋅

( )( )( )( )

222 2 1

2 21 12

0.84 μC 0.40 m4

3.36 μC 0.10 mE Q RE QR

= = =

The electric field on the surface of the second sphere is four times larger than the first sphere. This is the inverse of the ratio of their radii. ROUND: (a) 4

sphere 9.4 10 VV = ⋅

(b) 62 0.84 10 CQ −= ⋅

DOUBLE-CHECK: (a) The correct units of a voltage are volts. (b) The charge flow is non-zero and comparable to the total charge, as one would expect.

23.76. THINK: Determine the potential of an infinitesimally small piece dy along the y-axis on the x-axis. Then integrate to determine the potential. SKETCH:

RESEARCH: 0

1 ,4

dydVr

λπε

= 0

,L

V dV= 2 2 ,r x y= + ,Ayλ = 3.06 m,x = 4.0 cm,L =

7 28.0 10 C/mA −= ⋅

SIMPLIFY: ( )2 2 2 2

0 0 2 2 00 0 0 0

14 4 4 4

LL Ldy dy Ay A AV dV x y x L xr x y

λπε πε πε πε

= = = = + = + − +

CALCULATE:

( )( ) ( ) ( )− = ⋅ ⋅ + − = ⋅ 2 27 9 2 2 28.0 10 C/m 8.99 10 N m / C 0.03 m 0.04 m 0.03 m 1.438 10 VV

ROUND: = ⋅ 21.4 10 VV


922

DOUBLE-CHECK: As x gets larger, 2 2 0,x y x+ − ≈ as expected.

23.77. (a) Let 1 3.00 mCq = − and 2 5.00 mCq = be located at 1 2.00 mx = and 2 4.00 m,x = − respectively. There are three cases:

1 2

0 1 2

1( )4

q qV x

x x x xπε

= + − − for > 1 2, ,x x x 1 2

0 1 2

1( )4

q qV x

x x x xπε

= + − − for 1 2x x x< < and

1 2

0 1 2

1( )4

q qV x

x x x xπε

= + − − for < 1 2, .x x x

The three cases stem from − − >1 2, 0.x x x x

(b) Case 1 2, :x x x> 1 2 1 2 2 1tot

1 2 1 2

0 .q q q x q x

V xx x x x q q

+= = − =

− − + Case 1 2 :x x x< <

1 2 1 2 2 1

1 2 1 2

.q q q x q x

xx x x x q q

−= − =

− − − Case 1 2, :x x x< This case yields the same results as the first case.

Zeroes occur at the following points: ( )( ) ( )( )1 2 2 1

1 2

3.00 mC 4.00 m 5.00 mC 2.00 m11.0 m,

3.00 mC 5.00 mCq x q x

xq q

− − ++= = =

+ − +

( )( ) ( )( )1 2 2 1

1 2

3.00 mC 4.00 m 5.00 mC 2.00 m0.250 m.

3.00 mC 5.00 mCq x q x

xq q

− − −−= = = −

− − −

(c) .VEx

∂= −∂

( ) ( )

1 22 2

0 1 2

14

q qE

x x x xπε

= + − −

for > 1 2, ,x x x ( ) ( )

1 22 2

0 1 2

14

q qE

x x x xπε

= − + − −

for

1 2x x x< < and

( ) ( )1 2

2 20 1 2

14

q qE

x x x xπε

= − − − −

for < 1 2, .x x x

23.78. THINK: The forces acting on the charge are the coulomb and gravitational forces. For equilibrium, the total force must be zero. SKETCH:

RESEARCH: gravity ,F Mg= − coulomb ,F nqE= E = V/d, total gravity coulombF F F= +

SIMPLIFY: For equilibrium:

tot 0 .nqV MgdF Mg nqE Vd nq

= = = =

If the voltage is halved: 1 .2

MgdVnq

= The total force is then:

tot1 1 .2 2 2

nqV Mgd Mg gMa F Mg nq Mg ad nq d

= = − = − = − = −


923

If the voltage is doubled: 2 .MgdVnq

= The total force is then:

tot12 .nqV MgdMa F Mg nq Mg Mg a g

d nq d

= = − = − = =

CALCULATE: Not applicable. ROUND: Not applicable. DOUBLE-CHECK: If the voltage is halved, the particle goes down. If the voltage is doubled, the particle goes up. In both cases, the result makes sense.

23.79. THINK: (a) The potential is a function of distance but not direction. Since every segment, dq, is the same distance, R, from the origin, they have the same potential. (b) Same as part (a). (c) The electric field is dependent on direction, so this is not possible. SKETCH: A sketch is not necessary. RESEARCH:

(a) 0

14

qVRπε

=

(b) 0

1 ,4dqdV

Rπε =

V dV=

(c) Nothing to research. SIMPLIFY: (a) Nothing to simplify.

(b) 0

0 0

1 1 ,4 4

q dq qVR Rπε πε

= = which is the same result as part (a).

(c) Nothing to simplify. CALCULATE: Not applicable. ROUND: Not applicable. DOUBLE-CHECK: The fact that the result from (b) matches the prediction made in (a) supports the prediction.

23.80. THINK: (a) First determine the expression for the potential contribution. (b,c) Can be determined after determining the expression for part (a). SKETCH:


924

RESEARCH: (a) Let the center of the sphere be at the origin of coordinates, with the exterior charge at z R− on the positive z-axis. Let the image charge be at a coordinate z on that axis, with .z a< The requirement that the surface of the charge be equipotential with potential zero takes the form:

( ) ( )2 2 2 2 2 22 2 2 20 (1).

2 cos 2 cosQ q Q q

R a aR Z a azx y R z x y z Z θ θ= + = +

+ − + −+ + − + + −

( ), ,x y z is any point on the surface of the sphere (so 2 2 2 2x y z a+ + = ) with cos .z a θ= (b) Since the electric field at the exterior charge is the same whether the sphere or the image charge is present, the force on the exterior charge toward the sphere is the same as the image charge would exert.

( )2

2 2 20 0

1 14 4z z

qQ Q aRF e eR aR zπε πε

= = − − −

ze is the unit vector in the positive z-direction, as defined above. (c) The surface charge density, ( ),σ θ on the sphere is given by Gauss’ law applied to a “pillbox” partially

embedded at any point in the surface of the sphere: ( ) 0 ,rEσ θ ε= where rE is the radial component of the net electric field at the surface of the sphere. This can be determined from the contributions of the exterior and image charges via Coulomb’s law:

( )( )

( )( )

( )( )

( )( ) ( )

( )

3/2 3/22 2 2 20

2 2 2 2

3/2 3/22 2 2 20 0

14 2 cos 2 cos

1 / / 11 1 .4 42 cos 2 cos

x y z x y z

x y z r

Q xe ye z R e q xe ye z Z eE

R a aR Z a aZ

Q R a Qa R axe ye ze e

R a aR R a aR

θπε θ θ

πε πεθ θ

+ + − + + − = + + − + −

− − = + + = + − + −

SIMPLIFY: (a) Rearranging yields:

( ) ( )2 2

2 2 2 22 22 cos .q qR a z a a R z

Q Qθ

+ − + = −

Since the right side of this equation depends on ,θ while the left side does not, they are equal for all θ if and only if both are zero. This implies 2 2/ / .q Q z R= Therefore, ( ) ( )2 2 2 2 0.z R a R z a+ − + = The

quadratic formula gives two solutions for this .z R= Hence, ,q Q= − which is trivial and 2 /z a R= and / ,q Qa R= − the desired solution. Equation (1) requires that q be opposite in sign to Q.

(c) Using the coordinate and result of part (a), and the radial unit vector, ,re which is equal to

( ) /x y zxe ye ze a+ + at the surface of the sphere. Note that, as expected, the net electric field is in the radial

(normal) direction at the spherical surface. The surface charge density is therefore given by:

( )( )

( )

2 2

3/22 2

/ 1.

4 2 cos

Qa R a

R a aRσ θ

π θ

− =+ −

The total induced charge on the sphere can be determined by integrating this over the surface. CALCULATE: Not applicable. ROUND: Not applicable. DOUBLE-CHECK: Gauss’ law applied to a spherical “skin” around the conductor implies that the total surface charge is equal to the imagine charge, / .Qa R−

Chapter 24: Capacitors

925


In-Class Exercises

24.1. a 24.2. a 24.3. a 24.4. d 24.5. d 24.6. c 24.7. a 24.8. b 24.9. a 24.10. b 24.11. a 24.12. a 24.13. (a) True (b) False (c) True (d) False (e) False Multiple Choice

24.1. b 24.2. c 24.3. c 24.4. d 24.5. c 24.6. a 24.7. d 24.8. c 24.9. a 24.10. (a) F (b) F (c) T (d) F (e) T

Questions

24.11.

If two insulators were used the charge would not be able to flow into the insulators and no charge would be stored; thus, conductors must be used.

24.12. Work has to be done to separate a positively charged plate from a negatively charged plate. When the battery is disconnected, the charge on the plates has nowhere to go and must remain the same. The electric field from a plane of charge depends only on the charge, not upon the distance from the plane (ignoring edge effects) so the electric field will remain the same. The voltage difference between the plates will just be the product of the electric field with the separation distance (since the electric field is constant), so as you pull the plates apart you’ll be moving the same charge against an even voltage. When the battery remains connected, the voltage remains the same as the battery voltage. So as the plates are pulled apart, the electric field must decrease to make up for the increase in separation, which means the charge must flow off the plates (which it can do, because there’s a path to the battery). Thus the force becomes less and less with greater separation; a smaller charge against a smaller field. The work done in increasing the separation is less. Therefore, the work done is greater when the capacitor is disconnected from the battery.

24.13. Since capacitors can store charge and are found in a lot of electrical equipment, grounding is done to ensure the excess charge can be discharged safely.

24.14. A value of zero may be suitable since no charge will be stored on such a capacitor yielding a capacitance of zero. This is since the charge will flow right through the conductor like flowing through a wire and will not be create an electric field within the capacitor.

24.15. 22

,2 2

qC VUC

Δ= = 2

oldold

,2

qUC

= 2

newnew

,2

qUC

= 0old ,

AC

dε

= ( ) ( )

0 oldnew .

A C dC

d d d dε

= =′ ′+ +

The separation

distance is increased by .d′ 2

new oldnew

1 12

q d d dU UC d d

′ ′+ = = +

The energy stored has increased from the work in pulling the charges apart.


926

24.16. In order to increase the capacitance from 10.0 μF to 18.0 μF in a capacitor, you could add a dielectric in the capacitor with a dielectric constant of 1.80.

24.17. For two capacitors ( )

1

1 2series

1 2 1 2

1 1 ,C CC

C C C C

−

= + = + then

( )1 2 2

series 21 2 2 1

,1 /

C C CC CC C C C

= = <+ +

and ( )1 2 1

series 11 2 1 2

.1 /

C C CC CC C C C

= = <+ +

The resultant capacitance is always smaller than the smaller of the two values. In particular, if the difference between the two value is large (an order of magnitude or more), the resultant capacitance is less than but very close to the smaller of the two. For example, if we connect in series a capacitor

1 1 μFC = with a capacitor 2 10 μF,C = we get a capacitance of 0.91 μF.

24.18. Two capacitors are connected in series. Assume the potential difference 0V is due to a battery. The circuit is:

In series, the equivalent capacitance is eq 1 2

11 1 .C C C

= + The potential difference supplied by the battery is

0eq 1 2 1 2

1 1 ,q q qV qC C C C C

= = + = +

where 1

1

qVC

= and 22

.qVC

= Solving for 1V in the above yields:

0 1 1 02 2

.q qV V V VC C

= + = − Note that since 01 2

1 1 ,V qC C

= +

( ) ( )0 0 1 2

1 21 2

.1/ 1/

V V C Cq

C CC C= =

++

Then

( )0 1 2 0 10 1 2 0 1 0 21 0 0

1 2 2 1 2 1 2 1 2

1 .V C C V CV C C V C V C

V V VC C C C C C C C C

+ − = − = − = = + + + +

Similarly for 2 ,V

( )0 1 2 0 20 1 2 0 2 0 12 0 0

1 2 1 1 2 1 2 1 2

1 .V C C V CV C C V C V C

V V VC C C C C C C C C

+ − = − = − = = + + + +

24.19. (a) The limit is when the field reaches the dielectric strength of the material. The dielectric strength of air is given as 62.5 kV / mm 2.5 10 V/m.= ⋅ 2/E kq r= for a sphere, so

( )( )

− −⋅

⋅ = = ⋅ ≈ ⋅9 2 2

6 7 72

8.99 10 N m /C2.5 10 V/m 6.952 10 C 7.0 10 C.

0.0500 m

qq

(b) When the charge in the sphere exceeds the limit specified in (a), the charge on the sphere will create a strong enough electric field to create an ionized conductive channel of air. The charge will spark though the air discharging the sphere slightly.

24.20. (a) The energy stored on the capacitor is 2

0 .2

C VU =


927

(b) Due to the power supply, the potential difference across the capacitors’ plates remains constant as V when the dielectric material is inserted. To maintain this constant ,V the power supply must supply additional charge to the plates. The capacitance becomes 0 ,C Cκ= and the energy becomes

220 .

2 2C VCVU

κ= =

(c) The dielectric is pulled into the space between the plates. There is an applied electric field 0E

between the plates. When the dielectric is inserted, the molecules of the dielectric align with the field:

Then on each surface of the dielectric, there is an induced charge opposite to the charge on the adjacent plate. Since unlike charges attract, the dielectric is pulled into the space between the plates.

24.21. The capacitor is disconnected from the power supply; the charge Q on each plate remains constant while the dielectric is inserted, while the potential difference across the plates is reduced by a factor of .κ The

force with which the slab is pulled into the capacitor is 2

,2

d d QF Udx dx C

= − = −

Q is constant.

Consider two regions of the capacitor, one which is empty and one which contains dielectric material.

These two “pieces” are in parallel, so ( ) ( )0 0

empty dielectric ,L L x L x

C C Cd d

ε κε−= + = + where x is the depth

that the dielectric is inserted.

Then 2 2

2

1 .2 2

d Q Q dCFdx C dxC

= − = − −

Now

( ) ( ) ( )0 0 0 0 0 1 .L L x L x L L LdC d

dx dx d d d d dε κε ε κε ε κ −

= + = − + = −

( ) ( )22

0 02 1 1

22L V LQF

d dCε εκ κ

= − = −

24.22. Assume the coaxial capacitor contains a dielectric material of dielectric material of dielectric constant κ (as opposed to air). The capacitance of a cylindrical capacitor is:


928

( ) ( )( ) ( ) ( )

( )

0 0 0 0

2 1

0 0

2 2 2 2ln / ln / ln ln ln ln (1 /

2 2ln(1 / )ln ln ln(1 / )

L L L LC

r r R R R d R R d RR dL L

d RR R d R

πκε πκε πκε πκε

πκε πκε

= = = =− − − −−

= = −−− + −

Consider the series expansion for ( )ln 1 ,x− where 1,x ≤ 1:x ≠ ( )2 3

ln 1 ...2 3x xx x− = − − − − , which can

be approximated as ( )ln 1 x x− ≈ − for x close to zero. This approximation is valid, since the question states that d << R, so d / R is close to zero. Then:

( )0 02 2

.ln 1 /R

L LRC

ddπκε πκε

= − ≈−

Now consider the surface area of the cylinder: 2 .A LRπ= In the limit of small distance ,d this is the area

of both cylinders. Then we can make the replacement: 0 .d

AC

κε= This result is the formula for a parallel

plate capacitor, and in the limit of /d R approaching zero, this makes a great deal of sense.

24.23. The charge on each plate will be same. The battery will move charge from one plate to the other, keeping the overall charge on the device neutral.

24.24. The parallel plate capacitor is connected to a battery. As the plates are pulled apart: (a) The electrical potential on the plates does not change; therefore the charge on the plates would have to decrease. (b) The capacitance for the parallel plate capacitor is 0 / .C A dε= As the distance between the plates increases, C must decrease. By definition / ;C q V= since V remains the same, it must be that the charge on the plates decreases as they are pull apart. (c) The electric field between the plates of a parallel capacitor is uniform, and is equal to = /d.E V As the plates are pulled apart the electric field must decrease.

Problems

24.25. Assume the supercapacitor is made from parallel plates. The capacitance is 0 / .C A dε= Rearranging for

A yields: 0/ .A Cd ε= With 1.00 F,C = 31.00 mm 1.00 10 md −= = ⋅ and 120 8.85 10 F/m,ε −= ⋅ the area is

( )( )32

12

1.00 F 1.00 10 m1.13 km .

8.85 10 F/mA

−

−

⋅= =

⋅

24.26. The potential difference across the collinear cylinders is 100. V.V = The inner radius is 1 10.0 cm 0.100 m.r = = The outer radius is 2 15.0 cm 0.150 m.r = = The length of both cylinders is

40.0 cm 0.400 m.L = = By definition, .q CV= For a cylindrical capacitor, ( )0 2 12 / ln /r .C L rπε= For this system,

( )( )( )

1211

2 8.85 10 F/m 0.400 m5.488 10 F 54.9 pF.

ln 0.150/0.100C

π −−

⋅= = ⋅ ≈

Then ( )( )11 95.488 10 F 100. V 5.488 10 F 5.49 nC.q − −= ⋅ = ⋅ ≈ The electric field between the plates of a

cylindrical capacitor is 0/ 2 .E q rLπε= Themagnitude of the electric field just outside the inner surface is:

( )( )( )9

1 120 1

5.488 10 F 2466 V/m 2470 V/m.2 2 8.85 10 F/m 0.100 m 0.400 m

qEr Lπε π

−

−

⋅= = = ≈⋅

Its magnitude just inside the outer surface is:


929

( )( )( )9

2 120 2

5.488 10 F 1644 V/m 1640 V/m.2 2 8.85 10 F/m 0.150 m 0.400 m

qEr Lπε π

−

−

⋅= = = ≈⋅

24.27. For a spherical conductor, the capacitance is 04 .C Rπε= With 1.00 F,C = the radius must be

( )9 9

120

1.00 F 8.988 10 m 8.99 10 m.4 4 8.85 10 F/m

CRπε π −

= = = ⋅ ≈ ⋅⋅

24.28. The capacitance of a spherical capacitor made from two concentric conducting shells is: ( )πε= −s 0 1 2 2 14 / r .C r r r The capacitance of a parallel plate capacitor is: p 0 / .C A dε= The area A of the

parallel plate capacitor is equal to the area of the inner sphere in the spherical capacitor: 14 r .A π= With

( )= −2 1 ,d r r the fractional difference in capacitance between the two geometries is

( ) ( )( )

πε πε

πε

−− = = −

20 1 2 0 1s p 2

2p 10 1

4 /d 4 /d1.

4 /d

r r rC C rC rr

The capacitance of the parallel plate capacitor is ( )−2 1/ 1 100%r r smaller than that of the spherical capacitor.

24.29. The capacitance of a spherical conductor is: 04 .C Rπε= With the radius of the Earth being 66371 km 6.371 10 m,R = = ⋅ the Earth’s capacitance is:

( )( )6 12 4 44 6.371 10 m 8.85 10 F/m 7.0887 10 F 7.09 10 F.C π − − −= ⋅ ⋅ = ⋅ ≈ ⋅

24.30. By definition, capacitance is / .C q V= The capacitance of the spherical conductor is

( )1 2

02 1

4 .r

r rC

r=

−πε

The potential difference across the inner and outer spheres is 900. VV = when a charge of 86.726 10 Cq −= ⋅ is applied to them. The radius of the outer sphere is 2 0.210 m.r = The radius of the

inner sphere is:

( ) ( )1 20 2 1 0 1 2 2 1 0 2

2 1

2 21

0 20 2

4 4 4

44 1

r rC C r r r r Cr r C rr r

Cr rrr VC r

q

= − = = +−

= =+ +

πε πε πε

πεπε

( )( )( )( )

( )1 12

8

0.210 m0.160 m.

4 8.85 10 F/m 0.210 m 900. V1

6.726 10 C

rπ −

−

= ≈⋅

+⋅

24.31. THINK: The capacitor is a parallel plate capacitor of variable separation distance. The material between the two plates is air. When the initial separation is 0 0.500 cm 0.00500 md = = , the initial capacitance is

0 32.0 pF.C = (a) A battery is connected with a potential difference 9.0 V.V = Find the charge density σ on the left plate. When the separation is changed to ' 0.250 cm 0.00250 m,d = = find the new capacitance 'C and charge density, '.σ (b) The separation between the plates is again =0 0.00500 md and the battery is disconnected. The plates are moved to a separation of ' 0.00250 m.d = Find the new potential difference 'V between the plates.


930

SKETCH:

RESEARCH: By definition capacitance is / .C q V= Charge density is / .q Aσ = Note that as long as the battery is connected, the potential difference across the plates is constant. When the battery is disconnected, the total charge in the system must remain constant. SIMPLIFY:

(a) When the battery is first hooked up, the charge density is 0 0

0

C V VqA A d

εσ = = = (for a parallel plate

capacitor). When the separation is decreased from 0d to ',d the new capacitance is

0 0 0 0 0 .A Ad C d

Cd d d

ε ε′ = = =′ ′ ′

The new charge density is: 0' / 'V dσ ε= where V is the battery voltage. (b) Now that the battery is disconnected, q remains constant but V can change. When the plates are moved to a separation of d′ the new capacitance is 0 / ,A dε ′ which is C′ from above. The new potential difference is then 0/ / .V q C C V C′ ′ ′= = CALCULATE:

(a) ( )( )

σ−

−⋅

= = ⋅12

8 28.85 10 F/m 9.0 V

1.593 10 C/m ,0.00500 m

( )( )32.0 pF 0.00500 m

64.0 pF,0.00250 m

C′ = =

( )( )σ

−−

⋅′ = = ⋅

128 2

8.85 10 F/m 9.0 V3.186 10 C/m

0.00250 m

(b) ( )( )32.0 pF 9.0 V / 64.0 pF 4.5 VV ′ = = ROUND: For ,σ σ ′ and V ′ the precision is limited to two significant figures from .V 'C has three significant figures. (a) 8 21.6 10 C/m ,σ −= ⋅ 64.0 pF,C′ = 8 23.2 10 C/mσ −′ = ⋅ (b) 4.5 VV ′ = DOUBLE-CHECK: C is proportional to 1/ d for parallel plates; as d decreases C must increase. When q is constant C is proportional to 1/V, so as C increases, V decreases.

24.32. For equivalent capacitors of capacitance C in parallel, the equivalent capacitance is

eq1

.n

ii

C C nC=

= =

For a single capacitor, eq .C C= For two capacitors in parallel, eq 2 .C C= For for three capacitors in

parallel, eq 3 .C C= For equivalent capacitors of capacitance C in series, the equivalent capacitance is

eq1eq

1 1 .n

i i

n CCC C C n=

= = =

For two capacitors in series eq / 2.C C= For three capacitors in series, eq / 3.C C= Another combination is to have two capacitors in parallel and add the third in series with the first two, as shown:


931

In this case, the equivalent capacitance is eqeq

1 1 1 3 2 .2 2 3

C CC C C C

= + = = Lastly, there can be two

capacitors in series, with the third added in parallel with the first two, as shown:

In this case, the equivalent capacitance is eq3 .

2 2CC C C= + =

24.33. The capacitor can be treated like two capacitors in parallel where each has an area equal to half that of the original. One capacitor has the original plate spacing 1 1.00 mm 0.00100 m,d = = and the other has a plate

spacing 2 0.500 mm 0.000500 m.d = = The original area was 2 4 21.00 cm 1.00 10 m .A −= = ⋅ For capacitors in parallel, the equivalent capacitance is eq 1 2 .C C C= + For this system,

( )( )( )

( )( )( )

12 4 2 12 4 20 0

eq1 2

12

8.85 10 F/m 1.00 10 m 8.85 10 F/m 1.00 10 m/ 2 / 22 0.00100 m 2 0.000500 m

1.3281 10 F 1.33 pF.

A AC

d dε ε − − − −

−

⋅ ⋅ ⋅ ⋅= + = +

= ⋅ ≈

24.34. The capacitors have values 1 3.1 nF,C = 2 1.3 nFC = and 3 3.7 nF.C = The battery provides a voltage of 14.9 V.V = The circuit can be reduced to:

where 23 2 3C C C= + since 2C and 3C are in parallel. Now 1C and 23C are in series, and the equivalent capacitance is:

( )( )1 2 31 2 3

eqeq 1 23 1 2 3 1 2 31 2 3

1 1 1 1 1 .C C CC C C

CC C C C C C C C CC C C

++ += + = + = =

+ + ++

For capacitors in series, the charge on each capacitor is the total charge, eq .q C V= So, the charge on 23C is

.q For 2C and 3C in parallel, this means that 2 3 .q q q+ = Also, the potential difference across 2C and 3C is the same, so 2 3 .V V= Then

( )( )( )

( ) ( ) ( )

eq 12 3 2 2 3 3 2 2 3 2 2 2 3 2

2 3 2 3 1 2 3

2

3.1 nF 14.9 V5.7 V.

3.1 nF 1.3 nF 3.7 nF

C V C Vqq q q C V C V C V C V V C C VC C C C C C C

V

= + = + = + = + = = =+ + + +

= =+ +


932

24.35. The capacitors have values 1 3.5 nF,C = 2 2.1 nF,C = 3 1.3 nFC = and 4 4.9 nF.C = The battery has voltage 10.3 V.V = First, 3C and 4C are in parallel. Then 34 3 4C C C= + and the circuit becomes

Next, 1C and 34C are in series. Then 134 1 34 1 3 4

1 1 1 1 1 .C C C C C C

= + = ++

Then ( )1 3 4

1341 3 4

C C CC

C C C+

=+ +

and the

circuit becomes

Now 2C and 134C are in parallel. The equivalent capacitance of the circuit is:

( ) ( ) ( )( )1 3 4eq 2 134 2

1 3 4

3.5 nF 1.3 nF 4.9 nF2.1 nF 4.337 nF 4.3 nF.

3.5 nF 1.3 nF 4.9 nFC C C

C C C CC C C

+ += + = + = + = ≈

+ + + +

24.36. The capacitors have values 1 18.0 μF,C = 2 11.3 μF,C = 3 33.0 μFC = and 4 44.0 μF.C = The potential difference is 10.0 V.V = Capacitors 1C and 2C are in parallel, as are 3C and 4 .C Write 12 1 2C C C= + and

34 3 4 .C C C= + The circuit becomes

with 34C and 12C in series, the equivalent capacitance is

( )( )1 2 3 412 34eq

eq 12 34 12 34 1 2 3 4

1 1 1 .C C C CC C

CC C C C C C C C C

+ += + = =

+ + + +

The total charge required to charge the capacitors in the circuit is

( )( ) ( )( )( )1 2 3 4 4eq

1 2 3 4

18.0 μF 11.3 μF 33.0 μF 44.0 μF 10.0 V2.12 10 C.

18.0 μF 11.3 μF 33.0 μF 44.0 μFC C C C V

q C VC C C C

−+ ++ +

= = = = ⋅+ + + + + +

24.37. THINK: Six capacitors are arranged as shown in the question. (a) The capacitance of capacitor 3 is 3 2.3 nF.C = The equivalent capacitance of the combination of capacitors 2 and 3 is 23 5.000 nF.C = Find the capacitance of capacitor 2, 2 .C 3 2 and C C are in parallel. Use the formula for parallel capacitance. (b) The equivalent capacitance of the combination of capacitors 1, 2, and 3 is 123 1.914 nF.C = Find the capacitance of capacitor 1, 1.C 1 23and C C are in series. Use the formula for series capacitance. (c) The remaining capacitances are 4 51.3 nF, 1.7 nF,C C= = and 6 4.7 nF.C = Find the equivalent capacitance of the whole system, eq .C

(d) A battery with a potential difference of 11.7 VV = is connected as shown. Find the total charge q deposited on this system of capacitors. (e) Find the potential drop across capacitor 5 in this case. SKETCH: Consider the sketch in the question. Sketches of the simplified system are provided in the simplify step.


933

RESEARCH: For capacitors in series, the equivalent capacitance is eq1

1/ 1/ ,n

ii

C C=

= and the charge on

each is the same. For capacitors in parallel, the equivalent capacitance is 1

,n

eq ii

C C=

= and the potential

drop across each is the same. By definition, capacitance is / .C q V= Δ SIMPLIFY: (a) 23 2 3 2 23 3 C C C C C C= + = −

(b) 1

1123 1 23 123 23

1 1 1 1 1 CC C C C C

−

= + = −

(c) 4 5 6, , and C C C are in parallel. The equivalent capacitance between these capacitors is:

456 4 5 6 .C C C C= + +

Now, 123 456 and C C are in series. The equivalent capacitance of the entire circuit is: 1

eq123 456

1 1 .CC C

−

= +

(d) The total charge required to charge the capacitors is eq .q C V= Δ

(e) 4 5 6, , and C C C are in parallel. The voltage drop across each capacitor is the same: 4 5 6 .V V V= = The equivalent capacitors 123 456 and C C are in series. The charge on each 123 456and C C is the same, and they are equal to the total charge in the system: 123 456 .q q q= = Since 4 5 6, , and C C C are in parallel,

456 4 5 6 .q q q q q= = + + Then, ( )4 4 5 5 6 6 5 4 5 6q C V C V C V V C C C= + + = + + , ( )54 5 6

.qVC C C

=+ +

CALCULATE: (a) 2 5.000 nF 2.3 nF 2.700 nFC = − =

(b) 1

11 1 3.1011 nF

1.914 nF 5.000 nFC

− = − =

(c) −

= + + = = + =

1

456 eq1 11.3 nF 1.7 nF 4.7 nF 7.7 nF, 1.533 nF

1.914 nF 7.7 nFC C

(d) ( )( )1.533 nF 11.7 V 17.94 Cq = =

(e) ( )

( )5

17.94 C2.33 V

1.3 nF 1.7 nF 4.7 nFV = =

+ +

ROUND: (a) To 2 significant figures, 2 2.7 nF.C = (b) To 4 significant figures, 1 3.101 nF.C = (c) To 2 significant figures, eq 1.5 nF.C =


934

(d) To 2 significant figures, 18 C.q = (e) To 2 significant figures, 5 2.3 V.V = DOUBLE-CHECK: Note for equivalent capacitance 23 ,C where 2C and 3C are in parallel, 23 2 3, .C C C> Similarly, for equivalent capacitance 456 ,C where 4 5, ,C C and 6C are in parallel, 456 4 5 6, , .C C C C> Finally, the equivalent capacitance of the entire circuit eq 123 456, ,C C C< where the equivalent capacitances 123C and

456C are in series.

24.38. THINK: The capacitance values are 1 15.0 nF,C = 2 7.00 nFC = and 3 20.0 nF.C = The potential difference provided by the battery is 80.0 V.V = Note that 1C and 2C are in parallel with each other, while 3C is in series with the equivalent capacitor of 1C and 2 .C Find the magnitude and the sign of the charge 3lq on the left plate of 3 ,C the electric potential 3V across 3 ,C and the magnitude and sign of the charge 2rq on the right plate of 2 .C SKETCH:

RESEARCH: Capacitors 1C and 2C are in parallel; their equivalent capacitance is 12 1 2 .C C C= + Capacitor

3C is in series with 12 ;C the equivalent capacitance of the circuit is; therefore, 1

3 12eq

3 12 3 12

1 1 .C C

CC C C C

−

= + = +

Since 1C and 2C are in parallel the potential drop across them is equal, so 1 2 .V V= The potential drop across 3C and the equivalent capacitor 12C must sum to the potential drop across battery, 3 12 ,V V V+ = since they are in series. Finally, the total charge in the circuit is eq ,q C V= while the charge on a specific

capacitor iC is i i i .q C V= Since 3C and 12C are in series, the charges on these capacitors are equal to each other, and equal to the total charge in the circuit. The charges on 1C and 2C must sum to the total charge on their equivalent capacitor 12 ,C which is equal to the total charge in circuit .q SIMPLIFY: As explained above,

( )3 1 23 123 eq

3 12 1 2 3

.C C CC C

q q C V V VC C C C C

+= = = =

+ + +

At point ,A 3q is negative: the battery sets up an electric field in the wires. The field drives electrons from the negative end of the battery to the left of 3 .C Then,

( )3 1 23l

1 2 3

.C C C

q VC C C

+= −

+ +

3V is then 3 3 3/V q C= and 2q is 2 2 2 .q C V= 2V is found from 3 12 ,V V V+ = where 12 1 2 .V V V= = Then

2 3 ,V V V= − and ( )2 2 3 .q C V V= − At point ,B 2q is negative. The electric field pulls electrons from the left plate of 2C to the positive end of the battery so the net charge on the left plate of 2C is positive. The right plate must, therefore, be negatively charged. Then, ( )2, 2 3 .bq C V V= − −

CALCULATE: ( )( ) ( ) 7

3l

20.0 nF 15.0 nF 7.00 nF80.0 V 8.381 10 C

15.0 nF 7.00 nF 20.0 nFq −+

= − = − ⋅+ +


935

7

38.381 10 C 41.905 V

20.0 nFV

−− ⋅= =

( )( ) 72r 7.00 nF 80.0 V 41.905 V 2.6667 10 Cq −= − − = − ⋅

ROUND: To three significant figures, 73l 8.38 10 C,q −= − ⋅ 3 41.9 V,V = and 7

2r 2.67 10 C.q −= − ⋅ DOUBLE-CHECK: Note that 1 2 3 .q q q+ = 1q is found from ( )1 1 1 1 2 1 3 ,q C V C V C V V= = = − so

( )( ) 71 15.0 nF 80.0 V 41.905 V 5.7143 10 C.q −= − = ⋅

Then ( ) ( )7 7 71 2 5.7143 10 C 2.6667 10 C 8.381 10 C,q q − − −+ = ⋅ + ⋅ = ⋅ which is the magnitude of 3q that was

found above.

24.39. THINK: Fifty parallel plate capacitors are connected in series. The distance between the plates of the first capacitor is ,d between the plates of the second capacitor 2 ,d the third capacitor 3 ,d and so on. The area of the plates remains the same for all capacitors. Find the equivalent capacitance eqC in terms of 1C (the capacitance of the first capacitor). SKETCH:

RESEARCH: For capacitors in series, the equivalent capacitance is

i=1eq 1 2 3 i

1 1 1 1 1... .n

C C C C C= + + + =

The capacitance of a single parallel plate capacitor is: 0 / .C A dε= SIMPLIFY: The equivalent capacitance is:

( )3 501 2

eq 1 2 3 50 0 0 0 0 0

1 1 1 1 1... ... 1 2 3 ... 50 .d dd d d

C C C C C A A A A Aε ε ε ε ε= + + + + = + + + = + + + +

Since 1 ,d d= 2 2 ,d d= …, ,nd nd= it follows that: 50

i=1eq 0

1 i.dC Aε

= = Note 01

AC

dε

= and ( )50

i=1

1i ,

2n n +

=

and therefore ( )

( )1

eqeq 1

1 21 1 .2 1

n n CCC C n n

+ = = +

CALCULATE: With 50,n = ( )1 1

eq2

.127550 50 1

C CC = =

+

ROUND: The answer is precise. No rounding is required. DOUBLE-CHECK: For capacitors in series, the equivalent capacitance must be less than the value of the largest capacitor, in this case 1.C

24.40. THINK: A 1 5.00 nFC = capacitor initially charged to 1 60.0 VV = and a 2 7.00 nFC = capacitor charged to 2 40.0 VV = are connected to each other with the negative plate of 1C connected to the negative plate

2 .C Find the final charge on 2 .C SKETCH:

RESEARCH: For capacitors connected in parallel, the potential drop across each capacitor is the same, that is 1 2 .V V= In addition, charge must be conserved, that is the initial charge iq before the capacitors are


936

connected must equal the final charge fq when the capacitors are connected. Note 0 0 0/ .V q C= In general, the charge q on a capacitor is .q CV=

SIMPLIFY: i 1i 2i 1 1 2 2 ,q q q C V C V= + = + 1 f 1f 1 f 2 f 2 f 2, f

2 f 2

1 1 .C V Cq C V C V C V qC V C

= + = + = +

Then,

f i ,q q= ( )1 1 1 2 22, f 1 1 2 2 2, f

2 1 2

1 ./ 1

C C V C Vq C V C V qC C C

++ = + = +

CALCULATE: ( )( ) ( )( )

( ) ( )7

2, f

5.00 nF 60.0 V 7.00 nF 40.0 V 3.3833 10 C.

5.00 nF / 7.00 nF 1q −

+= = ⋅ +

ROUND: Rounding to three significant figures, 72, f 3.38 10 C 0.338 μC.q −= ⋅ =

DOUBLE-CHECK: The solution is reasonable given the magnitudes of the capacitors.

24.41. The charge on each plate has a magnitude of 60.0 μCq = and potential difference of 12.0 V.V = The

capacitance is; therefore, 6/ 60.0 μC /12.0 V 5.00 10 F.C q V −= = = ⋅ When the potential difference is ' 120. V,V = the potential energy stored in the capacitor is

( ) ( )( )2 261 1 5.00 10 F 120. V 0.0360 J.2 2

U C V −′= = ⋅ =

24.42. The potential difference across the defibrillator is 7500 V.V = It stores 2400 J.U = Generally

( ) 21/ 2 .U CV= Solving for C yields ( ) ( )22 52 / 2 2400 J / 7500 V 8.533 10 F 85 μF.C U V −= = = ⋅ ≈

24.43. The Earth has an electric field 150 N/C 150 V/m.E = = The electric energy density is

( ) ( )( )( )22 12 8 2 3 7 2 301/ 2 1/ 2 8.85 10 F / m 150 V/m 9.961 10 FV /m 1.0 10 FV /m .U Eε − − −= = ⋅ = ⋅ ≈ ⋅

24.44. THINK: The battery potential difference across two capacitors in series is 120. V.V = The capacitances are 3

1 1.00 10 μFC = ⋅ and 32 1.50 10 μF.C = ⋅ Find

(a) The total capacitance eqC of this circuit.

(b) The charge on each capacitor, 1q and 2 .q (c) The potential difference across each capacitor, 1V and 2 .V (d) The total energy stored in the circuit, .U SKETCH:

RESEARCH: (a) For a circuit of two capacitors connected in series, the equivalent capacitance is eq 1 21/ 1/ 1/ .C C C= + (b) For a circuit of two capacitors connected in series, the charge on each capacitor is the same, and equal to the total charge in circuit, that is 1 2 .q q q= = The total charge is found from eq .q C V= (c) For a circuit of two capacitors connected in series, the sum of the potential differences across each capacitor must be equal to the battery potential difference, that is 1 2 .V V V+ = In general, for each capacitor in series, i i i .q C V= (d) The total energy stored in the circuit will be the sum of the energy stored in each capacitor,

1 2 .U U U= + For each capacitor i ,C 2i i i/2 .U q C=


937

SIMPLIFY:

(a) 1 2eq

eq 1 2 1 2

1 1 1 C C

CC C C C C

= + =+

(b) 1 2 eqq q q C V= = =

(c) 11

1 1

;q qVC C

= = 22

2 2

q qVC C

= =

(d) 2 2 21 2

1 21 2 1 2

1 1 1 12 2 2

q q qU U UC C C C

= + = + = +

CALCULATE:

(a) ( )( )3 3

-4eq 3 3

1.00 10 μF 1.50 10 μF600. μF = 6.00 10 F 0.600 mF

1.00 10 μF 1.50 10 μFC

⋅ ⋅= = ⋅ =

⋅ + ⋅

(b) ( )( )1 2 0.600 mF 120. V 0.0720 Cq q q= = = =

(c) 1 3

0.0720 C 72.0 V;1.00 10 μF

V = =⋅

2 3

0.0720 C 48.0 V1.50 10 μF

V = =⋅

(d) ( )2

3 3

0.0720 C 1 1 4.32 J2 1.00 10 μF 1.50 10 μF

U

= + = ⋅ ⋅

ROUND: Since 1C has three significant figures, (a) eq 0.600 mFC =

(b) 1 2 0.0720 Cq q= = (c) 1 72.0 V;V = 2 48.0 VV = (d) 4.32 JU = DOUBLE-CHECK: It should be the case that ( ) ( )1 2 72.0 V 48.0 V 120. V.V V+ = + = This is correct. In addition, the potential energy in the circuit be equal to

( ) 2eq1/ 2 :C V ( )( )( )21/ 2 0.600 mF 120. V 4.32 J.U = =

24.45. THINK: Treat the neutron star as a spherical capacitor. The inner radius of the capacitor is the radius of the neutron star, 4

1 10.0 km 1.00 10 m.r = = ⋅ The outer radius is the radius of the neutron star and the

1.00 cm dipole layer. The charge density is ( )( )22 21.00 μC/cm 100. cm/m 0.0100 C/m .σ = = Find both

the capacitance C of the star and electrical energy U stored in the star’s dipole layer. SKETCH:

RESEARCH: The capacitance of a spherical capacitor is ( )( )0 1 2 2 14 / .C r r r rπε= − The total charge on the

dipole layer is 214 .q A rσ σ π= = Note since 1 2 ,r r≈ assume the areas of the inner and outer shells are the

same. The potential energy of a capacitor is ( ) 21/ 2 / .U q C=

SIMPLIFY: 1 20

2 1

4 ,r rC

r rπε=

−

( )222141 .

2 2

rqUC C

σ π= =


938

CALCULATE: ( )( )( )12 44 8.85 10 F / m 1.00 10 m 10000.01 m

1.113 F0.0100 m

Cπ −⋅ ⋅

= =

( ) ( )( )( )

222 4

130.0100 C/m 4 1.00 10 m

7.096 10 J2 1.113 F

Uπ ⋅

= = ⋅

ROUND: To 3 significant figures due to the thickness of the dipole layer, 1.00 FC = and 137.10 10 J.U = ⋅ DOUBLE-CHECK: There is an enormous amount of charge on the dipole layer, 71.30 10 C.q ≈ ⋅ Since

both C and U are proportional to ,q they should also be large (especially U, where 2U q∝ ).

24.46. THINK: A 30 4.00 10 nFC = ⋅ parallel plate capacitor is connected to a 12.0 VV = battery and charged.

Use what you know about capacitors in series. (a) Find the charge Q on the positive plate of the capacitor. (b) Find the electrical energy U stored in the capacitor. The capacitor is then disconnected from the 12.0 V battery and used to charge 3 uncharged capacitors, a 1 100. nFC = capacitor, a 2 200. nFC = capacitor, and a 3 300. nFC = capacitor, arranged in series. (c) Find the potential difference across each of the 4 capacitors, 0 ,V 1 ,V 2 ,V and 3 .V The capacitors are in series. (d) Determine the amount of electrical energy stored in the 0C capacitor that was transferred to the other 3 capacitors. SKETCH:

RESEARCH: In general, ,q CV= and for capacitors in series, 1eq

1 1 .n

i iC C=

=

(a) From above, the charge Q is 0 .Q C V=

(b) The electrical energy U is 20/ 2 .U Q C=

(c) For capacitors in series, the charge on each plate is the same, that is 1 2 3 .q q q= = The three capacitors can be replaced with an equivalent capacitor 123 ,C where 123C has a charge 123 1 2 3 .q q q q= = = Note the first capacitor 0C provides the voltage to charge the capacitors 1 ,C 2 ,C and 3C or rather 123 .C Some of the charge Q originally on 0C flows to the equivalent capacitor 123 .C By conservation of charge,

0 123 .Q q q= + Once 0q and 123q are determined, 0 ,V 1 ,V 2 ,V and 3V can be determined from .q CV= (d) The energy transferred to the other 3 capacitors is the sum of the energy on each capacitor. Use

2 / 2U q C= to determine the energy on each capacitor. SIMPLIFY: (a) 0Q C V=

(b) 20/ 2U Q C=

(c) From 0 123 ,V V=

0 123 123123 0

0 123 0

q q C

q qC C C

= =

where

123 1 2 3

1 1 1 1 .C C C C

= + +

From 0 123 ,Q q q= +

( )123

0 0 00 123 0

1 /

C QQ q q qC C C

= + =+

with 123 1 2 3 ,q q q q= = =


939

00

0

,q

VC

= 1231

1

,q

VC

= 1232

2

,q

VC

= and 1233

3

.q

VC

=

(d) 2 2

123 1231 2 3

1 2 3 123

1 1 12 2

q qU U U U

C C C C

Δ = + + = + + =

CALCULATE: (a) ( )( )3 5

0 4.00 10 nF 12.0 V 4.80 10 CQ C V −= = ⋅ = ⋅

(b) ( )( )

254

3

4.80 10 C2.88 10 J

2 4.00 10 nFU

−−

⋅= = ⋅

⋅

(c) 1

1231 1 1 54.5 nF;

100. nF 200. nF 300. nFC

− = + + =

( ) ( )( )

55

0 3

4.88 10 C 4.735 10 C1 /54.5 nF 4.00 10 nF

q−

−⋅= = ⋅+ ⋅

( )5 7123 1 2 3 3

54.5 nF 4.735 10 C 6.46 10 C4.00 10 nF

q q q q − − = = = = ⋅ = ⋅ ⋅

Then 5

0 3

4.735 10 C 11.84 V,4.00 10 nF

V−⋅= =

⋅

7

16.46 10 C 6.46 V,

100. nFV

−⋅= = 7

26.46 10 C 3.23 V

200. nFV

−⋅= = and

7

36.46 10 C 2.15 V.

300. nFV

−⋅= =

(d) The transferred energy is ( )

( )

276

6.46 10 C3.822 10 J.

2 54.5 nFU

−−

⋅Δ = = ⋅

ROUND: (a) 54.80 10 CQ −= ⋅ (b) 42.88 10 J 0.288 mJU −= ⋅ = (c) 0 11.8 V,V = 1 6.46 V,V = 2 3.23 V,V = 3 2.15 VV =

(d) The energy transferred is 63.822 10 J 3.82 μJ.−⋅ = DOUBLE-CHECK: Because 0C acts like a battery for 1 ,C 2C and 3C (is series), 0 1 2 3 :V V V V= + +

1 2 3 06.46 V 3.23 V+2.15 V 11.84 V .V V V V+ + = + = =

24.47. THINK: The circuit has 12.0 V,V = 1 500. pFC = and 2 500. pF.C = (a) Find the energy 0U delivered by the battery while the switch is closed to A and the capacitor 1C is fully charged. (b) Find the energy stored on 1C while the switch is closed to A and the capacitor 1C is fully charged. The potential difference across 1C is equal to the potential difference of the battery, .V (c) Find the total energy stored at 1C and 2C when the switch is thrown to .B The first capacitor 1C provides the voltage to charge the capacitor 2 ,C i.e. 1C acts as a battery. (d) Explain the energy loss, if there is any. Energy lost is the difference between the initial and final energies of the system. SKETCH:

RESEARCH: In general q CV= (a) The energy provided by the battery is given by .U QV=


940

(b) The energy stored in a capacitor is 2 / 2.U CV= (c) Then the new potential difference 1V across 1C must be equal to 2 ,V the potential difference across

2 ,C or 1 2 .V V= This implies that 1 1 2 2/ /q C q C= (since /V q C= ). By conservation of charge, the original charge 0q on 1C (before the switch was thrown to point B ) is 0 1 2 .q q q= + Each 1q and 2q can be

determined. Then the energy on each capacitor is ( )2 / 2 .U q C= (d) f iE E EΔ = − SIMPLIFY: (a) 2

1U QV C V= = as 0 1 .Q q C V= =

(b) 2

10 2

C VU =

(c) From 1 2 22 1

1 2 1

.q q Cq qC C C

= = Then 20 1 2 1

1

1 .Cq q q qC

= + = +

Also 0 1 .q C V= Then 1

12 1

.1 /

C VqC C

=+

21

11

,2qUC

= 22

222

qUC

=

(d) f i 1 2 0E E E U U UΔ = − = + − CALCULATE:

(a) ( )( )2 8500. pF 12.0 V 7.20 10 JU −= = ⋅

(b) ( )( )2 80 500. pF 12.0 V / 2 3.60 10 JU −= = ⋅

(c) ( )( )( ) ( )( )

91

500. pF 12.0 V3.00 10 C,

1 500. pF / 500. pFq −= = ⋅

+ 92

2 1 11

3.00 10 CCq q qC

− = = = ⋅

Then 22

1 222

qU UC

= = (since 1 2C C= and 1 2q q= ), ( )

( )

299

1

3.00 10 C9.00 10 J.

2 500. pFU

−−

⋅= = ⋅

(d) ( )9 8 82 02 2 9.00 10 J 3.60 10 J 1.80 10 JE U U − − −Δ = − = ⋅ − ⋅ = ⋅

Even though the battery supplies 87.20 10 J,−⋅ half of this is lost to heat in the system. Again, when 1C is connected to 2 ,C half of the energy is lost to heat. ROUND: The capacitors have 3 significant figures. The answers should be rounded to 3 significant figures as well. (a) 87.20 10 J 72.0 nJU −= ⋅ = (b) 0 36.0 nJU = (c) 1 2 9.00 nJU U= = (d) 18.0 nJEΔ = DOUBLE-CHECK: These answers are reasonable considering the initial values given.

24.48. THINK: (a) The energy density, which is related to the electric field, can be integrated to determine the total electrostatic potential energy. (b) The given formula can be integrated over the volume of the sphere to determine the gravitational potential energy. (c) The magnitudes of the values found in parts (a) and (b) can be used to determine what impact the electrostatic forces have on the structure of the Earth. SKETCH: Not required.


941

RESEARCH: (a) The energy density is given by ( ) 2

01/ 2 .u Eε= For the Earth of radius R, the electric field as a function of the radial position r is given by:

2

surface2 .

R EE

r=

The Earth can be treated as a conductor, so all excess charge resides at the surface of the Earth. Therefore, by Gauss’s Law 0E = for .r R< (b) As given in the question, the differential gravitational potential energy is ( )g / .dU Gm r dm= −

SIMPLIFY: (a) The electrostatic potential energy of the Earth is given by:

22 2 42 2 surface surfacee 0 02 20 0

2 4 2 2 4 2 30 surface 0 surface 0 surface

1 1sin 42 2

12 2 2 .

R R

RR

E R E RU d d r dr dr

r r

E R r dr E R E Rr

π πε ϕ θ θ ε π

πε πε πε

∞ ∞

∞∞ −

= =

= = − =

(b) The gravitational potential energy of the Earth is given by:

gGmU dU dm

r= = −

The mass is a function of the radius, 34 ,

3m rπρ=

and the differential mass element can be written as,

( )3 3 24 4 4 .3 3

dm dV d r d r r drρ ρ π πρ πρ = = = =

Therefore,

( ) ( ) ( )2 22 23 2 4 5 2 2 5

g 0 00

4 44 1 164 .3 3 3 5 15

RR RG GGU r r dr r dr r G R

rπ ρ π ρ

πρ πρ π ρ = − = − = − = −

CALCULATE:

(a) ( )( ) ( )3212 6 14e 2 8.854 10 F/m 150. V/m 6.371 10 m 3.237 10 JU π −= ⋅ − ⋅ = ⋅

(b) ( )( ) ( )2 52 11 3 2 3 3 6 32g

16 6.6742 10 m /kg s 5.515 10 kg/m 6.371 10 m 2.243 10 J15

U π −= − ⋅ ⋅ ⋅ = − ⋅

(c) ( )( )

3217

g e 14

2.243 10 J/ 6.930 10

3.237 10 JU U

− ⋅= = ⋅

⋅

The effects of the electrostatic forces on the Earth’s structure are insignificant compared to the effects of gravity. ROUND: (a) To three significant figures, 14

e 3.24 10 J.U = ⋅

(b) To four significant figures, 32g 2.243 10 J.U = − ⋅

(c) To three significant figures, 17g e/ 6.93 10 .U U = ⋅


942

DOUBLE-CHECK: The gravitational potential energy greatly exceeds the electrostatic potential energy, which is reasonable since the gravitational potential energy is proportional to 5 ,R while the electrostatic potential energy is proportional to 3 .R

24.49. In general, the energy stored in a capacitor is ( ) 21/ 2 .U CV= In terms of its dielectric strength max / mm,V

which yields ( )( ) ( )2 220 0 0max max

1 / mm / mm .2 2 2

A A AdU V V d V

d dκε κε κε

= = =

The ratio of MylarU to airU

is ( )( )

2Mylar maxMylar Mylar

2air air max air

/ mm.

/ mm

VUU V

=κ

κ From Table 24.1, this becomes:

( )( )( )( )

2Mylar 4

2air

3.1 280 kV / mm3.89 10 .

1 2.5 kV / mm

UU

= = ⋅

24.50. This set-up can be treated as two capacitors in parallel, one with the dielectric material and one with air. The total capacitance is dielectric air .C C C= + With air 1,κ = this becomes

( ) ( ) ( )2

0 0 0/ 2 / 21 .

2L L L L L

Cs s s

= + = +κε ε ε κ

24.51. The dielectric constant of air is 1.00059.κ = Its dielectric strength is / 2.5 kV / mm.V d = Treating the surface as having area ,A and assuming it is a plane, the charge is ( )0 air / .q CV A d Vε κ= = The charge

density of the surface is ( )0 air/ / .q A V dσ ε κ= = The maximum charge density is

( )( )( )12 8 5max 0 air

max

8.85 10 F / m 1.00059 2.5 kV/mm 2.2149 10 C / mm 2.2 10 C / m.Vd

σ ε κ − − − = = ⋅ = ⋅ ≈ ⋅

24.52. The thermo coax cable can be modeled as a cylindrical capacitor with 1 0.085 mmr = and 2 0.175 mm.r = With 9.7,k =

( )( )( )

( )( )12

100

2 1

2 8.85 10 F / m 9.727.473 10 F/m 750 pF/m.

ln / ln 0.175 mm/ 0.085 mmCL r r

ππε κ −−

⋅= = = ⋅ =

24.53. This system is treated as two capacitors in parallel, one with dimensions / 5L L⋅ and dielectric 1 ,κ the other with dimensions 4 / 5L L⋅ and dielectric 2 .κ Then

( ) ( )

( )( ) ( )

2 2 20 1 0 20 1 1 0 2 2 0 1 2

eq 1 2

21211

/ 5 4 / 5 45 5

8.85 10 F / m 0.100 m 4 5.0020.0 7.08310 F 70.8 pF.0.0100 m 5 5

L LA A LC C C

d d d d dε κ ε κε κ ε κ ε κ κ

−−

= + = + = + = +

⋅

= + = ≈

24.54. THINK: The capacitor has capacitance 4.0 nFC = and contains a sheet of Mylar with dielectric constant 3.1.κ = The capacitor is charged to 120 VV = and the power supply is then disconnected.

(a) Determine the work W required to completely remove the sheet of Mylar from the space between the two plates. (b) Determine the potential difference between the plates of the capacitor once the Mylar is completely removed.


943

SKETCH:

RESEARCH: (a) The work done is the change in potential energy: ,W U= Δ where 2 / 2 .U q C= Note that initially the capacitance is i air ,C Cκ= while the final capacitance is f air .C C= (b) The final potential is f f/ .V q C= Because the power supply is disconnected, the charge remains constant and is i i .q C V= SIMPLIFY:

(a) 2 2 2

f if i air

11 .2 2 2q q qW U UC C C

= − = − = − κ

Since q is conserved with the power supply being removed,

i i ,q C V= and 2 2i i

air

11 .2C VW

C κ = −

Since air i / ,C C κ= ( )2 2 2i i i i

i

11 1 .2 / 2C V C VWC

= − = − κ

κ κ

(b) air ii if i

f f air

.C VC VqV V

C C C= = = =

κ κ

CALCULATE:

(a) ( )( ) ( )

2

54.0 nF 120 V3.1 1 6.048 10 J

2W −= − = ⋅

(b) ( )( )f 3.1 120 V 372 VV = =

ROUND: To 2 significant figures, 56.0 10 JW −= ⋅ and f 370 V.V = DOUBLE-CHECK: When the dielectric material is removed and the power supply is disconnected, the capacitance must decrease while the charge stays constant. Since 1/ ,C V∝ the potential must increase.

24.55. THINK: A cylindrical capacitor is half-filled with a dielectric of constant .κ This can be treated as two cylindrical capacitors in parallel. It is connected with a battery of potential difference V across its two electrodes. Find the charge q deposited on the capacitor, and find the ratio of this charge to the charge

0q deposited on a completely empty capacitor connected in the same way across the same potential drop. SKETCH:

RESEARCH: The capacitance of a cylindrical capacitor is ( )0 2 12 / ln / .C L r rπε= For capacitors in parallel,

eq 1 2 .C C C= + In general, for a given potential, .q CV=


944

SIMPLIFY: When the capacitor is half-full, ( ) ( ) ( ) ( )0 0 0eq

2 1 2 1 2 1

2 / 2 2 / 21 .

ln / ln / ln /L L L

Cr r r r r r

πκε πε πεκ= + = + Then,

( ) ( )0

2 1

/ 21 .

l /nLq V

r rπε κ= + In the absence of a dielectric, ( ) ( )0 2 12 / ln / .C L r rπε= This gives the result:

( ) ( )0 0 2 12 / ln / .q LV r rπε= The ratio is ( ) ( )

( )0 2 1

0 0 2 1

1 / ln / 1.22 / ln /

LV r rqq LV r r

πε κ κπε

+ += =

CALCULATE: Not required. ROUND: Not required. DOUBLE-CHECK: Since the power source supplies a constant ,V the charge is not constant in the capacitor. It should be greater with the dielectric since q C∝ and dielectrics increase capacitance.

24.56. THINK: The dielectric slab has thickness .d The parallel plate capacitor with area 2 2100. cm 0.0100 mA = = is charged by a battery of 110. V.V = The plate separation distance is

250. cm 0.0250 m.d = = The dielectric constant is 2.31.κ = (a) Find the capacitance ,C the potential difference ,V the electric field ,E the total charge stored on the plate ,Q and electric energy stored U in the capacitor before inserting the dielectric material. (b) Find the above physical quantities, when the dielectric slab is inserted while the battery is kept connected. (c) Find the above physical quantities, when the dielectric slab is inserted after the battery was disconnected. SKETCH:

RESEARCH: (a) For a parallel plate capacitor, 0 / .C A dε= While connected to a battery, V across the capacitor is equal to that of the battery. The electric field is / .E V d= The total charge Q can be found from .Q CV= The electrical energy of a capacitor is ( )1/ 2 .U QV= (b) When the dielectric inserted the capacitance becomes 0' / .C A d Cκε κ= = With the battery still connected, the potential V stays constant but the charge changes; ' ' .Q C V= The electric field is / ;E V d= it does not change. The electrical energy becomes ( )' 1/ 2 ' .U Q V= (c) When the dielectric slab is inserted after the battery is disconnected the capacitance is still the same as in part (b). Now the potential does not remain constant, but the charge does; the charge is the same Q as in part (a), and the potential becomes " / '.V Q V= The electric field is " "/ ,E V d= and the electrical energy is ( )" 1/ 2 ".U QV= SIMPLIFY: (a) 0 / ,C A dε= batt ,V V= / ,E V d= battQ CV= and ( )1/ 2 .U QV= (b) ' ,C Cκ= batt ,V V= / ,E V d= batt' 'Q C V= and ( )1/ 2 ' .U Q V= (c) ' ,C Cκ= batt batt" / ' / ' / ,V Q C CV C V κ= = = " "/ ,E V d= battQ CV= and ( )1/ 2 ".U QV= CALCULATE:

(a) ( )( )12 2

128.85 10 F / m 0.0100 m

3.5417 10 F,0.0250 m

C−

−⋅

= = ⋅ 110. V,V = 3110. V 4.40 10 V/m,0.0250 m

E = = ⋅

( )( )12 103.5417 10 F 110. V 3.896 10 FQ − −= ⋅ = ⋅ and ( )( )( )10 81/ 2 3.896 10 J 110. V 2.143 10 J.U − −= ⋅ = ⋅


945

(b) ( )( )12 12' 2.31 3.5417 10 F 8.181 10 F,C − −= ⋅ = ⋅ 110. V,V = 34.40 10 V/m,E = ⋅

( )( )12 10' 8.181 10 F 110. V 8.999 10 CQ − −= ⋅ = ⋅ and ( )( )( )10 81/ 2 8.999 10 C 110. V 4.949 10 J.U − −= ⋅ = ⋅

(c) 12' 8.181 10 F,C −= ⋅ 110. V" 47.62 V,2.31

V = = 47.62 V" 1905 V/m,0.0250 m

E = = 103.896 10 CQ −= ⋅ and

( )( )( )10 91/ 2 3.896 10 C 47.62 V 9.276 10 J.U − −= ⋅ = ⋅

ROUND: Though ,V d and κ have 2 significant figures, A has three. Then: (a) 3.54 pF,C = 110. V,V = 34.40 10 V/m,E = ⋅ 390. pFQ = and 21.4 nJ.U = (b) ' 8.18 pF,C = 110. V,V = 34.40 10 V/m,E = ⋅ ' 900. pCQ = and 49.5 nJ.U =

(c) ' 8.18 pF,C = " 47.6 V,V = 3" 1.90 10 V/m,E = ⋅ 390. pCQ = and 9.28 nJ.U = DOUBLE-CHECK: Capacitance increases with a dielectric material, when the battery stays connected, q must increase. When the battery is disconnected before the dielectric is inserted, V must decrease.

24.57. THINK: A parallel plate capacitor has a capacitance of 120. pFC = and plate area of 2 2100. cm 0.0100 m .A = = The space between the plates is filled with mica of dielectric constant 5.40.κ =

The plates of the capacitor are kept at 50.0 V.V = I want to find: (a) The strength of the electric field mica, .E (b) The amount of free charge on the plates, .Q (c) The amount of induced charge on mica, ind .Q SKETCH:

RESEARCH: For a parallel plate capacitor, 0 / .C A dε κ= While connected to a battery, V across the capacitor is equal to that of the battery. The field in the mica is; therefore, just the field between the plates,

/ .E V d= The charge Q is .Q CV= The induced charge in the mica is found by considering net 0 induced .E E E= −

SIMPLIFY: (a) 0 0/ /C A d d A Cε κ ε κ= = and 0/ / .E V d VC Aε κ= = (b) Q CV=

(c) net ind ind net ,E E E E E E= − = − ind net

0 0 0 0 0

.Q QQ Q Q

A A A A Aε ε ε ε κε= − = − Then ( )ind 1 1/ .Q Q κ= −

CALCULATE:

(a) ( )( )

( )( )( )12 2

50.0 V 120. pF12550 V / m

8.85 10 F / m 0.0100 m 5.40E

−= =

⋅

(b) ( )( )120. pF 50.0 V 6.00 nCQ = =

(c) ( )ind16.00 nC 1 4.8888 nC

5.40Q = − =

ROUND: To three significant figures, 12.6 kV / m,E = 6.00 nCQ = and ind 4.89 nC.Q = DOUBLE-CHECK: The charge induced in the dielectric should be less than the charge on the capacitor plates.


946

24.58. THINK: Given a material with dielectric constant 3.40κ = and dielectric strength 7

max / 4.00 10 V/m,V d = ⋅ I want to design a capicitor with capacitance 47.0 pFC = , which can hold a charge of 7.50 nC.q = Let x and y be the dimensions of the parallel plates, and let z be the plate separation. SKETCH:

RESEARCH: 047.0 pF / ,C A zκε= = ,A xy= 7.50 nC ,q C V= = Δ 7/ 4.00 10 V / mE V z= Δ = ⋅ and

3.40.κ =

SIMPLIFY: .qVzE EC

Δ= = The dimension are minimized when ,x y=

1/220 0

0

.A x zCz x

C Cε εκ κ

κε

= = =

CALCULATE: ( )( )6

7

7.50 nC 3.99 10 m,4.00 10 V / m 47.0 pF

z −= = ⋅⋅

( )( )( )( )

1/26

312 3 -1 4 2

3.99 10 m 47.0 pF2.496 10 m

3.40 8.85 10 m kg s Ax

−−

− −

⋅ = = ⋅ ⋅

ROUND: 63.99 10 mz −= ⋅ and 32.50 10 m.x y −= = ⋅ DOUBLE-CHECK: Since the charge stored and the capacitance was very small, it is not surprising, though perhaps unrealistic, to have such small dimensions.

24.59. THINK: The plates on the parallel plate capacitor have a width 1.00 cm 0.0100 mW = = and a length 10.0 cm 0.100 m,L = = and therefore an area of 2 210.0 cm 0.00100 m .A = = The separation between the

plates is 0.100 mm 0.000100 m.d = = It is charged by a power supply at a potential difference of 31.00 10 V.V = ⋅ The power supply is then removed, and without being discharged, the capacitor is placed

in a vertical position above a container holding de-ionized water, such that the short sides of the plates are in contact with the water. Demonstrate that the water will rise between the plates, and determine the system of equations that would allow one to calculate the height to which the water rises between the plates. SKETCH:

RESEARCH: The capacitance of a parallel plate capacitor with air between the plates is air 0 / .C A dε= The capacitance of a parallel plate capacitor with de-ionized water between the plates is


947

water 0 air/ ,C A d Cκε κ= = where 80.4.κ = The charge on a capacitor is .q CV= The energy stored in a capacitor is ( )2 / 2 .U q C= The potential energy of a column of water with height h inside the parallel plate capacitor is g CM ,U mgh= where CMh is the height of the center of mass of the column of water. The mass

of the water is ,m Vρ= where 3 31.00 10 kg/m .ρ = ⋅

SIMPLIFY: With air, the capacitor has a capacitance of 0air .

ACd

ε= The charge on the capacitor is

air air .q C V= The energy stored in the capacitor is 2air

air .2q

UC

= Once the capacitor is charged and the battery

is removed, the charge on the capacitor stays the same. Since de-ionized water does not conduct electricity, the charge on each plate will stay the same even after the capacitor touches the water. By bringing the edge of the capacitor in contact with the free surface of the water in the tank, an upward force will act on the water. This can be proven from energy considerations. Assume that the water is indeed pulled upward between the plates until it completely fills the space between the plates (the water column height equals the length of the plates, L). The new capacitor with water as dielectric has a capacitance

0water air .

AC C

dκε

κ= = The energy stored by this capacitor would be: 2 2water air

water airwater air

1 .2 2q q

U UC Cκ κ

= = = The

energy stored by the capacitor with water as a dielectric is less than the energy stored by the capacitor with

air: water air air air air1 1 .U U U U U Uκκ κ

−Δ = − = − = The change (final minus initial) in the energy of the

system (capacitor) is negative, which means the system is doing work. This work is done against the force of gravity to pull the water upward between the plates. The potential energy of a column of water with

height equal to the length of the capacitor plate would be 2g CM

1 1 .2 2

U mgh VgL gWL dρ ρ= = = The work

done by the electric field in the capacitor is not enough to pull the water all the way up to a height L, but is enough to pull the water to some height h between the plates. The new capacitor is, in effect, a parallel combination of two capacitors: one with air, another with a dielectric (water). Their respective

capacitances are: 0 1 01

A WhCd d

κε κε= = and

( )00 22 .

W L hAC

d dεε −

= = The charge will no longer be

uniformly distributed on the plates. Rather, it will be redistributed such that the voltages on the parallel

capacitors are the same: 1 21 2

1 2

.q qV VC C

= = In addition, the total charge remains 1 2 air .q q q+ = The

energies stored in the two capacitors are 21

112

qUC

= and 22

22

.2qUC

= By conservation of energy, i fU U= , or

more specifically, air 1 2 ,2hU U U mg= + + where

2hmg represents the gravitational energy of the column of

water that is sucked upward between the plates of the capacitor. For a height h of the water between the plates, the mass of water is .m V dWhρ ρ= =

CALCULATE: ( )( )( )12 2 2

11air 4

8.85 10 F/m 1.00 10 m 10.0 10 m8.85 10 F,

1.00 10 mC

− − −−

−

⋅ ⋅ ⋅= = ⋅

⋅

( )( )11 3 8air 8.85 10 F 1.00 10 V 8.85 10 C,q − −= ⋅ ⋅ = ⋅

( )( )

285

air 11

8.85 10 C4.425 10 J,

2 8.85 10 FU

−−

−

⋅= = ⋅

⋅

57

water4.425 10 J 5.51 10 J

80.4U

−−⋅= = ⋅


948

Recall that water air :U U UΔ = − ( )5 51 80.4 4.425 10 J 4.372 10 J.80.4

U − −− Δ = ⋅ − ⋅

So water air ,U U< and the

capacitor is doing work against gravity to “suck” water up between the plates. So it is true that the water is drawn up to some height h between the plates of the capacitor. Note that if the water rose to height h L= , the gravitational potential energy stored in the system would be:

( )( )( )( ) ( )23 3 25

g

1.00 10 kg/m 9.81 m/s 0.0100 m 0.100 m 0.000100 m4.905 10 J.

2U −

⋅= = ⋅

So the work done by the capacitor’s electric field is not enough to suck up the water to the full height of L, since g water .U U> The parallel combination of the two “new” capacitors with water and air, respectively, has capacitances

( )( )( )

( )( )( ) ( )( )

12 28

1 4

12 210

2 4

80.4 8.85 10 F/m 1.00 10 m7.12 10 F

1.00 10 m8.85 10 F/m 1.00 10 m

8.85 10 0.100 m F,1.00 10 m

hC h

L hC h

− −−

−

− −−

−

⋅ ⋅= = ⋅

⋅⋅ ⋅ −

= = ⋅ −⋅

where h is measured in meters. By considering the new charge distribution and equivalent voltages on each 1C and 2 ,C there are two equations:

( ) ( ) ( )( )10 81 21 28 10

81 2

8.85 10 0.100 m 7.08 10 7.08 10 8.85 10 0.100 m

8.85 10 C

q q q h q hh h

q q

− −− −

−

= ⋅ − = ⋅⋅ ⋅ −

+ = ⋅

In addition, the energies stored in each capacitor is 21

112

qUC

= and 22

222

qUC

= , or 21

1 71.424 10qU

h−=⋅

and

( )22

2 9 .1.771 10 0.100 m

qUh−=

⋅ − The mass of the water between the parallel plates is

( )( )( ) ( )3 31.00 10 kg/m 0.000100 m 0.0100 m 0.00100 kg .m dWh h hρ= = ⋅ = The gravitational potential

energy of this water is: ( )( ) ( )

22 3 2

g

0.00100 kg 9.81 m/s/ 2 4.91 10 J .

2U mgh h h−= = = ⋅ By conservation

of energy,

( )

air 1 2

2 25 3 21 2

7 9

2

4.425 10 4.91 101.424 10 1.771 10 0.1 m

hU U U mg

q q hh h

− −− −

= + +

⋅ = + + ⋅⋅ ⋅ −

where the energies above are each measured in Joules. This provides the third equation needed to determine the height h. The system of equations contains the three unknowns, 1 2, , and q q h , where the charges are in Coulombs and the height in meters:

( ) ( )( )

( )

21 2

81 2

2 25 3 21 2

7 9

8.85 10 0.100 m 7.088.85 10

4.425 10 4.91 101.424 10 1.771 10 0.100 m

q h q hq q

q q hh h

−

−

− −− −

⋅ − =

+ = ⋅

⋅ = + + ⋅⋅ ⋅ −

ROUND: A computer algebra system can be used to solve the system. There are four solutions:

( )8 91 20.0956 m, 9.08 10 C, 2.32 10 C ,h q q− −= − = ⋅ = − ⋅

( )7 11 81 27.11 10 m, 5.04 10 C, 8.86 10 C ,h q q− − −= − ⋅ = − ⋅ = ⋅


949

( )3 5 51 21.27 10 m, 1.76 10 C, 1.75 10 C ,h q q− − −= − ⋅ = ⋅ = − ⋅

( )2 8 111 29.43 10 m, 8.84 10 C, 6.67 10 C .h q q− − −= ⋅ = ⋅ = ⋅

DOUBLE-CHECK: Note that if the height of the water column h were to equal the length of the capacitor plate, L (where 0.100 mL = ), it would defy the first equation in the system of equations above. This is consistent with an earlier finding, where it was shown that the final height of the water cannot be h L= since there is not enough energy stored in the capacitor’s electric field (with h L= ) to balance the gravitational potential energy in the water column of height h L= .

Additional Problems

24.60. This is like two parallel plate capacitors in parallel, each with an area ( ) 21/ 2 .A rπ= In parallel,

( ) ( )( ) ( )

( ) ( )

20 0 0 0

eq 1 2

2128

1 12

8.85 10 F/m 0.61 m1 11.1 2.980 10 F 30. nF.

2 0.0021 m

A A A rC C C

d d d d−

−

= + = + = + = +

⋅= + = ⋅ ≈

ε ε κ ε ε πκ κ

π

24.61. The largest potential difference that can be sustained without breakdown is about ( )( )2.5 kV/mm 15 mm 37.5 kV= . Next, consider the relationship between charge deposited and change in potential:

( ) ( )2

12 2 1 2 4 80 0.0025 m8.85 10 C N m 3.75 10 V 5.5 10 C.0.015 m

AQ CV V

dε − − − −

= = = ⋅ ⋅ = ⋅

24.62. The capacitances are given as 1 2.0 nFC = and 2 3 4.0 nF.C C= = The potential difference applied is 1.5 V.V = The potential difference on 1C is equal to that of the battery. Then 1 1 1q C V= 1C V=

( )( )2.0 nF 1.5 V 3.0 nC.= = Note 2C and 3C are in series. Their equivalent capacitance 23C is

( )( )2 323

2 3

4.0 nF 4.0 nF2.0 nF.

8.0 nFC C

CC C

= = =+

With 1C and 23C in parallel, 1 23 .V V V= = The charges for capacitors in series are equal; then 2 3 23q q q= = ( )( )23 2.0 nF 1.5 V 3.0 nC.C V= = =

24.63. The energy stored in a capacitor is ( ) 21/ 2 .U CV= Putting a Mylar insulator between the plates of a vacuum gap capacitor will increase the capacitance C by a factor of 3.1,κ = which is Mylar’s dielectric constant. Therefore, the percentage increase is

( ) ( ) ( )( )( ) ( ) ( )( )

2 2f i

2i

1/ 2 1/ 2100% 100% 1 100% 3.1 1 100% 210%

1/ 2

CV CVU UU CV

κκ

−−⋅ = ⋅ = − ⋅ = − ⋅ =

24.64. The capacitance of a parallel plate capacitor in air is given by 0 / ,C A dε= where A is the area of the plates and d is the separation between the plates. Changing the distance between the plates changes the capacitance. The potential energy of a capacitor is given by ( ) 21/ 2 .U CV= The work done is

( )

( )( )( ) ( ) ( )

2 2 2 20f i f i f i

f i

12 4 22 10

3 3

1 1 1 1 12 2 2 2

8.85 10 F/m 12.0 10 m 1 1 9.00 V 1.30 10 J2 2.75 10 m 1.50 10 m

AW U U U C V C V C C V V

d d

W

ε

− −−

− −

= Δ = − = − = − = −

⋅ ⋅ = − = − ⋅ ⋅ ⋅


950

24.65. The capacitance is 1.0 FC = for a square, parallel-plate capacitor. The separation is 0.10 mmd = 0.00010 m,= and is filled with paper of 5.0.κ = For a parallel plate capacitor 2

0 0/ / .C A d L dκε κε= =

Then ( )( )

( )( )120

0.00010 m 1.0 F1503 m 1.5 km.

5.0 8.85 10 F/mdCLκε −

= = = ≈⋅

24.66. The capacitance is 4.00 pFC = and the potential difference is 10.0 V.V = The plate separation is 3.00 mm.d =

(a) The charge is ( )( )4.00 pF 10.0 V 40.0 pC.Q CV= = =

(b) The energy stored is ( ) ( )( )( )221/ 2 1/ 2 4.00 pF 10.0 V 200. pJ.U CV= = =

(c) The area is ( )( ) ( )12 3 20/ 4.00 pF 0.00300 m / 8.85 10 F/m 1.36 10 m .A Cd ε − −= = ⋅ = ⋅

(d) The dielectric constant of polystyrene is 2.6.κ = Then ( )air 2.6 4.00 pF 10. pF.C Cκ= = =

24.67. First, simplify the circuit diagram to the following figure:

with 5 1 2C C C= + and 6 3 4 .C C C= + The total capacitance is given by ( ) 1

tot 5 61/ 1/ .C C C −= + Since the capacitors are in series, the charge on each capacitor is tot .Bq C V= The potential at point A is then given by 5/ .A D BV V V q C= + − Also, ( ) ( )5 1.0 mF 2.0 mF 3.0 mF,C = + = ( ) ( )6 3.0 mF 4.0 mF 7.0 mF,C = + =

tot 2.1 mF,C = and ( )( )2.1 mF 1.0 V 2.1 mC.q = = Therefore, the potential at point A is

( ) ( ) ( )= + − =0 V 1.0 V 2.1 mC / 3.0 mF 0.30 V.AV

24.68. The energy is given by 2 / 2 ,U q C= with 0 /C A d= κε for a parallel-plate capacitor. The energy stored is:

( ) ( )( )( )( )

24 32

12 3 20

4.20 10 C 1.30 10 m290 J.

2 2 7.0 8.85 10 F/m 6.40 10 mq dU

Aκε

− −

− −

⋅ ⋅= = =

⋅ ⋅

24.69. The capacitance is given by ( )( )( )

( )12 10 2

1308

9.1 8.85 10 F/m 1.00 10 m4.0 10 F.

2.00 10 mA

Cd

κε − −−

−

⋅ ⋅= = = ⋅

⋅

24.70. (a) Since the capacitors 1C and 2C have the same potential,

( )1 2 11 2

1 2 2

6.00 μF 40.0 μC 80.0 μC.3.00 μF

Q Q CqV Q QC C C C

= = = = =

(b) Since the total charge on 1C and 2C is equal to that on 3 ,C it is required that: 3 1 2 120. μC.Q Q Q= + = (c) The total voltage applied is:

31

1 3

80.0 μC 120. μC 37.3 V.6.00 μF 5.00 μF

QQC C

+ = + =

24.71. The capacitance of a cylindrical capacitor is given by:

( ) ( ) ( )( )( )12 3 -1 4 2

0

2 1

2 8.85 10 m kg s A 0.0724 m263 0.792 nF.

ln / ln 4.16 cm / 3.02 cmL

Cr r

πκ πε − −⋅= = =


951

24.72. (a) The capacitance is given by: ( )( )12 604 4 8.85 10 F/m 6.328 10 m 704 μF.C Rπε π −= = ⋅ ⋅ =

(b) ( )

( )

25214

7.8 104.3 10 J.

2 2 704 μFQU

C− ⋅

= = = ⋅

24.73. THINK: A parallel plate capacitor with an air gap is connected to a 6.00 V battery. The initial energy of the capacitor is i 72.0 nJ.U = After a dielectric material is inserted, the capacitor has an additional energy of 317 nJ. The final energy stored in the capacitor is f 72.0 nJ 317 nJ 389 nJ.U = + = SKETCH:

RESEARCH: The energy stored in a capacitor is given by ( ) 21/ 2 .U CV= The initial and final energy are

( ) 2i i1/ 2U C V= and ( ) 2

f f1/ 2 .U C V= SIMPLIFY: (a) Taking a ratio of fU and iU yields f i f i/ / .U U C C= Using f i ,C Cκ= the dielectric constant is found to be f i/ .U Uκ = (b) The charge in the capacitor is given by .Q CV= Using ( ) 21/ 2 ,U CV= it is found that the charge is

( )22 / 2 / .Q U V V U V= =

(c) The electric field inside a parallel plate capacitor is 0 0

2 .Q UEA AVκε κε

= =

(d) The electric field inside the capacitor after the dielectric material is inserted is ( )ff

f0 0

2 /2.

UUEAV AV

κκε ε

= = Using the result in (a) i f /U U κ= yields ( )f i i02 / .E U EAVε= = This means

that the field does not change. CALCULATE:

(a) The dielectric constant is 389 nJ 5.403.72.0 nJ

κ = =

(b) The charge in the capacitor after the dielectric material has been inserted is

( )9f

f

2 389 10 J20.129 μC.

6.00 VUQV

−⋅= = =

(c) The electric field inside the capacitor before the dielectric material is inserted is ( )

( )( )( )5i

i 12 2 2 3 20

2 72.0 nJ25.424 10 N/C.

8.85 10 C /N m 5.00 10 m 6.00 VUEAVε − −

= = = ⋅⋅ ⋅

(d) 5f i 5.424 10 N/CE E= = ⋅

ROUND: (a) 5.40κ = (b) f 0.129 μCQ = (c) 5

i 5.42 10 N/CE = ⋅

(d) 5f 5.42 10 N/CE = ⋅

DOUBLE-CHECK: The numerical results are reasonable.


952

24.74. THINK: Since charge is conserved, the initial charge on the capacitor charged by the battery is equal to the total charge on the capacitors that are connected. SKETCH:

RESEARCH: Since the charge is conserved, i 1 2 1 i 1 f f .Q Q Q C V C V CV= + = +

The energy stored in a capacitor is ( ) 21/ 2 .U CV= SIMPLIFY: Solving for C in the above equation yields ( )1 i f f/ .C C V V V= − The energy stored in the

second capacitor is ( ) 2f1/ 2 .U CV=

CALCULATE: ( )( )

( )8.00 μF 240. V 80.0 V

16.0 μF80.0 V

C−

= =

( )( )21 16.0 μF 80.0 V 51.2 mJ2

U = =

ROUND: Keeping only three significant figures gives 16.0 μFC = and 51.2 mJ.U = DOUBLE-CHECK: The equivalent capacitance of the two capacitors is eq 1 24.0 μF.C C C= + = The

charge on the equivalent capacitor is ( )( )eq f 24.0 μF 80.0 V 1.92 mC.Q C V= = = The initial charge on the

first capacitor is ( )( )i 1 i 8.00 μF 240. V 1.92 mC,Q C V= = = so charge is conserved, as expected.

24.75. THINK: In this problem, the work-energy relation is used. The dielectric constant of Mylar is 3.1.κ = Use the work-energy relation, and think about the initial and final energies of the capacitor. SKETCH:

RESEARCH: From work-energy relation, it is found that i f .W U U U= −Δ = − The initial and final

energies of the capacitor are ( ) 2i i1/ 2U CV= and

( )22 2i i i i i

2f i

1 1 1 .2 2 2

Q C V C VUC C κκ

= = =

SIMPLIFY: Therefore, ( ) ( )2i1/ 2 1 1/ .W CV κ= − The work done by the electric field is ,W FL= where L

is the length of the square plate, that is .L A= Using the capacitances of a parallel plate, 0 / ,C A dε= it is found that

( ) ( )2 20 0 02i ii

1 1 1 1 1 11 1 1 .2 2 2

A A AFL V F VV

d dL dε ε ε

κ κ κ = − = − = −

CALCULATE: Thus the net force done by the electric field is

( ) ( )12 2 2 4 2 2 9

3

1 18.85 10 C /N m 2.00 10 m 15.0 V 1 9.5 10 N.2 3.11.00 10 m

F− −

−

−

⋅ ⋅= − = ⋅ ⋅


953

ROUND: Keeping two significant figures gives 99.5 10 N.F −= ⋅ The positive sign of the force means that the direction of the force is the same direction as the motion of the dielectric material. DOUBLE-CHECK: The magnitude and direction of the force make sense, considering the scale.

24.76. THINK: This problem is similar to the motion of a projectile under a gravitational force. In this case the gravitational force is replaced by an electrostatic force. Let L be the length of the plates. SKETCH:

RESEARCH: The force acting on a proton is given by .F qE= For a parallel plate, the electric field is

0/ .E σ ε= Thus, 0/ .F qσ ε= The time required to reach the far edge of the capacitor is 0/ .t L V= Therefore, the deflection distance YΔ is

2 22

20 0 0

1 1 1 .2 2 2

qF L LY atm V m V

σε

Δ = = =

SIMPLIFY: Not applicable. CALCULATE: Putting in the numerical values gives:

( )( )( ) ( )( )

( )( )

219 6 2 2

227 12 2 2 6

1.602 10 C 1.0 10 C / m 2.0 10 m1 0.002168 m.2 1.67 10 kg 8.85 10 C / N m 1.0 10 m / s

Y− − −

− −

⋅ ⋅ ⋅ Δ = = ⋅ ⋅ ⋅

ROUND: 2.2 mm.YΔ ≈ DOUBLE-CHECK: This is reasonable.

24.77. THINK: A parallel plate capacitor with a squared area of side 10.0 cmL = and separation distance 2.5 mmd = is charged to a potential difference of 0 75.0 V,V = and then disconnected from the battery. I

want to determine the capacitor’s capacitance, 0 ,C and the energy, 0 ,U stored in it at this point. A dielectric with constant 3.4κ = is then inserted into the capacitor such that it fills 2/3 of the volume between the plates. I want to determine the new capacitance, new potential difference between the plates and energy of the capacitor, , C V′ ′ and .U ′ I want to determine how much work, if any, is required to insert the dielectric into the capacitor. SKETCH:


954

RESEARCH: (a) The capacitance of a parallel plate capacitor is given by 0 0 / .C A dε= The energy stored in the capacitor is ( ) 2

0 0 01/2 .U C V= (b) After a dielectric has been inserted, the capacitor can be treates as two capacitors in parallel, one with a dielectric, and one without. The new capacitance is obtained by adding the contributions of the two parts of the capacitor, i.e., 1 2 .C C C′ = + Since the charge on the capacitor is unchanged, and the potential is the same across both parts of the new capacitor, ( )0 0 0 0 / .C V C V V C C V′ ′ ′ ′= = The new energy stored on

the capacitor is ( ) 21/ 2 .U C V′ ′ ′= SIMPLIFY: (a) 0 0 / ;C A dε= ( ) 2

0 0 01/2U C V=

(b) The new capacitance is ( ) ( )1 2 0 0 01 21/3 2/3 .

3C C C C C Cκκ + ′ = + = + =

The new potential between the

plates is ( )0 0/V C C V′ ′= . The new energy stored in the capacitor is

( )( ) ( )

20 0

00 2

2 0 00

1 2 1 23 3 31 31/ 2

2 2 1 2 1 2

C VC C

C VU C V U

κ κ

κ κ

+ + ′ ′ ′= = = = + +

.

(c) By using the work energy relation, it is found that the applied work is ( )

0 0 0

2 13 .11 21 2

W U U U U Uκκκ

− ′= Δ = − = =− ++

Since κ is larger than 1, this means that the applied work is negative. Therefore, the external agent does not need to do work to insert the dielectric slab. CALCULATE: Substituting the numerical values yields,

(a) ( )( )212 2 2

0

8.85 10 C /N m 0.100 m35.42 pF,

0.0025 mC

−⋅= = ( )( )212 8

01 35.42 10 F 75.0 V 9.961 10 J2

U − −= ⋅ = ⋅

(b) ( ) ( )1 2 3.4

35.42 pF 92.08 pF,3

C+

′ = = ( )35.42 pF 75.0 V 28.85 V,92.08 pF

V ′ = =

( ) ( ) 883 3.83 10 J9.961 10 J1 2 3.4U −−′ = = ⋅⋅

+

(c) Not required. ROUND: Rounding all results to two significant figures gives (a) 0 35 pFC = and 7

0 1.0 10 J.U −= ⋅ (b) 92 pF,C′ = 29 VV ′ = and 83.8 10 J.U −= ⋅ (c) Not required. DOUBLE-CHECK: The answers are of reasonable magnitudes and their respective units make sense.

24.78. THINK: What would be the area of a parallel plate capacitor, with plate separation 1.0 mm,d = capable of storing the same amount of energy as a AAA battery, or 3400 J ? The potential difference of a AAA battery is 1.5 V.V = What would be the area of such a capacitor be if the potential difference is to be the maximum that can be applied without dielectric breakdown of the air between the plates?


955

SKETCH:

RESEARCH: The capacitance of the parallel plate capacitor is given by 0 / .C A dε= The energy that can

be stored in the capacitor is ( ) 21/ 2 .U CV= The dielectric strength of air is 2.5 kV/mm. SIMPLIFY: Substituting 0 /C A dε= into the expression for U and solving for A :

202

0

1 2 .2

A dUU V Ad V

εε

= =

CALCULATE: (a) Putting in the numerical values into the above expression gives

( )( )( )( )( )

311 2

212 2 2

2 1.0 10 m 3400 J3.413 10 m .

8.85 10 C / N m 1.5 VA

−

−

⋅= = ⋅

⋅

(b) Replacing V with the maximum voltage before dielectric breakdown occurs yields

( )( )( )( ) ( )( )( )

35 2

212 2 2 3

2 1.0 10 m 3400 J1.229 10 m .

8.85 10 C / N m 2.5 10 V/mm 1.0 mmA

−

−

⋅= = ⋅

⋅ ⋅

Note that ( )( )max 2.5 kV/mm 1.0 mm 2.5 kV.V = = ROUND: Rounding the results to two significant figures yields (a) 11 23.4 10 mA = ⋅ and (b) 5 21.2 10 m .A = ⋅ (c) These areas are in the order of many square km, making such a capacitor highly impractical. DOUBLE-CHECK: This makes sense.

24.79. THINK: Recall the equivalent capacitance of capacitors connected in series. The areas of the two capacitors in this question are the same. The charges are the same for both capacitors. SKETCH:

RESEARCH: An equivalent capacitance of two capacitors connected in series is eq 1 21/ 1/ 1/ .C C C= + The

capacitance of a parallel plate capacitor is given by 0 / .C A dε= Since 1C and 2C have the same area and the same plate separation, the capacitance of 2C is equal to 2 1 ,C Cκ= where κ is the dielectric constant

of material in 2 .C The equivalent capacitance, eqC = ( ) 11 21/ 1/C C −+ = ( ) 1

1 11/ 1/C Cκ −+ ( )1 / 1 .Cκ κ= + (a) The charge on the capacitors is .Q CV= (b) The total energy stored in the capacitors is ( ) ( )2 21/ 2 1/ 2 / .U CV Q C= = (c) The electric field inside 2C is given by ( ) ( )2 2 0/ / ,E Q A κε= and 2 1Q Q Q= =


956

SIMPLIFY:

(a)

( )0

1

1 1

AC dQ CV V V

εκκ

κ κ= = =

+ +

(b) ( ) ( )

( ) ( )

22 22 01

2 2 21 1

1 2 21 1

222 01

2 22 1

2 2 22 1

12 2 2 1 2 1

12 2 2 1 2 1

AC VVdQ C VU

C C

AC VVdQ C VU

C C

εκ κκ κ

κ κ

εκ κκ κ

κ κ κ

+ = = = =

+ +

+ = = = =

+ +

(c) ( ) ( )0

20 0 1 1

AVQ VEA d A d

κεκε κε κ κ

= = =+ +

CALCULATE: Substituting the numerical values into the above equations yields

(a) ( )( )( )

( )( )

12 410

1 2 3

7.0 8.854 10 F/m 1.00 10 m 96.0 V7.44 10 C

0.100 10 m 7.0 1Q Q

− −−

−

⋅ ⋅= = = ⋅

⋅ +

(b)

( ) ( )( )( ) ( )

( )

12 42 2

3

91 2

8.854 10 F/m 1.00 10 m7.0 96.0 V

0.100 10 m31.24 10

2 7.0 1U

− −

−

−

⋅ ⋅ ⋅ = = ⋅

+.

( )( )( ) ( )

( )

12 42

3

92 2

8.854 10 F/m 1.00 10 m7.0 96.0 V

0.100 10 m4.46 10 J

2 7.0 1U

− −

−

−

⋅ ⋅ ⋅ = = ⋅

+

The total energy is 1 2 31.24 nJ+4.456 nJ 35.70 nJ.U U U= + = =

(c) ( )( )2 3

96.0 V 120000 V/m0.100 10 m 7.0 1

E−

= =⋅ +

ROUND: Rounding all results to two significant figures gives the following answers. (a) 0.74 nCQ = (b) 36 nJU = (c) 5

2 1.2 10 V/mE = ⋅ DOUBLE-CHECK: Dimensional analysis confirms that all results are in the correct units.

24.80. THINK: The plates of the parallel plates capacitor consist of two metal discs of radius 1 4.00 cm 0.0400 m.R = = The separation between the plates is 2.00 mm 0.00200 m.d = = In part a) the

space between the plates is filled with air. In part b) a dielectric of out radius 1 ,R inner radius

2 2.00 cm 0.0200 m,R = = thickness 2.00 mm 0.00200 md = = and dielectric constant 2.00κ = is placed between the plates. In part B) the situation can be modeled as two capacitors in parallel. In part c) a solid disc of radius 1 ,R made of dielectric material having the dielectric constant 2.00κ = is placed between the plates. This situation can be modeled as two capacitors in series. The thickness of the dielectric layer is

/ 2.d


957

SKETCH:

RESEARCH: (a) The capacitance of the parallel plate capacitor in air is 1 0 / ,C A dε= while 2

1 .A Rπ=

(b) The equivalent capacitance of the capacitors in parallel is given by 2 i 2 ,C C C ′′′= + where

( )2 22 0 0 1 2/ /C A d R R dκε κε π′ ′= = − and 2

2 0 0 2/ / .C A d R dε ε π′′ ′′= =

(c) The equivalent capacitance 3C of the capacitors in series is ( ) 1

3 3 31/ 1/ ,C C C−

′′′= + where

( )23 0 1 / /2C R dκε π′ = and ( )2

3 0 1 / /2 .C R dε π′′ = SIMPLIFY: (a) 2

1 0 1R / .C dε π=

(b) ( )( )22 2

2 2 20 2 01 22 0 1 2 2

RR RC R R Rd d d

ε π ε πκε π κ−= + = − +

(c) Not necessary. CALCULATE:

(a) ( )( ) ( )212 2 2

111

8.85 10 C / N m 0.0400 m2.224 10 F

0.00200 mC

π−

−⋅

= = ⋅

(b) ( )( ) ( ) ( )( ) ( )( )

12 2 22 2 2 11

2

8.85 10 C / N m2.00 0.0400 m 0.0200 m 0.0200 m 3.892 10 F

0.00200 mC

π−

−⋅

= + + = ⋅

(c) ( )( ) ( )212 2 2

113

2.00 8.85 10 C / N m 0.0400 m8.897 10 F

0.00100 mC

π−

−⋅

′ = = ⋅


958

( )( ) ( )212 2 2

113 3

2.00 8.85 10 C / N m 0.0400 m4.485 10 F

0.00100 mC C

π−

−⋅

′′ ′= = = ⋅

( )( )11 1111

3 10

8.897 10 F 4.485 10 F2.990 10 F

1.244 10 FC

− −−

−

⋅ ⋅= = ⋅

⋅

ROUND: To 3 significant figures, (a) 11

1 2.22 10 FC −= ⋅ (b) 11

2 3.89 10 FC −= ⋅

(c) 113 2.99 10 FC −= ⋅

DOUBLE-CHECK: The value of a capacitor with a dielectric medium is greater than the value of the capacitor without a dielectric medium.

24.81. THINK: Capacitor 1 1.00 μFC = has an electric potential of 1 50.0 V.VΔ = Capacitor 2 2.00 μFC = has an electric potential of 2 20.0 V.VΔ = The two capacitors are connected positive plate to negative plate. Calculate the final charge, 1, fQ on capacitor 1C after the two capacitors have come to equilibrium. SKETCH:

RESEARCH: Because the capacitors are connected in such a way that the positive plate of each is connected to the negative plate of the other, they must be in series. Therefore the final charges on 1C and

2C must be equal. Because charge is conserved, the total initial charge must equal the total final charge. The initial charge of the system is i 1, i 2, iQ Q Q= + where 1, i 1 1Q C V= Δ and 2, i 2 2 .Q C V= Δ

SIMPLIFY: f i 1 1 2 2i i1 i2 1 1 2 2 i f f1 f2 f1, , , and .

2 2 2Q Q C V C VQ Q Q C V C V Q Q Q Q Q +

= + = + = = = = =

CALCULATE: ( )( ) ( )( )6 6

5f1

1.00 10 F 50.0 V 2.00 10 F 20.0 V4.50 10 C

2Q

− −−

⋅ + ⋅= = ⋅

ROUND: There were three significant figures provided in the question so the answer should be written as 5

1, f 4.50 10 CQ −= ⋅ or 45.0 μC. DOUBLE-CHECK: It is reasonable that there is less charge stored on capacitor 1C after it was connected to 2C because the potential across capacitor 1C would have to decrease in order for it to come into equilibrium with 2 .C

24.82. THINK: The spherical capacitor consists of two concentric conducting spheres of radius 1r and 2 ,r where

2 1.r r> The space between the spheres is filled with a dielectric material of electric permeability 010 .ε ε= The dielectric material starts at 1r and extends to radius ( )1 2 .R r R r< < The problem can be modeled as two spherical capacitors connected in series.


959

SKETCH:

RESEARCH: The equivalent capacitance for capacitors in series is eq 1 21/ 1/ 1/ .C C C= + The equation for

the first spherical capacitor is ( )( )1 1 14 / .C r R R rπε= −

SIMPLIFY: ( )eq 1 2 2 1 1 21/ 1/ 1/ / ,C C C C C C C= + = + substituting the values for 1C and 2C into the equation gives:

1 2

eq 1 2

1 .4 4R r r R

C r R r Rπε πε− −

= +

Recall that 010 ,ε ε= substituting this value into the equation gives: 2 2

21 2 2 1 2 1 2 12

1 2eq 0 1 0 2 0 0 1 2

1 22 1 2 1eq 0

2 1 2 10 1 2

10 101 1 1104 10 4 4 4 10

109 101 4 .9 104 10

r RR rR r r R R r r r R r r R r Rr R r RC r R r R r r R

r r RRr r r r R CRr r r r Rr r R

π ε πε πε πε

πεπε

− −− − − + −+= + = = + −

= = + −

In the limit of 1 ,R r= ( ) ( )( )2 2eq 0 1 2 1 2 1 0 1 2 2 14 10 / 10 10 4 / .C r r r r r r r r rπε πε = − = − In the limit of 2 ,R r=

( )eq 0 1 2 2 14 10 / .C r r r rπε = −

CALCULATE: Not required. ROUND: Not required. DOUBLE-CHECK: It is reasonable that the equivalent capacitance is greater in the limit 2 ,R r= because there would be more dielectric material present in the spherical capacitor.

24.83. THINK: This problem is similar to finding the maximum height in a projectile motion. However, in this case, the acceleration is due to the electric field of a parallel plate capacitor. Use the conservation of energy. SKETCH: (a) (b)


960

(c)

RESEARCH: (a) The energy required to reach the positive plate is given by .U q VΔ = Δ The energy of a proton is

2i (1/ 2) .K mv= In order for proton to reach the positive plate, the kinetic energy must be larger than ,UΔ

that is, i .K U> Δ (b) Not needed. (c) From conservation of energy, it is found that i i f f .K U K U+ = + Since at the maximum x displacement, the kinetic energy is zero, f 0.K = (d) Using conservation of energy, it is found that i i f f ,K U K U+ = + but in this case i f .U U= SIMPLIFY: (c) Using 2

i (1/ 2) xK mv= and ,U qV= the above equation becomes 2(1/ 2) 0 0 .xmv qV+ = + Using

i sin ,xv v θ= thus, the potential is ( )2 2sin / .2iV mv qθ= (d) It follows that i f .K K= Therefore, the speed of the proton as it reaches the negative plate is the same as the initial speed, 52.00 10 m/s.⋅ CALCULATE: (a) The kinetic energy of the proton is ( )( )2 1727 5(1/ 2) 3.34 10 J.1.67 10 kg 2.00 10 m/siK −−= = ⋅⋅ ⋅ The

energy required to reach the positive plate is ( )( ) 1719 4.806 10 J.300. V1.602 10U −−Δ = = ⋅⋅ Thus iK is less than .UΔ This means that the proton cannot reach the positive plate regardless of the angle .θ ROUND: Not needed. DOUBLE-CHECK: It is reasonable that the proton cannot reach the plate. Conservation of energy was used to balance the potential and kinetic energies, which is a key method of computing velocities.

24.84. THINK: The parallel p late capacitor has a dielectric, κ that is positioned between the plates. There is an air gap separating the dielectric material and the plates. The thickness of the dielectric material is 2 .d The thickness of the air gaps above and below the dielectric material are 1d and 3d respectively. The overall distance between the plates is .d SKETCH:

RESEARCH: The parallel plate capacitor can be modeled as three parallel plate capacitors in series. The equivalent capacitance of the capacitors in series is eq 1 2 31/ 1/ 1/ 1/ .C C C C= + + Two of the capacitors are air


961

filled and one has the material of dielectric, .κ The capacitance, C of each of the capacitors can be determined using the equations 1 0 1/ ,C A dε= 2 0 2/C A dε κ= and 3 0 3/ .C A dε= All three capacitors have the same plate area .A The total distance between the plates is 1 2 3 .d d d d= + +

SIMPLIFY: 3 2 1 3 1 2

eq 1 2 3 1 2 3

1 1 1 1 .C C C C C C

C C C C C C C+ +

= + + = Substituting in the values for 1 ,C 2C and 3C gives

( )

( )( )

20

2 3 1 3 1 2 1 2 31 2 3

3 2 3 1 3 1 2eq 0 00

1 2 3

111 1 ,

Ad d d d d d d d d

d d dd d d d d dC A A

Ad d d

κ κεκ κ

κ κε κ ε κκ

ε

+ + + += = = + +

but 1 3 2 ,d d d d+ = − substituting this into the equation gives:

( )2 2eq 0

1 1 d d dC A

κε κ

= − + ( )0

eq2 2

.A

Cd d d

ε κκ

=− +

CALCULATE: Not required. ROUND: Not required. DOUBLE-CHECK: It seems reasonable that the overall capacitance depends on the distance between the plates and the thickness of the dielectric material and not on the position of the material between the plates.


962

Chapter 25: Current and Resistance

In-Class Exercises

25.1. e 25.2. a 25.3. a 25.4. a 25.5. e 25.6. b 25.7. c 25.8. b 25.9. b Multiple Choice

25.1. d 25.2. d 25.3. c 25.4. c 25.5. c 25.6. a 25.7. c 25.8. a 25.9. c 25.10. b 25.11. d 25.12. d Questions

25.13. Subject to the applied potential and electric field ,E the electrons will accelerate indefinitely due to the electric force .F qE ma= = The drift velocity and current will increase indefinitely until some other effect takes over.

25.14. The voltage across the light bulb is constant. The resistance of a piece of metal (the filament in the bulb) is lower at low temperatures compared to higher temperatures. Since V iR= and V is constant, and the resistance is low, the current i must be large. When the light bulb is first turned on, the filament is cold, so the current is large. A large current increases the likelihood of the light bulb burning out.

25.15. They will be brighter if they are connected in parallel. In parallel, the light bulbs will pull twice the current from the battery, which is twice the power. In series, the circuit has twice the resistance, as it draws only half the current.

25.16. Resistors in parallel 1 2 2

parallel 21 2 1 2 2 1

1 11 /

R R RR RR R R R R R

= + = = <+ +

and 1 2 1parallel 1

1 2 1 2 1 2

1 1 .1 /

R R RR RR R R R R R

= + = = <+ +

The resultant resistance is always smaller than the smaller of the two values. In particular, if the difference between the two values is large (an order of magnitude or more), the resultant resistance is less than but very close to the smaller of the two. Example: if you connect in parallel a resistor 1 1 R = Ω and 2 10 ,R = Ω you get a resistance of 0.91 Ω . If 1 1 OhmR = and 2 1 k ,R = Ω you get a resistance of 0.999 .Ω

25.17. In calculating power, we can use any of the following three equivalent formulas: 2

2 .VP iV RiR

= = = For

resistors in series, the current is the same through all the resistors, so it makes sense to use 2 ,P Ri= and it is thus apparent that the higher the resistance, R of a resistor, the higher the power dissipated on that resistor. For resistors in parallel, the voltage across all the resistors is the same, so it makes sense to use

2

,VPR

= and it is thus apparent that the resistor with the lowest resistance will dissipate most power.

25.18. Consider the following diagram.

1 2

eqeq 1 2 1 2

1 1 1 ,R R

RR R R R R

= + =+

so

1 2 eq 1 1 2 2 eq V V V iR i R i R iR= = = = = 1 2 21 1 1

1 2 1 2

.R R Ri R i i i

R R R R= =

+ +


963

25.19. Since the resistors are in parallel, T

3T 2 3

21

1 1 1 1 1 1 1 1... ...,R R R R R R R R

= + + + = = + + + and 10 .R = Ω Let

1 1 .10

xR

= =Ω The series can be rewritten as:

2 3 2 3

T T

1 1... 1 1 ...,x x x x x xR R

= + + + + = + + + + but

( )2 3 11 ...

1x x x

x+ + + + =

− for 1 ,x < and 1 1,10

< so that means that

T T T

1 1 1 1 1 1 11 1 1 1 1 1

x xR x R x x R x

− ++ = = − = = − − − − T

1 1 1,xRx x−= = − which then gives the

formula T1 1 10 1 9 .R R Rx

= = − = Ω − Ω = Ω

25.20. The black wire, having lower resistance, will draw more power than the red wire since 2 ,P V R= where V is the battery voltage. Since the wire converts this electrical energy to thermal energy, the black wire will get hotter. Note that if the battery has significant internal resistance that will affect the temperature of the wires, but the black wire will still be hotter than the red wire.

25.21. No, ordinary incandescent light bulbs are not actually Ohm resistors. They can be operated over a wide enough range of currents, hence temperatures, that the temperature dependence of the lament resistance is significant. The resistance of an ordinary light bulb measured with an Ohmmeter at room temperature is substantially lower, that its resistance at an operating temperature of order 2000 K. By connecting light bulbs in series it is possible to operate them at a range of voltages, hence currents, wide enough to display this variation in resistance. The experiment is quite pretty as the light bulbs can be made to glow colors ranging from red through orange and yellow to white. A plot of V versus i for the light bulbs is not the straight line of an Ohm resistor it steepens noticeably the resistance increases as i increases.

25.22. (a) No assumptions can be made about the geometry and this is certainly not a steady state of equilibrium situation. However, if we consider a surface S surrounding the injection region as a Gaussian surface then the charge ( )Q t is given by Gauss’ Law ( ) ,

SQ t E dAε= ⋅

where the permittivity incorporates the

dielectric properties of the material. The material is ohmic so the electric field E drives current density .J Eσ= Hence, the above equation can be written ( )( ) / .

SQ t e J dAσ= ⋅

By the definition of J, the

integral here is the net rate of charge transport out of the volume surrounded by S. Charge conservation requires that this be equal to the rate of decrease of the charge within that volume (no charge can be gained or lost): / .

SJ dA dQ dt⋅ = −

This is the certainty equation for electric charge: it is similar to the

continuity equation of field mechanics, which expresses the conservation of field mass for particle number.

It has the advantage that it applies in every situation. Here it implies ( )/dQ e Qdt

σ= − is the desired

differential equation. (b) Students at this level should recognize immediately that the solution of a differential equation of this form is an exponential function. Explicitly, the equation implies

0

( )

0,

Q t t

Q

dQ dtQ e

σ′ ′= −′ or

0

( )ln ,Q t tQ e

σ = −

i.e. 0( ) exp ,tQ t Qe

σ = −

for all 0.t ≥ The charge in the injection region decays exponentially rapidly for a good conductor slowly for a poor one and the injected charge moves to the outer surface of the conductor.


964

(c) The proceeding result implies that the time required for the charge in the injection region to decrease

by half is 1/22ln .t εσ =

For copper, ( ) 181.678 10 mσ−−= ⋅ Ω at 20 C° and 0ε ε− by assumption

yielding

( )( )

−−

−

⋅= = ⋅

⋅ Ω

12 2 2Cu 191/2 17

8.85 10 C / N m ln21.03 10 s.

5.959 10 mt

This is less than the crossing time of light over a single atom so this calculation particularly the assumptions of Ohmic behavior and unit “dielectric constant” may not be very accurate in this case. It does indicate; however, that the evacuation of free charge from the interior of a good conductor is very rapid. For quartz the data is somewhat varied. The Handbook of Chemistry and Physics gives

( ) 131 10 mσ−

= ⋅ Ω at 20 C° for SiO2 and ( ) 03.75 4.1ε ε= − for fused quartz. A typical value for 1/2t

would be

( )( )( )

−

−−

⋅= =

⋅ Ω2

12 2 2

SiO1/2 113

3.9 8.85 10 C / N m ln2200 s

1 10 mt

over three minutes, some twenty one orders of magnitude longer. Reitz and Milford gives

( ) 1177.5 10 mσ −= ⋅ Ω for fused quartz.

This implies a value ( )( )

( )2

12 2 2

SiO 71/2 118

3.9 8.854 10 C / N m ln21.8 10 s,

1.3 10 mt

−

−−

⋅= = ⋅

⋅ Ω or about 210 days.

25.23. You can write the drift speed of electrons in a wire as ( ) .ivnqA

= For a wire connected across a potential

difference ,V you can find the current i in the wire by determining the resistance of the gold wire, which is just resisitivity, Au / ,R x Aρ= Δ where xΔ is its length. Therefore,

( ) resisitivity, Auresisitivity, Au

./

i V V VvnqA nqAR nq xnqA x A ρρ

= = = =ΔΔ

Thus, since none of the quantities in the equation above depend on ,A it has been shown that the speed of electrons does not depend on the cross-sectional area of the wire.

25.24. The brightness of a light bulb is proportional to its current, so to rank the brightness of the bulbs, you will need to find and rank the currents. The currents can be found by calculating the equivalent resistance for the different circuit elements. Bulbs 1 and 2 are in series, so 1 2 .i i= The equivalent resistance for the 2 bulbs is 2 .R The current through bulbs 1 and 2 is ( )= =1 2 / .2i i V R Bulbs 5 and 6 are in series, so 5 6 .i i= The equivalent resistance for the 2 bulbs is 2 .R The equivalent resistance for bulbs 4, 5, and 6 is

( )[ ] ( )−= =+1

456 2 / 3 .1/ 1/ 2R RR R Adding bulb 3 in series gives: ( )3456 5 / 3R R= and the current in bulb 3

is: ( )=3 3 / .5i V R The voltage across bulbs 4, 5 and 6 is then ( ) ( )3 / 5 2 / 5 .V V V− = This makes the currents in bulbs 4, 5 and 6: ( )=4 2 / 5i V R and ( )= =5 6 / .5i i V R Ranking the bulbs from dimmest to brightest: ( ) ( )5 6 4 1 2 3 .i i i i i i= < < = <

25.25. Conductor 1: length ,L= Radius ,R= Area ,A= Resistance R= and Voltage .V= Conductor 2: length ,L= Radius ,R= Area ,A= Resistance 2R= and Voltage .V= Power delivered is given as

= = = =2 2 2

11 2, , .

2 2PV V VP P P

R R R Therefore, the power delivered to the first would be twice that delivered

to the second.


965

Problems

25.26. The total charge in the Tevatron is .Q i t= Δ Now, / ,t L vΔ = where L is the beam circumference and v is the speed of the protons. 311 10 A,i −= ⋅

36.3 10 mL = ⋅ and 83.00 10 m/s.v c= = ⋅ This charge is made up of n protons:

( )( )( )( )

2 312

19 8

1.10 10 A 6.30 10 m 1.4 10 .

1.602 10 C 3.00 10 m/siL iLQ ne i t nv eC

−

−

⋅ ⋅= = Δ = = = = ⋅

⋅ ⋅

25.27. The area A is 2 6 23.14 10 mA rπ −= = ⋅ so the current density is 2 2/ 318.3 A/m 318 A/m .J i A= = ≈ The density of electrons is

62328 3

3

2.700 10 g1 electron 6.02 10 atoms 6.02 10 electrons / m ,atom 26.98 g m

n ⋅⋅ = = ⋅

and the drift speed is

( )( )2

38

28 19

318.3 A/m 3.30 10 m / s.6.02 10 electrons/m 1.602 10 A sd

Jvne

−−

= = = ⋅⋅ ⋅

25.28. THINK: The current is the same in both wires due to conservation of charge. This can be used to compute the ratio of current densities. The ratio of the drift velocities can then be computed by expressing the drift velocities in terms of the current density. The densities of charge carriers are charges per electron:

28 3Cu 8.50 10 mn −= ⋅ and 28 3

Al 6.02 10 m .n −= ⋅ The other values given in the question that will be needed

are 4Cu 5.00 10 m,d −= ⋅ and 4

Al 1.00 10 m.d −= ⋅ The lengths of the wires and the amount of current are not necessary to solve the question. SKETCH:

RESEARCH: / ,J i A= cross-sectional area,A = dJ nev= and 2 .A rπ= SIMPLIFY:

(a) ( )( )

22

AlCu Cu Al Al2 2

Al Al Cu CuCu

/ 2// / 2

dJ i A A dJ i A A dd

π

π= = = =

(b) ( )( )

Cu Cud,Cu Cu Ald

d,Al Al CuAl Al

/

/J n ev J nJv

ne v J nJ n e

= = =

CALCULATE:

(a) ( )( )

24Cu

24Al

1.00 10 m0.040000

5.00 10 m

JJ

−

−

⋅= =

⋅

(b) ( )28 3

d-Cu Cu Al28 3

d-Al Al Cu

6.02 10 m0.0400 0.028338.50 10 m

v J nv J n

−

−

⋅= = = ⋅


966

ROUND:

(a) Cu

Al

0.0400JJ

=

(b) d-Cu

d-Al

0.0283vv

=

DOUBLE-CHECK: The answers are dimensionless since they are ratios.

25.29. THINK: From the atomic weight and density of silver, the conduction electron density can be computed. Since both the current and the cross-sectional area of the wire are given, the current density can be computed and, using the calculated quantities, the drift speed of the electrons can be computed. Use the data: 20.923 mm ,A = 0.123 mA,i = 107.9 g/mol,M = 3

Ag 10.49 g/cm ,ρ = 23 1A 6.02 10 mol ,N −= ⋅

1 electron/atomN = and 91.602 10 C.e −= ⋅ SKETCH:

RESEARCH: (a) Ag A /n N N Mρ= (b) /J i A= (c) dJ nev= SIMPLIFY: (a) Not required. (b) Not required. (c) d /v J ne= CALCULATE:

(a) ( )( )3 23 1

22 31 10.49 g/cm 6.02 10 mol

5.853 10 cm107.9 g/mol

n−

−⋅

= = ⋅

(b) ( )( )

32

22 3

0.123 10 A 133.3 A/m0.923 mm 10 m/mm

J−

−

⋅= =

(c) ( )( ) ( )

28

d 322 3 2 19

133.3 A/m 1.421 10 m/s5.853 10 cm 10 cm/m 1.602 10 C

v −

− −= = ⋅

⋅ ⋅

ROUND: Three significant figures: (a) 22 35.85 10 cmn −= ⋅ (b) 2133 A/mJ = (c) 8

d 1.42 10 m/sv −= ⋅ DOUBLE-CHECK: These are reasonable values. Note that for part (a), only the composition and not the dimensions of the wire are relevant.

25.30. ( )( )

8Cu 23

10.9 m1.72 10 m 0.14 1.3 10 m / 2

LRA

ρπ

−

−= = ⋅ Ω = Ω

⋅


967

25.31. The resistances will be the same when their cross-sectional areas are the same.

( ) ( ) ( ) ( )2 22 2 2 2B

2B

1 1 1 13.00 mm 2.00 mm 9.00 mm 4.00 mm 5.00 mm 4 4 4 4

1 5.00 mm 1.12 mm2

R

R

π π π π π= − = − =

= =

25.32. The copper coil expands linearly with temperature. At 0 20. C 293.15 K,T = ° = it has resistance

0 0.10 .R = Ω The temperature coefficient of copper is 3 13.9 10 K .α − −= ⋅ At 100. C 173.15 K,T = − ° = ( )( ) ( ) ( )( )( )3 1

0 01 0.10 1 3.9 10 K 173.15 K 293.15 K 0.053 . R R T Tα − −= + − = Ω + ⋅ − = Ω

25.33. The area of 12 gauge copper wire is 2Cu 3.308 mm .A = The resistivity of copper and aluminum are,

8Cu 1.72 10 mρ −= ⋅ Ω and 8

Al 2.82 10 m.ρ −= ⋅ Ω In general the equation for resistance is / ,R L Aρ= meaning if the two wires have equal resistance per length ( ),L then

( )( )8 2Cu Al CuAl

Al 8Cu A Cu

2

l

2.82 10 m 3.308 mm 5.42 mm .

1.72 10 mA

AA Aρ ρρ

ρ

−

−

⋅ Ω= = = =

⋅ Ω This value corresponds to between 9 and 10 gauge wire.

25.34. Since the resistance is given as / ,R L Aρ= then the setup that maximizes L and minimizes A will give the largest resistance. This corresponds to choosing 3.00 cmL = and

( )( )= = 22.00 cm 0.010 cm 0.020 cm .A The resistivity is 2300 m.ρ = Ω Therefore, the maximum resistance is

( )( )( )

7max 2

2300 m 3.00 cm 100 cm 3.5 10 35 M .1 m0.020 cm

LRA

ρ Ω = = ⋅ = ⋅ Ω = Ω

25.35. THINK: The copper wire, 1 1 mL = and 1 0.5 mm,r = has an area of 1.A The wire is then stretched to

2 2 m.L = Since the overall volume ( )V AL= of the wire remains constant, if the wire doubles in length, the area must be halved. SKETCH:

RESEARCH: The resistance of the wire is i i i/ .R L Aρ= From the conservation of volume, it follows that

1 1 2 2 .V A L A L= = The fractional change in resistance is ( )2 1 1/ / .R R R R RΔ = −

SIMPLIFY: Since 2 12 ,L L= then ( )2 11/ 2 .A A= The change in resistance is then

( ) ( )( )

( )2 1 2 2 1 1 1 1 1 1 1 1 1 1

1 1 1 1 11 1

/ / 2 / 1/ 2 / 4 / /3.

/ //R R L A L A L A L A L A L AR

R R L A L AL Aρ

ρ− − − −Δ = = = = =

It is the same for

aluminum, independent of .ρ CALCULATE: Not required. ROUND: Not required. DOUBLE-CHECK: It would seem to make sense that the fractional change in resistance is the same for all materials, so having the equation independent of ρ makes sense.


968

25.36. THINK: The actual cross-sectional area of the resistor, 60 cmL = and 0 1 Ohm,R = is the difference in area from outer radius, 2.5 cm,b = and inner radius, 1.5 cm.a = The resistance of the resistor should vary linearly with temperature from 300 CT = ° to 0 20 CT = ° with 3 12.14 10 K .α − −= ⋅ SKETCH:

RESEARCH: The resistivity of the wire is given by / ,RA Lρ = where the area of interest is

( )2 2 .A b aπ= − The resistance varies with the temperature: ( )( )0 01 .R R T Tα= + − The percentage

change of resistance is found by ( )( )/ 100% .R RΔ SIMPLIFY:

(a) The resistivity is ( )2 2

0 .R b aRA

L L

πρ

−= =

(b) The fractional change in resistance is:

( ) ( ) ( )( ) ( ) ( )( )0 0 000

0 0

1% 100% 100% 100% .

R T T RR RR T TR R R

αα

+ − −−Δ = = = −

CALCULATE:

(a) ( ) ( ) ( )2 21.00 2.50 cm 1.50 cm

0.20944 m60.0 cm

πρ

Ω −= = Ω

(b) ( )( )( )3 1% 2.14 10 K 300. C 20. C 100% 59.92%RR

− −Δ = ⋅ ° − ° =

ROUND: (a) 0.209 mρ = Ω

(b) % 59.9%RR

Δ =

DOUBLE-CHECK: While the resistivity does seem high, it is expected since the wires are intended to carry a lot of power, so a high resistivity, and hence a high resistance, allows this for even a small current.

25.37. THINK: Since the current density across the junction is constant, ,J and they share the same cross-sectional area, they have the same current, .i At the junction, a positive charge will build up, and this means both electric fields, 1E and 2 ,E are pointing towards the junction. Gauss’s Law can then be used to determine the total charge built up on the interface. The electric fields are also related to the conductivities, 1σ and 2 .σ SKETCH:


969

RESEARCH: The conductivity is the inverse of resistivity, 1/ .σ ρ= The resistivity is related to electric

field by / .E Jρ = At the junction, Gauss’s law states 0/ .E dA q E= The current density ,J is / .J i A=

SIMPLIFY: From the conductivity and resistivity, the current density is ( )1/ 1/ / / ;E J J E J Eσ ρ σ= = = = therefore, 1 1 2 2E Eσ σ= or ( )1 2 1 2/ .E Eσ σ= Since 2E

is parallel

and 1E

is antiparallel to ,dA

( )2 1 0/ .E dA E E A q ε= − =

Solving this expression for q yields the following.

( ) σ σ σε ε ε ε εσ σ σ σ σ σ

= − = − = − = − = −

2 2 22 2 2 2

1 1 10 2 1 0

2 20

2 10 0

1 11 1i iq E E A E E A E E iJ E

This is what was required to be shown. CALCULATE: Not required. ROUND: Not required. DOUBLE-CHECK: The equation was verified, so it makes sense.

25.38. (a)

Since the potential across wire is 12.0 VVΔ = and the current is 33.20 10 A.i −= ⋅ Ohm’s Law states 3 / 12.0 V / 3.20 10 A 3750 .V iR R V t −Δ = = Δ = ⋅ = Ω

(b) Since the wire is 1000. kmL = long and has area of 24.50 mm ,A = the resistivity of it is

( )( ) ( )2 8/ 3750 4.50 mm / 1000. km 1.69 10 m,RA Lρ −= = Ω = ⋅ Ω therefore, the wire is most likely copper

( )6C 1.72 10 m .ρ −= ⋅ Ω

25.39. The current is 600. Ai = and the potential difference is 12.0 V.VΔ = Therefore, Ohm’s Law states ( ) / 12.0 V / 600. A 0.0200 .V iR R V iΔ = = Δ = = Ω

25.40. THINK: The resistance of the wire of radius 0.0250 cmr = and length 3.00 mL = is found by using its resistivity 81.72 10 m.ρ −= ⋅ Ω The potential drop is found using the current, 0.400 A,i = and Ohm’s Law. Assuming the electric field is constant, it is simply found through the potential drop over the length. SKETCH:

RESEARCH: The resistance is / .R L Aρ= The area of the wire is 2 .A rπ= The potential difference across the wire is .V iRΔ = The electric field across the wire is / .E V L= Δ SIMPLIFY:

(a) The resistance is 2 .L LRA r

ρ ρπ

= =

(b) The potential difference across the wire is .V iRΔ =

(c) The electric field across the wire is .VEL

Δ=


970

CALCULATE:

(a) ( )( )

( )

8

2

1.72 10 m 3.00 m0.2628

0.0250 cmR

π

−⋅ Ω= = Ω

(b) ( )( )0.400 A 0.2628 0.10512 VVΔ = Ω =

(c) 0.10512 V 0.03504 V/m 0.0350 V/m3.00 m

E = = =

ROUND: Rounding to three significant figures; (a) 0.263 R = Ω (b) 0.105 VVΔ = (c) 0.0350 V/mE = DOUBLE-CHECK: The wire loses very little potential over a long length. This means that this wire used in a standard circuit which would only be a few centimeters in length would only lose about 1 or 2 mV, making it a suitable material for a circuit.

25.41. THINK: The length and resistivity of the wire are 1.0 mL = and 81.72 10 m.ρ −= ⋅ Ω According to Table

25.2, the area of a 34 gauge wire is 20.0201 mm .A = Since the resistance changes linearly with the temperature, 0 20. CT = ° and 196 C,T = − ° the percentage change in resistance is proportional, by

3 13.9 10 K ,α − −= ⋅ to the temperature difference. Since the current is directly related to the resistance, the percentage change in resistance is related to the resistances themselves. Using the molar mass and density of copper, 0.06354 kgm = and 3 3

Cu 8.92 10 kg/m ,ρ = ⋅ the carrier density, ,n can be determined, which in turn allows velocity to be found. Use the value 0.1 V.VΔ = SKETCH:

RESEARCH: The resistance of the wire is / .R L Aρ= From Ohm’s Law, the potential drop across the

wire is .V IRΔ = The resistance changes linearly with temperature by ( )( )0 01 .R R T Tα= + − For a given

quantity, ,x the percentage change in it is therefore, ( )( )( )0 0/ / 100% .x x x x xΔ = − Current density is

2 / .J nev i A= = SIMPLIFY: (a) The original resistance is 0 / .R L Aρ= The cooled resistance is ( )( )0 01 ;R R T Tα= + − therefore,

( )( ) ( )( )0 0 000

0 0

1% 100% .

R T T RR RR T TR R R

αα

+ − −−Δ = = = − −

(b) The percentage change in current is found using the following equations.

( )( )( )( )

( )( )( ) ( )

0 0 00 0 0 0 00

0 0 0 0 0 0

0

0

1/ / 1/ 1// 1/ 1

100%1

R R T Ti i V R V R R R R R R Ri Ri i V R R RR R R T T

T T

T T

α

α

αα

− + −− Δ − Δ − − −Δ = = = = = =Δ + −

− −=

+ − (c) Assume each copper atom contributes just 1 e .− The molar volume of copper is Cu Cu / ;V mρ= therefore, A C / .n N mρ= The drift velocity is then, in general,


971

( )( )

( ) ( )( ) ( )( )

d

d0 0

d0 0

1

./ 1 1

iJ nevA

i V VvneA neAR neAR T T

V VvneA L A T T ne L T T

α

ρ α ρ α

= =

Δ Δ = = =

+ −

Δ Δ = =

+ − + −

At 0T T= and d / .V V ne Lρ= Δ CALCULATE: (a) ( )( )( )3 1% / 3.9 10 K 196 C 20. C 100% 84.24%R R − −Δ = − ⋅ − ° − ° = −

(b) ( )( )( )( )( ) ( )

3 1

3 1

3.9 10 K 196 C 20. C% / 100% 534.5%

1 3.9 10 K 196 C 20. Ci i

− −

− −

− ⋅ − ° − °Δ = =

+ ⋅ − ° − °

(c) ( )( )23 3 3

28 36.022 10 e 8.92 10 kg/m

8.4539 10 e / m0.06354 kg

n−

−⋅ ⋅

= = ⋅

At room temperature, ( )( )( )( )d 28 3 19 8

0.10 V 0.4293 mm/s.8.4535 10 e / m 1.602 10 C 1.72 10 m 1.0 m

V− − −

= =⋅ ⋅ ⋅ Ω

At temperature, 196 C,T = − ° ( )( )d 3 1

0.4293 mm/s 2.724 mm/s.1 3.9 10 K 196 C 20. C

V− −

= =+ ⋅ − ° − °

ROUND: (a) % / 84%R RΔ = − (decrease in resistance) (b) % / 530%I IΔ = (increase in current) (c) At room temperature, d 0.43 mm/s.V = At 77 K the speed is d 2.7 mm/s.V = DOUBLE-CHECK: Supercooling a resistor should greatly reduce the resistance and increase the current and drift velocity, which it does, so it makes sense.

25.42. If the current, 11 A,i = went entirely through the known resistor, 0 35 ,R = Ω the potential drop across it would be ( )( )0 11 A 35 385 V,V iRΔ = = Ω = which is too large, so the other resistor must be parallel to 0 .R

Therefore, by Ohm’s Law, 1 1 1

0 0

1 1 1 11 A 1 15.849 .120 V 35

iV i RR R V R

− − − Δ = + = − = − = Ω Δ Ω

Hence, 16 .VΔ = Ω

25.43. When the external resistor, 17.91 R = Ω is connected, it experiences a potential drop of 12.68 V,VΔ = so the current through the circuit is, by Ohm’s Law, / 12.68 V/17.91 0.70798 A 0.7080 A.i V R= Δ = Ω = = This is the same current running through the internal resistor, i ,R which is in series with ,R so since the battery has a total emf of emf 14.50 V,VΔ = the internal resistance is found using the following calculation.

( ) emfemf i i

14.50 V 17.91 2.5707 2.571 0.7080 A

VV i R R R R

iΔ

Δ = + = − = − Ω = Ω = Ω

25.44. The two resistors, 1 100. R = Ω and 2 400. ,R = Ω cause currents, 1 4.00 Ai = and 2 1.01 A,i = respectively. The currents they cause are the same through the internal resistor, i ,R and in both cases the emf of the battery, ,VΔ is the same. Since iR is in series with each of the other resistors, Ohm’s Laws says:


972

( ) ( ) ( )( )( ) ( )( )

( )

2 2 1 11 1 i 2 2 i 2 2 1 1 i 1 2 i

1 2

i

1.01 A 400. 4.00 A 100. 1.3378 1.34 .

4.00 A 1.01 A

i R i RV i R R i R R i R i R R i i Ri i

R

−Δ = + = + − = − =

−Ω − Ω

= = Ω ≈ Ω−

25.45. THINK: The resistance in each bulb is directly calculated by Ohm’s Law. Consider temperature effects on resistance to explain discrepancy with parts (a) and (b). Use the values: 1 6.20 V,VΔ = 1 4.1 A,i =

2 2.9 Ai = and 2 6.29 V.VΔ = SKETCH: (a) (b)

RESEARCH: By Ohm’s Law, 1 1V i RΔ = for one light bulb and ( )2 1V i R RΔ = + for both light bulbs. SIMPLIFY:

(a) 11 1

1

VV i R Ri

ΔΔ = =

(b) ( ) 22 2

2

2VV i R R Ri

ΔΔ = + =

CALCULATE:

(a) 16.20 V1.5122

4.1 AR = = Ω

(b) ( )6.29 V 1.084

2 2.9 AR = = Ω

ROUND: (a) 1.5 R = Ω (b) 1.1 .R = Ω (c) When two bulbs are put in series, it is expected that they glow dimmer than only one bulb. This would mean the one bulb would be hotter and thus have a larger resistance. DOUBLE-CHECK: Answer to part (c) helps to verify that the answers to parts (a) and (b) are reasonable.

25.46. Simplifying the circuit gives

( ) ( ) 1

eq 10.0 1/ 20.0 1/ 20.0 20.0 .R−

= Ω + Ω + Ω = Ω By Ohm’s Law, the current through eqR

is

( )eq 60.0 V/ 20.0 3.00 A.i = Ω = From the circuit setup, the current through eqR is the same as that through the 10.0 Ω resistor, which is 3.00 A.


973

25.47. THINK: The circuit can be redrawn to have the 10.0 Ω and 20.0 Ω resistors in series, both of which are parallel to the 30.0 Ω resistor, and then parallel again with the 40.0 Ω resistor. These resistors are then put in series with the 50.0 Ω resistor and the 60.0 V battery. SKETCH:

RESEARCH: Resistors in series combine as eq1

.n

ii

R R=

= Resistors in parallel combine as 1eq

1 1 .n

i iR R==

SIMPLIFY: The combined resistors in parallel become PR where

( )1

11 1 1P 1 2 3 4 P

1 2 3 4

1 1 1 R .R R R R RR R R R

−−− − −

= + + + = + + +

The equivalent resistance is eq 5 P .R R R= +

CALCULATE: 1

P1 1 1R 10.909091

10.0 20.0 30.0 40.0

− = + + = Ω Ω + Ω Ω Ω

eq 50.0 10.909091 60.909091 R = Ω + Ω = Ω

ROUND: The result should be rounded to three significant figures: eq 60.9 .R = Ω

DOUBLE-CHECK: If you add N equal resistors, R, in parallel, the equivalent resistance is R/N. Since the resistors in parallel are all about 30 Ω in each branch, the equivalent resistance should be about 10 Ω , which is close to the calculated answer of 11 .Ω . Therefore, the values of PR and eqR are reasonable.

25.48. THINK: When the switch is open, the current clearly breaks up into paths and the two left resistors, 1 3 3.00 ,R R= = Ω are in parallel with the two resistors, 2 5.00 R = Ω and 4 1.00 .R = Ω When the switch is

closed, it is not as obvious. Consider the potential drop from before the current splits to the switch arm. It should be the same regardless of which path is taken, likewise with the potential drop from the switch arm to after the current recombines. This means that the pairs of resistors 1R and 2 ,R and 3R and 4 ,R are connected in parallel. The pairs are subsequently connected in series with each other. SKETCH: (a) (b)


.n

ii

R R=


1 1 .n

i iR R=

=

The current is given by Ohm’s Law eq/ .i V R= Δ


974

SIMPLIFY:

(a) Equivalent resistance is ( ) ( )( ) 1

eq 1 3 2 4 eq1/ 1/ / .R R R R R i V R−

= + + + = Δ

(b) Equivalent resistance is ( )( ) ( )( )1 1

eq 1 2 3 4 eq1/ 1/ / .R R R R R i V R− −

= + + + = Δ

CALCULATE:

(a) 1

eq1 1 3.00 ;

3.00 3.00 5.00 1.00 R

− = + = Ω Ω + Ω Ω + Ω

therefore, 24.0 V 8.00 A.3.00

i = =Ω

(b) 1 1

eq1 1 1 1 2.625 ;

3.00 5.00 3.00 1.00 R

− − = + + + = Ω Ω Ω Ω Ω

therefore, 24 V 9.1429 A.2.625

i = =Ω

ROUND: (a) 8.00 Ai = (b) 9.14 Ai = DOUBLE-CHECK: Typically, when resistors are in parallel, they have a lower equivalent resistance than when in series and thus would yield a larger current, so it makes sense.

25.49. THINK: The circuit can be redrawn to have 3 2.00 R = Ω and 4 4.00 R = Ω in series, which are then parallel to 2 6.00 .R = Ω These are then in series with 1 6.00 R = Ω , 5 3.00 R = Ω , and the battery

12.0 V.V = Since 5R is in series with the equivalent resistance, the current through it is the same as the current through the whole circuit. Since 2 3 4 ,R R R= + the current through each branch is equal, and half of the total current. SKETCH:

RESEARCH: Resistors in series eq 1 2 .R R R= + Resistors in parallel ( ) 1eq 1 21/ 1/ .R R R −= + The current is

given by Ohm’s Law eq/ .i V R= Δ

SIMPLIFY:

(a) The resistors in parallel combine as 6 ,R where 1

62 3 4

1 1RR R R

−

= + + . The total equivalent

resistance is eq 1 6 5 .R R R R= + +

(b) The current through 5R is eq/ .i V R=

(c) The current through each branch is / 2i , so potential across 3R is 3 31 .2

V iRΔ =

CALCULATE:

(a) ( )

1

61 1 3.00 .

6.00 2.00 4.00 R

−

= + = Ω Ω + Ω eq 6.00 3.00 3.00 12.00 R = Ω + Ω + Ω = Ω

(b) ( ) ( )/ 1.00 A12.0 V 12.0 i = =Ω

(c) ( )( )

3

1.00 A 2.00 1.00 V

2V V

ΩΔ = Δ = =

ROUND: (a) eq 12.00 R = Ω


975

(b) 1.00 Ai = (c) 3 1.00 VVΔ = DOUBLE-CHECK: By considering the potential drop across 1 2 5, , and ,R R R the values are:

1 6 V,VΔ = 2 3 V,VΔ = and 5 3 V.VΔ = Hence, 1 2 5 12 V.V V VΔ + Δ + Δ = This value matches the total voltage provided by the battery. Using 3 4V VΔ + Δ instead of 2VΔ also gives 12 V, as expected.

25.50. THINK: The circuit can be redrawn to have 0R and 1R in series, which are then parallel to 1R . These are then in series with 1R . SKETCH:


.n

ii

R R=


1 1 .n

i iR R=

=

( )11 1 01 2

4 0 1 3 1 1 11 2 1 2 1 1 0

1 1 ;R R RR RR R R R R R R

R R R R R R R

− + = + = + + = + = + + + +

therefore,

( )( )2

21 1 00 1 0 1 1 0 1 1 0

1 0

2 ,2R R R

R R R R R R R R RR R+

− = − + = ++

and

2 2 2 2 21 0 0 1 1 0 1 1 0 0 12 2 3R R R R R R R R R R R+ + − = + = ( )1 01/ 3 .R R =

SIMPLIFY: Resistance 2R is 2 1 0 .R R R= + Resistance 3R is

1

31 1 0

1 1 .RR R R

−

= + + Equivalent resistance

0R is 1

0 1 3 11 1 0

1 1 .R R R RR R R

−

= + = + + + Thus, ( )

1

1 0 10 1

1 0 1 1 1 0

1 1 .R R R

R RR R R R R R

− + +

− = + = + +

( )( )2

2 2 2 21 1 00 1 0 1 1 0 1 1 0 0 1 0 1 1 1 0

1 0

2 2 00 1 1

2 22

3 .3

R R RR R R R R R R R R R R R R R R R

R RR

R R R

+− = − + = + + − = +

+

= =

CALCULATE: Not applicable. ROUND: Not applicable. DOUBLE-CHECK: If 1 0 ,R R= the total equivalent résistance would be 05 / 3.R If 1 0 / 2,R R= the total equivalent resistance would be 07 / 8.R Therefore, for eq 0 ,R R= 0 1 0/ 2 .R R R< < The answer satisfies this condition.

25.51. THINK: From the circuit, it is clear that resistors 1 5.00 R = Ω and 2 10.00 R = Ω are in series. Resistors

3 4 5.00 R R= = Ω are in parallel, with equivalent resistance 34 .R This is also true of resistors

5 6 2.00 ,R R= = Ω whose equivalent resistance is 56 .R This second pair of resistors are in turn connected in series with resistors 1R and 2 .R Ohm’s Law can be used to determine the current through the whole circuit which is the same as each resistor in series.


976

SKETCH:

RESEARCH: Equivalent resistances if resistors are in parallel are ( ) 1

34 3 41/ 1/ .R R R −= − Total current

through 4 resistors in series is ( )1 2 34 56/ .i V R R R R= Δ + + + The potential drop across a resistor is

i i .V iRΔ = SIMPLIFY: (a) The total current is ( )1 2 34 56/ ,i V R R R R= Δ + + + 1 1 ,V iRΔ = 2 2 ,V iRΔ = 3 4 34V V iRΔ = Δ = and

5 6 56 .V V iRΔ = Δ = (b) Current through 1R and 2R is .i Since 3 4R R= and 5 6 ,R R= the current splits evenly among them so

/ 2i i′ = through each of them. CALCULATE:

(a) ( ) ( )( ) 1

34 1/ 5.00 1/ 5.00 2.50 R−

= Ω − Ω = Ω ( ) ( )( ) 1

56 1/ 2.00 1/ 2.00 1.00 ,R−

= Ω − Ω = Ω

( )20.0 V / 5.00 10.00 2.50 1.00 1.08108 A,i = Ω + Ω + Ω + Ω = ( )( )1 1.08108 A 5.00 5.405 V,VΔ = Ω =

( )( )2 1.08108 A 10.0 10.81 V,VΔ = Ω = ( )( )3 4 1.08108 A 2.50 2.702 VV VΔ = Δ = Ω = and

( )( )5 6 1.08108 A 1.00 1.081 VV VΔ = Δ = Ω = (b) 1 2 1.08108 A,i i= = 3 4 5 6 1.08108 / 2 0.5405 Ai i i i= = = = = ROUND: (a) 1 5.41 V,VΔ = 2 10.8 V,VΔ = 3 4 2.70 VV VΔ = Δ = and 5 6 1.08 V.V VΔ = Δ = (b) 1 2 1.00 Ai i= = and 3 4 5 6 0.500 A.i i i i= = = = DOUBLE-CHECK: The sum of the 4 potential drops equal 20 V, so energy is conserved, so the answers make sense.

25.52. THINK: Ohm’s law can be used to relate the potential drop, 40.0 V,VΔ = to current 1 10.0 Ai = when resistors 1R and 2R are in series and to current 2 50.0 Ai = when the resistors are in parallel. This means there are two equations (the series and parallel configurations) and two unknowns (the resistors). Let 1R be the one that is larger, since the choice is arbitrary, and solve for it. SKETCH:

RESEARCH: For the series setup, ( )1 1 2 .V i R RΔ = + For the parallel setup, ( ) 1

2 1 21/ 1/ .V i R R −Δ = +

SIMPLIFY: The second resistor is, from series setup, ( ) ( )1 1 2 2 1 1 /V i R R R V i RΔ = + = Δ − and

1 2 .V R Ri

Δ = + From the parallel setup, ( )1 22 1 2 1 1

1 2 2 2

.R R V VV i R R R R

R R i i Δ ΔΔ = = + = − +

Therefore,

22 2

1 1 1 11 2 1

.V V R R P SR Ri i i

Δ Δ= − = −


977

With 2

1 2

,VPi i

Δ= and S = ΔV

i1

, 21 1 0R SR P− + =

2

14 .

2S S PR ± −

= 1R must be positive to get the

largest possible value, so 2

14 .

2S S PR + −=

CALCULATE: ( )

( )( )2

240.0 V3.2 ,

10.0 A 50.0 AP = = Ω and 40.0 V 4 .

10.0 AS = = Ω Therefore,

( ) ( ) ( )2

1

24 4 4 3.2 2.8944 .

2R

Ω + Ω − Ω= = Ω

ROUND: 1 2.89 R = Ω DOUBLE-CHECK: The other value for 1 ,R the negative in the quadratic, gives 1 1.11 ,R = Ω so eqR in

series and parallel is 4 Ω and 0.8 ,Ω respectively. ( )( )4 10.0 A 40 VΩ = and ( )( )0.8 50.0 A 40 V,Ω = which is consistent with emf voltage.

25.53. The voltage changes, from 0 110. VVΔ = to 1 150. V,VΔ = and the initial power is 0 100. W.P = Since the

resistance does not change, the power is, generally, 2 / ,P V R= Δ so the fractional change in power is

( )( )

22 2 21 0 1 0 1

2 2 20 0 0

150 V/ /% 1 1 0.8595.

/ 110 V

P P V R V R VPP V R V− Δ − Δ Δ

= = = − = − =Δ Δ

Therefore, % 86.0%P = (brighter).

25.54. (a) The average current, i, is simply the change in charge, 5.00 C,QΔ = over change in time,

0.100 ms.tΔ = ( ) ( )3/ 5.00 C / 0.100 10 s 50.0 kA.i Q t −= Δ Δ = ⋅ =

(b) If over the lightning bolt there is a 70.0 MVV = potential, the power is P iV= ( )( ) 1250.0 kA 70.0 MV 3.50 10 W= = ⋅

(c) The energy is the power times the change in time, ( )( )12 83.50 10 W 0.100 ms 3.50 10 J.E P t= Δ = ⋅ = ⋅

(d) Assuming the lightning obeys Ohm’s Law, the resistance is /R V i= Δ ( ) ( )70.0 MV / 50.0 kA= 31.40 10 .= ⋅ Ω

25.55. (a) If the hair dryer has power =1600. WP and requires a potential of =110. V,V the current supplied is then = = = =/ 1600. W /110. V 14.545 A 14.5 A.i P V i does not exceed 15.0 A, so it will not trip the circuit. (a) Assuming the hair dryer obeys Ohm’s Law, its effective resistance is given by

( ) ( )= Δ = = Ω/ 110. V / 14.545 A 7.56 .R V i

25.56. For a year of use, the time it is active is ( )( )( )1 year 365 days / year 24 hours / day 8760 h.tΔ = = The power of a regular light bulb is 100.00 W 0.10000 kW.P = = The power of the fluorescent bulb is

F 26.000 W 0.026000 kW.P = = Since it costs $0.12 / kWh to have each on, the cost of running each is:

( )( )( )$0.12 / kWh 0.10000 kW 8760 h $105.12,C = = ( )( )( )FC $0.12 / kWh 0.026 kW 8760 h $27.33.= =

25.57. To find the current through each, reduce circuit to


978

(a) The total current is eq/ 120 V / 288 0.42 A.i V R= Δ = Ω = i is current through the first resistor, so the

power over it is ( ) ( )221 0.42 A 192 33.3 W 33 W.P i R= = Ω = = The current through the two in parallel is

half of the original since they are the same resistance, so the power across each is 2 3P P=

( )( ) ( )( ) ( )= = Ω =2 21/ 2 1/ 4 0.42 A 192 8.5 W.i R

(b) The potential drop across each resistor is found by Ohm’s Law ( )( )Δ = = Ω =1 0.42 A 192 81 VV iR and

( ) ( )( )( )Δ = Δ = = Ω =2 3 1/ 2 1/ 2 0.42 A 192 40. V.V V iR

25.58. The current through the light bulb is / .i P V= The charge through the bulb is ,q i t= Δ so

( ) ( )1 A 60 min625 mAh 1.5 V1000 mA h

t 11 min.5.0 W

q VP tq i tV P

ΔΔ = Δ = Δ = =

Δ

25.59. THINK: The overall current through the resistor, ( ) which takes the values 1.00 , 2.00 and 3.00 ,R Ω Ω Ω is found using Ohm’s Law for when the load resistance is in series with the internal resistance,

i 2.00 ,R = Ω and the external emf, emf 12.0 V.V = I will determine an expression for the power across the load resistor and differentiate with respect to R, and solve this derivative equal to zero, in order to find a maximum in power. SKETCH:

RESEARCH: With R and iR in series, current through circuit is ( )emf i/ .i V R R= + Power through load

resistor is 2 .P i R= The power is maximized when / 0dP dR = and 2 2/ 0.d P dR <

SIMPLIFY: Power is ( )

2ememf f

i i

2

2 .V V

P R RR R R R

= + +

Therefore,

( ) ( ) ( )( )( )

( ) ( ) ( )

2 322 2emf emf i ii

2 2i iemf emf3 3 3

i i i

2

2 0,

dP dV V R R R R RR R RdR dR

R R R RRV VR R R R R R

− −− = = + − ++

+ − = − = = + + +

and hence iR R= is a critical point of P. The double-check step will verify that iR R= leads to a maximum.

CALCULATE: ( ) ( )( )

2

1 2

12.0 V 1.00 16.0 W,

1.00 2.00 P

Ω= =

Ω + Ω ( ) ( )( )

2

2 2

12.0 V 2.00 18.0 W

2.00 2.00 P

Ω= =

Ω + Ω and

( ) ( )( )

2

3 2

12.0 V 3.00 17.28 W.

3.00 2.00 P

Ω= =

Ω + Ω

ROUND: The values should be rounded to three significant figures each: 1 16.0 W,P = 2 18.0 WP = and

3 17.3 W.P =


979

DOUBLE-CHECK: The second derivative of P, 22

emf i2 4

22

))

(,

( i

V R Rd PdR R R+

−=

is clearly negative when

i ,R R= which verifies that iR R= yields a maximum for P.

25.60. THINK: Using the density and volume of water, 31000 kg/mρ = and 250 mL,V = the mass of the water can be determined. Along with the specific heat of water, ( )4.186 kJ/ kg K ,c = and the fact that the water goes from i 20 C 293 KT = ° = to f 100 C 373 KT = ° = the energy gained by the water can be determined. The change in energy over the time 45 stΔ = is equal to the power that the coil, 15 V,VΔ = dissipates. SKETCH:

RESEARCH: Power dissipated by coil is 2 / .P V R= Δ The energy gained by heating water, .Q mc TΔ = Δ The rate of energy gained by water is / ,P Q t= Δ Δ mass of water is .m Vρ= SIMPLIFY: Equating power dissipated by coil to energy rate gained by water gives the equation:

( )2f ic

.V T TV Q mc TP

R t t tρ −Δ Δ Δ= = = =

Δ Δ Δ

Therefore, ( )2

f i

.V tRVc T Tρ

Δ Δ=−

CALCULATE: ( ) ( )

( )( )( )( )( )2

3 3

15 V 45 s0.1209

1000 kg/m 0.25 L 4186 J/kg K 373 K 293 K 1 m /1000 LR = = Ω

−

ROUND: The values in the question have two significant figures, so round the answer to 1.2 m .R = Ω DOUBLE-CHECK: Since power is inversely proportional to resistance, the optimal way to heat using it would be to make the resistance as small as possible, so it makes sense.

25.61. THINK: Both the copper wire length ( 75.0 cm,l = diameter 0.500 mmd = and resistivity 8

C 1.69 10 mρ −= ⋅ Ω ) and silicon block (length 15.0 cm,L = width 2.00 mm,a = thickness ,b resistivity 4

S 8.70 10 mρ −= ⋅ Ω and resistance S 50.0 R = Ω ) can be thought of as resistors in series with a total

potential drop of 0.500 VVΔ = and a density of charge carriers of 23 31.23 10 m .−⋅ The current density, ,J can be used to determine the velocity of the carriers through the silicon. SKETCH:

RESEARCH: The resistance of the material is in general / .R L Aρ= The current through circuit is found by Ohm’s Law with copper wire and silicon block in series, ( )C S .V i R RΔ = − Area of the silicon block is

.A ab= The current density is d/ .J i A nev= = The time it takes to pass through silicon block is

d/ .t L vΔ = Power dissipated by silicon block is 2S S .P i R=


980

SIMPLIFY:

(a) Resistance of wire is ( )C

C 2 .1/ 4

lR

dρ

π=

(b) Current through circuit is C S

.ViR R

Δ=+

(c) Thickness of silicon block S SS

S

.L L

R bab aRρ ρ

= =

(d) Drift velocity of electrons v .d di iJ nevA abne

= = = So the time to cross the block is

d

.L Labnetv i

Δ = =

(e) 2S SP i R=

(f) Electric power is lost via heat. CALCULATE:

(a) ( )( )

( ) ( )

8

C 2

1.69 10 m 75.0 cm0.064553 64.553 m

1/ 4 0.500 mmR

π

−⋅ Ω= = Ω = Ω

(b) 0.500 V 0.009987 A 9.987 mA50.0 0.064533

i = = =Ω + Ω

(c) ( )( )

( )( )48.70 10 m 15.0 cm

0.001305 m 1.305 mm2.00 mm 50.0

b−⋅ Ω

= = =Ω

(d) ( )( )( )( )( )23 3 1915.0 cm 2.00 mm 1.305 mm 1.23 10 m 1.602 10 C

0.77243 s9.987 mA

t− −⋅ ⋅

Δ = =

(e) ( ) ( )2S 9.987 mA 50.0 0.004987 W 4.987 mWP = Ω = =

ROUND: (a) C 64.6 m R = Ω (b) 9.99 mAi = (c) 1.31 mmb = (d) 0.772 stΔ = (e) S 4.99 mWP = DOUBLE-CHECK: Drift velocity is / 20 cm/sL tΔ ≈ which is reasonable for such a small resistance. Also, the power lost is small, which is desirable in silicon, which is used in many electronic devices, so it makes sense.

Additional Problems

25.62. Electrical power is defined as 2P i R= or 2 /P V R= Δ or .P i V= Δ In the normal operation, the radio has

a resistance of ( ) ( )22 / 10.0 V / 3.33 30.0 Wr V P= Δ = = Ω and the current flowing through the radio is ( ) ( )/ / 3.00 A.30.0 W 10.0 Vi P V= Δ = = Now, if a 25.0 kV power supply is used, the required total

resistance, such that the current flowing through the radio is the same, is T / 25.0 kV / 3.0 A 8333.33 .R V i= Δ = = Ω Thus, the external resistance required is T .R R r= − The

number of resistors is T

1 1

8333.33 3.33 333 resistors.25

R rRNR R

− Ω − Ω= = = =Ω


981

All resistors are connected in series.

25.63. It is given 120 V,VΔ = 2.0 min 120 stΔ = = and 1 48 kJ.U = The power needed to cook one hot dog is

= Δ = ⋅ = ⋅4 21 1 / 4.8 10 J /120 s 4.0 10 W.P U t The current to produces this power is

= Δ = ⋅ =21 1 / 4.0 10 W /120 V 3.3 A.i P V The current to cook three hot dogs is ( )= = =13 3 3.3 A 10. A.i i

25.64. The aluminum wire and the copper wire dissipate the same power. Since the voltages across the wires are the same, this means that the resistances of the wires are the same. That is Al Cu ,R R=

Al Al Al Cu Cu Cu/ / .L A L Aρ ρ= Using the area of a circle 2 ,A rπ= it is found that 2 2Al Cu Al Al Cu Cu/ / .r r L Lρ ρ=

Thus, Al Cu Al Al Cu Cu/ .r r L Lρ ρ= Substituting the numerical values:

( ) ( )( )( )( )

8

Al 8

2.82 10 m 5.00 m1.00 mm 0.905 mm.

1.72 10 m 10.0 mr

−

−

⋅ Ω= =

⋅ Ω

25.65. The resistance of a cylindrical wire is / .R L Aρ= The length of the resistor is / .L RA ρ= Substituting the

numerical values yields ( )( ) ( )6 2 510.0 1.00 10 m / 1.00 10 m 1.00 m.L − −= Ω ⋅ ⋅ Ω =

25.66. Two resistive cylindrical wires of identical length are made of copper and aluminum. They carry the same current and have the same potential difference across their length. This means that they have the same resistance, that is, Cu Al ,R R= Al Al Al Cu Cu Cu/ / .L A L Aρ ρ= Since Al CuL L= and 2

Al AlA rπ= and 2Cu Cu ,A rπ= it

becomes 2 2Cu Al Cu Al/ /r r ρ ρ= or Cu Al Cu Al/ / .r r ρ ρ= Therefore, the ratio of their radii is

8Cu

8Al

1.72 10 m 0.781.2.82 10 m

rr

−

−

⋅ Ω= =⋅ Ω

25.67. (a)

The current flowing through the resistors is ( )1 2/ .i V R R= Δ + The power delivered to 1R is

( )2 21 1 1 2/ .P i R V R R R= = Δ + Substituting the numerical values gives:

( ) ( ) ( )2 29.00 V 200. / 200. 400. 0.0450 W.P = Ω Ω + Ω = (b)

The power delivered to 1R is 2

1P i R= or ( )221 9.00 V / 200. 0.405 W,P V R= Δ = Ω = ( )9.00 0.0450 W .P =

Therefore, the power delivered to the 200 Ω resistor by 9 V battery when the resistors are connected in parallel is 9.00 times greater than in series configuration.

25.68. The conductance of a wire is given by 1/ / .G R A Lρ= =

(a) From the electrical power 2 / ,P V R= Δ the conductance of the element is

( )1

2 2

1500 W 0.12 .110 V

PGV

−= = = ΩΔ

(b) Using 2 ,A rπ= the radius of the wire is 2 /r LGρ π= or / .r LGρ π= Substituting 89.7 10 m,ρ −= ⋅ Ω 3.5 mL = and 10.124 G −= Ω gives


982

( )( )( )8 19.7 10 m 3.5 m 0.124 0.12 mm.r

π

− −⋅ Ω Ω= =

25.69. The resistance of the light bulb is 21 1/ .R V P= Δ Consider each value in the problem to have three

significant figures. The power consumed by the bulb in a US household is

( )2 22

2 22 1

1

120. V 100. W 25.0 W.240. V

V VP P

R V Δ Δ = = = = Δ

25.70. (a) The minimum overall resistance is / 115 V / 200. A 0.575 .R V i= Δ = = Ω (b) The maximum electrical power is ( )( )200. A 115 V 23.0 kW.P i V= Δ = =

25.71. THINK: A battery with emf 12.0 V and internal resistance i 4.00 R = Ω is attached across an external resistor of resistance .R The maximum power that can be delivered to the resistor R is required. SKETCH:

RESEARCH: The power delivered to the resistor R is given by 2 .P i R= The current flowing through the

circuit is ( )i/ .i V R R= Δ + Therefore, the power is ( )2 2i/ .P V R R R= Δ + The maximum power delivered

to the resistor R is given when R satisfies / 0.dP dR = That is

( )( )

( )

22

2 3i i

20.

V RdP VdR R R R R

Δ −Δ= + =+ +

SIMPLIFY: Solving the above equation for R yields i 2 0R R R+ − = or i .R R= Thus, the maximum

power delivered to R is ( ) ( )22 2i i i/ 2 / .4P V R R V R= Δ = Δ

CALCULATE: Substituting the numerical values gives ( ) ( )( )212.0 V / 4 4.00 9.00 W.P = Ω =

ROUND: 9.00 WP = DOUBLE-CHECK: This is a reasonable amount of power for a 12 V battery to supply.

25.72. THINK: Calculate the resistance of a 10.0 m length of multiclad wire consisting of a zinc core of radius 1.00 mm surrounded by a copper sheath of thickness 1.00 mm. The wire can be treated as two resistors in parallel. SKETCH:

RESEARCH: The resistivity of zinc is 8Zn 5.964 10 m,ρ −= ⋅ Ω and the resistivity of copper is

8Cu 1.72 10 m.ρ −= ⋅ Ω Resistance is given by / .R L Aρ= The resistance of the zinc wire is therefore

2Zn Zn 1/ ,R L Rρ π= and the resistance of the hollow copper wire is ( )2 2

Cu Cu 2 1/ .R L R Rρ π π= − The net

resistance is Zn Cu1/ 1/ 1/ .R R R= +


983

SIMPLIFY:

ρ ρ

ρ ρ

− − ⋅ +

= + = = = + +

Cu Zn1 1

Cu Zn Cu Zn Cu Zneq

Cu ZnZn Cu Zn Cu Cu Zn

Cu Zn

1 1L L

R R R R A AR

L LR R R R R RA A

( )

( )

( )( )( )

ρρπ π ρ ρ

ρ ρ π ρ ρππ

⋅−

= =+ −+

−

CuZn2 2 2

1 2 1 Zn Cueq 2 2 2

Cu Zn Cu 1 Zn 2 122 212 1

LLr R R L

RL L r R R

RR R

CALCULATE:

( )( )( )( ) ( ) ( ) ( )( )

8 8

2 2e 28 3 3q

8 3

10.0 m 1.72 10 m 5.964 10 m

1.72 10 m 1.00 10 m 5.964 10 m 2.00 10 m 1.00 10 m

0.01664925

Rπ

− −

− − − − −

⋅ Ω ⋅ Ω=

⋅ ⋅ + ⋅ Ω ⋅ − ⋅

= Ω

ROUND: Keeping only three significant digits gives 0.0166 .R = Ω DOUBLE-CHECK: The combined resistance of the components of the wire is less than the resistance of either material alone, as expected for resistances in parallel.

25.73. THINK: The Stanford Linear Accelerator accelerated a beam of 142.0 10 electrons⋅ per second through a potential difference of 102.0 10 V.⋅ SKETCH: Not required. RESEARCH: Electrical current is defined by / ,i q t= i.e. the amount of charging passing per unit of time. The power in the beam is calculated by P i V= Δ and the effective ohmic resistance is / .R V i= Δ SIMPLIFY: (a) The electrical current in the beam is ( )/ / .i q t e n t= = (b) The power in the beam is .P i V= Δ (c) The effective resistance is / .R V i= Δ CALCULATE: (a) ( )( )19 10 51.602 10 C 2.0 10 V 3.20 10 Ai −= ⋅ ⋅ = ⋅

(b) ( )( )5 14 13.20 10 A 2.0 10 s 640 kWP − −= ⋅ ⋅ =

(c) ( )( )10 5 142.0 10 V 3.20 10 A 6.25 10 R −= ⋅ ⋅ = ⋅ Ω

ROUND: Keeping only two significant digits, the results become, (a) 53.2 10 Ai −= ⋅ (b) 640 kWP = (c) 146.3 10 R = ⋅ Ω DOUBLE-CHECK: Such large values are reasonable for a device meant to accelerate particles to relativistic speeds.

25.74. THINK: To solve this problem, the circuit needs to be simplified by finding equivalent resistances. Use the relationships of parallel and series resistors.


984

SKETCH:

RESEARCH: If two resistors in series, the equivalent resistance is eq A B ,R R R= + and if two resistors in

parallel, the equivalent resistances is eq A B1/ 1/ 1/R R R= + or ( )eq A B A B/ .R R R R R= +

SIMPLIFY: The equivalent resistance of two resistors in the circuit 1C is ( )eq 1 2 1 2/ .R R R R R= + The

equivalent resistance of three resistors in the circuit 2C is eq 3 eq eq 32R R R R R R= + + = +

( )1 2 1 2 32 / .R R R R R= + + Thus, the effective resistance is given by

1 2eff 3

eff 1 2

1 1 1 1 1 .2 2

R RR R RR R R R R

= + = = ++

The current flowing through 3R is the same as current through .R Therefore, emf emf

eff

.2

V Vi

R R= =

CALCULATE:

(a) ( )( )

eff

3.00 6.00 20.0 10.0 3.00 6.00 2

RΩ Ω Ω= + = ΩΩ + Ω

(b) ( )12.0 V 0.500 A

2 12.0 i = =

Ω

ROUND: Not required. DOUBLE-CHECK: Both of the calculated values have appropriate units for what the values represent.

25.75. THINK: For resistors connected in parallel, the potential differences across the resistors are the same. SKETCH:

RESEARCH: The equivalent resistance of two resistors in parallel ( )2 3 and R R is

( )( ) ( )= +eq 2 3 2 3/ .R R R R R

SIMPLIFY:

(a) The potential difference across 3R is bc ac ab 11 eq

1 .RV V V V iR V

R R

= − = − = +

(b) Since 1R and eqR are in series, the current flowing through 1R and eqR is 1 eq

.V ViR R R

= =+


985

(c) The rate thermal energy dissipated from 2R is 2

bc

2

.V

PR

=

CALCULATE: ( )( )

eq

3.00 6.00 2.00

3.00 6.00 R

Ω Ω= = Ω

Ω + Ω

(a) The potential difference across 3R is bc2.00 110. V 55.0 V.

2.00 2.00 V Ω = = Ω + Ω

(b) The current through 1R is 110. V 27.5 A.2.00 2.00

i = =Ω + Ω

(c) The thermal energy dissipated from 2R is ( )255.0 V

1.008 kW.3.00

P = =Ω

ROUND: Keeping three significant digits gives: (a) bc 55.0 VV = (b) 27.5 Ai = (c) 1.01 kWP = DOUBLE-CHECK: Each value has appropriate units for what is being measured.

25.76. THINK: When a potential difference V is applied across resistors connected in series, the resistors have identical currents. The potential difference across R1 is Vac = V . SKETCH:

RESEARCH:

Vac = Vab +Vbc ; The potential differences across R1 , R2 and R3 are Vab = i2R2 = VR2 / R2 + R3( ) and

Vbc = i2R3 = VR3 / R2 + R3( ), respectively. The currents are: i1 = V / R1 and i2 = i3 =V / R2 + R3( ). SIMPLIFY: Not required. CALCULATE: Substituting the values of the resistors and the potential difference across the battery yields (a) ac 1.500 V,V = ( )( ) ( )ab 1.500 V 4.00 / 4.00 6.00 0.600 V,V = Ω Ω + Ω = and bc ac ab 1.500 V 0.600 V 0.900 V.V V V= − = − = (b) ( )1 1.500 V / 2.00 0.750 Ai = Ω = and ( )2 3 1.500 V / 4.00 6.00 0.150 A.i i= = Ω + Ω = ROUND: Keeping three significant figures gives: (a) ac 1.500 V,V = ab 0.600 VV = and bc 0.900 V.V = (b) 1 0.750 Ai = and 2 3 0.150 A.i i= = DOUBLE-CHECK: The resistance through the right path is five times larger than the resistance through the center path. Therefore the current should be five times smaller in the right path, which the calculation shows before rounding for significant figures.

25.77. THINK: In order for a copper cable to start melting, its temperature must be increased to a melting point temperature ( )M 1358 K .T = The cable is insulated. This means the energy dissipated by the cable is used to increase its temperature. Use 2.5 m,L = and 12 V.V =


986

SKETCH:

RESEARCH: The resistance of the copper cable is / .R L Aρ= The energy dissipated by the cable is

( ) ( )2 2/ / .E V R t V A L tρ= Δ = Δ This energy must be equal to the amount of heat required to increase the

temperature of the cable from the room temperature to the melting point temperature, which is ( )M R ,Q cm T T cm T= − = Δ where c is the specific heat of copper.

SIMPLIFY: Using the mass of the copper cable given by D ,m ALρ= the time required to start melting the

cable is 22

DD 2 .

c L TV A t c AL T tL V

ρ ρρρ

ΔΔ = Δ Δ =

CALCULATE: ( )( )( )( ) ( )

( )

23 8

2

386 J/kg K 8940 kg/m 1.72 10 m 2.5 m 1358 K 300 K2.72 s

12 Vt

−⋅ Ω −Δ = =

ROUND: Rounding tΔ to two significant digits produces 2.7 s.tΔ = DOUBLE-CHECK: This is a short time interval, but the 12 V battery supplies a large voltage, and the insulation does not allow the heat of the wire to dissipate.

25.78. THINK: A piece of copper wire is used to form a circular loop of radius 10 cm. The cross-sectional area of the wire is 210 mm . The resistance between two points on the wire is needed. SKETCH:

RESEARCH: The resistance of a wire is given by / .R L Aρ= Thus, the resistances of each segment of the wire are 1 1 /R L Aρ= and 2 2 / .R L Aρ= SIMPLIFY: Since 2 13 ,L L= the resistance 2R is 2 1 13 / 3 .R L A Rρ= = Because 1R and 2R are in parallel, the effective resistance is R = R1R2 / R1 + R2( )= 3R1

2 / R1 + 3R1( )= 3/ 4( )R1 = 3/ 4( )ρL1 / A. Putting in

L1 = 1/ 4( )2πr gives R = 3 / 8( )ρπr / A.

CALCULATE: ( ) ( )8 45 2

0.100 m3 1.72 10 m 2.03 10 8 1.00 10 m

Rπ− −

−= ⋅ Ω = ⋅ Ω⋅

ROUND: 42.03 10 R −= ⋅ Ω

DOUBLE-CHECK: The resistance of L1 is R1 = ρL1 / A = 1.72 ⋅10−8 Ω m( )2π 10 ⋅10−2 m( )4 10 ⋅10−6 m2( )= 2.702 ⋅10−4 Ω.

The resistance of L2 is R2 = 3R1 = 8.105 ⋅10−4 Ω. Our result is less than R1 and R2 , because the two resistances are in parallel. So our answer seems reasonable.

25.79. THINK: Two conducting wires have identical length and identical radii of circular cross-sections. I want to calculate the ratio of the power dissipated by the two resistors (copper and steel). SKETCH: Not required. RESEARCH: The resistance of a wire is given by / .R L Aρ= The power dissipated by the wire is

2 2/ / .P V R V A Lρ= =


987

SIMPLIFY: Therefore, the ratio of powers of two wires is

copper copper copper

steel steel steel

2copper

2steel

/.

/P V A LP V A L

ρρ

=

Since copper steelL L= and copper steel ,A A= the ratio becomes coppersteel copper

steel

/ .PP

ρ ρ=

CALCULATE: copper

st

8

8eel

40.0 10 m 23.80951.68 10 m

PP

−

−

⋅ Ω= =⋅ Ω

ROUND: Rounding the result to three significant digits yields a ratio of 23.8:1. This is because copper is a better conducting material than steel. Moreover, the specific heat of copper is less than steel. This means that copper is less susceptible to heat than steel. DOUBLE-CHECK: Since the two wires have identical dimensions, and the power dissipated is inversely proportional to the resistivity of the wires, it is reasonable that the material with the higher resistivity dissipates the larger amount of power.

25.80. THINK: The resistance of a wire increases or decreases linearly as a function of temperature. SKETCH: Not required. RESEARCH: The resistance of the wire at temperature T is given by R = R0 1+α T − T0( )( ). The

resistance at temperature T is 2 / .R V P=

SIMPLIFY: The resistance at the temperature 0T becomes R0 = V 2 / P( )1+α T −T0( )( )−1.

CALCULATE: Substituting 20 C,T = ° 0 1800 C,T = ° 4 15 10 C ,α − −= − ⋅ ° 110 VV = and 40.0 WP = gives

( ) ( )( )( )2

14 1

0

110 V1 5 10 C 1800 C 20 C 2750 .

40 WR

−− −= − ⋅ ° ° − ° = Ω

ROUND: Rounding to one significant digit yields 0 3000 3 k .R = Ω = Ω DOUBLE-CHECK: Ohms are appropriate units for resistance. The calculation can be checked by rounding the values to the nearest power of 10. 0

3 10 C, T 1000 C, 10 CT α −−≈ ° ≈ ° ≈ − ° , 100 V, and 100 WV P≈ ≈ (note that P was rounded to 100 since log 40 1.5)> . Then,

R0 = 100 V /100 W( ) 1− 10−3 °C−1( )1000 °C − 0 °C( )( )≈1000 Ω.

The approximated value is the same order

of magnitude as the calculated value. This lends support to the calculation.

25.81. THINK: The energy dissipated in a resistor is equal to the energy required to move electrons along the direction of a current. The material may or may not be ohmic. The rate of energy dissipation in a resistor is equal to the amount of power required to push electrons. SKETCH:

RESEARCH: If the force on the electrons is F

and the average velocity of the electrons is ,v the required power is .P Fv= Since ,F qE= the power becomes .P qEv=


988

SIMPLIFY: Using the charge ,q neV= where V is a small finite volume, the required power is P nevVE= or .P E JV= (a) Therefore, the power dissipated per unit volume is / .P V E J=

(b) For an ohmic material the current density J

is related to E

by .J Eσ=

This equation yields the power dissipated per unit volume of ( ) 2/ EP V E Eσ σ= = or ( )( ) ( ) 2 2/ / 1/ .P V J J J Jσ σ ρ= = = CALCULATE: Not required. ROUND: Not required. DOUBLE-CHECK: Examine the two sides of the equation / .P V E J= The units of /P V are watts per

cubic meter. The units of the product E J are ( )( )2 3 3V/m A/m =V A/m W/m .= In (b), the units of

/P V are still watts per cubic meter. Since ,J Eσ=

the units of σ are 2A / mW . Therefore, the units of

( ) 2 21/ J Jσ ρ= are ( )( ) ( )( )22 2 2 2 4 3m W/A A/m = m W/A A /m = W/m . Thus, by dimensional analysis, the

computed equations are sensible.

Chapter 26: Direct Current Circuits

989

Chapter 26: Direct Current Circuits In-Class Exercises

26.1. b 26.2. c 26.3. c 26.4. c 26.5. d 26.6. e Multiple Choice

26.1. a 26.2. b 26.3. b 26.4. d 26.5. a 26.6. d 26.7. c 26.8. b 26.9. d, e & f

Questions

26.10. In diagram A) the voltmeter does not measure the voltage across the load resistor Load ,R but it measures the voltage across the series Load Ammeter .R R+ As long as the internal resistance of the Ammeter is much less than Load ,R the effect can be neglected. Similarly, the Ammeter does not measure only the current through the load resistor Load ,R but also the current through the Ammeter .R This means that the current measurement is affected by the value of Ammeter .R As long as AmmeterR is much less than LoadR the effect can be neglected.

In diagram B) the voltmeter measures the voltage across the load resistor. However, the current flowing through LoadR is affected by the internal resistance of the voltmeter. As a result the measured voltage is altered from the original value. As long as the load resistance of the voltmeter VoltmeterR is much larger than the load resistance Load ,R the effect can be neglected. The ammeter measures the net current flowing through the resistors Ammeter ,R LoadR and Voltmeter .R The effects of the internal resistors of the ammeter and the voltmeter are negligible if Load VoltmeterR R and Ammeter Load .R R

26.11. The capacitive time constant is given by τ =0 C.R Since the equivalent of two identical capacitors connected in series is ( )eqC 1/2 C,= the time constant is ( ) ( )τ τ= = =eq 0C 1/2 1/ 2 .R RC Therefore, the time constant decreases by a factor of 2.

26.12. The resistance in the two-point probe measurement is given by the resistance of the device and the wires since they are in series. The four-point measurement is designed such that the resistance of the wires is no longer apart of the measurement so the real potential is measured of the device. The four-point measurement, therefore, gives a better measurement of the real resistance.

26.13. For a capacitor, the rate of which it discharges is based on the current. If the resistance is high and the current is low then the discharge rate is slower. If the capacitor is large, then for a given voltage the amount of charge is higher and it will take a longer time for the capacitor to discharge. Hence, increasing R or C can increase the time constant.

26.14. The charge builds up on the capacitor. Thus, the emf of the capacitor balances the emf of the batteries. Summing around the circuit balances the emf, which must be zero regardless of the resistor. The current that satisfies this condition is zero.

26.15. An appropriate resistor R may be connected in series with the bulb and the battery, the value of the resistor can be solved by applying Kirchoff’s loop rule for the circuit containing ε (the voltage of the car battery), R and the bulb. bulb bulb0 /iR iR R i Rε ε− − = = − where = /i P V and = 2

bulb / .R V P

Therefore, ( ) ( )ε ε= − = − = − = Ω2 1.5 V 12.0 V 1.5 V 16 .

1.0 WV V VR VP P P


990

26.16. If the emf’s are doubled then the currents will also double as Kirchoff’s junction rule will still be satisfied as long as all the currents are doubled. Kirchoff’s Loop rule implies that if the potential drop across all the resistors is equal to the emf, then doubling the emf means that the currents must also double to account for the needed increases in the potential drop.

26.17. With the capacitors uncharged at 0,t = the potential difference across each capacitor at that instant is zero, just like the potential difference of a connecting wire. After a long time, the capacitors will be fully charged, and the potential across the two points will be such that the charge cannot flow between the two points across the capacitor. This has the same effect as open segments in the circuit.

26.18. When using a voltmeter, I want to measure the potential difference across a device. To do this, I set it up parallel to the component. Ammeters are instruments with very low resistance designed to measure the current. Think of this as a device submerged into a running stream. In order for the instrument to measure the flow, it has to be “in the stream”. Similarly, ammeters are in series, with the components they wish to measure.

26.19. This question provides an example of meter loading. In connecting an ordinary voltmeter and ammeter simultaneously to some component of a circuit, only two possible orientations are available: one can connect the ammeter in series with the parallel combination of the voltmeter and the component, or one can connect the voltmeter in parallel with the series combination of the ammeter and the component. In the first case, the ammeter measures not the current through the component but the current through the component and the voltmeter, which is slightly greater for any voltmeter with non-infinite resistance. In the second case, the voltmeter measures not the potential difference across the component, but the potential difference across the component and the ammeter, which will be slightly greater for any ammeter with non zero resistance. Simultaneous exact measurements of the current and voltage for the components alone are not possible with ordinary meters. The restriction “with ordinary meters” is reiterated here because it is possible to measure the voltage across a component, for example, without drawing current. This can be done via a “null measurement” such as is done with a potentiometer.

26.20. 2 /P V R= implies 2 / .R V P= A larger power rating implies a smaller filament resistance. Therefore, the answer is a 100 W bulb.

26.21. Since the constant is given by C,R the ratio of the times is equal to the ratio of the capacitances. The capacitances in each case are: series: ( )series 1 2 1 2/ ,C C C C C= + parallel: parallel 1 2 .C C C= + The ratio is then:

( )2 2 2parallel 1 2 1 1 2 2 1 2

series 1

1

2

2

1 2

1 2

1 2 2 1

2 2.C C CC C C C C C CC C C C C C CC

C

CC

C

+ + += = = = + +

+

+

The time to charge the capacitors in parallel is larger by a factor of 1 2

2 1

2C CC C

+ + (which is at least 2).

26.22. (a) The current at any time t is given by: τ−= /initiale

ti i where =initial /i V R and τ = .RC (b) The power of the battery is = .P Vi Integrated over all time, the power gives us the energy

( )2 /( ) 2

0 0 0t i t / e .t RCPd V d V R dt CV

∞ ∞ ∞ −= = =

(c) The power dissipation from R is = i .P R ( ) ( )2 2 /( ) 2

0 0t / e t 1/2 .t RCiRd V R d CV

∞ ∞ −= =

(d) Note that the energy provided by the battery less the energy dissipated by the resistor is the energy stored in the capacitor satisfying the law of conservation of energy.


991

Problems

26.23. The total resistance of the circuit is given by: total 1 2 .R R R= + The current is then total/ .I V R= Δ The potential drop across each resistor is then:

1 21 2

1 2 1 2

, .R R

V V V VR R R R

= Δ = Δ + +

The resistors in series construct a voltage divider. The voltage VΔ is divided between the two resistors with potential drop proportional to their respective resistances.

26.24. The current must be such that 2 10.0 W.P I R= =

( )

( ) ( )( )

22 2 2emf emf

emf

2 2 2 2

2

12.0 1.0 10.0 2 1.00 10.0 10.0 0 12.4 1.00.

V VI R P R r P rRP R PV

r R r RR R R R R

= = = + + + + = + + = − +

Solving this quadratic equation yields: 212.4 12.4 4.00 0.331

2R ± −= Ω = Ω or 12.1 .Ω

26.25. Kirchhoff’s Loop Rule around the upper loop and large loop yields ( ) ( )( ) ( ) ( )emf emf emf2.00 A 2.00 A 20.0 0 and 3.00 A 0 3.00 A ,V R V R V R− − Ω = − = =

respectively. Therefore, ( ) ( ) ( )( ) ( ) ( )( )

( )( )emf

3.00 A 2.00 A 2.00 A 20.0 0 1.00 A 2.00 A 20.0 40.0

3.00 A 40.0 120. V.

R R R R

V

− − Ω = = Ω = Ω

= Ω =

26.26. THINK: Close inspection of the diagram shows that there is no current flowing across the middle resistor is zero. This is because there is nothing different between the point above and below the middle resistor. That resistor can therefore be removed while changing nothing. SKETCH: The new diagram is then:

RESEARCH: The equivalent resistances of the top two resistors and the bottom two resistors are given by

top 2 ,R R R R= + = and bottom 2 ,R R R R= + = respectively. The system’s total equivalent resistance is given

by 1

eq1 1 .

2 2R

R R

− = +

SIMPLIFY: 1 1

eq1 1 2

2 2 2R R

R R R

− − = + = =

CALCULATE: Not required. ROUND: Not required. DOUBLE-CHECK: The total resistance is comparable to each of the individual resistors as one would expect.


992

26.27. THINK: (a) The dead battery is parallel to the starter and the live battery. (b) Kirchoff’s Laws can be used to find the currents. Use the data:

L D L S12.00 V, 9.950 V, 0.0100 , 1.100 , 0.0700 .DV V R R R= = = Ω = Ω = Ω SKETCH:

RESEARCH: Kirchoff’s Laws give: ( )

( )L D S

D L S

+ 1 1.1

i i ii i i

== −

( )( )

L L L S S

D D D S S

0 20 3

V i R i RV i R i R

− − =+ − =

SIMPLIFY: Substitute (1) into (2) and solve for D .i ( )L D S L S S+ 0V i i R i R− − = implies

( )L D L S L S 0,V i R i R R− − + = which in turn implies

( )L S L SD

L

V i R Ri

R− +

= (4)

( ) ( ) SSubstitute 4 into 3 and solve for .i ( )L s L S

D D S SL

0

V i R RV R i R

R − +

+ − =

implies

D L L DS

L D S D S L

V R V Ri

R R R R R R+

=+ +

(5)

( ) ( ) SSubstitute 1.1 into 3 and solve for .i ( )D L S D S S 0V i i R i R+ − − = implies

D L DS

D S

V i Ri

R R+

=+

(6)

( ) ( ) LSubstitute 6 into 2 and solve for .i D L DL L L S

D S

0V i R

V i R RR R

+− − = +

implies

L D L S D SL

L D L S D S

V R V R V Ri

R R R R R R+ −

=+ +

(7)

Substitute (5) and (7) into (1) and solve for L .i

CALCULATE: ( )( ) ( )( )

( )( ) ( )( ) ( )( )S

9.950 V 0.0100 12.00 V 1.100 149.938 A

0.0100 1.100 0.0700 1.100 0.0700 0.0100 i

Ω + Ω= =

Ω Ω + Ω Ω + Ω Ω

( )( ) ( )( ) ( )( )( )( ) ( )( ) ( )( )L

12.00 V 1.100 12.00 V 0.0700 9.950 V 0.0700 150.434 A

0.0100 1.100 0.0100 0.0700 1.100 0.0700 i

Ω + Ω − Ω= =

Ω Ω + Ω Ω + Ω Ω

D D150.434 A 149.938 A 150.434 A 149.938 A 0.496 Ai i= + = − = ROUND: Three significant figures: S 150. A,i = L 150. A,i = D 0.496 A.i = DOUBLE-CHECK: Inserting the calculated values back into the original Kirchoff’s equations;


993

( ) ( )( ) ( )( )L L L S S 12 V 150 A 0.01 150 A 0.07 0,V i R i R− − = − Ω − Ω = and

( ) ( )( ) ( )( )D D D S S 9.95 V 0.496 A 1.1 150 A 0.07 0,V i R i R+ − = + Ω − Ω = as required.

26.28. THINK: There is only one unknown, so one equation is sufficient to solve the problem. Use Kirchoff’s Loop Law to obtain the answer. SKETCH:

RESEARCH: Kirchoff’s Loop Law gives 1 1 1 2 0V i R V− + = for the first loop.

SIMPLIFY: 2 11

1

V Vi

R+

=

CALCULATE: 12.5 V 1.5 V 1.0 A

4.0 i += =

Ω

ROUND: 1 1.0 Ai = DOUBLE-CHECK: Consider the outside loop: 1 1 1 2 2 0.V i R i R− − = From the second loop,

2 2 2 2 2 20 / .V i R i V R− − = = So, ( )1 1 1 2 2 1 1 1 2 2 20 /V i R i R V i R V R R= − − = − − 1 1.0 A,i = as before.

26.29. THINK: Kirchoff’s Laws can be used to determine the currents. Use the values: A 6.0 V,V = B 12.0 V,V =

1 10.0 ,R = Ω = Ω2R 40.0 , and 3 10.0 .R = Ω SKETCH:

RESEARCH: 3 1 2 ,i i i= + A 1 1 3 3 ,V i R i R= + B 2 2 3 3 ,V i R i R= + − + − =A 1 1 2 2 B 0,V i R i R V and .P iV=

SIMPLIFY: ( ) A 2 3A 1 1 1 2 3 1

1 3

V i R

V i R i i R iR R

−= + + =

+

( )2

A 2 3 A 3 2 3B 2 2 1 2 3 B 2 2 2 3 B 2 2 2 3

1 3 1 3 1 3

V i R V R i R

V i R i i R V i R i R V i R i RR R R R R R

−= + + = + + = + − + + + +

A 3B

1 32 2

32 3

1 3

,

V RVR R

iRR R

R R

− + =

− + +

A 2 31

1 3

,V i R

iR R

−=

+ 3 1 2i i i= +

=A 1 A ,P i V B 2 BP i V=


994

CALCULATE: ( )( ) ( )2

2

6.0 V 10.0 10.0 12.0 V / 40.0 10.0 0.20 A

10.0 10.0 10.0 10.0 i

Ω Ω = − Ω − + Ω = Ω + Ω Ω + Ω

( )( )1 3

6.0 V 0.20 A 10.0 0.20 A, 0.20 A 0.20 A 0.40 A

10.0 10.0 i i

− Ω= = = + =

Ω + Ω

( )( )= =A 0.20 A 6.0 V 1.2 W,P ( )( )= =B 0.20 A 12.0 V 2.4 WP

ROUND: 1 2 3 A B0.20A, 0.20A, 0.40A, 1.2 W, and 2.4 W.i i i P P= = = = = DOUBLE-CHECK: The direction of 1i and 2i makes sense since they are in the direction of the driving force of the battery.

26.30. THINK: The circuit has three branches. Kirchoff’s Loop and junction laws can be used to find at least three linearly independent equations. Use the values: 1 5.00 ,R = Ω 2 10.0 ,R = Ω 3 15.0 ,R = Ω

emf ,1 10.0 V,V = and emf ,2 15.0 V.V = SKETCH:

RESEARCH:

emf ,1 emf ,2 2 2 1 1 0V V i R i R− + − = (1)

emf ,1 3 3 1 1 0V i R i R− − = (2)

1 2 3i i i+ = (3) SIMPLIFY: ( ) ( ) 1Substitute 3 into 2 and solve for :i ( )emf ,1 1 2 3 1 1 0V i i R i R− + − = implies

emf ,1 2 31

1 3

V i R

iR R

−=

+ (4)

( )( )( )( ) ( )

2

emf ,2 emf ,1 emf ,1 1 1 3emf ,1 2 3emf ,1 emf ,2 2 2 1 2

1 3 1 3 1 3 2

Substitute 4 into 1 and solve for :

/0 5

/

i

V V V R R RV i RV V i R R i

R R R R R R R

− + + − − + − = = + + +

CALCULATE: ( )( ) ( )

( )( ) ( )2

15.0V 10.0V 10.0V 5.00 / 5.00 15.0 0.54545 A

5.00 15.0 / 5.00 15.0 10.0 i

− + Ω Ω + Ω= =

Ω Ω Ω + Ω + Ω

( )21

10.0 V 15.0 0.09091 A

5.00 15.0 i

i− Ω

= =Ω + Ω

3 1 2 0.636363 Ai i i= + = ROUND: 1 0.0909 A,i = 2 0.545 A,i = 3 0.636 Ai =


995

DOUBLE-CHECK: The calculated values for the currents are all positive, which is consistent with the direction specified in the problem.

26.31. THINK: The circuit has six branches, and therefore six different currents. Kirchoff’s Laws can be used to find six equations to solve for six unknowns. SKETCH:

RESEARCH: There are four junctions, giving three linearly independent equations:

3 6 1

2 4 1

2 5 3

i i ii i ii i i

+ =+ =+ =

There are three loops that can be analyzed using Kirchoff’s loop rule. Analyzing each loop in the clockwise direction:

( )( )

1 3 1 2 4 3 5

emf ,1 6 6 7 emf ,2 3 5

emf ,1 2 4 4 2

00

0

i R R i R i RV i R R V i R

V i R i R

− + − − =− − + + + =

+ − =

The power supplied by each battery is given by .P Vi= SIMPLIFY: The most efficient method for solving a system of six equations and six unknowns is by matrix. Rearranging the equations:

( )( )

1 3 6

1 2 4

2 3 5

1 3 1 2 4 3 5

3 5 6 6 7 emf ,1 emf ,2

2 4 2 emf4 ,1

00

00

i i ii i i

i i ii R R i R i R

i R i R R V Vi R i R V

− + + =− + + =

− + =− + − − =

− + = −− = −

( )( )

emf,1 emf ,1 5

emf,2 emf ,2 6

P V V i

P V V i

=

=

CALCULATE:

1 1

2 2

3 3

4 4

5 5

6 6

12 6 5 0 0 0 0 0.25070 0 5 0 0 13 6 0.37460 6 0 6 0 0 6 0.1522

1 0 1 0 0 1 0 0.62541 1 0 1 0 0 0 0.2224

0 1 1 0 1 0 0 0.40

i ii ii ii ii ii i

− − − − − − − − −

= = −

−

− 30


996

The current through resistors 1 3 and R R is 1 0.2507 Ai = clockwise. The current through resistor 4R is

2 0.3746 Ai = counterclockwise. The current through resistor 5R is 3 0.1522 Ai = counterclockwise. The current through resistor 2R is 4 0.6254 Ai = clockwise. The current through resistors 6 7 and R R is

6 0.4030 Ai = clockwise. Current 5 0.2224 Ai = clockwise does not flow through any resistor.

( ) ( )( )( ) ( )( )

emf,1

emf,2

6.00 V 0.2224 A 1.3344 W

12.0 V 0.4030 A 4.8360 W

P V

P V

= =

= =

ROUND: 1 0.251 Ai = clockwise

2 0.375 Ai = counterclockwise

3 0.152 Ai = counterclockwise

4 0.625 Ai = clockwise



( )( )

emf,1

emf,2

1.33 W

4.84 W

P V

P V

=

=

DOUBLE-CHECK: The currents through the batteries are flowing from lower to higher potential, as expected.

26.32. THINK: When the potential difference between a and b is zero, no current will flow. The potential difference will be zero when the ratio of the resistances above the ammeter is equal to the ratio of the resistances below the ammeter. Use 1 25.0 cmL = and 2 75.0 cm.L = SKETCH:

RESEARCH: The current is zero when 1

2 1

2

L11 L L

L

,xx

RR R R R RR R

= = 1 100. .R = Ω 1L 1L /AR ρ= and

2L 2L /A.R ρ=

SIMPLIFY: ( ) ( )1 2 1L /A L /A ,xR Rρ ρ= ( )1 2 1L /LxR R=

CALCULATE: ( )( )75.0 cm / 25.0 cm 300. 100. xR = = ΩΩ ROUND: 300. xR = Ω DOUBLE-CHECK: xR is comparable to 1R as one would expect.

26.33. THINK: Suppose the total equivalent resistance of the ladder up to some arbitrary point is given by L .R Since the ladder is infinite, it does not matter what point on the ladder is chosen for the analysis, and


997

adding one more segment to the end will not change the equivalent resistance of the network. SKETCH:

RESEARCH: The ladder consists of the array with resistance LR plus another segment with resistance .R′ R′ contributes one resistor of resistance, R, in parallel with the array, and two resistors of resistance, R, in series with the array. The total resistance is now L L L2 .R R R R R′ = + =

SIMPLIFY: 1

LL L

L L

1 12 2 2RRR R R R R R

R R R R

−

= + = + + = + +

( ) ( ) 2 2LL L L L L L L

L

2 2 2 2 0RRR R R R R R R R RR R RR R

R R= + + = + + − − =

+

CALCULATE: Solving the quadratic equation for LR gives ( )L 1 3 .R R= +

ROUND: Since no values are given in the question, it is best to leave the answer in its precise form,

( )L 1 3 .R R= +

DOUBLE-CHECK: Consider the first rung of three resistors in series. The equivalent resistance is eq 3 .R R= Now, add another rung of three resistors. One resistor is in parallel with the first rung, and two

resistors are in series with the first rung. The equivalent resistance is now 1

eq,11 1 112 2.75 .

3 4R R R R

R R

− = + + = =

Adding another rung gives 1

eq,24 1 412 2.7333 .

11 15R R R R

R R

− = + + = =

Repeating the process,

eq,3153 2.7324 ,56

R R R= = eq,4571 2.73206 ,209

R R R= = …, ( )1 3 .nR R→ + This verifies the value found in the

solution.

26.34. THINK: This is a very famous and very tricky problem. Superposition can be used to find the answer. I will inject 1 Amp into A as specified below and extract 1 Amp from B as specified below.


998

SKETCH:

1 Amp extracted from B:

RESEARCH: The superposition of the two cases has 1 Amp entering A and leaving B. The superposition of the currents indicates that ( )1/2 A passes through the resistor ABR , showing the effective resistance between the two points is /2.R SIMPLIFY: Not required. CALCULATE: Not required. ROUND: Not required. DOUBLE-CHECK: It makes sense that the effective resistance is less than R since there are other pathways for the current to flow.

26.35. Let Ai be the maximum current (i.e. full scale value) the ammeter can measure without the shunt. If the shunt is to extend the full scale value by a factor tot A/ ,N i i= then

shuntA shunt A

A

1.i

i i Ni Ni

+ = = −

Since the ammeter and shunt have the same voltage across them,


999

i,AAi,A A shunt shunt shunt i,A

shunt

.1

RiR i R i R R

i N= = =

−

To allow a current of 100 A, the resistance of the shunt resistor must be ( )

( )shunt

1.00 1.0010.1 m .

99.0100. A /1.00 A 1R

Ω Ω= = = Ω

−

The fraction of the total current flowing through the ammeter is ( )( )

A

tot

1.00 A0.0100.

100. Aii

= =

The fraction of the total current flowing through the shunt is ( )( )

shunt

tot

1.00 A1 0.990.

100. Aii

= − =

26.36. The voltage across the device must be smaller than the voltage across the device and the resistor by a factor of N. ( ) = +1,V 1,V series .N V V V Since =1,V series :i i

( ) ( ) = = + = + = −

1,V series series series1,V 1,V 1,V series 1,V

1,V series 1,V 1,V

1 1V V R R

N V V V N R N RR R R R

Numerical Application: ( ) 6 6series 100. 1 1.00 10 99.0 10 99.0 M .R = − ⋅ Ω = ⋅ Ω = Ω The 1.00 V potential drop

across the voltmeter is 1.00% of the total power. The other 99.0 V potential drop occurs across the added series resistor and is 99.0% of the total.

26.37. The sketch illustrates the case of measuring ab .V

The total resistance is ( ) ( )i i/ .RR R R R + + The total current is

( )( )5

5 75i

5 7i

6.0000 V 3.0149 10 A.1.0000 10 1.00 10 1.0000 10 1.0000 10 1.00 10

ViRR R

R R

−= = = ⋅⋅ Ω ⋅ Ω+ + ⋅ Ω

+ ⋅ Ω + ⋅ Ω

The potential across the voltmeter is

( ) ( )( )5 75i

ab 5 7i

1.0000 10 1.00 10 3.0149 10 A 2.985 V 2.99 V,1.0000 10 1.00 10

RRV i

R R− ⋅ Ω ⋅ Ω= = ⋅ = =

+ ⋅ Ω + ⋅ Ω

bc ab6.0000 V 6.0000 V 2.985 V 3.015 V 3.02 V.V V= − = − = = Increasing iR will reduce the error since the voltmeter will draw less current.

26.38. (a) The current is to be 10.0 mA for a voltage of 9.00 V. ( )emf / 9.00 V/ 10.0 mA 900. R V i= = = Ω (b) The current is 2.50 mA. The resistance is variable .R R+ The current is given by


1000

( ) ( )( )emf emf variablevariable emf

variable

9.00 V 2.50 mA 900. V 896 .

2.50 mAV V iR

i i R iR RR R i

− Ω−= + = = = = Ω

+

26.39. THINK: (a) The total resistance must first be determined in order to find the current. Since the resistors are in series, the same current flows through both of them. (b) The current that flows through the circuit is the result of the equivalent resistance including the ammeter. The same current flows through the 1.00 kΩ resistor and the parallel combination of resistor and ammeter. Of the current flowing through this combination, the majority will flow through the lower resistance, i.e., the ammeter. The fraction of the current that goes through the Ammeter can be calculated using the resistances. SKETCH: (a) (b)

RESEARCH: (a) eq 2 ,R R= 1.00 k ,R = Ω a eq/ ,i V R= V 12.0 V=

(b) The current that flows through the circuit is beq

,ViR

= where Aeq

A

,RR

R RR R

= + +

A 1.0 .R = Ω The

current flowing through the resistor/ammeter combination is split into two parts. R 1 / ,i V R= Δ and

Amm 2 A/ .i V R= Δ SIMPLIFY:

(a) a 2ViR

=

(b) ( ) ( )b Amm2 Amm

AmmAmm AmmAmm Amm Amm AmmAmm

Amm

2i RRV RRV Vi

R R RR R R R R RRRRR R

Δ= = = ⋅ = ++ + + +

CALCULATE:

(a) ( )a 3

12.0 V 6.00 mA2 1.00 10

i = =⋅ Ω

(b) ( )12.0 V 0.01198 A

1.00 k 2.0 =

Ω + Ω

ROUND: (a) a 6.00 mAi = (b) Amm 0.012 Ai = DOUBLE-CHECK: The ammeter measures the current across the other resistor acting like a short across the first resistor, as would be expected.


1001

26.40. THINK: (a) I need to find the total resistance and then find the potential drop in each resistor. (b) When a voltmeter is connected across one of the resistors, the combination of the resistor and the voltmeter will have an equivalent resistance slightly different from that of the resistor alone. This will cause a change in the potential drop across the resistor/voltmeter combination. I need to calculate the new potential drop. SKETCH: (a) (b)

RESEARCH: (a) Since they are identical and in series, the resistors have the same potential drop of / 2.V (b) The total resistance is now given by ( )total voltmeter voltmeter/ .R R R R R R= + + The potential drop across the

voltmeter is then ( )( )voltmeter voltmeter voltmeter total/ 1/ .V V R R R R R= + SIMPLIFY: Not required. CALCULATE: (a) 12.0V / 2 6.00V= (b) ( )( ) ( )total 100. k 10.0 M 100. k / 10.0 M 100. k 199.009901 kR = Ω + Ω Ω Ω + Ω = Ω

( )( ) ( ) ( )voltmeter total12.0 V 10.0 M 100. k / 10.0 M 100. k 1/ 5.97 VV R = Ω Ω Ω + Ω =

The percentage change is ( )6.00 V 5.97 V /6.00 V 0.500%.− = ROUND: The percentage change is 0.500%. DOUBLE-CHECK: It make sense that the voltmeter will reduce the voltage since any voltmeter (with non-infinite resistance) will draw a small amount of current.

26.41. The equation for the charge of a capacitor in an RC circuit over time is ( ) /inital .tI t I e τ−= Use the

equations: ,RCτ = 100. 200. 300. ,R = Ω + Ω = Ω 10.0 mF,C = ( )

initial

ln / ,I t

tI

τ

= −

( )

initial

ln ,I tI

τ

−

and

( )( ) 5.00 mC300. 10.0 mF ln 8.99 s.100. mC

t = − Ω =

26.42. The circuit can be easily simplified to


1002

where eq 2.00 k 6.00 k 8.00 kR = Ω + Ω = Ω and ( ) ( ) 1

eq 1/ 2.00 μF 1/ 6.00 μF 1.50 μF.C−

= + = The time

constant is then ( )( )eq eq 8.00 k 1.50 μF 12 ms.T R C= = Ω = The initial charge of the 2.00 μF capacitor,

with initial potential 10.0 VV = is ( )( ) 50 2.00 μF 10.0 V 2.00 10 C.q CV −= = = ⋅ The charge decays as

( ) /0 .tq t q e τ−= When 1/ 2 ,t T= the charge left is ( ) 1/2 6

0 01/ 2 0.606 6.06 10 C.q t q e q− −= = = ⋅

26.43. Since the position of the resistor with respect to the capacitor is irrelevant, the circuit is simplified to:

The maximum charge of the capacitor is ( )( ) 4

0 C 20.0 μF 12.0 V 2.40 10 C.q V −= Δ = = ⋅ In general, the

capacitor charges as ( ) 0 1 .t

RCq t q e−

= −

When ( ) ( ) 0 :1/ 2q t q=

( )0 01 1 11 ln ln 2 .2 2 2

t tRC RCq q e e t RC RC

− − = − = = − =

Therefore, ( )( ) ( )3.00 20.0 μF ln 2 41.6 μs.t = Ω =

26.44. By Ohm’s law, the power, 1.21 GW,P = is related to potential, 12.0 V,V = and resistance, ,R by

( )22 2 12.0 V 119 n .

1.21 GWV VP RR P

= = = = Ω

The time to charge the capacitor, 1.00 ,C F= to 90.0% is ( ) 0 01 0.900 0.100.t t

RC RCq t q e q e− −

= − = =

Therefore, ( ) ( )( ) ( ) 9ln 0.100 119 n 1.00 F ln 0.100 274 10 s 274 ns.t RC −= − = − Ω = ⋅ =

26.45. THINK: The charge on the capacitor, 90.0 μF,C = decays exponentially through the resistor, 60.0 .R = Ω The energy on the capacitor is proportional to the square of the charge, so the energy also

decays exponentially. If 80.0% of the energy is lost, then 20.0% is left on the capacitor.


1003

SKETCH:

RESEARCH: The charge on the capacitor is given by ( ) /0 .t RCq t q e−= The energy on the capacitor is given

by ( ) ( )2 .E t q t= To determine the time when there is 20.0% energy remaining, consider the equation:

( ) ( )0.200 0 .E t E= SIMPLIFY: Determine time, t:

( ) ( ) ( )

( ) ( )

2 2 /2 20 0

2 /

0.200 0 0.200 2 2 2

2 0.200 ln 0.200 ln 0.200 .2

t R C

t RC

q t q e qE t E

C C Ct RCe t

RC

−

−

= = = =

− = = = −

CALCULATE: ( ) ( ) 360.0 90.0 μF

ln 0.200 4.3455 10 s2

t −Ω= − = ⋅

ROUND: t = 4.35 ms

DOUBLE-CHECK: After t = 4.35 ms, the charge on the capacitor is 0.451 of the maximum charge. This value squared gives 0.203, which is 20.0% with rounding error considered.

26.46. THINK: After sufficient time, the potential on both plates (area 2A 2.00 cm= and separation 0.100 mmd = ) will be 60.0 V.VΔ = Since the capacitors are in series, the total charge on each will be the

same. The potential drop across a capacitor is needed to find its electric field. The second capacitor has dielectric constant 7.00κ = and dielectric strength S 5.70 kV/mm.= SKETCH:

RESEARCH: The capacitance of the air filled capacitor is 1 0A/ ,C dε= an that with the dielectric is

2 0A/ .C dκε= The charge on a capacitor is .Q C V= Δ The energy stored in a capacitor is ( )2 / 2 .U Q C= The electric field inside a capacitor is / .E V d= SIMPLIFY: (a) Equivalent capacitance is

11 10 0

eq1 2 0 0

A A1 1 11 .A A 1

d dCC C d d

ε ε κε κε κ κ

−− − = + = + = + = +

Charge on the first capacitor is 1 eq .Q Q C V= = Δ

(b) Charge on the second capacitor is 2 eq .Q Q C V= = Δ

(c) The total energy on both plates is 2 22eq 2

eqeq eq

1 .2 2 2

C VQU C VC C

Δ= = = Δ


1004

(d) The potential drop across the second capacitor is ( )2 2 2 0/ / .AV Q C Qd κεΔ = = The electric field across

it is then ( )2 2 0/ / .AE V d Q κε= Δ = CALCULATE:

(a) ( )( )( )12 2 2 4 2

11eq 4

8.854 10 C / N m 2.00 10 m7.00 1.54945 10 F7.00 1 1.00 10 m

C− −

−−

⋅ ⋅ = = ⋅ + ⋅

( )( )11 101 1.54945 10 F 60.0 V 9.2967 10 CQ − −= ⋅ = ⋅

(b) 102 9.2967 10 CQ −= ⋅

(c) ( )( )211 81 1.54945 10 F 60.0 V 2.789 10 J2

U − −= ⋅ = ⋅

(d) ( )( )( )

10

2 12 2 2 4 2

9.2967 10 C 74998 V/m7.00 8.854 10 C / N m 2.00 10 m

E−

− −

⋅= =⋅ ⋅

ROUND: (a) 10

1 9.30 10 CQ −= ⋅

(b) 102 9.30 10 CQ −= ⋅

(c) 82.79 10 JU −= ⋅ (d) 2 75.0 kV/mE = DOUBLE-CHECK: Numerically, 2 7.5VVΔ = and 1 0/ A 52.5,V Qd εΔ = = so 1 2 60V ,V V VΔ + Δ = = Δ which means energy was conserved. Also, since 2E S< (dielectric strength), this capacitor is clearly viable, so it makes sense.

26.47. THINK: Since the dielectric material ( 2.5,κ = 50.0 μmd = and 124.0 10 mρ = ⋅ Ω ) acts as the resistor and it shares the same cross sectional area as the capacitor, 0.050 μF,C = a time constant, ,τ should be independent of the actual capacitance and resistance, and only depend on the material. SKETCH:

RESEARCH: The capacitance is 0A / .C dκε= The resistance is / A.R dρ= The time constant is .RCτ =

SIMPLIFY: The time constant is 00 .

AdRCA d

κερτ κρε = = =

CALCULATE: ( ) ( )( )12 12 2 22.5 4.0 10 m 8.85 10 C / N m 88.5 sτ −= ⋅ Ω ⋅ =

ROUND: 89 sτ = DOUBLE-CHECK: While this value seems relatively high, it is nonetheless perfectly reasonable. The high resistivity and greater than 1 dielectric material, both imply bigger R and ,C so a high τ is reasonable.

26.48. THINK: Since the current varies with time due to the charging of the capacitor, 2.00 mF,C = the energy lost due to heat from the resistor, 100. ,R = Ω is found by integrating the power dissipation of the resistor


1005

over time. When the capacitor is fully charged it has the same potential as the battery, 12.0 V.VΔ = SKETCH:

RESEARCH: When the capacitor is fully charged, the energy stored in it is ( ) 21/ 2 .U C V= Δ The power

across the resistor is 2 .P I R= The current decreases exponentially by ( ) /0 ,ti t i e τ−= where RCτ = and

0 / .i V R=

SIMPLIFY: Energy across the capacitor: ( ) 21/2 .CU CV= Energy dissipated through the resistor:

( ) ( )( ) ( )2 2

2 2 2 / 2 / 2 /

0 0 0 0 0.t t t

R oV VU P t dt I t Rdt R I e dt R e dt e dtR R

τ τ τ∞ ∞ ∞ ∞ ∞− − − = = = = =

Therefore, 2 2 2

02 /

0

0 .2 22

tR

V V VU eeR R R

τ τ ττ ∞− − = = − =−

Therefore, ,RCτ = ( )2

212 2R

V RCU CV

R= = and .C RU U=

CALCULATE: ( )( )( )21/2 2.00 mF 12.0 V 0.144 JC RU U= = = ROUND: 0.144 J,C RU U= = the same energy for both. DOUBLE-CHECK: The energy stored in capacitor is same as energy lost to heat by the resistor. This makes sense if I consider that the total internal energy should stay the same. Therefore, energy lost by resistor is energy gained by capacitor, so energy is conserved.

26.49. THINK: Normally, to be fully discharged the time needs to go to infinity. After 2.0 ms,tΔ = the capacitor should be as close to fully discharged as possible. A good standard of discharge is when the final charge is less than 0.01% which roughly corresponds to a time of 10 ,τ where τ is the time constant of the circuit. From ,τ the capacitance, ,C can be determined and using 5.0 JE = the potential difference on the plates is found. 10.0 k .R = Ω SKETCH: Not required. RESEARCH: The time constant is approximated as ( )1/10 ,tτ = Δ an is also .RCτ = The energy stored in

the capacitor is ( ) 21/2 .E C V= Δ

SIMPLIFY: The capacitance is 10

tCR Rτ Δ= = the potential difference is then 2 .EV

CΔ =

CALCULATE: 82.0 ms 2.0 10 F 0.020 μF,10 10.0 k

C −= = ⋅ =⋅ Ω

( )2 5.0 J

22361 V0.020 μF

VΔ = =

ROUND: 0.020 μF,C = 22 kVVΔ = DOUBLE-CHECK: If instead I chose to be only 99% discharged, corresponding to only 5 ,t τΔ = the potential across the capacitor would be about 16 kV, which is also high, so our choice is reasonable.

26.50. THINK: (a) When switch 1S is closed, the current flows solely through resistors 1 100. R = Ω and 3 300. R = Ω which are in series with a battery emf 6.00 V.V =


1006

(b) When switch 2S is closed, the current splits between 1 100. R = Ω in one branch and 2 200. R = Ω with a capacitor 4.00 mFC = in the other branch. Initially there is no charge on the capacitor so there is no potential drop across it, meaning it does not initially contribute to the current. These branches are then in series with resistor 3 300. R = Ω and battery emf 6.00 V.V = (c) The capacitor, 4.00 mF,C = will charge but only through resistor 2 200. ,R = Ω so as to give a time constant .τ As it charges over 10.0 min 600. s,t = = the current through that branch will decrease exponentially. (d) When the capacitor, 4.00 mF,C = is fully charged, no current flows through that branch. This means that initially, the battery emf 6.00 V,V = is in series with resistors 1 100. R = Ω and 3 300. .R = Ω The initial potential in the capacitor must still equal the potential drop across resistor 1.R When switch 1S is opened, the capacitor begins to discharge though resistors 1R and 2 200. .R = Ω As capacitor discharges, the current will decrease exponentially to f 1.00 mAi = . SKETCH: (a) (b)

(c) (d)

RESEARCH: (a) The equivalent resistance is eq 1 3 .R R R= + By Ohm’s Law, the current through circuit is 1 emf eq/ .i V R=

(b) The equivalent resistance of resistors 1 and 2 is ( ) 112 1 21/R 1/ .R R −= + The total equivalent resistance is

then eq 3 12 .R R R= + By Ohm’s law, the current through circuit is 2 emf eq/ .I V R=

(c) As the capacitor charges, the current through it decrease as ( ) /0

tCi t i e τ−= where, 2 .R Cτ = The current

through resistor 1R is Ri and the total current out of the battery is .R Ci i i= +

(d) Potential drop across 1R initially is 1 1 1 .Ci R V V= Δ = Δ Current decays exponentially as ( ) /0 ,ti t i e τ−=

where ( )1 2R R Cτ = + and by Ohm’s law, ( )0 1 2/ .Ci V R R= Δ + SIMPLIFY:

(a) emf emf1

eq 1 3

V Vi

R R R= =

+

(b) 1

1 212

1 2 1 2

1 1 ,R R

RR R R R

−

= + = + emf emf

2eq 3 12

.V V

iR R R

= =+

(c) Calculate ( )Ci t and infer i from it.


1007

(d) Initial current is 1 10

1 2 1 2

CV R ii

R R R RΔ

= =+ +

when ( ) f .i t i= Therefore,

( ) ( )f 1 2 f 1 2/ /1 1f

1 2 1 1 1 1

ln .t ti R R i R RR ii e e t

R R R i R iτ τ τ− −

+ += = = − +

CALCULATE:

(a) 16.00 V 0.0150 A 15.0 mA

100. 300. i = = =

Ω + Ω

(b) ( )( )

12

100. 200. 66.67 ,

100. 200. R

Ω Ω= = Ω

Ω + Ω 2

6.00 V 0.01636 A 16.36 mA300. 66.67

i = = =Ω + Ω

(c) ( )( )200. 4.00 mF 0.800 s.τ = Ω = ( )600. s

750.0.8 sC 0 0 0 A.i t i e i e

−−= = ≈ Regardless of what 0i is after

10.0 min, the current through that branch is effectively 0.0 A. Therefore, 15.0 mA.Ri i= = Since there is no current through the capacitor, the circuit is equivalent to having switch 2S open, as in part (a) so current through battery is then the same as in part (a).

(d) ( )( )100. 200. 4.0 mF 1.20 sτ = Ω + Ω = and ( ) ( )( )( )( )

1.00 mA 100. 200. 1.20 s ln 1.9313 s

100. 15.0 mAt

Ω + Ω= − = Ω

ROUND: (a) 1 15.0 mAi = (b) 2 16.4 mAi = (c) 15.0 mAi = (d) 1.93 st = DOUBLE-CHECK: (a) This is a reasonable value for current. (b) Since I added a resistor in parallel, the overall resistance is expected to decrease, and hence the current increase, so it makes sense. (c) From part (b), I saw the overall current was greater than in part (a). If a piece of the current found in part (b) dies off, the final current should be smaller, so it makes sense. (d) Since I know the current should reduce to zero in 600. s,t = then reducing by 80.0% in only 1.93 s is very reasonable so it makes sense.

26.51. THINK: The capacitor, 2.00 μF,C = charges via the battery, 10.0 V,VΔ = through resistor, 1 10.0 ,R = Ω so the resistors, 2 4.00 R = Ω and 3 10.0 ,R = Ω can be simplified to be in parallel. After a long time, the capacitor becomes fully charged and no current goes through it. The potential drop across it is then the same as the drop across 2R and 3 .R The energy of the capacitor is proportional to the square of the potential drop across it. The total energy lost across 3R is determined by integrating the power across it over time.


1008

SKETCH:

RESEARCH: The current through the circuit after a long time is ( )emf 1 23/ .i V R R= + Resistors in parallel

add as ( ) 11 123 2 3 .R R R

−− −= + The potential drop across the capacitor is C 23 .V iRΔ = The energy in the

capacitor is given by ( )2C / 2.E C V= Δ When the switch is open, the current through 3R is 3 C 2/ .i V R= Δ

The current across 3R varies as ( ) 23/3 3 .t R Ci t i e−= The power across 3R is given by ( )2

3 3 3 .P i t R= The energy

across 3R is given by ( )3 30.E P t dt

∞=

SIMPLIFY:

(a) The potential drop across the capacitor is given by: emf 23C 23

1 23

.V R

V iRR R

Δ = =+

(b) The energy in the capacitor is given by ( )2C

1 .2

E C V= Δ

(c) The energy across 3R is given by: ( ) 23 23

22 / 2 /2 C

3 3 3 30 0 03

t R C t R CVE P t dt R i e dt e dt

R∞ ∞ ∞− −Δ

= = =

23

2 2 22 /C 23 C 23 C 23

3 3 30

0 .2 2 2

tt R C

t

V R C V R C V R Ce

R R R

=∞−

=

Δ Δ Δ = − = + =

CALCULATE:

(a) ( ) ( )( ) 11 123 10.0 4.00 2.86 ,R

−− −= Ω + Ω = Ω ( )( )

C

10.0 V 2.86 2.22 V

10.0 2.86 V

ΩΔ = =

Ω + Ω

(b) ( )( )2 61 2.00 μF 2.22 V 4.938 10 J2

E −= = ⋅

(c) ( ) ( )( )

( )

2

63

2.22 V 2.86 2.00 μF1.411 10 J

2 10.0 E −Ω

= = ⋅Ω

ROUND: (a) C 2.22 VVΔ = (b) 4.94 μJE = (c) 3 1.41 μJE =


1009

DOUBLE-CHECK: The energy across 2R is ( )22 C 23 2/ 2 3.527 μJ.E V R C R= Δ = The result makes sense

because energy is conserved: 2 3 .E E E+ =

26.52. THINK: (a) The capacitor, 15 μFC = and 100.0 V,CVΔ = is fully discharged when the charge is less than 0.01%, which roughly corresponds to a time of 10 ,τ where τ is the time constant of the circuit. The resistor in question is a cube of gold of sides 2.5 mml = and resistivity 82.44 10 m.Rρ −= ⋅ Ω (b) The capacitor, 15 μFC = and 100.0 V,CVΔ = is fully discharged so that all the initial stored energy has gone to heating the resistor. The resistor in question is a cube of gold of size 2.5 mm,l = density

3 319.3 10 kg / mDρ = ⋅ and specific heat 129 J / kg C.c = ° Assume the cube is initially at room temperature, i 20.0 C.T = ° SKETCH: (a) (b)

RESEARCH: (a) The resistance of the cube is / A.RR Lρ= The time constant is 10 .t τ=

(b) The energy of the capacitor is ( ) 21/ 2 .C CU C V= Δ The energy gained by the gold block increases its temperature as .Q mc T= Δ Mass of gold is .Dm Vρ= The energy the cube gains is same energy the capacitor dissipates, .CU Q= SIMPLIFY: (a) The time for discharge is ( ) ( )210 10 10 / A 10 / 10 / .R R Rt RC L C l l C C lτ ρ ρ ρ= = = = =

(b) To find the final temperature: ( ) ( )2 1/ 2 .CQ U mc T C V= Δ = Δ Therefore,

( ) 2f i

12D cVc T T C Vρ − = Δ

2

f i3 .2

c

D

C VT T

l cρΔ

= +

CALCULATE: (a) ( )( ) ( )8 910 2.44 10 m 15 μF / 2.5 mm 1.464 10 st − −= ⋅ Ω = ⋅

(b) ( )( )

( )( ) ( )( )2

f 34 3

15 μF 100.0 V20.0 C 21.928 C

2 1.93 10 kg/m 2.5 mm 129 J/ kg CT = + ° = °

⋅ °

ROUND: (a) 1.5 nst = (b) f 22 CT = °


1010

DOUBLE-CHECK: (a) The calculated value has appropriate units for time, and the magnitude of the value is reasonable for a discharge time. (b) The temperature of the gold cube does not change appreciable, which would be desirable for real circuits, so it makes sense.

26.53. THINK: Consider any given rung on the ladder to have a total equivalent capacitance of 0 .C Next, determine the equivalent capacitance, 1 ,C of 0C with the next rung and two legs. Since the ladder is infinite, it should not matter where on the ladder the analysis is performed. If 0C is the equivalent capacitance of all the capacitors beyond some point, then adding another set of capacitors to the mix should not affect anything and 1C should equal 0 ,C giving a recursive relation in C and thus the total equivalent capacitance, in terms of C, can be determined. SKETCH:

RESEARCH: Capacitors add in series as 1 1 1eq 1 2 .C C C− − −= + Capacitors in parallel add as eq 1 2 .C C C= +

SIMPLIFY: 0C is parallel to C, which gives 0 .C C C′ = + C′ is in series with 2 C’s, which gives:

1 0 0

1 2 1 1 .C C C C C

= + =+

Therefore, ( ) ( ) 2 200 0 0 0 0 0 0 0 0

0

2 2 2 2CC

C C C C C CC C C C C C C CC C CCC C

+ = + + = + + + = ++

2 20 0 2 2 0.C CC C + − =

CALCULATE: Using the quadratic equation: 2 2

02 4 8 12 1 3 .

4 2 4 2C C C CC C C

− ± + − ±= = − ± =

03 1 ,2

C C −=

since 0C must be positive.

ROUND: Not necessary. DOUBLE-CHECK: Consider the first rung of three capacitors in series. The equivalent of these is C/3. Adding another rung of three capacitors puts one capacitor in parallel with C/3 and then two capacitors in series with this to get:

43 3C CC + = and then

12 3 4 .4 11

CC C

− + =


1011

Adding another rung performs the same operation as before to get:

4 1511 11

C C C+ = and then 12 41 56 .

56 153C

C C

− + =

Continuing on gives: 15 5641 41

C C C+ = and then 12 153 209 ,

209 571C

C C

− + =

209 780571 571

C C C+ = and then

12 571 0.36602534 ,780

CC C

− + =

0.36602534 1.36602534C C C+ = and therefore

12 1 0.366025399 .1.36602534

CC C

− + =

The series converges around 0 0.366025 .C C= The solution is

( )0 3 1/ 2 0.366025 ,C C C= − = which is the same as the above result. Therefore, by continuously adding

rungs to the ladder, it converges to the previous result.

Additional Problems

26.54. (a) If the switch is closed for a long time, the capacitor is fully charged and there is no current through that branch. Therefore, the current through The 4.0 Ω resistor is 0 A.i = (b) With no current through 2 ,R the potential drop across it is 2 0 V.VΔ = The two resistors, 1 6.0 R = Ω and 3 8.0 ,R = Ω are in series with each other, so the current through them is

( ) ( ) ( )1 3/ / 0.714 A.10.0 V 14.0 i V R R= Δ + = =Ω The potential drop across the 6.0 Ω resistor is

( )( )1 1 0.714 A 6.0 4.286 V,V iRΔ = = Ω = and across the 8.0 Ω resistor is 3 3V iRΔ =

( )( )0.714 A 8.0 5.714 V.= Ω = Therefore, 1 4.3 V,VΔ = 2 0.0 VVΔ = and 3 5.7 V.VΔ = (c) The potential on the capacitor is the same as the potential drop across the 8.0 Ω resistor since they are parallel, so 3 5.7 V.CV VΔ = Δ =

26.55. (a) The maximum current through the ammeter is A 1.5 mA.i = The ammeter has resistance = Ω1 75 .R The current through a resistor is given by = / ,i V R where V is the potential difference across the resistor. Since current flows through the path of least resistance, when a shunt resistor of small resistance shuntR is connected in parallel with the ammeter, most of the current flows through the shunt resistor. The shunt resistor carries most of the load so that the ammeter is not damaged.

From Kirchoff’s rules = +shunt Ai i i and =shunt shunt A 1.i R i R Therefore,

= + = +

A 1 1

A Ashunt shunt

1 .i R R

i i iR R

For known current Ai (measured by ammeter) and known resistances 1R and shunt ,R the new maximum current i can be calculated. Note that A .i i> A shunt resistor is added in parallel with an ammeter so the current can be increased without damaging the ammeter.


1012

(b) From Kirchoff’s rules shown above, ( )( ) −Ω

= = = = ⋅ Ω = Ω− −

3A 1 A 1shunt

shunt A

1.5 mA 75 7.5 10 7.5 m .

15 A 1.5 mAi R i R

Ri i i

26.56. The potential on the capacitor, 150. μF,C = when it is fully charged is 200. V.VΔ = The potential

decreases exponentially as it discharges through 1.00 M ,R = Ω by ( ) / .t RCV t Ve−Δ = When

( ) 50.0 V,V tΔ = ( ) ( ) 150.0 V 200. V .4

t tRC RCV t e e

− −Δ = = = Therefore, the result is

( )ln 4.00 207.94 s 208 st RC= = = or 3.47 min.

26.57. The capacitor, ,C discharges through the bulb, f 2.5 k ,R = Ω in 0.20 ms.DtΔ = The charging time is 0.80 ms.CtΔ = For simplicity assume the charging and discharging time are the time constants of the

circuits. Therefore, τ −ΔΔ = = = = = ⋅ =

Ω8

ff

0.20 ms 8.0 10 F 80. nF,2.5 k

dd d

tt R C C

R and

τΔ

Δ = = = = = Ω0.80 ms 10. k .C 80. nF

CC C

tt RC R

26.58. The potential, emf ,V of the battery is the same with ammeter, 0 53 ,R = Ω as without. The external resistance 1130 ,R = Ω has a current of 5.25 mAI = with ammeter, so by Ohm’s law

( ) ( ) ( )( )0emf 0

5.25 mA 53 1130 5.4962 mA 5.50 mA.

1130 i R R

V i R R i R iR+ Ω + Ω

′ ′= + = = = = =Ω

26.59. THINK: When the switch is set to X for a long time, the capacitor, 10.0 μF,C = charges fully so that it has the same potential as the battery, C emf 9.00 V.V VΔ = = After placing the switch on Y, the capacitor discharges through resistor 2 40.0 R = Ω and decreases exponentially for both immediately ( )0 st = and

1.00 mst = after the switch. SKETCH:

RESEARCH: By Ohm’s law, the current initially is 0 2/ ,Ci V R= Δ where emf .CV VΔ = Current decays

exponentially as ( ) /0 ,ti t i e τ−= where .RCτ =

SIMPLIFY: (a) Initial current is 0 2 emf 2/ / .Ci V R V R= Δ =

(b) After 1 ms,t = current is ( ) /0 0 .

tt RCi t I e i eτ

−−= =

CALCULATE:

(a) 09.00 V 0.225 A40.0

i = =Ω


1013

(b) ( ) ( ) ( )( )1.00 ms

40.0 10.0 μF1.00ms 0.225 A 0.01847 Ai e−

Ω= =

ROUND: (a) 0 225 mAi = (b) ( )1.00 ms 18.5 mAi = DOUBLE-CHECK: After only 1.00 ms, the current decreases by almost 90.0%, which would make this a desirable circuit, so it makes sense.

26.60. THINK: Since the two resistors, 2.2 k ,R = Ω the two capacitors, 3.8 μF,C = and the battery,

emf 12.0 V,V = are all in series, the order doesn’t matter, so equivalent resistance and capacitance are used to determine the time constant, .τ The current then decreases exponentially from it’s initial current to

=f 1.50 mAi in time .t SKETCH:

RESEARCH: The equivalent resistance is 2 ,eqR R R R= + = and the equivalent capacitance is

( ) 11/C+1/ /2.eqC C C−= = Initial potential in capacitor, emf .CV VΔ = By Ohm’s law, the initial current is

0 / .C eqi V R= Δ

SIMPLIFY: The current at time t is ( ) ( )( )/ C C C/220 .

2 2eq eq

t t tR Ct CR RC

eq

V V Vi t i e e e e

R R Rτ

− − −− Δ Δ Δ

= = = =

( ) Cf f

2

tRCV

i t i i eR

−Δ= = f

C

2

tRCRi

eV

−

=Δ

f

C

2 ln

Rit RC

V

= − Δ

CALCULATE: ( )( ) ( )( )2 2.2 k 1.5 mA2.2 k 3.8 μF ln 0.004998 s

12.0 Vt

Ω= − Ω =

ROUND: 5.0 mst = DOUBLE-CHECK: The initial value of the current was about 2.7 mA. The circuit decays to about half its original current in roughly 5 ms, which makes this a desirable circuit, so it makes sense.

26.61. The charge on the capacitor increases exponentially with a time constant, 3.1 s.τ = Since the amount of energy in the capacitor is proportional to the square of the charge, the energy also increases exponentially.

The charge on the capacitor is given by ( ) ( )/0 1 .tq t q e τ−= − The energy on the capacitor is given by

( ) ( )2 / 2 .E t q t C= The time to get to half of the maximum energy is given by ( ) max / 2,E t E= where 2

max 0 / 2 .E q C= This gives:


1014

( ) ( ) ( )22 /220 /0

max

11 1 1 2 4 2 2 2

1 1 ln 1 (3.1 s)ln 1 3.8 s.2 2

tt

q eq tqE t E e

C C C

t

ττ

τ

−−

−= = = = − =

= − − = − − =

26.62. THINK: When the switch is closed for a long time, the capacitors, 1 1.00 μFC = and 2 2.00 μF,C = are fully charged so no current flows through them, and thus the current only flows through the two resistors,

1 1.00 kR = Ω and 2 2.00 k ,R = Ω driven by a battery emf 10.0 V.V = At this point, the potential drop across each resistor is equal to the potential on its complementary capacitor. Since the capacitors are in series, they have the same charge on them. SKETCH: (a) (b)

RESEARCH: The current, by Ohm’s law is found in both cases as ( )emf 1 2/ .i V R R= + When the switch is closed, the potential drop across capacitor jC is / (for 1,2).j j j jV Q C IR jΔ = = = When switch is open,

charge on each plate is emf .eqQ C V=

SIMPLIFY: (a) The charges on the capacitor are given by: ( )emf 1 2/ / .j j j j j j j jQ C iR Q iR C V R C R R= = = Δ +

( )1 emf 1 1 1 2/Q V R C R R= + and ( )2 emf 2 2 1 2/ .Q V R C R R= +

(b) The charge on each capacitor is ( ) ( )1emf 1 2 emf 1 2 1 2 emf1/ 1/ / .eqQ C V C C V C C C C V−= = + = +

CALCULATE:

(a) ( )( )( ) 6

1

10.0 V 1.00 k 1.00 μF3.33 10 C

1.00 k 2.00 kQ −Ω

= = ⋅Ω + Ω

and ( )( )( ) 5

2

10.0 V 2.00 k 2.00 μF1.33 10 C

1.00 k 2.00 kQ −Ω

= = ⋅Ω + Ω

(b) ( )( )( ) 610.0 V 1.00 μF 2.00 μF

6.67 10 C1.00 μF 2.00 μF

Q −= = ⋅+

ROUND: (a) 1 3.33 μC,Q = and 2 13.3 μCQ = (b) 6.67 μCQ = DOUBLE-CHECK: More charge builds up when the capacitors have their own resistor than when they are paired together. This is because even though the potential drop across them is the same in both cases, when the switch is open, the overall capacitance of the circuit is less than the sum of two. So a smaller C gives a smaller ,Q so it makes sense.

26.63. THINK: From Kirchoff’s rules, an equation can be obtained for the sum of the three currents, 1 ,i 2i and

3 ,i and two equations for the two inner loops of the circuit. This will yield 3 equations for 3 unknowns


1015

(the currents) and can be solved by simple substitution. Once the currents are known, the power over each resistor is found via Ohm’s law. 1 10.0 ,R = Ω 2 20.0 ,R = Ω 3 30.0 ,R = Ω 1 15.0 VV = and 2 9.00 V.V = SKETCH:

RESEARCH: Looking at point A, the 3 currents all flow into it, so 1 2 3 0.i i i+ + = Going clockwise in each loop (upper and lower) yields 2 equations 1 1 2 2 2 0i R i R V− + − = and 2 2 2 1 3 3 0.V i R V i R− + + = The power

across a resistor is 2 .P i R= SIMPLIFY: Since all resistances are in ohms and all voltages are in volts the equations can be rewritten for simplicity with only the magnitudes of the values, knowing the final answer will be in amperes.

1 2 3 1 2 30 ,i i i i i i+ + = = − − and 1 1 2 2 2 0i R i R V− + − = 1 210.0 20.0 9.00 0i i − + − = which implies

( ) ( )2 3 2 2 3 2 310.0 20.0 9.00 30.0 10.0 9.00 9.00 10.0 / 30.0 .i i i i i i i− − − + = + = = − Therefore,

2 2 2 1 3 3 0V i R V i R− + + =

2 3

2 3

33

3 3

3

3

9.00 20.0 15.0 30.0 24.0 20.0 30.0

9.00 10.020.0 30.0 24.0

30.020.06.00 30.0 24.0

3236 18 3

27.055.0

i ii i

ii

i i

i

i

− + += −

−− =

− − =

= −

= −

Therefore, ( )

2

9.00 10.0 27.0 / 55.0 51.0 ,30.0 110.

i− −

= = and 151.0 27.0 3.00 .110. 55.0 110.

i = − − = The power across each is

then 21 1 1 ,P i R= 2

2 2 2P i R= and 23 3 3 .P i R=

CALCULATE: ( )2

13.00 A 10.0 0.00744 W,110.

P = Ω =

( )2

251.0 A 20.0 4.299 W110.

P = Ω =

and

( )2

327.0 A 30.0 7.230 W.55.0

P = − Ω =

ROUND: 1 7.44 mW,P = 2 4.30 WP = and 3 7.23 W.P = DOUBLE-CHECK: Looking back at the values for current, it is found that

1 2 33.00 51.0 27.0A A A 0,110. 110. 55.0

i i i+ + = + − =


1016

which is what would be expected. Going from left to right on each branch gives

1 13.00 V,11.0

i R− = − 2 2 23.00 V11.0

V i R− = − and 3 3 13.00 V.11.0

i R V− − = −

So the potential drop across each branch in parallel is the same, so the answers make sense.

26.64. THINK: From Kirchhoff’s rules, an equation can be obtained for the sum of the three currents, 1 ,i 2i and

3 ,i and two equations can be obtained for the two inner loops of the circuit. This will yield 3 equations for 3 unknowns (the currents) and can be solved by substitution. Once the currents are known, the voltage drop over resistor 2 is found via Ohm’s law. 1 30.0 ,R = Ω 2 40.0 ,R = Ω 3 20.0 ,R = Ω enf,1 12.0 VV = and

emf,2 16.0 V.V = SKETCH:

RESEARCH: By the choice of directions of currents, at point A, the currents sum as 2 1 3 0.i i i− − = Going clockwise in the upper and lower loops gives 2 equations: 1 1 1 2 2 0V i R i R− + + = and

2 2 3 3 2 0.i R i R V− − − = Potential drop across resistor 2 is 2 2 .V i RΔ = SIMPLIFY: Since all resistances are in Ω and all voltages are in V, I can rewrite the equations for simplicity with only the magnitudes of the values, knowing out final currents are in A. It follows that

2 1 3 2 1 30 ,i i i i i i− − = = + and therefore, 21 2 1

12.0 40.012.0 30.0 40.0 ,

30.0ii i i −

= + =

22 3 3

16.0 40.016.0 40.0 20.0 ,

20.0ii i i +

= − − = − and

2 22 1 3 2 2 2

2 2

12.0 40.0 16.0 40.0 12.0 16.0 40.0 40.0 2 10 30.0 20.0 30.0 20.0 30.0 20.0 5 3

13 2 6 .3 5 65

i ii i i i i i

i i

− += + = − = − − − = − −

= − = −

The potential drop across it is 2 2 .V i RΔ =

CALCULATE: ( )6 A 40.0 3.6923 V65

V Δ = Ω =

ROUND: 3.69 VVΔ = DOUBLE-CHECK: Going back to equation for 1i and 3 ,i I get 1 34 / 65 Ai = and 3 8 /13 A,i = − where

1 3 26 / 65 A ,i i i+ = − = so the currents are consistent. Calculating the potential drop across the upper and lower branches gives 1 1 1 3.6923 VV i R− = and 2 3 3 3.6923 V,V i R+ = so each branch has the same potential drop across it, so it makes sense.


1017

26.65. (a) From equation 24.10 of the textbook, the capacitance of a spherical capacitor is ( )04 / ,C ab b aπε= − where 1.10 cmb = and 1.00 cm.a = Since it is connected in series with resistor 10.0 MR = Ω and emf voltage, emf 10.0 V,V = the time constant is

( )( )( )( )( )

( )12 2 2

404 8.8542 10 N m /C 10.0 M 1.10 cm 1.00 cm4

1.22 10 s.1.10 cm 1.00 cm

RbaRC

b a

ππετ−

−⋅ Ω

= = = = ⋅− −

(b) The charge on the capacitor still grows exponentially: ( ) ( )/0 1 ,tq t q e τ−= − where 0 emf .q CV=

Therefore,

( ) ( )( )

( )( )( )( )( )

( )( )4

0.1 ms0 emf

12 2 2 0.100 ms1.22 10 s

40.100 ms 1

4 8.8542 10 N m /C 10.0 V 1.10 cm 1.00 cm1

1.10 cm 1.00 cm

68.5 μC

V baq e

b a

e

τπε

π−

−

−−

⋅

= − −

⋅ = − −

=

26.66. THINK: Since the circuit has three branches, four equations (one for each branch and the equation for the current at a junction) can be written down simply by inspection. However, a deeper analysis is required to fully understand the evolution of the circuit. For example, as the capacitor, 30.0 μF,C = charges, the current through it, ( )1 ,i t starts at some maximum and decays to zero. When this happens, the

other currents, ( )2i t and ( )3i t must become equal. Even though there are two branches with batteries,

emf,2 80.0 VV = and emf,3 80.0 V,V = and resistors, A 40.0 R = Ω and B 1.0 ,R = Ω and C 20.0 R = Ω and

D 1.0 ,R = Ω respectively, the capacitor effectively sees two resistors in parallel to charge through. All three currents and the potential across each branch are time dependent. SKETCH:

RESEARCH: Starting at junction X and going clockwise around the three loops gives three equations for the potential along them. Also, junction X gives an equation relating the currents.

( ) ( )C 3 3 emf,3 0V t R i t V−Δ + − = (1)

( ) ( )C emf,2 2 2 0V t V R i t−Δ + − = (2)

( ) ( )2 2 emf,2 3 3 emf,3 0R i t V R i t V− + − = (3)

( ) ( ) ( )2 1 3i t i t i t= + (4)

where 2 A BR R R= + and 3 C D .R R R= + The charge on the capacitor is ( ) ( )/max 1 ,tQ t Q e τ−= − where

e .qR Cτ = The equivalent resistor is 2R and 3R in parallel. The current through the capacitor decays as

( ) ( ) /1 1 0 ,ti t i e τ−= where ( )1 0i is the initial current through the capacitor.


1018

SIMPLIFY: The time constant is given by: 1

2 3e

2 3 2 3

1 1 .qCR R

R C CR R R R

τ−

= = + = +

Consider the voltage drop

from junction X to Y. Along each branch it must be equal, so for the bottom two branches: ( ) ( ) ( ) ( )2 2 emf,2 emf,3 3 3 emf,3 emf,2 2 2 3 3 .R i t V V R i t V V R i t R i t− = − + = + Using equation (4), and substituting

in for ( )2i t and ( )3i t gives:

( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )emf,2 emf,3 2 1emf,2 emf,3 2 1 3 3 3 2 1 2 3 3 3

2 3

,V V R i t

V V R i t i t R i t R i t R R i t i tR R

+ − + = + + = + + = +

( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )emf,2 emf,3 3 1emf,2 emf,3 2 2 3 2 1 3 1 2 3 2 2

2 3

.V V R i t

V V R i t R i t i t R i t R R i t i tR R

+ + + = + − = − + + = +

When ,t → ∞ ( )1 0,i t → so the steady state current is given by: ( ) ( ) emf,2 emf,32 3 s

2 3

lim lim .t t

V Vi t i t i

R R→∞ →∞

+= = =

+

At all times, the voltage drop from X to Y is the same along any branch. Now that the steady state current is reached, the voltage drop across the capacitor can be determined and thus the maximum charge and initial current on the capacitor. Compare ( )CV tΔ = ∞ to both branches:

( ) ( )C C,max 3 3 emf,3 3 s emf,3V t V R i t V R i VΔ = ∞ = Δ = = ∞ − = − or

( )C,max emf,2 2 2 emf,2 2 s .V V R i t V R iΔ = − = ∞ = −

emf,2 emf,3 emf,2 emf,3C,max 3 emf,3 emf,2 2

2 3 2 3

emf,2 emf,3 emf,2 emf,32 3 2 33 emf,3 emf,2 2

2 3 2 3 2 3 2 3

3 emf,2 2 emf,3

2

V V V VV R V V R

R R R RV V V VR R R R

R V V RR R R R R R R R

R V R VR

+ + Δ = − = − + +

+ + + + − = − + + + +

−

+3 emf,2 2 emf,3

3 2 3

.R V R V

R R R−

=+

The maximum charge is given by: ( )3 emf,2 2 emf,3

max C,max2 3

.C R V R V

Q C VR R

−= Δ =

+ In general, ( )Q t is related to

( )1i t by: ( ) ( ) ( )/ / /max1 max max

11 0 .t t tdQ t Qdi t Q e Q e edt dt

τ τ τ

τ τ− − − = = − = − − =

( ) ( )3 emf,2 2 emf,3 emf,2 emf,3max 2 31

2 3 2 3 2 3

0C R V R V V VQ R R

iR R CR R R Rτ

− += = = − +

Now that ( )1 0i is determined, ( )2i t and ( )3i t can be expressed in simpler terms:

( ) ( ) ( )

( )

/3 emf,2 2 emf,3 3emf,2 emf,3 3 1 emf,2 emf,3

22 3 2 3 2 3

3 2 emf,2 emf,3 /s

2 3

/ /

/.

t

t

R V R V R eV V R i t V Vi t

R R R R R RR R V V

i eR R

τ

τ

−

−

−+ + += = +

+ + + −

= + +

( ) ( ) ( )

( )

/2 emf,2 2 emf,3 3emf,2 emf,3 2 1 emf,2 emf,3

32 3 2 3 2 3

2 3 emf,3 emf,2 /s

2 3

/ /

/

t

t

R V R V R eV V R i t V Vi t

R R R R R RR R V V

i eR R

τ

τ

−

−

−+ − += = −

+ + + −

= + +

The potential across the capacitor for any given times is then given by:

( ) ( ) ( ) ( )3 emf,2 2 emf,3/ /maxC

2 3

1 1 .t tQ t R V R VQV t e e

C C R Rτ τ− −−

Δ = = − = −+


1019

Therefore, in addition to the previous four equations, there are six additional ones. 2 3

2 3

CR RR R

τ =+

(5)

emf,2 emf,3s

2 3

V Vi

R R+

=+

(6)

( ) emf,2 emf,3 /1

2 3

tV Vi t e

R Rτ−

= −

(7)

( ) ( )3 2 emf,2 emf,3 /2 s

2 3

/ tR R V Vi t i e

R Rτ−

−= +

+ (8)

( ) ( )2 3 emf,3 emf,2 /3 s

2 3

/ tR R V Vi t i e

R Rτ−

−= +

+ (9)

( ) ( )3 emf,2 2 emf,3 /C

2 3

1 tR V R VV t e

R Rτ−−

Δ = −+

(10)

CALCULATE: 1 40.0 1.0 41.0 ,R = Ω + Ω = Ω 2 20.0 1.0 21.0 R = Ω + Ω = Ω

( )( )( ) ( )1

11 4 130.0 F 41.0 21.0 4.16613 10 s 2400.31 s

41.0 21.0 μ

τ−

−− − − Ω Ω

= = ⋅ = Ω + Ω

s80.0 V 80.0 V 2.580645161 A,41.0 21.0

i += =Ω + Ω

emf,2 emf,3

2 3

80.0 V 80.0 V 1.858304297 A41.0 21.0

V VR R

− = − = −Ω Ω

( ) ( )3 2 emf,2 emf,3

2 3

/ 21.0 / 41.0 80.0 V 80.0 V0.6294256 A

21.0 41.0 R R V V

R R− Ω Ω −

= = −+ Ω + Ω

( ) ( )2 3 emf,3 emf,2

2 3

/ 41.0 / 21.0 80.0 V 80.0 V1.228878648 A

21.0 41.0 R R V V

R R− Ω Ω −

= =+ Ω + Ω

( ) ( )3 emf,2 2 emf,3

2 3

21.0 80.0 V 41.0 80.0 V25.8069516 V

21.0 41.0 R V R V

R RΩ − Ω−

= = −+ Ω + Ω

ROUND: The initial four equations, rounded to three significant figures are:

(1) ( ) ( )12400 s79.98 V 0.03 V 80.0 V 0,te−−

+ − = (2) ( ) ( )12400 s80.0 V 79.98 V 0.011 V 0,te−− − + =

(3) ( ) ( )12400 s159.96 V 0.041 V 160.0 V 0te−−

+ − = and

( ) ( ) ( ) ( ) ( ) ( )1 1 12400 s 2400 s 2400 s0.629 A 1.86 A 1.23 A .t t te e e− − −− − −

− = − + DOUBLE-CHECK: The initial four equations within rounding are still valid, so the values of the coefficients are correct. Checking ( )2 0i and ( )3 0i using equations (8) and (9) gives:

( ) ( ) ( )( )

emf,2 emf,3 3 emf,2 2 emf,3 3 emf,2 3 2 emf,22

2 3 22 3 2

/ / 1 /0

1 /V V R V R V R V R R V

iR R RR R R

+ + − += = =

+ + and

( ) ( ) ( )( )

emf,2 3 2 emf,2 2 emf,3 3 emf,3 2 3 emf,33

2 3 33 2 3

/ / 1 /0 .

1 /V V R V R V R V R R V

iR R RR R R

+ Δ − − += = =

+ +

These results satisfy ( ) ( ) ( ) ( ) ( )1 2 3 emf,2 2 emf,3 30 0 0 / / .i i i V R V R= − = − Also, consider that initially the emf,2V

battery “sees” only 2R first (likewise for battery emf,3V and 3 ),R so the initial current is simply emf,2 2/V R (or emf,3 3/ ),V R so the equations for the currents make sense.


1020

26.67. THINK: The capacitor of capacitance is 10.0 μF,C = is charged through a resistor of resistance 10.0 ,R = Ω with a battery, emf 10.0 V.V = It is discharged through a resistor, 1.00 .R ′ = Ω For either

charging or discharging, it takes the same number of time constants to get to half of the maximum value. The energy on the capacitor is proportional to the square of the charge. SKETCH:

RESEARCH: The capacitor’s charge is given by ( ) ( )/

0 1 .tq t q e τ−= − In general, the energy on the

capacitor is given by ( ) ( )2 / 2 .E t q t C= The time constant is either RCτ = or .R Cτ ′ ′= SIMPLIFY:

(a) When ( ) 0 / 2,q t q= then: ( ) ( )/ /0 0

1 1 11 1 ln ln2.2 2 2

t tq t q q e e tτ τ τ τ− − = = − = − = − =

(b) If ( ) 01 ,2

q t q= the energy is: ( ) ( ) ( )22 20 0

max

/ 2 1 1 .2 2 4 2 4

q t q qE t E

C C C

= = = =

(c) The time constant for discharging is .R Cτ ′ ′=

(d) The capacitor discharges to half the original charge in ( )ln 2 .t τ ′= CALCULATE: (a) ( )ln2.00, or 0.693t τ τ= (b) 1.00:4.00. (c) ( )( )1.00 10.0 μF 10.0 μsτ ′ = Ω =

(d) ( ) ( )10.0 μs ln 2.00 6.93 μst = = ROUND: (a) 0.693τ (b) 1.00:4.00. (c) 10.0 μsτ ′ = (d) 6.93 μst =

DOUBLE-CHECK: In general, charge decreases exponentially as ( ) /0 .tq t q e τ−= For 10 μsτ ′ = and

6.93 μs,t = the charge is ( ) 6.93/100 06.93 μs 0.497 ,q q e q−= = which is about half the original charge.

26.68. THINK: From Kirchhoff’s rules, an equation can be obtained for the sum of the three currents, 1 ,i 2i and

3 ,i and two equations can be obtained for the two inner loops of the circuit. This will yield 3 equations for 3 unknowns (the currents) and can be solved by simple substitution. Once the currents are known, the voltage drop over resistor 2 is found via Ohm’s law. 1 3.00 ,R = Ω 2 2.00 ,R = Ω 3 5.00 ,R = Ω emf,1 10.0 VV = and emf,2 6.00V.V =


1021

SKETCH:

RESEARCH: By the choice of directions of currents, at point A, the currents sum as 1 2 3 0.i i i− − = Going clockwise in the upper and lower loops gives 2 equations: 1 1 emf,1 3 3 0i R V i R− + − = and

3 3 2 2 emf,2 0.i R i R V− + = Potential drop across resistor 2 is 2 2 .V i RΔ = The power across the third resistor is 23 3 .P i R=

SIMPLIFY: From equation of currents: 1 2 3 3 1 2 .i i i i i i= + = − From the upper loop:

( ) ( )emf,1 3 3 1 1 1 emf,1 1 3 3 1 / / .V i R i R i V R i R R− = = − From the lower loop:

( ) ( )emf,2 3 3 2 2 2 emf,2 2 3 3 2 / / .V i R i R i V R i R R+ = = + Therefore,

emf,1 emf,2 emf,1 emf,23 3 3 3 33 1 2 3 3 3 3

1 1 2 2 3 1 2 1 21

emf,1 emf,2emf,1 emf,23 3 3 3 3 33 3

1 23 1 2 1 2 3 1 2

emf,1 e3

1 3

V V V VR R R R Ri i i i i i i

R R R R R R R R R

V VV VR R R R R Ri i

R RR R R R R R R R

V Vi

R R

−

= − = − − − = − − +

− + + = − = + +

= −

1mf,2

2 3 3 1 2

1 1 1R R R R R

− + +

CALCULATE:

(a) ( )( ) ( )( )

1

310.0 V 6.00 V 1 1 1 0.06452 A

5.00 3.00 2.00 3.00 5.00 2.00 5.00 i

− = − =+ + Ω Ω Ω Ω Ω Ω Ω

(b) ( ) ( )20.06452 A 5.00 0.02081 WP = Ω = ROUND: (a) 64.5 mAi = (b) 20.8 mWP = DOUBLE-CHECK: Going back to equation for 1i and 2 ,i the currents can be calculated as 1 3.2258 Ai = and 2 3.1613 A.i = Their difference is 1 2 0.0045 A,i i− = which is also 3 ;i therefore, the current is correct and it makes sense.

26.69. THINK: From Kirchhoff’s rules, an equation can be obtained for the sum of the five currents, 1 ,i 2 ,i 3 ,i

4i and 5 ,i and two equations can be obtained for the two inner loops of the circuit. Since the voltage drop across resistor 3 is zero, the current through that branch is also zero. Ohm’s law allows an equation for the ratios of the resistors. Once 2R is known, the current through it is obtained by equation the potential drop across both 1R and 2R is equal to the emf voltage. 1 8.00 ,R = Ω 4 2.00 ,R = Ω 5 6.00 R = Ω and


1022

emf 15.0 V.V = SKETCH:

RESEARCH: By the choice of directions of currents, two equations arise 1 2 3i i i= + and 5 3 4 .i i i= + Since the current through 3R is zero, 3 3 3 0.V i RΔ = = Then, two sets of potential drops are equal: 1 1 4 4i R i R= and

2 2 5 5 .i R i R= The potential across 1R and 2R is emf 1 1 2 2 .V i R i R= + SIMPLIFY: Since 3i is zero, the current becomes 1 2i i= and 4 5 .i i= Dividing the potential drops across each resistors yields

1 51 1 4 4 1 42

2 2 5 5 2 5 4

.R Ri R i R R R R

i R i R R R R= = =

The current through it is 2 1 ;i i i= = therefore, ( ) ( )emf 1 1 2 1 emf 1 2 / .V i R R i V R R= + = +

CALCULATE: ( )( )

2

8.00 6.00 24.0 ,

2.00 R

Ω Ω= = Ω

Ω 1215.0 V 0.46875 A

8.00 240 i i= = =

Ω + Ω

ROUND: 2 24.0 ,R = Ω 2 469 mAi = DOUBLE-CHECK: Solving for ( )4 5 emp u s/ 1.875 Ai i V R R= = + = for a total current of T 2.34375 Ai =

coming out of the battery. Since there is no current in 3 ,R the circuit is just 1R and 2R in parallel with

4R and 5R to gives ( ) ( ) 1

eq 1 2 4 51/ 1/ ,R R R R R−

= + + + and produces a current of

emp eq T/ 2.34375 ,i V R i= Δ = = so it makes sense.

26.70. THINK: From Kirchhoff’s rules, equations can be obtained for the sum of the five currents, 1 ,i 2 ,i 3 ,i 4i and 5 ,i and three equations for the three inner loops of the circuit. This will yield 5 equations and 4 unknowns (the currents) and can be solved by simple substitution. 1 1.00 ,R = Ω 2 2.00 ,R = Ω

3 3.00 ,R = Ω 4 4.00 ,R = Ω 5 5.00 ,R = Ω emf,1 12.0 VV = and emf,2 6.00 V.V = SKETCH:


1023

RESEARCH: By the choice of directions of currents, at point A, 1 2 3 4 0,i i i i− − − = and at point B,

2 4 5 0.i i i+ − = By going clockwise in each loop yields 3 equations: emf,1 1 1 3 3 0,V i R i R− − =

2 2 emf,2 4 4 0i R V i R− − − = and 4 4 emf,2 5 5 3 3 0.i R V i R i R− + − + = Potential drop across resistor 2 is 2 2 .V i RΔ =

The power across the third resistor is 23 3 .P i R=

SIMPLIFY: (a) Using the equation 5 2 4 ,i i i= + the other 4 can be simplified to: 1) 1 2 3 4 0i i i i− − − = ; 2)

1 1 3 3 emf,1R i R i V+ = ; 3) 2 2 4 4 emf,2R i R i V− + = ; 4) ( )5 2 3 3 4 5 4 emf,2 .R i R i R R i V− + + = (b) Since the resistance are in Ω and all voltages are in V, the equations can be rewritten for simplicity with only the magnitude of the values, knowing that the final currents are in A, which yields: 1)

1 2 3 4 0i i i i− − − = ; 2) 1 33 12.0i i+ = ; 3) 2 42 4 6.00i i− + = ; 4) 2 3 45 3 9 6.00.i i i− + =

( ) ( ) ( )( ) ( ) ( )

( ) ( ) ( )( )( ) ( ) ( ) ( )

2 3 4

3 4

3 4

4

2 1 : 4 12.0 22 2 3 : 8 6 30.0 3

5 2 4 : 23 4 54.0 423 / 8 2 4 : 85.0 / 4 32.25 4 .

i i i AA i i A

A i i AA A i B

− + + = ≡+ + = ≡

− + − + = − ≡+ = ≡

Therefore, ( ) 44 129 / 85.0,B i = ( ) ( )3 43 30.0 6 / 8 222 / 85.0,A i I = − =

( ) 2 3 42 12.0 4 3.00 / 85.0,A i i i = − − = and ( ) 1 2 3 41 354 / 85.0.i i i i = + + =

CALCULATE: 4129 A 1.5176 A85.0

i = =

ROUND: 4 1.52 Ai = DOUBLE-CHECK: If these currents are used to calculate the potential drop from ,A B→ you get

3 3 5 5 0.071 V,i R i R− + = 4 4 2 0.071 V,i R V− + Δ = and 2 2 0.071 V,i R− = so the potential drops are all the same, so it makes sense.

26.71. THINK: Consider any square on the grid to have been reduced so that every side has a capacitance, ,C′ which is the equivalent of all the capacitors above, below and along each side. Since the grid is infinite, then no side has more capacitors than any other, so all four are reduced to the same capacitance. Next, consider the same analysis except for only three sides, so that one side is still of capacitance, C, while the others are .C′ When those four sides are reduced to one equivalent capacitance, the result should be equal to the original value of .C′ This is because the grid is infinite and adding an extra square to the already reduced side should affect nothing, resulting in the same capacitance, giving a recursive relation in C, and thus the total equivalent capacitance, in terms of C, can be determined. SKETCH:


1024

RESEARCH: Capacitors in series add as 1 1 1eq 1 2 .C C C− − −= + Capacitors in parallel add as eq 1 2 .C C C= +

SIMPLIFY: When all four sides are reduced to ,C′ the equivalent capacitance (across A to B) is: 11 1 1 4 .

3CC C

C C C

− ′ ′′ ′= + + + = ′ ′ ′

Looking at when one side is reduced using the other three reduced gives 0C as: 1

01 1 1 .

3CC C C

C C C

− ′ = + + + = + ′ ′ ′

Since 0 :C C′= 2 3 3

CC C C C′′ ′= + = and 3 .

2C C′ = Therefore, the total equivalent capacitance is:

4 4 3 4 2 .3 3 2 2CC C C C

′ ′′ = = = =

CALCULATE: Not applicable. ROUND: Not applicable. DOUBLE-CHECK: Consider an intersection on the grid. If a voltage was applied to this point, it would see equal capacitance (since it is infinite) in all four directions, meaning it would contribute an equal charge, q, to each direction. If the same voltage with opposite polarity was applied to any adjacent intersection, it would see a –q along each direction. This means the capacitor that joins the two intersections is actually double the charge on one, meaning the potential sees an effective capacitance twice the size of any one capacitor, so an equivalent capacitance of 2C is correct.

( )0 / 3 3 / 2 .C C C C C′ ′= + = Therefore the total equivalent capacitance is

( ) ( )( )4 / 3 4 / 3 3 / 2 2 .C C C C′′ ′= = =

Chapter 27: Magnetism

1025


In-Class Exercises

27.1. a 27.2. e 27.3. d 27.4. d 27.5. d Multiple Choice

27.1. b 27.2. c 27.3. e 27.4. e 27.5. a 27.6. a 27.7. a,c,d,e are true; b is false 27.8. b

Questions

27.9. (a)

(b)

(c)

(d)

27.10. Zero. The force acting on a charged particle in a magnetic field is .F qv B= × By definition of the cross-

product (and confirmed by experiment), this force is always perpendicular to the velocity of the particle at any point in the magnetic field. Thus, the work done by the magnetic field on the charged particle is zero. The effect of this force on the particle is that it changes the direction of the particle’s velocity, but not its magnitude. Hence, the uniform circular motion the particle has in the magnetic field (the cyclotron motion).


1026

27.11. A = Parabolic (electric field). B = Circular (magnetic field). The forces acting on a charged particle under either an electric field or a magnetic field is F qE=

or ,F qv B= ×

respectively.

27.12.

(a) The direction of the force acting on a charge moving in a magnetic field is given by the right-hand rule. If the fingers point in the direction of ,v then to produce a force in the negative x-direction, the magnetic field has to act out of the page, in positive z-direction. (b) Yes, it does change. For the negatively charged electron, the field must point into the page, in negative z-direction. The direction of the force depends on the charge.

27.13. A magnetic potential is used to represent magnetic fields in regions of zero current density in some applications, but the construction is not as useful as its electrical counterpart. This is because the electric potential represents a potential energy (per unit charge), which is part of a conserved total energy. It keeps track of the work done by the electric field on a charge moving in that field, and can thus be used to analyze the dynamics of charged particles. But the magnetic field never does any work on a charged particle, as the magnetic force is perpendicular to the particle’s velocity. There is no work for a magnetic scalar potential energy to track. It represents no contribution to a conserved total energy, and hence, does not enter into any dynamics. It is more useful in advanced treatments of electromagnetic theory to represent the magnetic field as the curl of a vector potential: ,B A= ∇×

for a suitable vector field, .A

27.14. This is possible if the direction of the current is parallel or anti-parallel to that of the magnetic field. In such a case, 0.dF idL B= × =

27.15. Yes, it is possible. In order for this to work, the force due to the electric field, ,F qE=

has to be perpendicular to the velocity vector at all times. One way to achieve this is to have a the electric field from a point particle, say a proton, for which the electric field points in radial outwards direction. An electron with suitable initial velocity can then make circular orbits around the proton. For these the speed does not change. If the electric field is replaced with a uniform magnetic field, the speed of a charged particle never changes. Note that in both cases described here, only the speed is constant, but the direction of the velocity vector changes. (In the case that the initial velocity vector is parallel or anti-parallel to the magnetic field even the direction stays constant.)

27.16. The charged particle will move in a helix around the magnetic field lines. Its motion in the z-direction is unaffected by the magnetic field, and therefore the time required involves determining the component of the initial velocity in the z-direction, which is simply v multiplied by the cosine of the angle. Thus, the time

required is ,cosz

z ztv v θΔ ΔΔ = = where zΔ is the extent of the region along the z-direction.


1027

27.17.

The magnetic force acts in a direction perpendicular to both the velocity and magnetic fields. Since these are both in the horizontal plane, the force acts into or out of the page. The right-hand rule shows that a positive charge experiences a net force outwards. Thus, for the negatively charged electron, the force is directed inwards.

27.18. Recall that a velocity selector works with perpendicular magnetic and electric fields. At the Earth’s surface, there is an approximately perpendicular relation between the electric and magnetic fields. Thus, on a line perpendicular to E and B, charged particles will travel without deflection if they have the correct velocity. This velocity has a magnitude E/B. It is known that the Earth’s magnetic field is approximately 0.3 gauss or

−⋅ 53 10 T. Therefore, the value of E/B at the Earth’s surface is of the order:

− = ⋅⋅

65

150 N 5 10 m/s.3 10 T

When pointed west, magnetically speaking the beam would be un-deflected. Once you are facing West, North is on your right.

27.19. A cyclotron has both electric and magnetic fields. It is the alternating electric field which does the work to

increase the particle’s kinetic energy. Although the magnetic field does not do any work (it does not change the particle’s kinetic energy), it nevertheless plays an important role in keeping the particle in a circular orbit. As the electric field accelerates the particle, the radius of the circular orbit increases so that the particle follows a spiral trajectory. The alternating electric field and the static uniform magnetic are crucial for the operation of the cyclotron as a particle accelerator.


1028

Problems

27.20.

B ,F qv B= × B sin ,F q vB θ= 90θ = ° and ,q e= so, B .F evB= Inserting the values gives:

( )( )( )19 5 14B 1.602 10 C 4.0 10 m/s 0.40 T 2.6 10 N.F − −= ⋅ ⋅ = ⋅

27.21. B ,F qv B= × B sin ,F q vB θ= 90θ = °, 2q e= − B B 2F F evB = = +

( )18

5B19 5

3.0 10 N 9.4 10 T2 2 1.602 10 C 1.0 10 m/sFBev

−−

−

⋅= = = ⋅⋅ ⋅

27.22. THINK: The particle moves in a straight line at constant speed. Thus, the net force must be zero. The electric force is the negative of the magnetic force. Using the right-hand rule, the direction of the magnetic field can be determined. For the magnitude, set the magnitude of the net force to zero. 10.0 μC, q =

300. m/sv = and E = 100. V/m. SKETCH: (a) (b)

RESEARCH: ( ) 0F qE q v B= + × =

SIMPLIFY: ( ) 0 q E v B v B E+ × = × = −

(a) ( )ˆˆ ˆ ˆ Ev B Ey vz B Ey B xv

× = − × = − = − (by the right-hand rule)

(b) ( )ˆ ˆ ˆ v B Ez vz B Ez× = − × = −

There is no solution. z B×

is either zero, or a vector in the xy-plane. CALCULATE:

(a) 100. V/m 1 T 0.3333 T;300. m/s 3

B = − = − = −

so ˆ0.333 T.B x= −

(b) No solution. No magnetic field will keep the particle moving at a constant speed in a straight line. ROUND: (a) = −

ˆ0.333 TB x

(b) Not applicable. DOUBLE-CHECK: No Lorenz force can counteract an electric force in z-direction, if the particle is also traveling in z-direction, because the Lorenz force is always perpendicular to the velocity vector.


1029

27.23. THINK: First determine the components of the force. Once the components are determined, the magnitude and the direction of the force can be found. = 20.0 μC,q v = 50.0 m/s, = 0.700 TzB and

= 0.300 T.yB

SKETCH:

RESEARCH: ,F qv B= × ˆˆ ˆ ,x y z× = ˆˆ ˆ,y z x× = ˆ ˆ ˆ,z x y× = A B B A× = − ×

SIMPLIFY: ( ) ( ) ( )ˆ ˆ ˆ ˆˆ ˆ ˆ ˆ ˆ ˆ ˆz y z y z y y zF qvx B z B y qv B x z B x y qv B y B z F y F z= × + = × + × = − + = +

2 2 2 2y z z yF F F qv B B= + = +

1 1 1tan tan tany yz

z zy

B BFB BF

θ − − − = = =

CALCULATE: ( )( ) ( ) ( )2 26 420.0 10 C 50.0 m/s 0.700 T 0.300 T 7.616 10 NF − −= ⋅ + = ⋅

1 0.300 Ttan 23.200.700 T

θ − = = °

ROUND: 47.62 10 NF −= ⋅

and the direction of the force is in the yz-plane, 23.2θ = ° above the negative

y-axis. DOUBLE-CHECK: These results are reasonable. The Right Hand Rule dictates that the direction of the magnetic force be in the -y, +z-plane.

27.24. THINK: The only force acting on the particle is the magnetic force. The components of this force can be determined, and then the points where all the components vanish can be determined.


1030

SKETCH:

RESEARCH: 0,F qv B= × = ˆˆ ˆ ,x y z× = ˆˆ ˆ,y z x× = ˆ ˆ ˆ,z x y× = A B B A× = − ×

SIMPLIFY: ( ) ( ) ( ) ( )( ) ( )

0 0

0

ˆ ˆˆ ˆ ˆ ˆ ˆ0ˆ ˆ 0

F qv xx x ay y xy b z qv x ay x y xy b x z

qv x ay z b xy y

= = − + − = − × + − × = − + − =

0 x ay x ay − = =

( ) 20 0 0 bb xy b ay y b ay ya

− = − = − = = ± x ay ab = = ±

Only the positive values are needed since it only here where the field is non-zero. So, F = 0 at the point

( ) ( ), , / .x y ab b a=

CALCULATE: Not necessary. ROUND: Not necessary. DOUBLE-CHECK: For the given field, the results are correct. However, it turns out that the B

field

given cannot actually be physically created. The fundamental equations which govern the behavior of all electric and magnetic fields are called Maxwell’s equations, named after James Clerk Maxwell, who first unified them. One of these equations states that:

ˆ ˆ ˆ 0.yx zBB B

B x y zx y z

∂∂ ∂∇ = + + =

∂ ∂ ∂

This is not obeyed by the given B

field. Interestingly, 0B∇ = exists because as far as is now known,

magnetic monopoles do not exist, only dipoles. This is in contrast to electric fields where monopoles (isolated positive and negative charges) do exist. In order to create the B

field given in this problem,

magnetic monopoles are required to exist.

27.25. 21 2 2

eVK eV mv eV vm

= = =

( )( )314

19

2 9.11 10 kg 400. V2 1 2 1 3.37 10 T0.200 m 1.602 10 C

mv m eV mVBer er m r e

−−

−

⋅= = = = = ⋅

⋅

27.26.


1031

35θ = °, 0.040 T ,B B= =

54.0 10 m/s ,v v= ⋅ = sinxv v vθ ⊥= = (perpendicular to B

), cosyv v vθ= =

(parallel to B

).

(a) ( )( )( )

( )( )

31 55

19

9.1 10 kg 4.0 10 m/s sin353.3 10 m

1.602 10 C 0.040 Tmvrq B

−−⊥

−

⋅ ⋅ °= = = ⋅

⋅

(b) The time it takes to travel 2π radians around the circle is:

2 2 2 .mvr mt

v v q B q Bπ π π⊥

⊥ ⊥

= = =

During this time it is moving forward with speed ,v and will move a distance, d, given by:

( )( )( )( )

31 54

19

2 9.1 10 kg 4.0 10 m/s cos352 2 cos 2.9 10 m.1.602 10 C 0.040 T

v m mvd v tq B q B

ππ π θ−

−−

⋅ ⋅ °= = = = = ⋅

⋅

27.27. By Newton’s second law, the quantity /dp dt is equal to the net force on the particle, exerted by the electric and magnetic fields. By the Work Energy Theorem, the quantity dK/dt is the rate at which work is done on the particle by the electric field. The magnetic force is always perpendicular to the particle’s velocity and does no work. Hence, these quantities can be written:

( ),dp q E v Bdt

= − + × .dK qE v

dt= −

Slightly modified, these relationships can be put into a form that transforms simply from one reference frame to another according to Einstein’s Special Theory of Relativity (see Chapter 35). They can be used to show that in a world governed by Einsteinian dynamics, the simplest force law that can be written is the combined electromagnetic force law above.

27.28. ( )

( )

28 63

19

1.88 10 kg 3.0 10 m/s7.0 10 m 7.0 mm

1.602 10 C 0.50 Tmvrq B

−−

−

⋅ ⋅= = = ⋅ =

⋅

27.29.

The net force is directed toward the center of the circle. From the right-hand rule, a positive charge requires the magnetic field to be oriented into the plane, in the negative z-direction. Since an electron is negatively charged, it can be concluded that the field points out of the page, along the positive z-direction. The magnitude is given by:

( )( )31 12 1

19

9.1 10 kg 1.2 10 sˆ 6.816 T 6.8 T.

1.602 10 Cq B mB B zm q

ωω− −

−

⋅ ⋅= = = = =

⋅

27.30. ,mvrq B

= 21 2 2

KK mv vm

= = 2 2 m K Kmrq B m q B

= =

The mass, charge and fields are the same for the two particles.


1032

⋅ = = = = ⋅

211 1

22 22

2 4.00 10 eV 1.412.00 10 eV2

q BK mr Kr q B KK m

So, the 400 eV particle travels in an orbit of radius 1.41 times that of the radius of the 200 eV particle.

27.31. THINK: The proton moves through a magnetic field. The component of the velocity parallel to the field is unchanged. The component perpendicular, however, will create a circular motion. The velocity of the proton is ( ) 5ˆˆ ˆ1.0 2.0 3.0 10 m/sv x y z= + + and the field is ˆ0.50 T.B z= SKETCH:

RESEARCH: The radius of the circular motion in a magnetic field is:

.mvrq B

=

SIMPLIFY: The speed in the xy-plane is 2 2 .xy x yv v v= + The radius of the circle is: 2 2 .x ymr v v

q B= +

CALCULATE: ( )( ) ( ) ( )27 2 25 5 3

19

1.6726 10 kg 1.0 10 m/s 2.0 10 m/s 4.6688 10 m1.6022 10 C 0.50 T

r−

−−

⋅= ⋅ + ⋅ = ⋅⋅

( ) ( )2 25 5 51.0 10 m/s 2.0 10 m/s 2.23607 10 m/sxyv = ⋅ + ⋅ = ⋅

ROUND: The values are given to two significant figures. The proton will follow a helical path with a velocity of 53.0 10 m/s⋅ along the z-axis, with the circular motion in the xy-plane having a speed of

52.2 10 m/s⋅ and a radius of 4.7 mm. DOUBLE-CHECK: The angular velocity is on the same order of magnitude as the original velocity. Dimensional analysis confirms the units are correct.

27.32. THINK: The copper sphere accelerates in the region of the electric field, and gains an amount of kinetic energy equal to the potential difference times the charge on the sphere. At this speed, the sphere enters the magnetic field, which curves its path. The sphere has a mass of 63.00 10 kgm −= ⋅ and a charge of

45.00 10 C.−⋅ The potential difference is V = 7000. V, and the magnetic field is B = 4.00 T, perpendicular to the direction of the particle’s initial velocity.


1033

SKETCH:

RESEARCH: The kinetic energy will be 2 / 2 .KE mv PE qV= = = The radius of the path in the magnetic field is / .r mv qB=

SIMPLIFY: The velocity is 2 2 /v qV m= or 2 / .v qV m= The radius is then:

2 1 2 .mv m qV mVrqB qB m B q

= = =

CALCULATE: ( )( )6

4

2 3.00 10 kg 7000. V1 2.29128 m4.00 T 5.00 10 C

r−

−

⋅= =

⋅

ROUND: The least precise values have three significant figures, so the radius of the copper’s path in the magnetic field is 2.29 m. DOUBLE-CHECK: The very large potential difference accelerates the sphere to a high speed, therefore a large radius of curvature is reasonable.

27.33. THINK: The particles will move in circular, clockwise paths (in the direction of )v B× within the

magnetic field. The radius of curvature of the path is proportional to the mass of the particle, and inversely proportional to the charge of the particle. Both particles move at the same speed within the same magnetic field. The radii, charges and masses of the particles can then be compared. SKETCH:

RESEARCH: The radius is related to the mass and charge of the particles by / .r mv q B= At the instant

that the particles enter the magnetic field, the magnetic force acting on them is .BF qv B= × For the


1034

particles to travel in a straight line, the force on the particles due to the electric field must oppose the force due to the magnetic field: .E BF F qE qv B= − = − ×

SIMPLIFY: Since the velocity and the magnetic field is the same for both particles, 1 1 2 2

1 2

.r q r qv

B m m= = The

ratio of the masses is:

( )1 1 1

2 2 2

1 .2 2 4

m r q Rqm r q R q

= = =

For the particles to travel in a straight line, ( ) ( )ˆ ˆ ˆ

ˆ.qE qv B qvB x z qvB y

E vBy= − × = − × = − −

=

Therefore, the electric field must have magnitude E vB= and point in the positive y-direction in order for the particles to move in a straight line. CALCULATE: Not applicable. ROUND: Not applicable. DOUBLE CHECK: Since the mass increases with radius and charge, it makes sense that the particle with the smaller charge and radius has the smaller mass.

27.34. THINK: The question goes through the elements of a mass spectrometer. A source of gold and molybdenum emits singly ionized atoms at various velocities toward the velocity filter. The velocity filter described in the diagram uses a magnetic field, 1B

, and an electric field E

to select the velocity of the

exiting particles. It can be shown that the velocity of the ions must be 1 / .v B E= If the velocities are smaller, the force due to the electric field will dominate pushing the particles off the path through the filter. If the velocity is greater than 1 /B E then the force due to the magnetic field dominates and will also push the ion off its path. The particles exit the filter with the selected velocity and enter the mass spectrometer. The magnetic field inside the mass spectrometer curves the path of the entering particles based on their mass and charge. Thus, particles of different charges and masses can be separated. The gold and molybdenum have beams in the mass spectrometer with diameters of 2 40.00 cmd = and

1 19.81 cm,d = respectively. The mass of the gold ion is −= ⋅ 25gold 3.27 10 kg.m Both ion types have charges

of 191.6 10 C.q e −= = ⋅ The electric and magnetic field within the velocity filter are = ⋅ 4 ˆ1.789 10 V/mE y

and 1 ˆ1.00 T,B z=

respectively. SKETCH:


1035

RESEARCH: a)

(b) For the ions to pass through the velocity filter, the force due to the electric field must cancel the force due to the magnetic field: e 0 .F qE qv B= = The forces due to an electric and magnetic field are given by

eF qE=

and B 0 ,F qv B= × respectively.

(c) The radius of the ion’s path in a magnetic field is given by ( )/ .R mv q B= (d) The mass of the molybdenum can be determined by setting the velocity, charge and magnetic field equal to each other for each ion and comparing. SIMPLIFY: (b) The velocity of the exiting particle is then 0 1/ .v E B= This velocity does not depend on any parameters of the ions. (c) The radius of the circular path is ( )0 2/ .r mv q B=

(d) 0 Au+ Mo+

2 Au+ Mo+

.v r r

q B m m= = Solving for the mass of the molybdenum gives: Mo+

Mo+ Au+Au+

.r

m mr

=


1036

CALCULATE:

(b) 4

40

1.789 10 V/m 1.789 10 m/s1.00 T

v ⋅= = ⋅

(d) ( )25 25Mo+

19.81 cm/2 3.27 10 kg 1.6194 10 kg40.00 cm/2

m − −= ⋅ = ⋅

ROUND: The values are reported to three significant figures. (b) The velocity filter allows particles traveling 41.79 10 m/s⋅ to exit the filter. This value does not depend on the type of ion. It is based on the fact that the particle is charged and only depends on the fields. (c) The equation for the radius of the semi-circular path is / .r mv q B=

(d) The mass of the molybdenum ion is 251.62 10 kg.−⋅ DOUBLE-CHECK: The calculated velocity has appropriate units, and the actual mass of molybdenum is about 251.64 10 kg.−⋅ These facts help to support the answers as reasonable.

27.35. THINK: This question explores the function of an accelerator. The 3 +He ion source ejects particles into region 1. The magnetic field in this region bends the path of the 3 +He particle into a semicircular path. Particles then enter the 3rd region containing an electric field that accelerates the particles toward region 2. In region 2, the path of the particles is bent into a semicircle again. This time the particles do not pass through and electric field. The particle then enters region 1 again. The path is bent again and the particle accelerates once more before it exits the accelerator. The 3 +He ions have a mass of 275.02 10 kgm −= ⋅ and

a charge of 191.60 10 C.q e −= = ⋅ The ions start with a kinetic energy of:

( )−−

⋅ ⋅= = = ⋅

3 1916

4.00 10 eV 1.60 10 J4.00 keV 6.40 10 J.

1.00 eVK

The magnetic field in region 1 is 1 1.00 T.B = In region 2, the magnetic field, 2 ,B is unknown. The 3rd region contains an electric field of E = 60.0 kV/m and has a length of l = 50.0 cm = 0.500 m. The distance between the source and the aperture is d = 7.00 cm = 0.0700 m. SKETCH:

RESEARCH: The kinetic energy is equal to 2 / 2.KE mv= The radius of the path of a charged particle in a magnetic field is / .r mv q B= The force on the particle in region 3 is e ,F qE= which must equal F = ma. With the acceleration of this region, the velocity at which it exits region 3 can be determined from:

2 2f 0 2 .v v ad= +

SIMPLIFY: Let 0 ,v 1v and 2v be the velocity of the ion after it is ejected from the source, region 3 and

region 3 the second time, respectively. The velocity after the source is given by 0 2 / .v KE m= The


1037

acceleration of region 3 is ma = qE or a = qE/m. The velocity, 1 ,v is then 2 21 0 02 2 / .v v al v qEl m= + = +

Similarly, the velocity, 2 ,v is given by: 2 2 2 2

2 1 1 0 02 2 / 2 / 2 / 4 / .v v al v qEl m v qEl m qEl m v qEl m= + = + = + + = + The radius of the path, the first time the ion enters region 1 is 1 0 1/ .R mv qB= The radius of the path in region 2 is 2 1 2/ .R mv qB= The radius of the path the second time it goes through region 1 is

3 1 1/ .R mv qB= For the particle to exit the aperture, a distance d = 7 cm from the ion source:

0 0 11 1 11 2 3

1 2 1 1 2

2 2 2 2 2 .mv v vmv mv vmR R R dqB qB qB q B B

+− + = − + = − =

Solve for 2B to determine the magnetic field required in region 2:

( ) ( ) ( )

0 1 0 112 1 2

1 2 1

0 1 1 1 12 1 2

1 0 1 1 0 1 1

2 2

2 .

2 / / 2 2

v v v vv qd qdB v BB B m B m

v v v mB vqdB v BB m v v B qd m m v v qdB

+ + − = − =

+ − = = = + − + −

Region 1 must have dimensions larger than 3 1 1/ .R mv qB= The velocity of the ions as they exit the

accelerator is ( )22 0 4 / .v v qEl m= +

CALCULATE: ( )−

−

⋅= =

⋅

16

0 27

2 6.40 10 J504995 m/s

5.02 10 kgv

( ) ( )( )( )− −

− −

⋅ ⋅ ⋅= + =

⋅ ⋅

16 19 3

1 27 27

2 6.40 10 J 2 1.60 10 C 60.0 10 V/m 0.500 m1472186 m/s

5.02 10 kg 5.02 10 kgv

( ) ( )( )( )− −

− −

⋅ ⋅ ⋅= + =

⋅ ⋅

16 19 3

2 27 27

2 6.40 10 J 4 1.60 10 C 60.0 10 V/m 0.500 m2019822 m/s

5.02 10 kg 5.02 10 kgv

( )( )( )( )( ) ( )( )( )

27

2 27 19

2 5.02 10 kg 1.00 T 1472186 m/s1.70866 T

2 5.02 10 kg 504955 m/s 1472186 m/s 1.60 10 C 0.0700 m 1.00 TB

−

− −

⋅= =

⋅ + − ⋅

( )( )( )( )

271

3 191

5.02 10 kg 1472186 m/s0.046190 m

1.60 10 C 1.00 Tmvx RqB

−

−

⋅= = = =

⋅

ROUND: The values are reported to three significant figures. (a) The magnetic field of region 2 is 1.71 T. (b) The region must have dimensions greater than 4.62 cm. (c) The velocity the ions leave the accelerator is 62.02 10 m/s.⋅ DOUBLE-CHECK: Dimensional analysis confirms all the answers are in the correct units. These results are reasonable.

27.36. The force on a wire of length l and current i in a magnetic field B is given by sin .F li B liB θ= × =

The magnitude of the magnetic field is:

( )( )0.500 N 0.02083 T 20.8 mT.

sin 2.00 m 24.0 A sin30.0FB

li θ= = = ≈

°

27.37. The force on the wire is ( )net B g ˆ ˆ ˆ ˆ ˆ ˆ.F ma F F iL B mgy iLB x z mgy iLBy mgy= = + = × + = − × − + = − + For the

conductor to stay at rest, a = 0 or mg = iLB. The suspended mass is then:


1038

( )( )( )( )2

20.0 A 0.200 m 1.00 T0.408 kg.

9.81 m/siLBm

g= = =

27.38.

For the wire to levitate, the force of the magnetic field must equal the force of gravity on the wire:

2 .iLB mg Vg r Lgρ ρπ= = = The current required to levitate the wire is:

( ) ( ) ( )( )( )

23 223

8940 kg/m 0.000500 m 9.81 m/s1.38 10 A.

0.0001 T/G0.500 Gr giB

πρπ= = = ⋅

27.39. THINK: First, the relationship between the current in the sheet and the magnetic field must be established. The force on the sheet can then be determined. The sheet has length, L = 1.0 m, width, w = 0.50 m, and thickness, t = 1.0 mm = 0.0010 m. The magnetic field, B = 5.0 T, is perpendicular to the sheet and the current flowing through it. The current is 3.0 A.i = SKETCH:

RESEARCH: The force on a wire carrying current in a magnetic field is .F iL B= ×

SIMPLIFY: Imagine that the sheet is constructed of many wires of length, L, carrying a charge dq . The

infinitesimal force on the sheet due to the wire is .dF dqL B= ×

Since L is perpendicular to B, .dF dqLB= The infinitesimal current is equal to the current density times the differential area, .dq jdA= The current density is equal to the total current divided by the cross sectional area, / / .j i A i wt= = The infinitesimal force is then / .dF dqLB jdALB iLBdA wt= = = Integrating over the area gives:

0 0.

w ti iF LB dx dy LBwt iLBwt wt

= = =

This is the same result as that for a wire. CALCULATE: ( )( )( )3.0 A 1.0 m 5.0 T 15.0 NF = = ROUND: The result is reported to two significant figures. The force on the sheet is 15 N. This is the same as the force on a wire of the same length with the same current and magnetic field.


1039

DOUBLE-CHECK: The force on the sheet is the same as the force on a wire. This is expected since only the magnitude of the current matters in a wire (the size of the wire is not relevant).

27.40. THINK: For the rod to remain stationary, the forces of the magnetic field and gravity along the plane of the incline must cancel. SKETCH:

RESEARCH: The force due to the magnetic field is B .F iL B iLB= × =

Along the plane of the incline, the

force is B B cos cos .xF F iLBθ θ= = The force due to gravity along the surface of the incline is g sin .xF mg θ=

SIMPLIFY: Equating these forces gives the current:

cos siniLB mgθ θ= or tan .mgiLB

θ=

The current must go out of the page in the side view of the system, by the right-hand rule. CALCULATE: Not applicable. ROUND: Not applicable. DOUBLE CHECK: According to the derived expression, the strength of the current required to hold the wire stationary decreases as the magnetic field increases, which is logical.

27.41. THINK: The loop will experience a torque in the presence of a magnetic field as discussed in the chapter. The torque is also equal to the moment of inertia of the loop times its angular acceleration. The loop is square with sides of length, d = 8.0 cm and current, i = 0.15 A. The wire has a diameter of 0.50 mm (which corresponds to a radius of r = 0.25 mm) and a density of 38960 kg/m .ρ = The magnetic field is B = 1.0 T and points °35.0 away from the normal of the loop.

SKETCH:

RESEARCH: The torque on the loop is sin .iABτ θ= The moment of inertia for a rod about its center is = 2 /12.I Md The moment of a rod about an axis along its length is 4 / 4.I r Lρπ= The parallel axis

theorem states = + 2off center cm .I I Md

SIMPLIFY: The moment of inertia of the loop is:

( )

ρπ ρπ

ρπ ρπ ρπ ρπ ρπ ρπ

= + + + + +

= + + = + + = + = +

2 22 2 4 4

2 4 2 2 4 2 2 4 2 3 4

1 1 1 112 12 4 2 4 2

1 1 1 1 1 1 4 1 2 1 .6 2 2 6 2 2 6 2 3 2

d dI Md Md r d M r d M

Md r d Md Md r d r d d r d r d r d


1040

The angular acceleration is sinI iABτ α θ= = or sin / .iAB Iα θ=

( ) ( )θ θα

ρπ ρπ = =

+ +

2

2 2 2 2 2

sin sin 2 / 3 / 2 2 / 3 / 2

id B iBr d d r d r d r d

CALCULATE: ( )( )( )

( ) ( ) ( ) ( ) ( ){ }απ

°= = +

22 23

0.15 A 1.0 T sin35.0 916.94 rad/s8960 kg/m 0.00025 m 2 0.080 m / 3 0.00025 m / 2 0.080 m

ROUND: The values are given to two significant figures, thus the loop experiences an initial angular acceleration of 3 20.92 10 rad/s .α = ⋅

DOUBLE-CHECK: One Tesla represents a magnetic field of large magnitude, resulting in a correspondingly large acceleration. This result is reasonable.

27.42. THINK: The rail-gun uses a magnetic force to accelerate a current carrying wire. The wire has a radius of 45.1 10 mr −= ⋅ and a density of 38960 kg/m .ρ = The current through the wire is 41.00 10 A⋅ and the

magnetic field is B = 2.0 T. The wire travels a distance of L = 1.0 m before being ejected. SKETCH:

RESEARCH: The force on the wire is .F iLB= The velocity of the wire is given by 2 20 2 .v v ad= +

SIMPLIFY: The wire accelerates at a rate of:

2 2 .F iLB iLB iBam m r L rρπ ρπ

= = = =

The ejected velocity is:

ρπ= =2

2

22 iBLv aLr

or ρπ

= 2

2 .iBLvr


( ) ( )4

23 4

2 1.00 10 A 2.0 T 1.0 m2337 m/s

8960 kg/m 5.1 10 mv

π −

⋅= =

⋅

ROUND: The velocity is reported to two significant figures, like the given values. The wire exits the rail-gun at a speed of 2.3 km/s. DOUBLE-CHECK: This result is very fast, about 7 times the speed of sound. This Navy has used this technology to accelerate 7 lb objects to this speed. As a comparison, this is double the speed of a bullet from a conventional rifle. A bullet weighs 55 g.

27.43. THINK: Using integration, the force on each segment of the loop can be found. From this, the net force can be found.


1041

SKETCH:

RESEARCH: In the differential limit, the magnetic force on current carrying wire in a magnetic field is:

.BdF i dL B= ×

SIMPLIFY: The force on wire 1 is:

( )/2 /2 /2 /220 0 0,1 /2/2 /2 /2

ˆ ˆ ˆ ˆ 0.2

l l l l

B ll l l

iB iB iBF i dx B dx zx xz xdxy x

a a a −− − − = × = × + = − = =

The force on wire 2 is:

( ) ( ) ( ) ( )/2 /2 /20 0 0 0,2 /2/2 /2

ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ .l l l

B ll l

iB iB iB iB lF dy zx xz zz xx dy zz xx y zz xx

a a a a−− −= × + = − + = − + = − +

Along wire 2, / 2x l= and 0,z = so 2

0,2 ˆ.

2BiB l

F xa

=

Wire 3 is similar to wire 1, so ,3 ,1 0.B BF F= = The force on wire 4 is:

( ) ( ) ( ) ( ) ( )2/2 /20 0 0 0 0

,4 /2/2ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ0 / 2 .

2l l

B ll

iB iB iB l iB l iB lF dy zx xz zz xx y zz xx z l x x

a a a a a− − = − × + = − = − + = − + =

The net force is: 2

0net ,2 ,4 ˆ.B B

iB lF F F x

a= + =

CALCULATE: Not applicable. ROUND: Not applicable. DOUBLE CHECK: It is reasonable that the force is directly proportional to the magnetic field strength

0 ,B the side length l of the loop, and the current i.

27.44. The loop experiences a torque of:

( )( )( )3 6 9sin 20.0 2.00 10 A 0.0800 m 0.0600 m 50.0 10 T 9.60 10 N m.NiABτ θ − − −= = ⋅ ⋅ = ⋅

By the right-hand rule, the torque is in the positive y-direction. To hold the loop steady, a torque of the same magnitude must be applied in the negative y-direction.

27.45. The torque on the loop due to the magnetic field is sin .iAB NiABτ θ= Ν = This is equal to the applied torque, .rFτ = Equating the torques gives the magnetic field:

( ) ( )2

1.2 N 0.1353358 T 0.14 T.120 0.49 A 0.048 m

rF rF FBNiA Ni rNi r π ππ

= = = = = =

27.46. The torque on the pencil is:

( ) ( )τ θ π θ π − = = = ° = ⋅

2 230.00600 msin sin 20 3.00 A 5.00 T sin60.0 7.35 10 N m.

2 2dNiAB Ni B


1042

27.47. THINK: The loop feels a torque if it carries a current in the presence of a uniform magnetic field. The maximum torque occurs when the magnetic moment of the loop is perpendicular to the magnetic field. The loop has a radius of r = 0.500 m, density of 38960 kg/m ,ρ = and the wire has a cross-sectional area of

5 21.00 10 m .A −′ = ⋅ A potential difference of 0.012 VV = is applied to the wire. The loop is in a magnetic field of 0.25 T.B = SKETCH:

RESEARCH: The resistivity of the wire is given by 9R 16.78 10 m.ρ −= ⋅ Ω The current is found using

V = iR. The magnetic moment of the loop is given by .iAnμ = The torque is equal to ,Iτ α= where I is

the inertial moment of the loop about its diameter. The inertial moment is given by 2 .I r dm= The

torque due to the magnetic field is given by .Bτ μ= ×

SIMPLIFY: The mass of the loop is given by 2 22 2 .M V rA A y xρ ρ π πρ= = = + The inertial moment of one quarter of the loop is:

( )

32 2

1/4 0 02 2 2 2

2 2 2 2 2 2 3

0

2 2

1 1 12 0 2 .2 3 6 3

r r

r

xdx x dxI x dm Ax Ar x r x

A r x r x A r r Ar

π πρ ρ

π ρ πρ πρ

−= = = − − −

= − − − + = − =

The total inertial moment is four times this magnitude: 3tot

4 .3

I Arπρ= The torque and thus the angular

acceleration is maximized when the magnetic moment is perpendicular to the magnetic field. The torque is then given by 2

max sin .B B B iAB i r Bτ μ μ θ μ π= × = = = = The angular acceleration is:

2max

max 3

3 .44 / 3

i r B iBI ArAr

τ παρπρ

= = = The current is: R R2.

l rRiV AV AV

ρ π ρ= = = The angular acceleration is:

( )R Rmax 2

3 2 / 3.

4 2r AV B B

Ar A Vπ ρ πρα

ρ ρ= =

CALCULATE: ( )( )

( )( ) ( )

92

max 23 5 2

3 16.78 10 m 0.250 T1.8386 rad/s

2 8960 kg/m 1.00 10 m 0.012 V

πα

−

−

⋅ Ω= =

⋅

ROUND: The result is reported to two significant figures since the potential difference is the least precise value (given to two significant figures). The maximum angular acceleration is 2

max 1.8 rad/s .α = DOUBLE-CHECK: This is a reasonable value for an angular acceleration of a copper wire in a small magnetic field.


1043

27.48.

The torque on the coil as a function of θ is ( )sin 90 cos .B B B= × = ° − =τ μ μ θ μ θ The magnetic

moment of the coil is .NiAμ = Assume the coil contributes little to the inertial moment of the galvanometer. Assume the mass is distributed evenly through the rod. The torque on the rod due to gravity is:

sin sin ,r F LF LMgτ θ θ= × = =

where r is the distance to the center of mass of the rod. Equating the two torques gives:

1cos sin tan tan .B NiAB NIABB LMgLMg LMg LMgμμ θ θ θ θ −

= = = =

27.49. Assume the electron orbits the hydrogen with speed, v. The current of the electron going around its orbit is:

./ 2

q e ev evit d v d rπ

= = − = − = −

The magnetic moment of the orbit is: ( )2 2 1 .2 2eviA i r r evr

rμ π π

π= = = − = − Angular momentum is given

by L = rp = rmv. Using the angular momentum, the moment is: 1 1 / 2 .2 2 2

m ermv eLerv e rvm m m

μ = − = − = − = −

27.50. THINK: The magnetic field produces a torque on the coil. This stretches the spring until it creates a torque equal but opposite to the torque due to the magnetic field. The ring has a diameter of d = 0.0800 m and carries a current of 1.00 A.i = The spring constant is k = 100. N/m and the magnetic field is B = 2.00 T.


1044

SKETCH:

RESEARCH: The torque due to the spring is s .r Fτ = ×

The torque due to the magnetic field is sin .iABτ θ=

SIMPLIFY: The angle θ is given by: 2sin ./ 2L L

d dθ Δ Δ= = The torque due to the spring is

s s sin / 2.r F dk Lτ θ= × = Δ Equating this to the torque due to the magnetic field gives:

( )22 / 42cos / 2 sin .2

i d BiAB i dBdk L iAB Ldk dk k

π πθ θΔ = Δ = = =


( )1.00 A 0.0800 m 2.00 T

0.002513274 m2 100. N/m

Lπ

Δ = =

ROUND: The values are given to three significant figures, thus the extension is 2.51 mm.LΔ = DOUBLE-CHECK: This result is reasonable.

27.51. THINK: The coil experiences a torque due to the magnetic field. The coil, however, is hinged along one of its lengths. The torque then can be determined in the normal way. The force on each segment is calculated to determine the torque on the coil. The coil has N = 40, a width of w = 16.0 cm, a height of h = 30.0 cm and carries a current of 0.200 A. The magnetic field is ( )ˆˆ0.065 0.250 T.B x z= +


1045

SKETCH:

RESEARCH: The force on a length of wire carrying current in a magnetic field is .F Nil B= ×

The torque

is given by .r Fτ = ×

SIMPLIFY: The force on the segment a-b is ( )ab ab ˆ ˆˆ ˆ ˆ ˆ.x z z zF Nil x B x B z NiwB x z NiwB y= × + = × = −

The force on the segment b-c is:

( ) ( ) ( ) ( )bc bc ˆ ˆ ˆ ˆˆ ˆ ˆ ˆ ˆ ˆ ˆ .x z x z x z x zF Nil y B x B z Nih B y x B y z Nih B z B x Nih B z B x = × + = × + × = − + = − +

The net force on the coil is zero since the magnetic field does no work. The coil is free to rotate about the y-axis. The only force that can contribute is bc :F bc ˆˆ ˆ ˆ ˆ.z x xr F wx F wx Nih B x B z wNihB yτ = × = × = × − =

The door rotates counterclockwise when looking from the top. CALCULATE: ( )( )( )( )ab ˆ ˆ40 0.200 A 0.160 m 0.250 T 0.320 NF y y= − = −

( )( )( ) ( ) ( ) ( ) ( )bc ˆ ˆ ˆˆ ˆ ˆ40 0.200 A 0.300 m 0.065 T 0.250 T 0.156 0.600 N 0.600 0.156 NF z x z x x z = − + = − + = −

( )( )( )( )( ) ˆ ˆ40 0.200 A 0.160 m 0.300 m 0.065 T 0.02496 Nmy yτ = = ROUND: (a) The force on segment a-b is = −

ab ˆ0.320 NF y

(b) The force on segment b-c is ( )= −

bc ˆˆ0.60 0.156 NF x z or bc 0.62 NF = directed 15° from the x-axis toward the negative z-axis. (c) The total force is net 0.F = (d) The torque on the coil is 0.025 Nmτ = and rotates along the y-axis in counterclockwise fashion. (e) The coil rotates in a counterclockwise fashion as seen from above. DOUBLE-CHECK: This result is reasonable.

( )( )( )sin 40 0.200 A 0.160 m 0.300 m ,NiABτ θ= = ( ) ( ) ( )2 20.065 T 0.250 T sin 14.6 0.025 N m+ ° =

27.52. The Hall voltage is given by: V .iBneh

Δ = The carrier density of the electron sheet is:

( )( )( )( )

624 3

19 9 3

10.0 10 1.00 T9.19 10 e/m .

V 1.60 10 C 10.0 10 m 0.680 10 ViBn

eh

−

− − −

⋅= = = ⋅

Δ ⋅ ⋅ ⋅


1046

27.53. THINK: The question asks for the carrier density of the thin film, and the nature of the carriers. The film has a thickness of 1.50 μm.h = The current is i = 12.3 mA and the voltage reads V = -20.1 mV. The magnetic field is B = 0.90 T.

SKETCH:

RESEARCH:

(a) Due to the magnetic force, the charge carriers are accumulated on the front side of the sample. Since the polarity of the Hall potential is negative, the charge carriers are holes.

(b) The Hall voltage is given by HV .iBneh

Δ =

SIMPLIFY:

(b) The charge carrier density is H

.V

iBnhe

=Δ

CALCULATE:

(b) ( )

( )( )( )3

24 36 19 3

12.3 10 0.90 T2.29478 10 e/m

1.50 10 m 1.6 10 C 20.1 10 Vn

−

− − −

⋅= = ⋅

⋅ − ⋅ − ⋅

ROUND: (b) The values are given to two significant figures or greater, so the carrier density of the film is

24 32.3 10 e/m .n = ⋅ DOUBLE-CHECK: This is a reasonable value for a carrier density.

Additional Problems

27.54. The radius of the proton’s path is: 22 2 .1/

mrfmv m r m rrqB qB T qB f qB

ππ π = = = =

The radius of the path and

its frequency are: mvrqB

= and ,2qBf

mπ= respectively. In the cyclotron:

( )( )( )( )

27 8

19

1.67 10 kg 2.998 10 m/s / 20.17 m,

1.602 10 C 9 Tr

−

−

⋅ ⋅= =

⋅

( )( )( )

19

27

1.602 10 C 9 T1.4 MHz.

2 1.67 10 kgf

π

−

−

⋅= =

⋅

In the Earth’s magnetic field:

( )( )( )( )( )

27 8

19

1.67 10 kg 2.998 10 m/s / 231 km,

1.602 10 C 0.5 G 0.0001 T/Gr

−

−

⋅ ⋅= =

⋅

( )( )( )

19 4

27

1.602 10 C 0.5 10 T0.76 kHz.

2 1.67 10 kgf

π

− −

−

⋅ ⋅= =

⋅

27.55. The force on the wire is: ( )( )( ) ( )θ −= × = = ° − ° = ≈ ⋅

2sin 3.41 A 0.100 m 0.220 T sin 90.0 10.0 0.07388 N 7.39 10 N.F iL B iLB


1047

27.56. The radius of a charged particle’s path in a magnetic field is / .r mv q B= For this electron, the radius of its path is:

( )( )( )( )

31 7

19 4

9.109 10 kg 6.00 10 m/s6.82 m.

1.602 10 C 0.500 10 Tr

−

− −

⋅ ⋅= =

⋅ ⋅

27.57. The force on a current carrying wire in a magnetic field is sin .F ilB θ= To determine the minimum current, set 90 :θ = °

( )5

4

1.0 N 232558 A 2.3 10 A.0.100 m 0.43 10 T

FilB −

= = = ≈ ⋅⋅

The minimum current required for the wire to experience a force of 1.0 N is 52.3 10 A.i = ⋅

27.58. For the ball’s path to have a radius, r, the magnetic field must be:

( )( )( )( )

35.00 10 kg 3000. m/s 0.500 T.

15.0 C 2.00 mmv mvr Bq B q r

−⋅= = = =

Since the particle has a positive charge, the field must point downward, by the second right-hand rule, for it to travel clockwise.

27.59. For the particle to travel in a straight path, the force due to each field must cancel: E B .F qE F qvB= = = For this to occur, the speed of the charged particle must be:

45

3

1.00 10 V/m 2.00 10 m/s,50.0 10 T

EvB −

⋅= = = ⋅⋅

in the positive x-direction.

27.60. The torque on a current carrying coil in a magnetic field is:

( )( ) ( ) ( )( )23

3

sin

100 100. 10 A 0.100 m 0.0100 T sin30.0

0.0015707 N m 1.57 10 N m.

NiABτ θ

π−

−

=

= ⋅ °

= ≈ ⋅

27.61. (a) The electron must travel in a circular path with a radius of 60.0 cm, as shown in the figure below.

By the right-hand rule, B must be in the negative z-direction.

( )( )( )( )

31 56

19e

9.11 10 kg 2.00 10 m/s1.89 10 T

1.60 10 C 0.600 mmvBq r

−−

−

⋅ ⋅= = = ⋅

⋅

(b) The magnetic force is perpendicular to the motion and does no work.


1048

(c) Since the speed of an electron does not change, the time the electron takes to travel a quarter-circle is given by:

( )( )

65

3.14159 0.600 m4.71 10 s.

2 2 2.00 10 m/srtv

π −= = = ⋅⋅

27.62. THINK: First, determine the current using Ohm’s law and then determine the force on the wire. SKETCH:

RESEARCH: Ohm’s law is V = iR. Use the values: V = 12.0 V, B = 5.00 T, and 3 .R = Ω .F iL B= ×

In this case, since the top and bottom part of the loop have currents traveling in opposite directions, their forces will cancel. Only the right side of the loop will contribute to the force. SIMPLIFY: F = iLB, to the right (from the right-hand rule), / .V iR i V R= = Substitute the expression for I into the expression for F to get: / .F VLB R=

CALCULATE: ( )( )12.0 V 1.00 m 5.00 T 20.0 N3.00

F = =Ω

ROUND: F = 20.0 N to the right. DOUBLE-CHECK: The final expression makes sense since it is expected that if a larger voltage is applied, a larger force is attained.

27.63. THINK: As the alpha particle enters the region of the magnetic field, its motion will be deflected into a curved path. The radius of curvature is determined by the mass, charge, and initial velocity, and by the strength of the field. All quantities are given except for the velocity. The particle’s velocity can be determined by employing the law of conservation of energy. The period of revolution can be determined from the particle’s radius of curvature and velocity. 276.6 10 kg,mα

−= ⋅ 2 ,q eα = + 0.340 T,B =

and 2700 V.VΔ =

SKETCH:

RESEARCH: The radius of curvature is given by / .r mv q B= By conservation of kinetic energy,

212

q V mv= . The period of revolution is given by 2 rTvπ= .

SIMPLIFY: ( )19

2 527

2 2 1.6 10 C 270021 5.1 10 m/s2 6.6 10 kg

q Vq V mv v

m

−

−

⋅= = = = ⋅

⋅


1049

CALCULATE: ( )( )27 5

19

6.6 10 kg 5.1 10 m/s0.0309 m

2 1.6 10 C 0.340 Tr

−

−

⋅ ⋅= =

⋅

( )19

527

2 2 1.6 10 C 27005.12 10 m/s

6.6 10 kgv

−

−

⋅= = ⋅

⋅

( ) 75

2 0.031 m3.82 10 s

5.12 10 m/sT

π −= = ⋅⋅

ROUND: Rounding to two significant figures, 0.031 m,r = 55.1 10 m/s,v = ⋅ and 73.8 10 s.T −= ⋅ DOUBLE CHECK: All calculated values have the correct units. The numerical values are appropriate to the scale of the particle.

( )192 5

27

2 2 1.6 10 C 270021 5.1 10 m/s2 6.6 10 kg

q Vq V mv v

m

−

−

⋅= = = = ⋅

⋅

( )( )27 5

19

6.6 10 kg 5.1 10 m/s0.031 m

2 1.6 10 C 0.340 Tr

−

−

⋅ ⋅= =

⋅

It is also known that ( ) 7

5

2 0.031 m2 3.8 10 s.5.1 10 m/s

rTv

ππ −= = = ⋅⋅

27.64. THINK: The electric field component and the vertical component must cancel each other. SKETCH: Not necessary. RESEARCH: It is required that ( ) 0v B E v B B E B E B× = − × = − =

(since for any vector,

( ) 0).A B B× =

But since E

is not perpendicular to ,B

0.E B ≠ Note that ˆ150 N/CE z= −

and

( )ˆ ˆ50.0 20.0 μT.B y z= −

This scenario cannot occur. SIMPLIFY: Nothing to simplify. CALCULATE: No calculations are necessary. ROUND: There are no values to round. DOUBLE-CHECK: No Lorenz force can counteract an electric force in z-direction, if the particle is also traveling in z-direction, because the Lorenz force is always perpendicular to the velocity vector.

27.65. THINK: Determine the velocity in terms of the mass and see how this changes the answer. SKETCH: Not necessary. RESEARCH: / ,v qBr m= B = 0.15 T, r = 0.05 m and 276.6 10 kg.m −= ⋅ SIMPLIFY: It is not necessary to simplify.

CALCULATE: ( )( )( )19

527

1.6 10 C 0.15 T 0.05 m1.8 10 m/s

6.6 10 kgv

−

−

⋅= = ⋅

⋅

Note that for 3 ,4

m m′ = 4 ;3 / 4 3qBrv vm

′ = = the velocity increases by a factor of 4 / 3.

ROUND: 52 10 m/sv = ⋅ DOUBLE-CHECK: Since there is an inverse relationship between v and m, it makes sense that decreasing m by a factor of 3/4 increases v by factor of ( ) 13 / 4 4 / 3.− =

27.66. THINK: First determine the radius of curvature and then determine the amount the electron deviates over a distance of = 1.00 m.l Use 0.300 TB = as the Earth’s magnetic field. Convert the energy into Joules.


1050

SKETCH:

The dashed line is the path of the electron.

RESEARCH: Geometry gives 2 2 .d r r l= − − 2 ,mv p mErqB qB qB

= = = ( )197500 eV 1.609 10 J/eV .E −= ⋅

SIMPLIFY: It is not necessary to simplify.


( )( )31 3 19

19 4

2 9.11 10 kg 7.50 10 eV 1.602 10 J/eV9.7354 m

1.602 10 C 0.300 10 Tr

− −

− −

⋅ ⋅ ⋅= =

⋅ ⋅

( ) ( ) ( )2 29.7354 m 9.7354 m 1.00 m 0.051495 m,d = − − = upward from the ground.

ROUND: To three significant figures, the answer should be rounded to: d = 0.0515 m upward. DOUBLE-CHECK: Note that d << 1.00 m, as is expected since the magnetic field of the Earth is fairly weak.

27.67. THINK: Since the electric field from the plates will cause the proton to move in the negative y-direction, the magnetic field must apply a force in the positive y-direction. It is determined from the right-hand rule that B

must be in the negative z-direction.

SKETCH:

RESEARCH: 61.35 10 m/s,v = ⋅ V = 200. V and 335.0 10 md −= ⋅ , and / .E V d= It is required that

.vB E=

SIMPLIFY: E VBv vd

= =


36 3

200. V4.23 10 T

1.35 10 m/s 35.0 10 mB −

−= = ⋅

⋅ ⋅

ROUND: To three significant figures, 3 ˆ4.23 10 TB z−= − ⋅ DOUBLE-CHECK: The final expression for B makes sense. If the applied voltage is larger, one needs a larger magnetic field.

27.68. THINK: Determine the radius of curvature. This distance will allow the electron to be trapped in the field.


1051

SKETCH:

RESEARCH: The magnitude is given by B 0 ,F qv B= in the positive y-direction (by the right-hand rule).

0mvd r

q B= =

SIMPLIFY: 00

q Bdv

m=

CALCULATE: Not applicable. ROUND: Not applicable. DOUBLE-CHECK: It makes sense that for larger magnetic fields and for larger widths, the escape velocity is larger.

27.69. THINK: Equilibrium occurs when the net torque on the coil is zero. Use the values: 2 ,A d= 0.200 m,d = i= 5.00 A, m = 0.250 kg, B = 0.00500 T, and N = 30.

SKETCH:

RESEARCH: ( )g sin / 2 ,mg dτ θ= B cos .NiABτ θ= It is required that g B .τ τ=

SIMPLIFY: 2g B

1 sin cos 2

mgd Nid Bτ τ θ θ= =

1sin 2 2 2 tan tancos

NdiB NdiB NdiBmg mg mg

θ θ θθ

− = = =


1052

CALCULATE: ( )( )( )( )

( )( )1

2

2 30 0.200 m 5.00 A 0.00500 Ttan 6.9740

0.250 kg 9.81 m/s− = = °

θ

ROUND: The number of turns is precise, so it does not limit the precision of the answer. The rest of the values are given to three significant figures of precision, so it is appropriate to round the final answer to:

6.97θ = °. DOUBLE-CHECK: It makes sense that θ is inversely proportional to m, since the less the coil weighs, the more vertical it must be.

27.70. THINK: It can be deduced from symmetry that the net force is in the negative y-direction. Therefore, .B yF F=

SKETCH:

RESEARCH: From the book, sin .BF iLB θ= The objective is to sum up the forces due to each point of the semi-circle. This means I will integrate BF over the length of the wire.

SIMPLIFY: ( )1

0 0 0sin sin sin

l

y lF iLB dL iLB Rd iLBR d

π πθ θ θ θ θ= = = in the negative y-direction.

CALCULATE: ( )0cos 1 1 2yF iLRB iLRB iLRBπθ= = − − − = in the negative y-direction.

ROUND: Not applicable.

DOUBLE-CHECK: Note that 0

cos 0.xF iR Bdπ

θ θ= = This confirms the analysis based on symmetry.

27.71. THINK: The magnetic force on the proton due to the presence of the magnetic field will affect only the component of the proton’s velocity that is perpendicular to the magnetic field. SKETCH:

RESEARCH: (a) When the proton enters the magnetic field, the component of its velocity that is parallel to the field will be unaffected, so the proton advances along the z-axis at a constant speed. The component of the particle’s velocity that is perpendicular to the field will be forced into a circular path in the xy-plane. Thus,


1053

the trajectory of the proton will be a helix, as shown in the figure. The magnetic force on the proton is given by:

( ) .BF qv B q v v B qv B qv B qv B⊥ ⊥ ⊥= × = + × = × + × = ×

(b) The magnitude of the magnetic force is given by .BF qv B qv B⊥ ⊥= × = By Newton’s Second Law,

2 / .BF ma qv B mv r⊥ ⊥= =

(c) The period of the circular motion in the xy-plane projection is given by 2 .rTvπ

⊥

= The frequency is

given by 1/ .f T= (d) The pitch of the motion is .p v T= SIMPLIFY: (b) The radius of the trajectory projected onto the xy-plane is given by

( )osin 90 cos .mvmv mvr

qB qB qB⊥

+= = =

θ θ

(c) 2 2 .mv mT

v qB qB⊥

⊥

= =

π π

(d) ( )sin .p v Tθ= CALCULATE:

(b) ( )( ) ( )

( )( )

27 6

19

1.67 10 kg 1.00 10 m/s cos 60.010.44 mm.

1.60 10 C 0.500 Tr

−

−

⋅ ⋅ °= =

⋅

(c) ( )

( )( )

277

19

2 1.67 10 kg1.312 10 s,

1.60 10 C 0.500 TT

π −−

−

⋅= = ⋅

⋅ ( )

67

1 7.624 10 Hz.1.312 10 s

f−

= = ⋅⋅

(d) ( ) ( )( )6 71.00 10 m/s sin 60.0 1.312 10 s 113.6 mm.p −= ⋅ ° ⋅ =

ROUND: Rounding to three significant figures, (b) 10.4 mm (c) 71.31 10 s,T −= ⋅ 67.62 10 Hzf = ⋅ (d) 114 mm DOUBLE CHECK: All calculated values have correct units. The magnitudes are appropriate for subatomic particles.


1054

Chapter 28: Magnetic Fields of Moving Charges In-Class Exercises

28.1. c 28.2. a 28.3. a 28.4. a 28.5. b 28.6. e 28.7. d 28.8. d 28.9. a Multiple Choice

28.1. b 28.2. c 28.3. c 28.4. a 28.5. d 28.6. a 28.7. c 28.8. c 28.9. a 28.10. d Questions

28.11. The wires are twisted in order to cancel out the magnetic fields generated by these wires.

28.12. Since the currents running through the wire generate magnetic fields, these fields may overpower the magnetic field of the Earth and make the compass give a false direction.

28.13. No, an ideal solenoid cannot exist, since we cannot have an infinitely long solenoid. To a certain extent, yes, it renders the derivation void. However, the derivation is an approximation and is an important theoretical example.

28.14. In Example 28.1, the right hand rule implies that the magnetic dipole of the loop points out of the page. Application of the right hand rule to the straight wire tells us that the magnetic field produced by wire points out of the page. Assume the angle between the dipole moment and the field remains fixed. Since the dipole strength is also constant, the only quantity left to vary is field strength. If the potential energy is to be reduced, the loop must move towards a region of smaller magnetic field strength. That is, the loop must move away from the straight current-carrying wire.

28.15. By Coulomb’s Law, the electric force between the particles has magnitude ( )e2 2

0/ 4 .F q dπε= For the

magnetic force, the version of the Biot-Savart Law given in the text can be adapted to describe the magnetic field produced by a moving particle via the replacement

( ) ( ) s / / Idl dq dt dl dq dl dt qv with q charge and sv , the velocity of the source particle. The

magnetic field produced by one particle at the location of the other can be written as ( )30 / 4B qvd dμ π=

with ,v common velocity and ,d the separate of the particles. The magnitude of the magnetic force one

particle is given by ( )( ) ( )3 2 2 2e 0 0/ 4 ( ) / / 4 .F qvB qv qvd d q v dμ π μ π= = ⋅ = Since the vectors ,v d and

v d⋅ are mutually perpendicular (the site of the angle between any two of them is unity) the ratio of forces is 2

m e 0 0/F F vμ ε= which also 2 2m e/ / ,F F v c= where c is the speed of light.

28.16. The field is given by Ampere’s law ( )( ) 0 enc2 / 2 .B a b iπ μ+ = Current density is then given by:

( )( )2 2/J i b aπ= −

The area of interest is:

( )( )

2

0

/ 2

(2 ) / 2

a b A

B a b AJ

π

π μ

+ = + =

Chapter 28: Magnetic Fields of Moving Charges

1055

( )

( ) ( )

220

2 2

22

02 2

( ) 2

2 .

a b iB aa b b a

a b ai

a b b a

μ ππ π

μπ

+ = − ⋅ + −

+ − = ⋅

+ −

28.17. The magnetic field at point P would be zero. The contribution from part A would be zero since P lies along the axis of A. The currents through B and C points in opposite directions and yield magnetic fields that cancel out at P .

28.18. Ampere’s law states that, 0 ,C

Bdl iμ= but since B is constant the integral must be zero. If so, i is zero

everywhere and consequently 0J = everywhere.

28.19. (a) Since molecular hydrogen is diamagnetic, the molecules must have no intrinsic dipole moment. Since the nuclear spins cannot cancel the electron spins, the electron spins must be opposite to cancel each other. (b) With only a singly electron, the hydrogen atoms must have an intrinsic magnetic moment. Atomic hydrogen gas, if it could be maintained, would have to exhibit paramagnetic or ferromagnetic behavior. But ferromagnetism would require inter atomic interactions strong enough to align the atoms in domains, which is not consistent with the gaseous state. Hence one would expect atomic hydrogen to be paramagnetic.

28.20. The saturation of magnetizations for paramagnetic and ferromagnetic materials is of comparable magnitude. In both types of materials the intrinsic magnetic moments of the atoms arise from a few unpaired electron spins. Magnetization effects in ferromagnetic materials are more pronounced at low applied fields because the atoms come pre-aligned in their domains, but once both types of atoms have been forced into essentially uniform alignment, the magnetization they produce is comparable. For either type of material maximum magnetizations of order 2 66 310 A m / m 10 A/m=

magnetic dipole moment

per unit volume are typical.

28.21. The wire carries a current which produces a magnetic field. This magnetic field will deflect the electron by the Lorentz force in the left direction.

Problems

28.22. The force of wire 1 on wire 2 is ( ) ( )1 2 2 2 0 1 0 1 2/ 2 / 2 .F i LB i L i d i i L dμ π μ π→ = = = Since 1 22 ,i i=

( )21 2 0 1 / .F i L dμ π→ = Solving for the current 1i gives

( )( )( )( )

61 2

1 70

0.0030 m 7.0 10 N0.23 A.

4 10 T m/A 1.0 mdF

iL

ππμ π

−

→−

⋅= = =

⋅

The current on the other wire is 2 0.46 A.i =

28.23. The magnetic field created by the wire is given by the Biot-Savart Law ( )0 / 2B i rμ π= . The force on the

electron is given by the Lorentz force ( )0 / 2 .F qvB qv i rμ π= = The acceleration of the electron is 19 5 7

12 2031

(1.602 10 C)(4.0 10 m/s)(4 10 T m/A)(15 A)4.2 10 m/s

2 2 (9.109 10 kg)(0.050 m)qv iFa

m mrπμ

π π

− −

−

⋅ ⋅ ⋅= = = = ⋅

⋅

The direction of the acceleration is radially away from the wire.

28.24. The magnitude of the magnetic field created by a moving charge along it is line of motion is zero. By the Biot-Savart Law,


1056

0 02 2

ˆˆ0,

4 4qdv rids rB

r rμ μπ π

××= = =

since the angle between the angle between the velocity and the position vector r is zero. The situation is the same for an electron and a proton.

28.25. The field along the axis of a current loop of radius R as measured at a distance x from the center of the loop is

( )2

03/22 2

.2

i RBx R

μ=

+

The current of the loop must be

( ) ( ) ( )( )( )

( )3/22 23/2 6 62 2

5 92 27 6

0

2 2.00 10 m 6.38 10 m26.00 10 T 7.14 10 A.

4 10 T m/A 2.00 10 m

x R Bi

Rμ π−

−

⋅ + ⋅+ = = ⋅ = ⋅⋅ ⋅

28.26. The magnitude of the magnetic field at a perpendicular distance from a current-carrying wire is

0 .2

iB

rμπ ⊥

=

The average magnetic field will occur at the average distance from the wire to the ammeter. The maximum distance is from the wire to the corner of the ammeter and the minimum distance is from the wire to the midpoint of a side of the ammeter. The average distance is half of the sum of these distances:

, ave2 1 ,4

r l⊥

+=

where l is the side length of the ammeter. Therefore, the current through the wire is:

( ) ( )( )( )( )( )

4ave, ave ave

70 0

2 2 1 2 2 1 0.0300 m 3.00 G 10 T/G227.2 A.

4 4 4 10 T m/A

lBr Bi

π ππμ μ π

−⊥

−

+ += = = =

⋅

28.27. THINK: A force due to the magnetic field generated by a current carrying wire acts on a moving particle. In order for the net force on the particle to be zero, a second force of equal magnitude and opposite direction must act on the particle. Such a force can be generated by another current carrying wire placed near the first wire. Assume the second wire is to be parallel to the first and has the same magnitude of current. The wire along the x-axis has a current of 2 A oriented along the x-axis. The particle has a charge of 3 μCq = − and travels parallel to the z-axis through point ( , , ) (0,2,0).x y z =


1057

SKETCH:

RESEARCH: The magnetic field produced by the current is given by the ( )0 / 2 .B i rμ π= The force on the particle is given by the Lorentz force, 0 .F qv B= SIMPLIFY: If the wires carry the same current then the new wire must be equidistant from the point that the particle passes through the xy-plane. Only then will the magnetic force on the particle due to each wire be equal.

By the right hand rule, the currents will be in the same direction. This means that 1 2 .r r=

CALCULATE: The requirement 1 2r r= means that the second wire should be placed parallel to the first wire (parallel to the x-axis) so that it passes through the point ( , , ) (0,4,0).x y z = ROUND: Not necessary. DOUBLE-CHECK: It is reasonable that two wires carrying the same current need to be equidistant from a point in order for the magnitude of the force to be the same.

28.28. THINK: The current through the wire creates a magnetic field by the Biot-Savart Law. The straight part of the wire only creates a magnetic field at points perpendicular to it. Therefore this part of the wire can be ignored. The magnetic field at the center of the semicircle is created by the charge moving through the semicircle. SKETCH:

RESEARCH: The Biot-Savart Law can be employed in the form 02

sin4

idB dsr

μ θπ

= . Going around the

semicircle, the angle φ can be related to the current element by .ds rdφ= SIMPLIFY: Performing the integration gives

0 0 0020

0

sin sinsin sin .4 4 44

i i iiB Rdr rr r

ππμ μ θπ μ θθ μ θφ φ

π ππ = = = =

The angleθ between the current and the radial vector r is 90° for the loop, thus ( )0 / .4B i rμ=

CALCULATE: ( ) ( )

( )7

54 10 T m/A 12.0 A

3.76991 10 T4 0.100 m

Bπ −

−⋅

= = ⋅

ROUND: The values are given to three significant figures, thus the magnetic field produced by the wire is 53.77 10 TB −= ⋅ and points into the page.

DOUBLE-CHECK: The magnetic field is very small, as would be expected from a real-world point of view.


1058

28.29. THINK: Each of the wires creates a magnetic field at the origin. The sum of these fields and the Earth’s magnetic field will produce a force on the compass, causing it to align with the total field. The wires carry a current of 1 2 25.0 A.i i= = The Earth’s magnetic field is 5

Eˆ2.6 10 T.B y−= ⋅

SKETCH:

RESEARCH: The magnetic field produced by a wire is ( )0 / 2 .B i dμ π=

SIMPLIFY: The magnetic field of wire 1 is ( )1 0 1 1ˆ( ) / 2 .B i y dμ π= −

Wire 2 produces a magnetic field

of ( )2 0 2 2ˆ / 2 .B i x dμ π=

The sum of the magnetic fields is 0 1 0 2

net 1 2 E E

1 2

ˆ ˆ .2 2

i iB B B B y x B

d dμ μπ π

= + + = − + +

CALCULATE: ( )( )

( )( )( )

( )7 7

5net

5 6

4 10 T m/A 25.0 A 4 10 T m/A 25.0 Aˆ ˆ ˆ2.6 10 T

2 0.15 m 2 0.090 mˆ ˆ5.5555 10 T 7.3333 10 T

B y x y

x y

π ππ π

− −−

− −

⋅ ⋅= − + + ⋅

= ⋅ − ⋅

The direction of the field is 6

15

7.3333 10 Ttan 7.5196 .5.5555 10 T

θ−

−−

− ⋅= = − ° ⋅

ROUND: The angle is accurate to two significant figures. The compass points 7.5° below the x-axis. DOUBLE-CHECK: This is a reasonable answer. The compass points towards the east if y is north.

28.30. THINK: The coil will levitate if the force from the magnetic field cancels the force of gravity. The coils have radii of 20.0 cm.R = The current of the bottom coil is i and travels in the clockwise direction. By the right hand rule the top coil has a current of the same magnitude, moving in a counter clockwise direction. The mass of the coils is 0.0500 kg.m = The distance between the coils is 2.00 mm.d = SKETCH:

RESEARCH: The force of gravity is g .F mg= The magnetic force on the top coil due to the bottom coil is

( ) ( )B 0 1 2 0 1 2/ 2 2 / 2 .F i i L d i i R dμ π μ π π= =

SIMPLIFY: Equating the two forces give 2

0 1 2 02.

2i i R i R

mgd d

μ π μπ

= = The amount of current is 2

0

mgdiRμ

= or

0

.mgdiRμ

=


( )( )

2

7

0.0500 kg 9.81 m/s 0.00200 m62.476 A

4 10 T m/A 0.200 mi

π −= =

⋅


1059

ROUND: Reporting to 3 significant figures, the current in the coils is 62.5 A and travel in opposite directions. DOUBLE-CHECK: Dimensional analysis provides a check:

( )

2 2

2 2

kg m/s m kg m m A A s mkg m m A A

m/A m N/ A m m m s kg m m m si

T

= = = = .

28.31. THINK: The current carrying wires along the x- and y-axes will each generate a magnetic field. The superposition of these fields generates a net field. The magnitude and direction of this net field at a point on the z-axis is to be determined. SKETCH:

RESEARCH: Both currents produce a magnetic field of ( )0 / 2 .B i rμ π=

The magnetic field produced by

the wire along the x-axis gives ( ) ( )1 0 / 2 .B i y bμ π= −

The wire along the y-axis creates a magnetic

field of ( )2 0 / 2 .B ix bμ π=

SIMPLIFY: The total magnetic field is then 0net 1 2 .

2i

B B B x yb

μπ

= + = −

The magnitude of the field

is ( )220 0 021 1 .

2 2 2i i i

Bb b b

μ μ μπ π π

= + − = = The direction of the field is 1 0 0tan2 2

i ib b

μ μθπ π

− −=

in the x-y

plane at a height of b.

CALCULATE: 1tan 1 45− = − ° in the x-y plane at point b.

ROUND: Not applicable. DOUBLE CHECK: Both the right hand rule and the symmetry of the problem indicates that the net field should be in the fourth quadrant.

28.32. THINK: The loop creates a magnetic field at its center by the Biot-Savart Law. The loop has length 0.100 ml = and carries a current of 0.300 A.i =

SKETCH:


1060

RESEARCH: The Biot-Savart Law states 02

sin .4

i dsdBr

μ θπ

= ⋅ The angle θ is found by using the equations:

sin / ,d rθ = 2 2 ,r s d= + and / 2.d l=

SIMPLIFY: The field due to one side of the loop is ( )

0 02 3/22 2

.4 4

i idd dsdB dsr s d

μ μπ π

= ⋅ =+

Since there are

four sides, the total loop is four times this value. The total magnetic field is then

( ) ( )0 0

3/2 3/202 2 2 2

0 0 0 02 2 2 2 2

0

24 44 4

2 2 2 8022

d d

d

d

id idds dsB dBs d s d

sid i i idd ldd s d d d

μ μπ π

μ μ μ μπ π ππ

−= = ⋅ =

+ +

= = − = = +

CALCULATE: ( )( )

( )7

68 4 10 T m/A 0.300 A

3.394 10 T0.100 m

Bπ

π

−−

⋅= = ⋅

ROUND: To 3 significant figures, the magnetic field at the center of the loop is 63.39 10 T.B −= ⋅ DOUBLE-CHECK: The current is small, so the magnetic field it generates is expected to be small. This is a reasonable value.

28.33. THINK: In order for wire 1 to levitate, the forces on it must cancel. Both wire 2 and 3 will create magnetic fields that will interact with wire 1. Both wires create forces with horizontal and vertical components. The horizontal components will add deconstructively. The vertical components however will add constructively. Therefore, only the vertical components need be calculated. Wires 2 and 3 each carry a current of 600. A.i = All three wires have a linear mass density of 100. g/m.λ = The wires are arranged as shown in the figure. SKETCH:

RESEARCH: The force of gravity on the wire is g .F mg= The force between two wires carrying current is

( )21 0 1 2 / 2 .F i i L dμ π=

SIMPLIFY: The vertical component of the magnetic force for one wire is 0 3 1 0 3 131 .

2 ( / 2)i i L i i L

Fh h

μ μπ π

= = The

total force due to the wires is then 0 3 1B 31

22 .

i i LF F

hμπ

= = Equating this to the force of gravity gives:

0 3 12.

i i Lmg Lg

hμλπ

= = Solving for the current 1i gives: 10 3

.2

h gii

π λμ

=

CALCULATE: ( )3 2

1 7

(0.100 m)(100. 10 kg/m)(9.81 m/s ) 204.375 A2 4 10 T m/A (600. A)

i ππ

−

−

⋅= =⋅

ROUND: The current of wire 1 required to levitate is 1 204 A.i = DOUBLE-CHECK: The current in wire 1 is on the same order of magnitude as the other currents. This is a reasonable answer.


1061

28.34. THINK: The net field is a superposition of the fields created by the top wire, the bottom wire and the loop. The wires are 2.00 cm apart and carry a current of 3.00 A.i = The radius of the loop is 1.00 cm.r = SKETCH:

RESEARCH: The magnetic field produced by an infinite wire is ( )0 / 2 .B i rμ π= A semi infinite wire is

half this value, ( )0 / 4 .B i rμ π= A full loop produces a magnetic field of ( )0 / 2 .B i rμ= The half loop

produces half of this, ( )0 / 4 .B i rμ= SIMPLIFY: By the right hand rule, the magnetic field points into the page. The magnetic field is the sum of all the fields.

0 0 0 0net top bottom loop

2 14 4 4 4

i i i iB B B B

r r r rμ μ μ μπ π π

= + + = + + = +

CALCULATE: ( )( )

( )7

4net

4 10 T m/A 3.00 A 2 ˆ1 1.54 10 T4 0.0100 m

B zπ

π

−−

⋅ = + = − ⋅

ROUND: To 3 significant figures, the magnetic field at the origin is 4 ˆ1.54 10 T .z−− ⋅ DOUBLE-CHECK: The field due to a single infinite wire similar to the wires in the problem would be

( )( ) ( )( )7 54 10 T m/A 3.00 A / 2 0.0100 m 6.00 10 T,B π π− −= ⋅ = ⋅ which is similar to the result. Therefore,

the result is reasonable.

28.35. THINK: The wire creates a magnetic field that produces a Lorentz force on the moving charged particle. The question asked for the force if the particle travels in various directions. The velocity is 3000 m/s in various directions. SKETCH:

RESEARCH: The magnetic field produced by an infinite wire is ( )0 / 2 .B i dμ π= By the right hand rule

the field points in the positive z-direction. The force produced by the magnetic field is .F qv B= ×

SIMPLIFY: The force is given by ( )0 0ˆ ˆ ˆ2 2q i q i

F qv B v z v n zd d

μ μπ π

= × = × = ⋅ × where n is the direction of the

particle.


( ) ( ) ( )7

29.00 C 4 10 T m/A 7.00 A

ˆ ˆ ˆ ˆ3000. m/s 1.89 10 N2 2.00 m

F n z n zπ

π

−−

⋅= ⋅ × = ⋅ ×

Note that ˆ ˆ ˆ,x z y× = − ˆ ˆ ˆ,y z x× = and ˆ ˆ 0.z z− × =

ROUND: The force should be reported to 3 significant figures.


1062

(a) The force is 2 ˆ1.89 10 N F y−= − ⋅

if the particle travels in the positive x-direction.

(b) The force is 2 ˆ1.89 10 N F x−= ⋅

if the particle travels in the positive y-direction. (c) The force is 0F = if the particle travels in the negative z-direction. DOUBLE-CHECK: The right hand rule confirms the directions of the forces for each direction of motion of the particle.

28.36. THINK: The wire produces a magnetic field that creates a force on the loop. The wire has current of w 10.0 Ai = and is 0.50 md = away from the bottom wire of the loop. The loop carries a current of

l 2.00 Ai = and has sides of length 1.00 m.a = SKETCH:

RESEARCH: The force on two wires carrying a current is ( )0 1 2 / 2 .F i i dμ π= The torque is given by

.r Fτ = ×

SIMPLIFY: The forces on part ② and ④ cancel each other. The force on ① is ( )1 0 w l / 2F i i dμ π= and

points towards the long wire. The force on ③ is 3 0 w l / 2 ( )F i i d aμ π= + and points away from the long

wire. The total force is then 0 w l1 3net

1 12i i

F F Fd d a

μπ

= + = − +

and points towards the long wire. Because

the force and the length between the axis of rotation are parallel there is no torque on the loop.

CALCULATE：( ) π

π

−−

⋅ = − = − ⋅

76

net

4 10 T m/A (10.0 A)(2.00 A) 1 1 5.33333 10 N2 0.50 m 1.50 m

F y

ROUND: The force is reported to two significant figures. (a) The net force between the loop and the wire is 6 ˆ5.3 10 N.F y−= − ⋅ (b) There is no torque on the loop. DOUBLE-CHECK: The force between the long wire and the lower arm of the loop is attractive, because the currents are in the same direction. The currents of the long wire and the upper arm of the loop are in opposite directions, therefore the force is repulsive. Since the lower arm is closer to the long wire, the attractive force dominates, and the net force is in the negative y direction, as calculated.

28.37. The magnetic field at the center of the box is the sum of the fields produced by the coils. A coil produces a

magnetic field of ( )

20

3/22 2.

2

NiRB n

x R

μ=

+

The magnetic field produced by the coil on the x z− plane is

( ) 7 2

5xz 3/22 2

4 10 T m/A (30)(5.00 A)(0.500 m)( ) 6.66 10 T

2 (0.500 m) (0.500 m)B y y

π −−

⋅= + = ⋅

+


1063

The magnetic field produced by the other coil has the same magnitude but points in the negative x-direction. Therefore 5

tot 6.66 10 T[ - ].B x y−= ⋅ + The magnitude of the field is 52 6.66 10 T−⋅ ⋅ or 59.42 10 T−⋅ . The direction of the field is at an angle of 45° from the negative x-direction towards the

positive y-axis.

28.38. Each side of the loop will create the same magnetic field at the center of the loop. The total field is 4 times the field of one side. The field at the center is given by the Biot-Savart Law:

0 03 2 sin .

4 4i ids rdB ds

r rμ μ θπ π

×= =

Since sin / ,d rθ = the differential element of magnetic field is

0 03 2 2 3/2 .

4 4 ( )i idd dsdB ds

r d sμ μπ π

= =+

Integration gives

0 0 0 0 02 2 3/2 2 2 3/2 2 2 20

0

2.

4 4 2 2( ) ( ) 2 2 2

dd d

d

sid id id i ids ds dBdd s d s d dd d s

μ μ μ μ μπ π π π π−

= = = = = + + +

The total field is then 0 0 0tot

2 2 2 24 .

( / 2)i i i

B Bd L Lμ μ μ

π π π= = = =

28.39. Using Ampere’s Law, the magnetic field at various points can be determined. 0 enclosed .B ds iμ= For the

cylinder, assuming the current is distributed evenly, 0 enc2B r iπ μ= or ( )0 enc / 2 .B i rμ π= The field at

a 0r r= = is zero since is does not enclose any current a 0.B = The field at br r R= < is

( )7260 enc 0 b 0 b

b tot 2 2 2b b

4 10 T m/A (1.35 A)(0.0400 m)1.08 10 T.

2 2 2 2 (0.100 m)i r ir

B ir r R R

πμ μ π μπ π π π π

−−

⋅ = = = = = ⋅

Note that i is

equal to the fraction of total area of the conductor’s cross section and the total current. The field at

cr R= is ( )( )

( )7

60 enc 0 totc

c

4 10 T m/A 1.35 A2.70 10 T.

2 2 2 0.100 mi i

Br R

πμ μπ π π

−−

⋅= = = = ⋅ The field at dr R> is

( )( )( )

760 enc

dd

4 10 T m/A 1.35 A1.69 10 T.

2 2 0.160 mi

Br

πμπ π

−−

⋅= = = ⋅ By inspection it can be seen that the magnetic field

at br , cr and dr the magnetic field will point to the right.

28.40. The loop creates a magnetic field of l 0 / (2 )B i Rμ= at its center and is directed upwards. Out of the page. Both wires contribute a magnetic field of w 0 / (2 )B i Rμ π= pointing out of the page. The total fields is then

0tot l w

22 (1 ),2

iB B B

Rμ

π= + = + and points out of the page.


1064

28.41.

The current that flows through a ring of radius r which lies in the region a r b< < is given by

2 2 20 0 0 02 | ( ).

r raa

i J dA J d J J r aπρ ρ πρ π= = = = − To find the magnetic field employ Ampere’s Law

0 enclosed .B ds iμ⋅ = For a cylinder this becomes ( ) 0 enc2B r iπ μ= or 0 enc / (2 ).B i rμ π= If r a< then

0r aB < = , thus if a r b< < then 2 2enc 0 ( )i J r aπ= − and

2 2 2 20 0 0 0( ) ( )

.2 2a r b

J r a J r aB

r rμ π μ

π< <

− −= = If r b>

then 2 2enc 0 ( )i J b aπ= − and

2 20 0 ( )

.2r b

J b aB

rμ

>

−= Note that if r b= then

2 20 0 ( )

.2a r b r b

J b aB B

bμ

< < >

−= =

28.42.

The current within a loop of radius Rρ ≤ is given by

3320 0 0

0 00 0 0

2 2 2( ) 2 ( ) 2 | .

3 3r r r rJ J J rr ri J r dA J r r dr J r dr r dr

R R R Rπ π ππ π

′ ′′ ′ ′ ′ ′ ′ ′= = = = = =

The magnetic field is given by Ampere’s Law

( ) 0 enc0 enc2 .

2i

B ds B r i Br

μπ μ

π⋅ = = =

The magnetic field in the region r R< is 3 2

0 0 0 02.

2 3 3J r J r

Br R R

μ π μπ

= =

The magnetic field in the region

r R> is 3 2

0 0 0 02.

2 3 3J R J R

Br R r

μ π μπ

= =


1065

28.43. THINK: To find the magnetic field above the center of the surface of a current carrying sheet, use Ampere’s Law. The path taken should be far from the edges and should be rectangular as shown in the diagram. The current density of the sheet is 1.5 A/cm.J = SKETCH:

RESEARCH: The direction of the magnetic field is found using the right hand rule to be +x above the surface of the conductor. Ampere’s Law states 0 enclosed2 .B ds B r iπ μ= =

SIMPLIFY: Note that sections 1 and 3 are perpendicular the field. 0B ds⋅ = for these two sections. If the path of 4 and 2 has a length of L, then by Ampere’s Law, 1 2 0 enclosed 0 .B ds B L B L i JLμ μ⋅ = + = =

By

symmetry 1 2 .B B= Thus, 1 02B Jμ= or 1 0 / 2.B Jμ=

CALCULATE: ( )7

51

4 10 T m/A (1.5 A/cm)(100 cm/m)9.42478 10 T

2B

π −−

⋅= = ⋅

ROUND: The magnetic field is accurate to two significant figures. The magnetic field near the surface of the conductor is 5

1 9.4 10 T.B −= ⋅ DOUBLE-CHECK: The form for the magnetic field is similar to that of a solenoid. It is divided by a factor of 2, which makes sense when considering the setup of a solenoid. The form of the equation is similar to that of question 28.12. This makes sense because the magnetic field inside a solenoid is generated by a current carrying wire on both sides of the Amperian loop, whereas the field generated by the flat conducting surface originates on one side of the Amperian loop only. In effect, the flat conductor can be seen as similar to half a solenoid, flattened out. See figure 28.21 in the text for a visual.

28.44. THINK: The magnetic field is the sum of the field produced by the wire core CB and the sheath S .B The wire has a radius of 1.00 mm.a = The sheath has an inner radius of 1.50 mmb = and outer radius of

2.00 mm.c = The current of the outer sheath opposes the current in the core. SKETCH:

RESEARCH: The current density of the core is 2C / ( )J i aπ= and the current density of the sheath is

2 2S /[ ( )].J i c bπ= − − The enclosed current is calculated by enclosed .i JdA=

The magnetic field is derived

using Ampere’s Law: 0 enclosed .B ds iμ⋅ =

SIMPLIFY: When the radius is within the core , r a≤ , the magnetic field is 2

0 enc 0 0 2 0 0

2 20

02 2

2

22

r

r aiB ds B r i JdA rd dra

i r ria a

ππ μ μ μ θ

πμ π μπ

≤⋅ = = = =

= =


1066

or 2

0 02 2 .

2 2r ai irrBr a a

μ μπ π≤ = = If the radius is between the core and the sheath, ,a r b< <

02a r bB ds B r iπ μ< ≤= = or 0 / (2 ).a r bB i rμ π< ≤ = Within the sheath, ,b r c< < the magnetic field is

2 2 22

0 enc 0 02 2 2 2 2 20

22 | 12( ) ( )

r rb r c C bb

i i r r bB ds B r i i rd dr i ic b c b c b

π ππ μ μ θ μ μππ< <

− −= = = + = − = − − − −

2 20

2 212b r c

i r bBr c b

μπ< <

−= − −

If the radius is outside of the cable, ,r c≥ then the magnetic field is 0 enc 02 ( ) 0r cB ds B r i i iπ μ μ≥⋅ = = = − =

or 0.r cB ≥ = In summary the magnetic fields of various regions are 2 2

0 0 02 2 2, , 1 , 0.

2 22r a a r b b r c r cir i i r bB B B B

r ra c bμ μ μ

π ππ≤ < ≤ < < ≥

−= = = − = −

CALCULATE: In order to graph the behavior of the magnetic field as a function of the radius, set the

magnetic field in units of 0 .2u i

aπ

ROUND: There is no need to round. DOUBLE-CHECK: Note that the magnetic field outside of the coaxial cable is zero. These cables are used when equipment that is sensitive to magnetic fields needs current.

28.45. THINK: Ampere’s Law can be used to determine the magnitude of the magnetic field in the two regions. SKETCH: A sketch is included at the end of the SIMPLIFY step, once the two equations have been found. RESEARCH: The current with the conductor is given by ( ) .i J r dA= The magnetic field is found using

Ampere’s Law 0 enclosedB ds iμ= or 0 enc

2i

Br

μπ

= .

SIMPLIFY:

( ) ( )( )( )

/ /0 0 00 0

/ 0/0

2 / /0

( ) 2 ( ) 2 2 ( ) |

( ) ( 0) 2

(1 ) 2

r r r R r R r

r R R

r R r R

i J r dA J r r dr J r e dr J R R r e

R R r e R R e J

R e Rre J

π π π

π

π

′ ′− −

− −

− −

′ ′ ′ ′ ′ ′= = = = − +

= − + − − +

= − −

If r R< then 2 / 2 /0 0 00( ) 2 [ ( ) ].

2r R r R

r RJ

B R R R r e J R R R r er r

μ μπ

π− −

< = − + = − + If r R> then

22 / 2 2 1 2 10 0 0 0 0

0( ) 2 2 [1 2 ]2

R Rr R

J J RB R R R R e J R R e e

r r rμ μ μππ

− − −> = − + = − = − .


1067

CALCULATE: There are no values to substitute. ROUND: There are no values to round. DOUBLE-CHECK: Note that the two computed formulas agree when .r R=

28.46. The magnetic field in a solenoid is given by the equation:

( ) ( )7 30

10004 10 T m/A 2.00 A 6.28 10 T.0.400 m

B niμ π − − = = ⋅ = ⋅

28.47. The magnetic field in a solenoid is given by 0 .B inμ= Let the magnetic field of solenoid b be 0 .bB inμ= The magnetic field of solenoid a is ( ) ( ) ( ) ( )0 04 / 3 4 / 3 4 / 3 .a bB i N L in Bμ μ= = = The ratio of solenoid A magnetic field to that of solenoid B is 4:3.

28.48. The magnetic field at a point 1.00 cmr = from the axis of the solenoid will be the sum of the field due to the solenoid and the field produced by the wire. The solenoid has a magnetic field of S 0 SB i nμ= along the axis of the solenoid.

The wire produces a field which is perpendicular to the radial vector of ( )w 0 w / 2 .B i rμ π= The magnitude of the field is then

2 2 2 2tot S w 0 S w( ) ( / 2 )B B B i n i rμ π= + = +

( ) ( )( )( ) ( )( )

22

7 -1 4tot

10.0 A4 10 T m/A 0.250 A 1000 m 3.72 10 T.

2 0.0100 mB π

π− −

= ⋅ + = ⋅

28.49. (a) The magnetic field produced by the wire is

( ) ( )( ) ( )( )7 50 / 2 4 10 T m/A 2.5 A / 2 0.039 m 1.3 10 T.B i rμ π π π− −= = ⋅ = ⋅

(b) The magnetic field of the solenoid is

( )( )7 20

324 10 T m/A 2.5 A 0.010 T 1.0 10 T.0.01 m

B inμ π − − = = ⋅ = = ⋅

This field is much larger for the solenoid than the wire.

28.50. The magnetic field of a loop is 2

02 2 3/2 .

2 ( )i RB

x Rμ

=+

Therefore a coil of N loop produces a field of2

02 2 3/2 .

2 ( )iN RB

x Rμ

=+

Let / 2x R= gives

3/22 2

0 0 02 2 3/2 3 3/2

4 .2 5 2( ) 2 (5 / 4)iN iN iNR RB

Rx R Rμ μ μ = = = +

The field at the center of the coils is then

( )73/2 3/20 6

tot

4 10 T m/A (0.123 A)(15)4 42 2.21 10 T.5 5 (0.750 m)

i NB B

R

πμ −−

⋅ = = = = ⋅


1068

28.51. THINK: If the perpendicular momentum of a particle is not large enough, its radius of motion will not be large enough to enter the detector. The minimum momentum perpendicular to the axis of the solenoid is determined by a condition such that the centripetal force is equal to the force due to the magnetic field. SKETCH:

RESEARCH: Since the particle originates from the axis of the detector, the minimum radius of the circular motion of the particle must be equal to the radius of the detector as shown above. The magnetic force on the particle is .F qvB= Centripetal acceleration is 2 / .Ca v r= The magnetic field due to the solenoid is 0 .B inμ=

SIMPLIFY: Using Newton’s Second Law, the momentum is 2 / .qvB mv r mv p qrB= = = Therefore, the minimum momentum is 0 .p qrinμ=

CALCULATE: Substituting the numerical values yields.

( )( )( )( )( )7 19 2 1 194 10 T m/A 1.602 10 C 0.80 m 22 A 550 10 m 1.949 10 kg m/sp π − − − −= ⋅ ⋅ ⋅ = ⋅

ROUND: Rounding the result to two significant figures gives 191.9 10 kg m/s.p −= ⋅ DOUBLE-CHECK: This is a reasonable value.

28.52. The magnetic potential energy of a magnetic dipole in an external magnetic field is given by .U Bμ= −

Therefore, the magnitude of the difference in energy for an electron “spin up” and “spin down” is

up down 2 .U U U BμΔ = − = This means the magnitude of the magnetic field is / 2 .B U μ= Δ

( )25

24 2

9.460 10 JPutting in the numerical values gives 0.05094 T.2 9.285 10 A m

B−

−

⋅= =⋅

28.53. The energy of a dipole in a magnetic field is .U Bμ= − The dipole has its lowest energy min .U Bμ= −

and its highest energy max .U Bμ= The energy required to rotate the dipole from its lowest energy to its highest energy is 2 .U BμΔ = This means that the thermal energy needed is UΔ which corresponds to a temperature / 2 / .B BT U k B kμ= Δ = Substituting the numerical values of the dipole moment of hydrogen atom and 0.15 TB = yields

( )( )( )

24

23

2 9.27 10 J/T 0.15 T0.20 K.

1.38 10 J/KT

−

−

⋅= =

⋅


1069

28.54.

The magnetic permeability of aluminum is ( )Al 01 .μ χ μ= + Applying Ampere’s Law around an Amperian loop of radius r gives

enc(2 ) .B ds B ds B r iπ μ= = =

The current enclosed by the Amperian loop is 2

enc 2 .ri iR

ππ

= Therefore, the magnetic field inside a wire is

given by 2 .2

irBR

μπ

= This means the maximum magnetic field is located at the surface of the wire where

the magnitude is .2

iBR

μπ

= Thus, the maximum current is

( )( )( )

( )( )( )3

maxmax 5 7

Al 0

2 1.0 10 m 0.0105 T252 A.

1 1 2.2 10 4 10 T m/A

RBi

ππχ μ π

−

− −

⋅= = =

+ + ⋅ ⋅

28.55. The magnitude of the magnetic field inside a solenoid is given by m 0 ( / ).B in i N Lμ κ μ= = Thus the relative magnetic permeability mκ is given by the equation:

( ) ( )( ) ( ) ( )

2

m 70

2.96 T 3.5 10 m54.96 55.

4 10 T m/A 3.0 A 500BL

iNκ

μ π

−

−

⋅ ⋅= = = ≈

⋅ ⋅ ⋅

28.56.

The magnetic permeability of tungsten is ( )W 01 .μ χ μ= + Applying Ampere’s Law around an Amperian loop of radius r gives

enc(2 ) .B ds B ds B r iπ μ= = =

The current enclosed by the Amperian loop is 2

enc 2 .ri iR

ππ

= Therefore, the magnetic field is

( ) ( )( )( )( )( )

5 7 3W 0 3

2 23

1 6.8 10 4 10 T m/A 3.5 A 0.60 10 m12.9 10 T.

2 2 1.2 10 m

iB r

R

πχ μπ π

− − −−

−

+ ⋅ ⋅ ⋅ += = = ⋅ ⋅

28.57. THINK: To determine the magnetic moment, the effective current of the system is needed. This implies the speed of the ball is required.


1070

SKETCH:

RESEARCH: The ball travels in a circular orbit and it travels a distance of 2 Rπ in time T , where T is the time for one revolution. The effective current is given by / .i q T= Since 2 / ,T R vπ= this becomes

( )/ 2 .i qv Rπ= The effective magnetic moment is ( )2 / 2 / 2.iA qv R R qvRμ π π= = = From the centripetal

force, it is found that the speed is 2 / / .mv R F v FR m= =

SIMPLIFY: Combining the above results yields 1 .2

FRq Rm

μ =

CALCULATE: Putting in the numerical values gives

( ) ( )( ) ( )6 5 225.0 N 1.00 m1 2.00 10 1.00 m 1.118 10 A m .2 0.200 kg

Cμ − −= ⋅ = ⋅

ROUND: Keeping 3 significant figures gives 5 21.12 10 A m .μ −= ⋅ DOUBLE-CHECK: This magnetic moment is appropriately small for a small charge moving at a low velocity.

28.58. THINK: The magnetic field due to a proton is modeled as a dipole field. Using the value of the magnetic field, the potential energy of an electron spin in the magnetic field is .U Bμ= − ⋅

SKETCH:

RESEARCH: The electron field due to an electric dipole is given by ( )30/ 2 .E P Rπε=

The

corresponding magnetic field is obtained by replacing ( )01/ 4πε with ( )0 / 4μ π and P

with .μ

Thus, ( )30 / 2 .B Rμ μ π=

SIMPLIFY: The energy difference between two electron-spin configurations is anti parallel

03

0

03

0

( ) ( )

2 22

ee

Pe e

e P

U U U

B B

Ba

a

μ μμ μμ μπ

μ μ μπ

Δ = −

= − − − −

= =

=

CALCULATE: Inserting all the numerical values yields


1071

( )( )( )( )

7 24 2625 6

311

4 10 T m/A 9.27 10 J/T 1.41 10 J/T3.528 10 J 2.204 10 eV.

5.292 10 mU

π

π

− − −− −

−

⋅ ⋅ ⋅Δ = = ⋅ = ⋅

⋅

ROUND: Rounding the result to three significant digits produces 62.20 10 eV.U −Δ = ⋅ DOUBLE-CHECK: This is reasonable. A small difference in potential is expected for these small particles.

28.59. THINK: The classical angular momentum of rotating object is related to its moment of inertia. To get the magnetic dipole of a uniformly changed sphere, the spherical volume is divided into small elements. Each element produces a current and a magnetic dipole moment. The dipole moment of all elements is then added to get the net dipole moment. SKETCH:

RESEARCH: (a) The classical angular momentum of the sphere is given by ( ) 22 / 5 .L I mRω ω= = (b) The current produced by a small volume element dV is / (2 ).i dVρ ω π= Thus the magnetic dipole

moment of this element is 2( sin ) .2

dVd rρωμ π θπ

= Summing all the elements gives

( )( )22 2 2

0 0 0sin sin .

2R r r dr d d

π π ρωμ θ θ θ φ=

(c) The gyromagnetic ratio is simply the ratio of the results from parts (a) and (b): / .e Lγ μ= SIMPLIFY:

(b) 4 3

0 0

3 4

0 015 3 5 5cos 2

cos 01

2 sin2

sin

4(1 cos ) cos5 3 5 3 5

R

R

r drd

d r dr

R x R Rd x

π

π

π

ρωμ π θ θ

ρπω θ θ

ρπω θ θ ρπω ρπω−

= ⋅

= ⋅

= − − = − + =

Since 34 ,3

R qρ π = the magnetic moment becomes 2 / 5.q Rμ ω=

(c) Taking the ratio of the magnetic dipole moment and the angular momentum yields: 2

2

5 .2 25

e

q Rq

L mmR

ωμγ

ω= = = Substituting q e= − gives: ( )/ 2 .e e mγ = −

CALCULATE: Not required ROUND: Not required DOUBLE-CHECK: The magnetic dipole and the angular momentum should both be quadratic in R, so it is logical that the ratio of these two quantities is independent of R.


1072

Additional Problems

28.60.

The magnitude of magnetic field due to one of the coils is 2

0 11 2 2 3/2 .

2 ( )iN RB

x Rμ

=+

Since 1 2 ,B B= the net

magnetic field is 2

01 2 2 2 3/2 .

( )iNR

B B Bx Rμ

= + =+

Putting in 0.500 m, 2.00 m, 7.00 Ax R i= = = and

50N = yields ( )( )( )( )

( ) ( )

274

3/22 2

4 10 T m/A 7.00 A 50 2.00 m2.01 10 T.

0.500 m 2.00 mB

π −−

⋅= = ⋅

+

28.61.

Since the horizontal distance between points A and B is large compared to d, the magnetic field at point B can be approximated by two parallel wires carrying opposite currents. By the right hand rule, the magnetic field at point B is directed into the page from both currents. Since point B is a distance of d/2 away from each wire, the magnitude of magnetic field at point B is twice that at point A. So, the strength of the magnetic field at point B is ( )2 2.00 mT 4.00 mT.B = =

28.62.

Applying the right hand rule gives the direction of the magnetic field due to the wire at the compass needle in the westward direction. The magnitude of wireB is

( )70

4 10 T m/A 500.0 A8.33 μT.

2 2 12.0 mI

Bd

πμπ π

−⋅ ⋅= = =

⋅

The deflection of the compass needle is wire

Earth

8.33 μTarctan arctan 11.8 .40.0 μT

BB

δθ

= = = °

The deflection is

westward.


1073

28.63. The magnetic dipole moment is defined as 2 .iA i Rμ π= = This means the current that produces this

magnetic dipole moment is ( )2/ .i Rμ π= Substituting the numerical values gives the current of 22 2

9 96 2

8.0 10 A m 4.07 10 A 4.1 10 A.(2.5 10 m)

iπ

⋅= = ⋅ ≈ ⋅⋅

28.64. The potential energy of a current loop in a magnetic field is given by .U Bμ= − The magnitude of the

magnetic dipole moment is 2 .iA i Rμ π= = The direction of the magnetic dipole moment can be determined using the right hand rule. In this case, the magnetic dipole is in the positive z-direction.

Therefore, it follows that ( )22 3 20.10 A 0.12 m 4.5 10 A m .i R z zμ π π −= = ⋅ ⋅ = ⋅ The energy is given by

( ) ( )3 2 34.5 10 A m 1.5 T 6.8 10 J.U B z zμ − −= − = ⋅ − = ⋅ If the loop can move freely, the loop will rotate such

that its magnetic dipole moment aligns with the direction of the magnetic field. This means the magnetic dipole moment is 3 24.5 10 ( ) A m .zμ −= ⋅ −

Thus the minimum energy is 3 2 34.5 10 A m ( ) ( 1.5 T) 6.8 10 J.U z z− −= − ⋅ − − = − ⋅

28.65. The magnitude of magnetic field inside a solenoid is given by ( )0 0 / .B in i N Lμ μ= = Simplifying this, the

number of turns of the wire is ( )0/ .N BL iμ= Putting in the numerical values, 0.20 A, 0.90 mi L= = and

35.0 10 TB −= ⋅ yields( )( )

( )( )

3

7

5.0 10 T 0.90 m17904 18000 turns.

4 10 T m/A 0.20 AN

π

−

−

⋅= = ≈

⋅

28.66.

Applying Ampere’s Law around a loop as shown in the figure gives 0 enc .B ds B ds iμ= = Thus, the

magnetic field is ( )0 enc / 2 .B i rμ π= The enclosed current is given by encenc

total

A ii i

A= − when encA is cross

sectional area of the shield that is enclosed by the loop and totalA is the cross sectional area shield. This

means the areas are 2 2enc ( )A r aπ= − and 2 2

total ( ).A b aπ= − Thus the magnetic field inside the shield is 2 2 2 2

0 02 2 2 21 .

2 2i ir a b rBr rb a b a

μ μπ π

− −= − = − −

28.67. THINK: The torque due to the current in a loop of wire in a magnetic field must balance the torque due to weight.


1074

SKETCH:

RESEARCH: The torque on a current loop in a uniform magnetic field is given by

B ( ) .B iN A B iNA z Bτ μ= × = × = − ×

Using Newton’s Second Law, the torque due to the weight is found to

be ( ) ( ) W 0.5 0.5 ( ).r T ax mg z amg x zτ = × = × − = − ×

SIMPLIFY: Since the system is in equilibrium, the net torque must be zero: B w 0.τ τ τ= + = Thus,

B w

0.5 ( ) 0.5 ( ).iNAz B amg x z amg z x

τ τ= −

− × = × = − ×

This means that the magnetic field vector is in positive .x Substituting B Bx=

gives 0.5 .iNAB amg=

After simplifying and using A ab= , 1 1 .2 2 ( ) 2

amg amg mgB x x xiNA iN ab iNb

= = =

CALCULATE: Substituting the numerical values produces ( )( ) 20.0500 kg 9.81 m/s

0.0245 T.2(1.00 A) 50 (0.200 m)

B x= =⋅ ⋅

ROUND: Three significant figures yields, 24.5 mT.B =

DOUBLE-CHECK: The magnetic force must be in the positive z-direction to balance gravity. By the right hand rule, it can be seen that the magnetic field must point in the positive x-direction for this to occur. This is consistent with the result calculated above. The result is reasonable.

28.68. THINK: In this problem, the net magnetic field due to two parallel wires is determined by adding the contributions from the wire. SKETCH:

RESEARCH: The magnitude of the magnetic field of a long wire is given by ( )0 / 2 .B i Rμ π= The net

magnetic field is net 1 2 .B B B= +

Because of the symmetry of this problem, the y-component of the magnetic fields cancel out and only the x-component remains. Thus, the net magnetic field becomes


1075

0

ˆ ˆ ˆsin sin 2 sin

ˆsin

B B x B x B xi

B xR

θ θ θμ θ

π

= − − = −−

=

SIMPLIFY: Since 2 2( / 2)sin ,R d

Rθ −

= the magnitude of the magnetic field simplifies to

220

2 4i dB R

Rμ

π= −

CALCULATE: Inserting the numerical values of the parameters gives

( )( )( )

( ) ( )27 222 5

22

4 10 T m/A 10.0 A 20.0 10 m12.0 10 m 1.843 10 T.

412.0 10 mB

π

π

− −− −

−

⋅ ⋅= ⋅ − = ⋅

⋅

ROUND: Keeping three significant figures, 18.4 μT.B = DOUBLE-CHECK: The magnetic field due to one wire at the same position is 16.7 μT. Therefore, it is reasonable that the answer for two wires is slightly larger than this, considering that the y-components cancel out.

28.69. THINK: In this problem the force on a particle due to a magnetic field must balance the force due to gravity. SKETCH:

RESEARCH: The force acting on the particle due to the magnetic field is B sin .F qvB θ= Since the angle

between v

and B

is 90.0 ,° the force due to the magnetic field becomes B .F qvB= This force must balance the gravitational force which is given by g .F mg= Therefore B gF F= or .qvB mg=

SIMPLIFY: The magnetic field due to the current in the wire is ( )0 / 2 .B i dμ π= The change of the particle

is then found to be ( ) ( )0/ 2 / .q mg vB mg d v iπ μ= = CALCULATE: Inserting the numerical values gives a charge of

( )( )( )( )

6 24

7

1.00 10 kg 9.81 m/s 2 (0.100 m)4.905 10 C.

1000. m/s 4 10 T m/A (10.0 A)q

π

π

−−

−

⋅= = ⋅

⋅

ROUND: Rounding the result to 3 significant figures gives 44.91 10 C.q −= ⋅ DOUBLE-CHECK: Dimensional analysis confirms the calculation provided the answer in the correct

units: ( )

2 2kg m/s m kg m/s A m A s C .

m/s T m/A A m/sm/s N/ A mq

= = = = =

28.70. THINK: The torque on a loop of wire in a magnetic field is given by ,Bτ μ= ×

where μ is the magnetic dipole moment of the wire.


1076

SKETCH:

RESEARCH: (a) Using the right hand rule, the direction of current is counterclockwise as seen by an observer looking in the negative μ direction as shown in the above figure.

(b) Using the magnetic dipole moment ˆ,iNAnμ =

the torque on the wire is ˆ ,iNAn Bτ = ×

where n is a

unit vector normal to the loop. Since ˆ sinn B B θ× =

and 2A Rπ= , the magnitude of the torque is 2 sin .iN R Bτ π θ=

SIMPLIFY: From the equation, the number of turns needed to produce τ is 2 .sin

NiR B

τπ θ

=

CALCULATE: (b) Substituting the numerical values of the parameters yields

( )( )( ) ( ) ( )

1 22

3.40 N m49.98 50. turns.

5.00 A 5.00 10 m 2.00 T sin 60.0N

π −= = =

⋅ °

(c) Replacing the values of the above R with 22.5 10 mR −= ⋅ gives the number of turns ( )

( )( ) ( ) ( )2 22

3.40 Nm100. turns.

5.00 A 2.50 10 m 2.00 T sin 60.0N

π −= =

⋅ °

ROUND: Not needed. DOUBLE-CHECK: Since N is inversely proportional to 2R , the ratio of the results in (b) and (c) is

( )2211 2

2 22 1 1

/ 2 1 50 .4 200

RN RN R R

= = = =

28.71. THINK: Assuming the inner loop is sufficiently small such that the magnetic field due to the larger loop is same across the surface of the smaller loop, the torque on the small loop can be determined by its magnetic moment. SKETCH:

RESEARCH: The torque experienced by the small loop is given by .Bτ μ= ×

The magnetic field in the

centre of the loop is given by 0 1 ˆ.2

iB y

Rμ

=

The magnetic dipole moment of the small loop is

22 2 2

ˆ.i A i r xμ π= =

SIMPLIFY: Combining all the above expressions yields the torque.


1077

( )2 2

2 0 1 0 1 2 0 1 22

ˆ ˆ ˆ ˆ2 2 2

i i i r i i ri r x y x y

R R Rμ πμ πμτ τ π

= = × = × =

CALCULATE: Putting in all the numerical values gives

( )( )

7 27

4 10 T m/A (14.0 A)(14.0 A)(0.00900 m)1.254 10 N m.

2 0.250 m

π πτ

−−

⋅= = ⋅

ROUND: Rounding to 3 significant figures gives, 71.25 10 N m.τ −= ⋅

DOUBLE-CHECK: The units are correct: 2 2T m/A A A m N A m

N m .m A m

τ = = =

28.72. THINK: Two parallel wires carrying currents in the same direction have an attractive force. Two parallel wires carrying currents in opposite directions have a repulsive force. SKETCH:

RESEARCH: By considering the direction of emf potentials, the currents in the wires have the same direction. Therefore the force between the wires is attractive. The force between the two wires is given by

0 1 2 .2i i L

Fa

μπ

=

SIMPLIFY: The currents through the wires are given by emf,11

1

Vi

R= and emf,2

22

.V

iR

= Thus, the force

becomes 0 emf,1 emf,2

1 2

.2V V L

FaR R

μπ

= Solving for 2R gives: 0 emf,1 emf,22

1

.2

V V LR

aR Fμ

π=

CALCULATE: Substituting the numerical values gives

( )( )( )( )( )( )( )

7

2 5

4 10 T m/A 9.00 V 9.00 V 0.250 m5.063 .

2 0.00400 m 5.00 4.00 10 NR

π

π

−

−

⋅= = Ω

Ω ⋅

ROUND: Rounding the result to 3 significant figures gives 2 5.06 .R = Ω DOUBLE-CHECK: To 1 significant figure, the value of 2R is the same as 1.R This is reasonable.

28.73. THINK: To solve this problem, the forces due to an electric field and a magnetic field are computed separately. The forces are added as vectors to get a net force. SKETCH: (a) (b)


1078

RESEARCH: Using the right hand rule and since the charge of proton is positive, the directions of forces are shown above. The magnitude of the electric force on the proton is ,EF qE= and the magnitude of the magnetic force is .BF qvB= SIMPLIFY:

(a) The acceleration of the proton is net ( ).F qvB qE qa vB Em m m

−= = = −

(b) The acceleration of the proton if the velocity is reversed is net ( ).B E

F qvB qE qa F F vB Em m m

− −= = − − = = − +

CALCULATE: Substituting the numerical values yields the acceleration

(a) ( )( )( )19

10 227

1.60 10 e 200. m/s 1.20 T 1000. V/m 7.28 10 m/s1.67 10 kg

a−

−

⋅= − = − ⋅⋅

(b) ( )( )( )19

11 227

1.60 10 e 200. m/s 1.20 T +1000. V/m 1.19 10 m/s1.67 10 kg

a−

−

⋅= − = − ⋅⋅

ROUND: (a) 10 27.28 10 m/sa = − ⋅ (b) 11 21.19 10 m/s− ⋅ DOUBLE-CHECK: It is expected that the result in (b) is larger than in (a). This is consistent with the calculated values.

28.74. THINK: The net acceleration of a toy airplane is due to the gravitational acceleration and the magnetic field of a wire. However for this problem, the gravitational force is ignored. SKETCH:

RESEARCH: Using a right hand rule, the magnetic force on the plane is directed toward the wire. The net acceleration of the plane due to the magnetic field is / / .Ba F m qvB m= =

SIMPLIFY: Substituting the magnetic field of the wire ( )0 / 2B i dμ π= yields 0 .2qv i

amdμ

π=

CALCULATE: Putting in the numerical values gives the acceleration: ( ) ( )

( )( )3 7

5 236 10 C (2.8 m/s) 4 10 T m/A (25 A)

1.674 10 m/s .2 0.175 kg 0.172 m

aπ

π

− −−

⋅ ⋅ ⋅ ⋅= = ⋅

ROUND: Rounding the result to two significant figures gives 5 21.7 10 m/s .a −= ⋅ DOUBLE-CHECK: It is expected that the result will be much less than the value of the gravitational acceleration.

28.75. THINK: To do this problem, the inertia of a long thin rod is required. The torque on a wire is also needed. The measure of the angle θ is 25.0 ,° and the current is 2.00 A.i = Let 4 20.200 10 mA −= ⋅ and

29.00 10 T.B −= ⋅


1079

SKETCH:

RESEARCH: The magnetic dipole moment of the wire is given by ˆ.NiAnμ =

(a) The torque on the wire is .Bτ μ= ×

The magnitude of this torque is sin sin .B NiABτ μ θ θ= = (b) The angular velocity of the rod when it strikes the bell is determined by using conservation of energy, that is, i fE E= or i i f f .U K U K+ = + SIMPLIFY: (a) sin sin .B NiABτ μ θ θ= = (b) Since i 0,K = the final kinetic energy is

( ) ( )f i f

21 cos cos 0 cos 1 cos2

K U U

I B B B B Bω μ θ μ μ θ μ μ θ

= −

= − + ° = − + = −

Thus the angular velocity is ( ) 2

2 (1 cos ) 2 (1 cos ) ,1/12

B NiABI mL

μ θ θω − −= = using 21 ,12

I mL= the inertia of a

thin rod. CALCULATE: Putting in the numerical values gives the following values. (a) ( )( )( )( ) ( )4 2 2 470 2.00 A 0.200 10 m 9.00 10 T sin 25.0 1.06 10 N mτ − − −= ⋅ ⋅ ° = ⋅

(b) ( )( )( )( )( )

( )( )( )

1/24 2 2

2

2 70 2.00 A 0.200 10 m 9.00 10 T 1 cos25.01.72 rad/s

1/12 0.0300 kg 0.0800 mω

− − ⋅ ⋅ − ° = =

ROUND: Rounding to 3 significant figures yields 41.06 10 N m,τ −= ⋅ 1.72 rad/s.ω = DOUBLE-CHECK: The torque should have units of Newton-meters, while the angular velocity should have units of radians per second.

28.76. THINK: Using a right hand rule, the sum of the magnetic fields of two parallel wires carrying opposite currents cannot be zero between the two wires. SKETCH:

RESEARCH: The magnitude of the magnetic field of a long wire is ( )0 / 2 .B i Rμ π= Since 1 2i i< and 1i is in an opposite direction to 2 ,i using the right hand rule, it is found that the location of the zero magnetic field must be to the left of the left-hand wire, as shown in the figure. Assuming the location is a distance x

to the left of the left-hand wire, then the net magnetic field is 0 2 0 1net 2 1 0.

2 ( ) 2i i

B B Bx d x

μ μπ π

= − = − =+

SIMPLIFY: Solving for x yields


1080

2 1 12 1 1

2 1

.i i i dxi i x i d x

x d x i i= = + =

+ −

Since 2 12 ,i i=

1

1 1

.2

i dx di i

= =−

CALCULATE: Not required. ROUND: Not required. DOUBLE-CHECK: This result is expected since the ratio of 2 1/ 2.i i = This means the ratio of distances is

2

1

2 2d dd d

= = also.

28.77. THINK: In order for a coil to float in mid-air, the downward force of gravity must be balanced an upward force due to the current loop in the magnetic field. SKETCH:

RESEARCH: By using right-hand rule 1, the direction of the forces can be determined. For the y-component yB of the magnetic field the force due to the current is in the radial direction of the coil.

Therefore, this component cannot be responsible for levitating the coil. For the x-component xB of the magnetic field, with a counterclockwise current as viewed from the bar magnet, the resulting force is in the y-direction, towards the bar magnet (see figure on right). This is the correct direction for balancing the weight of the coil. The magnitude of the y-component of the force on an element dl is

sin sin .xydF Ni dl B NiB dlθ θ= × =

Thus the total magnetic force on the current loop is 2

0sin .

R

yF NiB dlπ

θ= Newton’s Second Law requires that .yF mg=

SIMPLIFY: The integral simplifies to: 2 sin .yF RNiBπ θ= Therefore,

2 sin .2 sin

mgRNiB mg iRNB

π θπ θ

= =

CALCULATE: Substituting in the numerical values yields

( )( )

3 2

2

10.0 10 kg (9.81 m/s )4.416 A.

2 5.00 10 m 10.0(0.0100 T)sin(45.0 )i

π

−

−

⋅= =

⋅ °

ROUND: To 3 significant figures, the current is 4.42 A,i = counterclockwise as viewed from the bar magnet. DOUBLE-CHECK: It takes large currents to generate strong magnetic forces. A current of 4 A is realistic to levitate a 10 g mass.


1081

28.78. THINK: In this problem, Ampere’s Law is applied on three different circular loops. SKETCH:

RESEARCH: Loops 1 ,L 2L and 3L are Amperian loops.

(a) For distances ,r a< applying Ampere’s Law on the loop 1 ,L gives 0 enc(2 ) .B ds B r iπ μ⋅ = = Since

enc 0,i = the field is also zero, 0.B = (b) For distances r between a and ,b applying Ampere’s law on the loop 2L yields

0 enc(2 ) .B ds B r iπ μ⋅ = =

The enclosed current is given by enc enc /i A i A= or ( )( )

2 2

enc 2 2.

r ai i

b a

π

π

−=

−

(c) For distances ,r b> applying Ampere’s Law on 3L gives 0 enc 0 ,2 2

i iB

r rμ μ

π π= since enc .i i=

SIMPLIFY: Thus, the magnetic field is ( )( )

2 20

2 2.

2

r aiB

r b aμπ

−=

−

CALCULATE: Putting in the numerical values gives (a) 0B =

(b) ( )

( )( ) ( )( ) ( )

2 277

2 22

4 10 T m/A (0.100 A) 6.50 cm 5.00 cm2.21 10 T

2 6.50 10 m 7.00 cm 5.00 cmB

π

π

−−

−

⋅ ⋅ − = = ⋅ ⋅ −

(c) ( )

( )7

72

4 10 T m/A (0.100 A)2.22 10 T

2 9.00 10 mB

π

π

−−

−

⋅ ⋅= = ⋅

⋅

ROUND: Keeping 3 significant figures yields the following results for (b) and (c). Note that the value found in (a) is precise. (a) 0B = (b) 72.21 10 TB −= ⋅ (c) 72.22 10 TB −= ⋅

DOUBLE-CHECK: The units of the calculated values are T, which is appropriate for magnetic fields.

28.79. THINK: To solve this problem, the current enclosed by an Amperian loop must be determined. SKETCH:

RESEARCH: Applying Ampere’s Law on a loop as shown above gives 0 enc(2 ) .B ds B r iπ μ⋅ = =

enci is the

current enclosed by the Amperian loop, that is 2

enc 0 0( ) ( ) .

ri J r dA J r r dr d

πθ′ ′ ′= =


1082

SIMPLIFY: Since ( )J r is a function of r only, the above integral becomes enc 02 ( ) .

ri J r r drπ ′ ′ ′=

Substituting 0( ) (1 / )J r J r R= − yields 2 2 3 2 3

enc 0 0 000

2 2 2 .2 3 2 3

rr r r r r ri J r dr J J

R R Rπ π π

′ ′ ′′ ′= − = − = −

Thus, the magnetic field is 2 3 2

0 00 0

2.

2 2 3 2 3J r r r rB J

r R Rμ π μ

π

= − = −

CALCULATE: Not required. ROUND: Not required. DOUBLE-CHECK: The form of the answer is reasonable.

28.80. THINK: The maximum torque on a circular wire in a magnetic field is when its magnetic moment is perpendicular to the magnetic field vector. SKETCH:

RESEARCH: The torque on the circular wire is given by .Bτ μ= ×

The magnitude of the torque is

sinBτ μ θ= where θ is the angle between μ

and B

. SIMPLIFY: (a) The maximum torque is when 90 ,θ = ° that is, .Bτ μ= Using 2 ,iA i Rμ π= = the torque becomes

2 .i R Bτ π= CALCULATE: Inserting the numerical values gives the torque:

( ) ( )22 3 4(3.0 A) 5.0 10 m 5.0 10 T 1.18 10 N m.τ π − − −= ⋅ ⋅ = ⋅

ROUND: Keeping only two significant figures yields 41.2 10 N m.τ −= ⋅ (b) The magnetic potential energy is given by cos .U Bμ θ= − The maximum and the minimum potential energies are when 180θ = ° and 0 ,θ = ° that is, maxU Bμ= + and min .U Bμ= − CALCULATE: Since the values of Bμ is the same as in (a), the range of the potential energy is

4 4max min 2 2 1.2 10 J 2.4 10 J.U U U Bμ − −Δ = − = = ⋅ ⋅ = ⋅

ROUND: Not needed. DOUBLE-CHECK: The change in potential is a change in energy, so it is appropriate that the final answer have joules as units.

Chapter 29: Electromagnetic Induction

1083


In-Class Exercises

29.1. c 29.2. a 29.3. c 29.4. c 29.5. a 29.6. d 29.7. a 29.8. e Multiple Choice

29.1. d 29.2. c 29.3. a 29.4. a 29.5. a 29.6. c 29.7. d 29.8. b

Questions

29.9. A refrigerator’s electrical circuit contains a motor with a large number of winding coils, making it highly inductive. The electromagnetic induction due to the coil can create a large voltage, on the order of kV between the prongs. This voltage is great enough to ionize the air and the process of ionization produces light, creating a visible spark.

29.10. Large machinery and motors often convert electrical energy to mechanical energy or vice-versa to complete a task. The conversion from electrical energy to mechanical energy requires the creation of magnetic fluxes. Changes in the magnetic flux reaching a pacemaker, due to movements of the machine or the person, will create currents in the circuitry of the pacemaker, changing its behavior which can be dangerous.

29.11. As the metal moves through the non-uniform magnetic field, it experiences a changing magnetic flux. The flux induces an emf in the metal, if it is a conductor, and produces eddy currents. Lenz’s law states that the induced currents create a force to oppose the movement of the metal through the field. This action is analogous to the drag force or force of friction used to create the damping of a harmonic oscillator.

29.12. Lenz’s law requires that as the magnet moves down the cylinder, a current is produced in the aluminum cylinder, which in turn creates a magnetic field that opposes the magnet’s motion. The force of the currents on the magnet is proportional to the velocity of the magnet. Thus, the magnet will continue to accelerate until it reaches a terminal speed that creates a force equal and opposite to the force of gravity.

29.13. (a) The currents produced in the aluminum, by induction, create a force that opposes the motion of the magnet. The magnet falling in the glass tube does not create a current since glass is an insulator. Thus, the magnet in the glass tube falls faster since there is no magnetic field produced to oppose the force of gravity. (b) Because the glass has nearly infinite resistance, no eddy currents are created as the magnet passes through it. The aluminum being a good conductor does produce eddy currents as the magnet falls through it. Thus, the aluminum tube has a larger eddy current.

29.14. (a) The B field inside the solenoid is uniform and equal to i 0 .B niμ= Outside the solenoid, the field is zero, o 0.B = The B field through the ring is only that of the field inside the solenoid of radius, a. The flux

is then 2 2 20 0 .BA ni a n a Ctμ π μ πΦ = = = Thus, the emf is 2 2

ind 0/ 2 .V d dt n a Ctμ πΔ = Φ =

(b) The magnitude of the electric field is then 1 0 0 1sin / 2 .B V t bRμ ω= 2 202 2rE V n a Ctπ μ π= Δ = or

2 20 .na Ct

Er

μ=

(c) The ring is not necessary for the induced electric field to exist. The solenoid will produce a magnetic field from the current being passed through the wire inducing an electric field on each concurrent loop of wire.


1084

29.15. Lenz’s law requires that the induced current opposes the change in the magnetic field. Therefore, the B field created by the induced current is downward. To produce a magnetic field in this direction, the current must flow clockwise as seen from above.

29.16. The area of the loop perpendicular to the field is given by ( )2 cos .A L tω= The potential difference is:

( ) ( )( ) ( )( ) ( )2 2 2in,L cos sin sin .

d ABd dA dV B B L t BL t BL tdt dt dt dt

ω ω ω ω ωΦΔ = − = − = − = − = − − =

29.17. The emf produced by a loop is given by ind ,V vBLΔ = where L is the length of the moving conductor. By converting L into the differential, dr, and integrating from the center of the disk to the edge for the emf of

the disk gives ind 0.

RV vBdrΔ = The velocity of an element, dr, is given by .v rω= The emf is then:

2ind 0

1 .2

RV r Bdr R Bω ωΔ = =

29.18. Separation of charge due to the magnetic force, ,qv B× engenders a compensating electric field of

magnitude .E v B vB= × = The corresponding potential difference across height, l, is:

( )( )( )1.80 m 2.00 m/s 25.0 T 90.0 V.V lE lvB= = = = In equilibrium this drives no current. However, such a large magnetic field offers further hazards due to any metal objects about the man’s body and to stress on blood vessels, which are carrying conducting fluids in motion like iron.

29.19. The flux through the inside copper cylinder is constant during the process, so: 2 2

i f i i f f i i f f .B A B A B r B rπ πΦ = Φ = = The final magnetic field is given by:

2

if i

f

.r

B Br

=

If the initial B field is 1.0 T and the radius compresses by a factor of 14, then final field is given by:

( ) ( ) ( ) = = = = ⋅

22 2 2i

f i ii

14 14 1.0 T 2.0 10 T./14r

B B Br

Experimental magnetic fields are typically lower than 10 T. This is a huge magnetic field.

29.20. Lenz’s law requires that the induced current opposes the change in the magnetic field. Therefore, the B field created by the induced current is downward. To produce a magnetic field in this direction, the current must flow clockwise as seen from above.

29.21. The inductance of a solenoid is given by 20 .L n lAμ= Let d denote the length of the wire. The number of

turns in each case is / 2 .N d rπ= The inductance is then:

( )2 20 0 0 0 0

1 .2 2

dL n lA n nl A nNA n r ndrr

μ μ μ μ π μπ

= = = = =

For both solenoids, the number of turns per unit length is equal, and the distance of the wire is the same. Therefore, the ratio of the inductances is:

01

2

/ 2 1 .2 / 2 2

ndrLL nd r

μμ

= =

Thus, the inductance of the second solenoid is twice that of the first solenoid.

Problems

29.22. The magnetic flux through the coil is given by:


1085

( ) ( ) ( )θ πΦ = = ° − ° =2 2cos 20 5.00 T 0.400 m cos 90 25.8 21.9 T mNBA

29.23. The potential difference around the loop is: ( ) ( )π π −ΔΦ ΔΦ Δ Δ − = − ≈ − = − = − = − = − = ⋅ Δ Δ Δ Δ

22 5emf

0 T 1.20 T0.0100 m 1.89 10 V.20.0 s

ABd B BV A rdt t t t t

Note that the area of the ring is perpendicular to the field. Thus, the normal of the area is parallel to the field and cos 1.θ =

29.24. The voltage across the loop is given by:

( ) ( ) ( )3

2 2Bind

1.50cos cos cos cos40.0 4.50 .

d td d dBV NAB NA NA NL tdt dt dt dt

θ θ θΦ= − = − = − = − = − °

The current induced if the loop passes a resistance of 3.00 R = Ω is:

( ) ( )( ) ( ) ( )( )

2 22 2ind cos40.0 4.50 8 0.200 m cos 40.0 4.50 2.00 s

1.47 A.3.00

NL tVi

R R° °

= = − = − =Ω

29.25. Because the magnetic field is perpendicular to the normal of the loop, there is no flux through the loop: cos90 0.ABΦ = ° =

Since there is no flux through the loop, there is no induced voltage: ( )0

0.ddV

dt dtΦ= − = − =

29.26. THINK: The change in the area of the loop creates a change in the magnetic flux through the loop. The change in flux produces a current. The loop has a resistance of 30.0 R = Ω and a radius which changes from i 20.0 cmr = to f 25.0 cmr = in 1.00 s. The magnetic field of the Earth is about 54.26 10 T.−⋅ SKETCH:

RESEARCH: The flux through the loop is B cosAB θΦ = or B ,ABΦ = since the B field is perpendicular to the surface of the loop. The induced potential difference is given by ind B / .V d dt= − Φ This potential must also satisfy .V iR= SIMPLIFY: The induced current in the loop is:

2 22 2ind B f i1 1 .

V d r rdAB B dA B d r B dr BiR R dt R dt R dt R dt R dt R t

π π π Φ − = = − = − = − = − = − ≈ − Δ

CALCULATE: ( ) ( ) ( )2 25

74.26 10 T 0.250 m 0.200 m

1.00374 10 A30.0 1.00 s

iπ−

− ⋅ − = − = − ⋅ Ω

ROUND: The induced current in the loop is 71.00 10 A.i −= − ⋅ DOUBLE-CHECK: This current is very small, as one would expect. The negative sign indicates that the direction of the induced current is such that the magnetic field due to the induced current opposes the change in magnetic flux that induces the current.

29.27. THINK: The current in the outer loop generates a magnetic field. Because the magnitude of the current in the outer loop changes with time, the magnetic field it generates also changes. The changing magnetic field, in turn, induces a potential difference and thus a current in the inner loop. Let I be the current in the outer loop and i be the induced current in the inner loop.


1086

SKETCH:

RESEARCH: The current through the large loop is 0 1sin / .I V t Rω= This creates a magnetic field at the center of the loop of:

01 .

2I

Bb

μ=

Since the radius of the inner loop is much smaller than the radius of the outer loop, the magnetic field through the inner loop is 1 0 0 1sin / 2 .B V t bRμ ω= This magnetic field creates a flux of:

22 0 0

1 11

sin .2B

a VB A B a t

bRμ ππ ωΦ = = =

The induced potential across the inner loop is then: 2

0 0ind

1

sin .2

B a Vd dV tdt dt bR

μ π ω Φ

Δ = − = −

This voltage corresponds to a current in the inner loop of: 2

ind 0 0

2 2 1

1 sin .2

V a Vdi tR R dt bR

μ π ω Δ

= = −

SIMPLIFY: The potential difference induced in the inner loop is:

( )2 2 2

0 0 0 0 0 0ind

1 1 1

sin sin cos ,2 2 2

a V a V a Vd dV t t tdt bR bR dt bR

μ π μ π μ π ωω ω ω

Δ = − = − = −

and the induced current in the inner loop is:

( )2 2 2

ind 0 0 0 0 0 0

2 2 1 1 2 1 2

1 sin sin cos .2 2 2

V a V a V a Vd di t t tR R dt bR bR R dt bR R

μ π μ π μ π ωω ω ω Δ

= = − = − = −

CALCULATE: Not applicable. ROUND: Not applicable. DOUBLE CHECK: The time dependence on the current for the outer loop and inner loop is shown in the plot below. For example, for / 2tω π< (taking positive values to be the counterclockwise direction) if the current in the outer loop is moving counterclockwise and increasing then the current in the inner loop is increasing in the clockwise direction. This is consistent with Lenz’s Law.


1087

29.28. THINK: The varying current, 1i , through the outer solenoid creates a varying magnetic field, 1 ,B within the coil. This varying B field creates a flux in the inner solenoid, which in turn creates an induced emf. SKETCH:

RESEARCH: The magnetic field generated by the outer solenoid is given by 1 0 1 0 0 cos .B ni ni tμ μ ω= = The flux generated in the inner solenoid is given by 2 1.A BΦ = The induced emf in the inner solenoid is given

by 1ind 2 .

dBdV Adt dtΦΔ = − = −

SIMPLIFY: ( ) ( )1ind 2 2 0 0 2 0 0 2 0 0cos cos sin .

dB d dV A A ni t A ni t A ni tdt dt dt

μ ω μ ω μ ω ωΔ = − = − = − = This

corresponds to a current of ( )2 2 0 0/ sin / ,i V R A ni t Rμ ω ω= Δ = in the inner solenoid. The current of the inner solenoid induces a B field of:

( ) 2 20 2 0 0 0 2 0

2 0 2

sin sin.

n A ni t n A i tB ni

R Rμ μ ω ω μ ω ωμ= = =

CALCULATE: Not applicable. ROUND: Not applicable. DOUBLE CHECK: The induced magnetic field of the inner solenoid must oppose the change in flux of the outer solenoid. It can be seen from the expressions for 2B and 1B the two fields will always have opposite directions, satisfying this requirement.

29.29. (a) The decreasing B field creates a changing flux through the loop, confined to the area of the dotted circle of radius, r = 3.00 cm. The varying flux creates an emf of:

( ) ( )22 2 2B f i

ind .d r Bd ABd B BdB BV r r r

dt dt dt dt t t

ππ π πΦ − ΔΔ = − = − = − = − ≈ − = − Δ Δ

This corresponds to a current of:

( )ππ − − = = − = − = = Δ Ω

22f i

0.0300 m 1.00 T 2.00 T 0.00707 A 7.07 mA0.200 2.00 s

B BV riR R t

(b) The B field points into the page, thus a decrease in the B field will induce a current corresponding to a B field which points into the page. By the right-hand rule, the induced current flows clockwise.

29.30. The airplane’s wings are approximated by a straight wire. The voltage across a wire moving in a B field is:

( )( )( )4mach3 3 340. m/s 10.0 m 0.500 10 T 0.510 V.V vLB v LB −= = = ⋅ =

29.31. THINK: As a conductor travels through a magnetic field, perpendicular to the ground, of intensity 0.426 G,B = it creates a voltage difference between its ends. The length of metal of interest is L = 5.00 m

and rotates at 41.00 10 rpm.⋅


1088

SKETCH:

RESEARCH: The voltage across a wire moving in a magnetic field is ind .V vLBΔ = Each element of the blade travels at a different speed, .v rω= To calculate the voltage, the length must be divided into pieces of length, dl, which travel at .v lω= The value should be integrated over the total length, from 0 to L.

SIMPLIFY: 2

0 0

12

L LV vBdl l Bdl BLω ω= = =

In terms of the blade’s rpm, the voltage is ( ) 22 rpm1 .

2 60 sV BL

π =


424

1.00 10 rpm0.426 10 T 5.00 m 0.55763 V

60 s/rpmV

π−

⋅ = ⋅ =

ROUND: The voltage across the hub of the helicopter’s blade to its edge is ind 0.558 V.VΔ = DOUBLE-CHECK: This result is a comparable voltage to the one induced by the airplane’s wing span traveling at mach 3, V = 0.5 V.

29.32. THINK: The expanding loop creates a changing flux through the loop. Lenz’s law implies that the changing flux induces a current in the loop. This is similar to increasing the magnetic field within the loop. To counteract the increase in flux, the current must create a magnetic field opposite to the B field. By the right-hand rule, the current must flow clockwise. The radius of the loop expands by 0 ,r r vt= + where

0 0.100 mr = and v = 0.0150 m/s. The resistance of the wire is = Ω12.0 .R The B field has a uniform value of 0 0.750 TB = upward. The problem asks for the induced current at the time, t = 5.00 s. SKETCH:

RESEARCH: The flux through the loop is 2

B .AB r BπΦ = = The induced current of the loop is / ,i V R= where the voltage is given by B / .V d dt= − Φ SIMPLIFY: The induced current in the wire is:

( ) ( ) ( ) ( )2 02B0 0

21 1 2 .B r vt vdV d B d Bi r r vt v r vt

R R dt R dt R dt R Rππ π+Φ = = − = − = − + = − = − +

CALCULATE: The magnitude of the induced current at t = 5.00 s is: ( ) ( ) ( )( )

π = − + = Ω

2 0.750 T0.0150 m/s 0.100 m 0.0150 m/s 0.0010308 A.5.00 s12.0

i

ROUND: =1.03 mAi at 5.00 s, travelling clockwise through the loop.

DOUBLE-CHECK: ( )2 2

2

T m V s m ATm/s m m/s s A

s m V si

= + = = = Ω Ω


1089

29.33. THINK: Terminal velocity will be reached when the force due to the changing magnetic flux cancels the weight of the bar. SKETCH: A sketch is not necessary.

RESEARCH: Bind term ,dV Bv

dtωΦ

Δ = − = ind ,V

iR

Δ= B ,F iBω= B gravityF F mg= =

SIMPLIFY: 2 2

ind termterm 2 2

V B v mgRiB mg B mg mg vR R B

ωω ωω

Δ= = = =

CALCULATE: No calculations are necessary. ROUND: Rounding is not necessary. DOUBLE-CHECK: It makes sense the large m is, the higher termv has to be to compensate for the greater gravitational force.

29.34. THINK: (a) The change in area causes an induced voltage. (b) After finding the induced voltage, the induced current can be determined. (c) The induced current will cause a force opposite to the direction of motion (from Lenz’s law) which requires extF compensating for it. (d) Determine extW and extP from ext .F SKETCH: Provided with the question. RESEARCH:

(a) Bind

d dAV B BvLdt dtΦ

Δ = − = =

(b) ind ,ViR

Δ= in the clockwise direction.

(c) B ind extF i LB F= = (d) ext ext ,W F y= Δ extP Fv= (e) 2

ext R indP P i R= = SIMPLIFY: (a) indV BvLΔ =

(b) indBvLi

R=

(c) 2 2

B extL B vF F

R= =

(d) 2 2

ext ,L B vW yR

= Δ 2 2 2

extL B vP

R=

(e) 2 2 2

RL B vP

R=

CALCULATE: Not necessary. ROUND: Not necessary. DOUBLE-CHECK: (e) This is due to the law of conservation of energy. The work done has to go somewhere, and in this case is dissipated by the resistor as heat.

29.35. THINK: The current in the wire will cause a magnetic field. The changing current will cause a changing flux through the loop, inducing a potential. SKETCH: Provided with question.


1090

RESEARCH: For a wire: 0 2 .4

iBr

μπ =

Bind ,dV

dtΦ

Δ = ( )= +2.00 A 0.300 A/s ,i t A = 7.00 m by 5.00 m,

B .B dAΦ =

SIMPLIFY: ( ) ( ) ( )μ μ μπ π π

Φ = = =

8 m 0 0 0

B 1 m

2 8.00 mln ln8.005.00 m 5.00 m 5.00 m4 2 1.00 m 2i ii dr

r

( ) ( ) ( ) ( )( )μ μπ π

Φ Δ = = =

0 0Bind ln8.00 ln8.00 0.300 A/s5.00 m 5.00 m2 2

d diVdt dt

CALCULATE: ( ) ( )( )ππ

−− ⋅Δ = = ⋅

77

ind4 10 H/m ln8.00 0.300 A/s 6.238 10 V5.00 m 2

V

ROUND: −Δ = ⋅ 7ind 6.24 10 VV

DOUBLE-CHECK: It makes sense that the larger the rate of change of the current, the larger the induced voltage.

29.36. THINK: (a) By the right-hand rule, the flux is into the page. Since the square is moving away from the wire, the flux is decreasing. Lenz’s law states that the current is moving clockwise. (c) The top and bottom parts have the same contributions and cancel each other. SKETCH: (b)

RESEARCH: Use =2 20.0 cmx and =1 10.0 cmx as the end points. 0 2 ,4

iBr

μπ =

B ,B dAΦ =

ind ,dVdtΦΔ = − ind

ind ,V

iR

Δ= r = 10.0 cm, i = 1.00 A, v = 10.0 cm/s, = Ω0.0200 ,R L = 10.0 cm,

left i 1nd ,F i LBx= right i 2nd ,F i LBx= and net right left .F F F= −

SIMPLIFY: ( ) ( )2

12 1

0 0B

2 ln ln4 2

x vt

x vt

iL dr L i x vt x vtr

μ μπ π

+

+ Φ = = + − +

2

0Bind

12Lid v vV

dt x vt x vtμπ

ΦΔ = − = − − + +


1091

( ) ( )

( )

( )

indnet ind ind 1 ind 1 1

0

2 2 2

22

2

2

2

2

01

1

2 20 0 1

12 2 2

02

1

1 1 22 4

1 12 2

1 12

VF i LBx i LBx i LB x x LB x x

RLi iv L x x

R x vt x vt r

L i v x xR x vt x vt r

L i v x xx vt x vtR

μ μπ π

μ μπ π

μπ

Δ= − = − = −

= − − − + + −

= − − + + −

= − − + + 1

r

CALCULATE: At time t = 0:

( ) ( ) ( ) ( )( )

2 2

2

16

27

net

0.100 m 1.00 A 4 10 H/m 0.100 m/s 20.0 cm 10.0 cm1 120.0 cm 10.0 cm 10.0 cm2 0.0200

1.00 10 N

Fπ

π

−

−

⋅ − = − − Ω

= ⋅

ROUND: −= ⋅ 16net 1.00 10 NF

DOUBLE-CHECK: It makes sense that for larger velocities and currents through the wire, the induced force is larger. This is in some ways analogous to how a car traveling faster than another has a larger drag force.

29.37. ( ) ( )cos 2 ,t BA ftπΦ = ( )ind 2 sin 2 .dV fBA ftdt

π πΦΔ = = − The maximum occurs when ( )sin 2 1.ftπ =

πΔ = =ind,max 2 110. V.V fBA Substitute the values to obtain: ( )( )π π

= = =2

110. V 110. V 17.5 Hz.2 2 1.00 T 1.00 m

fBA

29.38. THINK: First relate the magnetic flux to the angular speed and then determine the maximum angular speed. Use the values 0.87 T,B = = 20.0300 m .A SKETCH: A sketch is not necessary.

RESEARCH: For a single loop: ( ) ( )cos .t BA tωΦ = ( )ind sindV BA tdt

ω ωΦΔ = − =

ind,max 170 V ,V BAωΔ = = since the maximum occurs when ( )sin 1.tω =

SIMPLIFY: ind,maxVBA

ωΔ

=

CALCULATE: ( )ω = =2

170 V 6513 Hz0.87 T 0.0300 m

ROUND: 6500 Hzω = DOUBLE-CHECK: It is reasonable that the higher the applied voltage, the higher the angular speed.

29.39. THINK: First determine an expression for the magnetic flux, and then use Faraday’s law to determine the induced voltage. SKETCH: A sketch is not necessary. RESEARCH: 4

Earth 0.300 G 0.300 10 T,B −= = ⋅ ( )B cos ,NBA tωΦ = 2 ,A rπ= r = 0.250 m, 51.00 10 ,N = ⋅

( )2 150. Hz ,ω π= ind Bind

1 ,V d

iR R dt

Δ Φ = = −

Bind,peak

peak

1 ,d

iR dt

Φ = −

1500. R = Ω

SIMPLIFY:

(a) ( )( ) ( )ind1 sin sin ;NBAi NBA t tR R

ωω ω ω = − − =

The peak occurs at ( )sin 1:tω = ind,peak .NBAiR

ω=


1092

(b) ( )avg ind,peak0.7071 ,i i= 2avg avgP i R=

CALCULATE:

(a) ( )( )( ) ( )

( )

24

ind,peak

51.00 10 0.300 10 T 0.250 m 2 150. Hz0.3701 A

1500. i

π−⋅ ⋅= =

Ω

(b) ( )avg 0.7071 0.3701 A 0.2617 A,i = = ( ) ( )2avg 0.2617 A 1500. 102.7 WP = Ω =

ROUND: (a) ind,peak 0.370 Ai =

(b) avg 0.2617 A,i = avg 102.7 WP =

DOUBLE-CHECK: The answer seems reasonable since there are a very large number of turns for the generator turning at a very fast rate.

29.40. First solve for n: ( )( )0 6 2 20

0.025 T 33158.1.2566 10 m kg s A 0.60 A

BB ni ni

μμ − − −

= = = =⋅

( )( )( )22 6 2 21 0 2 1 200 1.2566 10 m kg s A 33158 0.034 m 0.0302 H,M N n rπμ π − − −= = ⋅ =

( ) ( )( )2 20 1+ 2.4 si t i t−=

( )( )( )( )( )20.0302 H 2 0.60 A 2.4 s 2.0 s 0.17 VdiV Mdt

−= − = − = −

The results match those of the example.

29.41. The potential across an inductor is given by: ind ,diV Ldt

Δ = − where didt

is the slope.

29.42. THINK: The potential difference induced in the solenoid is due to the changing current in the coil. Using

the mutual inductance of the solenoid due to the coil, the potential difference induced in the solenoid can be calculated. Assume the magnetic field of the short coil is uniform. This is not strictly accurate, but necessary to answer the question and will give a reasonable approximation. SKETCH:

RESEARCH: The mutual inductance between the coil and the solenoid is

s c s

c

,N

Mi

→Φ=


1093

where sN is the number of turns in the solenoid, c s→Φ is the flux in the solenoid resulting from the magnetic field through the coil, and ci is the current in the coil. The flux is given by

0c2

s .in rπμ→Φ =

ind ,diV Mdt

Δ = s 30,N = n = 60/cm = 6000/m, r = 0.0800 m, = 2.00 A .12.0 s

didt

SIMPLIFY: s2

ind 0diV N n rdt

πμΔ =

CALCULATE: ( )( ) ( )( )π π − − Δ = ⋅ = ⋅

27 4ind

2.00 A30 6000 / 4 10 H/m 0.0800 m 7.57986 10 V12.0 s

V m

ROUND: −Δ = ⋅ 4ind 7.58 10 VV

DOUBLE-CHECK: It makes sense that for larger changes in current, larger potential differences are induced.

29.43. (a) τ = = =ΩL

1.00 H 1.00 μs1.00 M

LR

(b) ( ) ( )L/emf 1 .tVi t e

Rτ−= − At t = 0, ( ) 0.i t = At = 2.00 μs,t ( ) ( ) ( )( )−= − =

Ω2.00 μs / 1.00 μs10.0 V 1 8.65 μA.

1.00 Mi t e

At steady state. :t → ∞ ( )∞ = =emf 10.0 μ .V

i AR

29.44. For an RL circuit: ( ) ( )/emf 1 ,tVi t e

Rτ−= − where τ = = 0.0250 s.L

R

( ) ( ) ( ) ( )( )/

emf emf

0.300 A 120. 1 ln 1 ln 1 0.0576 s0.0250 s 40.0 V

ti t R i t Re t t

V Vτ τ−

Ω= − − − = = − = −

29.45. The potential drop is the sum of the potential drop across the resistor and the inductor:

( )( ) ( )( )3.0 A 3.25 0.440 3.6 A/s 11 V.diV iR Ldt

Δ = + = Ω + =

29.46. THINK: In a circuit containing only a resistor, the current would be established almost instantaneously. However, with the RL circuit, the current must increase exponentially from zero to the steady state.

emf 1 218 V, 6.0 , 5.0 H.V R R L= = = Ω = SKETCH: Provided with question. RESEARCH: (a) The inductor functions as an open-circuit, so emf 2/ 18 V / 6.0 3.0 A.i V R= = Ω = (b) The inductor acts as an open-circuit, so there is no current across it and hence no current across 1.R (c) The current across 2R is given by Ohm’s Law, emf / .i V R= (d) The potential difference across a resistor is also given by Ohm’s Law, .V iRΔ = (e) Same as (d). (f) The sum of the voltages around any loop is zero. (g) The rate of current change across 1R is the same as that of L. SIMPLIFY: (a) 2/i V R= (b) Not applicable. (c)

2R emf 2/i V R=

(d) 1 1R R 1V i RΔ =


1094

(e) 2 2R R 2V i RΔ =

(f) 1 1emf L R L emf R0 V V V V V V− − = = −

(g) LL

Vdi diV Ldt dt L

= =

CALCULATE: (a) 18 V / 6.0 3.0 Ai = Ω = (b)

1R 0i =

(c) 2R 18 V / 6.0 3.0 Ai = Ω =

(d) ( )( )1R 0 A 6.0 0VΔ = Ω =

(e) ( )( )2R 3.0 A 6.0 18 VVΔ = Ω =

(f) L 18 V 0 18 VV = − =

(g) 18 V 3.6 A/s5.0 H

didt

= =

ROUND: Not necessary. The values are already to the correct number of significant figures. DOUBLE CHECK: The branch of the circuit which contains only a resistor and a source of emf behaves as a simple resistor circuit, with the current being established almost instantaneously. For the branch of

the circuit which contains a resistor and an inductor, equation 29.29 states ( ) ( )/ /emf 1 .t L RVi t e

R− = −

When ( )0, 0,t i t= = as found above.

29.47. THINK: After a long time, the inductor acts like a short-circuit. The circuit is in steady state, so the current is no longer changing. emf 1 218 V, 6.0 , 5.0 H.V R R L= = = Ω = SKETCH: An equivalent sketch when the circuit is in steady-state is as follows.

RESEARCH: The current from the battery is given by emftot

net

,V

iR

= where 1

net2 1

1 1 .RR R

−

= +

The current

through each resistor is given by Ohm’s Law, / .i V R= The sum of the potentials around any loop must be zero:

1 2emf R emf R L0, 0.V V V V V+ = + + =

SIMPLIFY:

(a) ( )emftot 1 2

1 2

Vi R R

R R= +

(b) 1

1

RR

1

Vi

R=

(c) 2

2

RR

2

Vi

R=

(d) 1 1emf R R emf0 V V V V+ = = −

(e) 2 L 2 Lemf R R emf emf+ 0 diV V V V V V V L

dt+ = = − − = − −

(f) LdiV Ldt

=


1095

(g) 1R L Ldi di Vdt dt L

= =

CALCULATE:

(a) ( )( ) ( )tot18 V 6.0 6.0 6.0 A

6.0 6.0 i = Ω + Ω =

Ω Ω

(b) 1R

18 V 3.0 A6.0

i = =Ω

(c) 2R

18 V 3.0 A6.0

i = =Ω

(d) 1R 18 VV = −

(e) ( )( )2R 18 V 5.0 H 0 18 VV = − − = −

(f) ( )L 5.0 H 0 0V = =

(g) 1R 0 05.0 H

didt

= =

ROUND: Not necessary. DOUBLE CHECK: Evaluating the loop containing the inductor using equation 29.29 shows that after a

long time, ( )( )/ /emf emf2

2 2

( ) 1 ,t L RV Vi t e

R R−= − = as found above. Kirchoff’s rules can be used to show that

1 2R R ,i i= also as found above.

29.48. THINK: As the current begins to flow through the circuit, the self induced potential difference in the inductor opposes the change in current. As the change in current decreases, the self induced potential difference also decreases until the current reaches the steady state given by Ohm’s Law, emf / .i V R= When the switch is opened, the current will continue to flow, at a decreasing rate, through the loop composed of

3 , ,R L and 2R until the energy which has been stored in the inductor is dissipated. SKETCH: (a) (b)

(c)

RESEARCH: (a) Immediately after the switch is closed, the inductor is like an open-circuit. Clearly,

3R 0,i = and

1 2 3R 1 R 2 R 3 0.V i R i R i R+ + + = so 2 1R R

1 2

.Vi iR R

= =+

(b) After a long time, the inductor acts like a short-circuit. 2 3tot 1

2 3

,R R

R RR R

= ++

1tot

VV V RR

′ = −


1096

(c) When the switch is opened, 2 3L R R .i i i= = In fact, the equivalent resistance of this circuit is 2 3R R R′ = +

and the circuit can be redrawn accordingly. Since the current in an inductor cannot change

instantaneously, from part (b): ( )1 tot

initial3 3

/,

V R V RViR R

−′= = 2 3

tot 12 3

R RR R

R R= +

+ and .L

Rτ =

′ The

current for an RL circuit is ( ) ( )/initial .ti t i e τ−= Immediately after opening the switch, 0t ≈ and

( ) ( )00 initial initial .i t i e i= =

SIMPLIFY:

(a) 1 2 2 1 3R 1 R 2 R R R

1 2

0 0 , 0VV i R i R i i iR R

+ + + = = = =+

(b) 1

2 3R

2 3tot 1 2 1 3 2 31

2 3

( )V R RV ViR RR R R R R R RR

R R

+= = =

+ +++

2

2 3 1 2 1 3 2 3 1 2 1 3 31R

2 2 2 1 2 1 3 2 3 2 1 2 1 2 2 3 1 2 1 2 2 3

( ) ( ) ( ) ( )( ) ( )

V R R V R R R R R R V R R R R V RRV ViR R R R R R R R R R R R R R R R R R R R R R

′ + + + − += = − = = + + + + + +

3

2 3 1 2 1 3 2 3 1 2 1 31 2R

3 3 3 1 2 1 3 2 3 3 1 2 1 2 2 3 1 2 1 2 2 3

( ) ( ) ( ) ( )( )

V R R V R R R R R R V R R R RR V RV ViR R R R R R R R R R R R R R R R R R R R R R

′ + + + − += = − = = + + + + + +

(c) ( ) ( )/initial

ti t i e τ−= , where 1R 0,i = ( ) ( ) ( )

3 2

1 tot 2R R initial

3 1 2 1 3 2 3

/,

V R V R V Ri i i t i

R R R R R R R−

= − = = =+ +

.

CALCULATE: Not applicable. ROUND: Not applicable. DOUBLE CHECK: When the switch is closed and the current is increasing according to

( )/emf( ) 1 ,tVi t e

Rτ−= − as ,t → ∞ emf( ) ,

Vi t

R= which agrees with the result.

29.49. The energy density is given by 2B 0/ 2 .u B μ= Determine the volume that gives B 1 J:Vu =

( ) ( )( )

0 0 3 32 25

2 1 J 2 1 J1.0 10 m .

5.0 10 TV

Bμ μ

−= = = ⋅

⋅

This volume is equivalent to a 10 m by 10 m by 10 m cube. This is a fraction of the size of a house.

29.50. (a) The magnetic energy density is given by: ( )μ μ

= = = ⋅22 6 3B

0 0

1 1 3.00 T 3.58 10 J/m .2 2

u B

(b) The total energy is given by B B .U Vu= 2 ,V R Lπ= R = 0.500 m, L = 1.50 m.

( ) ( )( )π = ⋅ = ⋅2 6 3 6B 0.500 m 1.50 m 3.58 10 J/m 4.22 10 JU

29.51. (a) ( )2 210 26 3B

0 0

1 1 4.0 10 T 6.4 10 J/m2 2

u Bμ μ

= = ⋅ = ⋅

(b) The associated mass density is then: 26 3

9 3Brest2 8 2 2

6.37 10 J/m 7.07 10 kg/m(2.998 10 m/s )

uc

ρ ⋅= = = ⋅⋅

29.52. THINK: The emf potential and the resistance can be used to find the maximum current. Then the energy stored in the magnetic field of the inductor at one fourth of this current can be found. The equation for the rise in current as a function of time can be used to find the time for the circuit to reach a current of one


1097

fourth of its maximum value. The inductance of the inductor is = 40.0 mH,L the resistance of the resistor is = Ω0.500 ,R and the emf potential is =emf 20.0 V.V SKETCH:

RESEARCH: (a) At steady-state,

emfmax .

Vi

R=

The time of interest is when max / 4.i i= Use the equation 2B

1 .2

U Li=

(b) ( ) ( )RL/max max

11 ,4

ti t i e iτ−= − = RLLR

τ =

SIMPLIFY:

(a) 2 2

emf emfB 2

1 1 12 4 32

V LVU L

R R = =

(b) ( )RL/max max RL

RL

1 3 3 31 ln ln ln4 4 4 4

t t Li i e tR

τ ττ

− = − = − = − = −

CALCULATE:

(a) ( )( )

( ) = = Ω

2

2

0.0400 H 20.0 V1 2.00 J32 0.500

U

(b) = − = Ω

0.0400 H 3ln 0.0230 s0.500 4

t

ROUND: (a) U = 2.00 J (b) t = 0.0230 s DOUBLE-CHECK: It makes sense that the time it takes to reach one fourth of the maximum value is comparable to the time constant, RL .τ

29.53. THINK: The equation for the rate of energy production due to a potential across a resistance can be used to determine the heat generated. The induced potential can be found by using Faraday’s Law. Then the rise in temperature due to this heat can be found for the ring of mass = 0.0150 kgm and specific heat

capacity o129 J/kg C.c = The strength of the magnetic field is B = 0.0800 T, the radius of the ring is r = 0.00750 m, the time change between maximum and zero magnetic flux is Δ = 0.0400 s,t and the resistance of the ring is 661.9 10 .R −= ⋅ Ω SKETCH:


1098

RESEARCH: The induced potential in the ring is given by: ind .d BAVdt tΦΔ = − = −

Δ The rate of energy

production as heat is given by

( )2ind .

VP

RΔ

=

The power produced multiplied by the time difference is equal to the heat generated: .P t Q mc TΔ = = Δ

SIMPLIFY: The temperature rise is

( ) ( )22 22ind

BrV t t BATmcR mcR t mcR t

πΔ Δ Δ Δ = = − = Δ Δ


( )( )( )( )π

−−

Δ = = ⋅ °⋅ Ω

22

5o 6

0.0800 T 0.00750 m4.1715 10 C

0.0150 kg 129 J/kg C 61.9 10 0.0400 sT

ROUND: To three significant figures, the temperature rise is −Δ = ⋅ °54.17 10 C.T DOUBLE-CHECK: It makes sense that for larger fields, TΔ is larger, and for larger masses, TΔ is smaller since it would take more work to heat up the ring. As expected, the temperature increase is quite small.

29.54. THINK: Consider the energy of the dipole before and after the flip and relate this to the work done. SKETCH: A sketch is not necessary. RESEARCH: When the dipole is in alignment: .U NiAB= − When the dipole is anti-parallel to the field:

.U NiAB= SIMPLIFY: The work done must therefore be 2 .W U NiAB= Δ = CALCULATE: No calculations are necessary. ROUND: Rounding is not necessary. DOUBLE-CHECK: Larger fluxes (larger NAB) yield more work for the power supply.

29.55. THINK: Determine the energy density of the electric field and the magnetic field separately. SKETCH: A sketch is not necessary.

RESEARCH: 2B

0

1 ,2

u Bμ

= 2E 0

1 ,2

u Eε= 00 ,

k EB

ω×

=

0 0

,k

ωμ ε

=

( ) ( )0, cos ,E x t E k x tω= • −

( ) ( )0, cos .B x t B k x tω= • −

SIMPLIFY:

( )( )

2 2 222 210 0 02B

0 2 2 2 2 22 2E 0 0 0 0 0 0 00 0 0

cos1 1 1 1 12 2 cos

B k x t k E k EB BuE

u E E k x t E k E

ωε

μ μ ε μ ε μ ε ωω

− • − × × = = = = = • −

Note that k

is perpendicular to 0E

so 2 2 2

0 0 ,k E k E× =

so the above expression becomes B

E

1.uu

=

CALCULATE: No calculations are necessary. ROUND: Rounding is not necessary. DOUBLE-CHECK: This result shows that the energy in this type of wave is partitioned equally between the electric and magnetic fields.


1099

Additional Problems

29.56.

The induced voltage is given by: ( )( )ΦΔ = − = = = =ind

2.00 V2.00 V 20.0 m/s.0.100 m 1.00 T

dV vLB vdt

29.57. The potential difference is given by Faraday’s law:

( ) ( )π π −ΦΔ = − = = = = ⋅22 3ind 0.0400 m 1.50 T/s 7.54 10 Vd dB dBV A R

dt dt dt

Note that the radius of the coil is irrelevant.

29.58. The inductor cannot have the current jump instantaneously. From Kirchoff’s loop law: emf

emf 0 .Vdi diV L

d L dθ θ− = =

Integrate both sides to get: ( ) emf C.V

i t tL

= + Since ( )0 0,i = C = 0. The expression is then ( ) emf .V

i t tL

=

29.59. The energy stored in a solenoid is given by 2B / 2.U Li= The energy is dependent only on the magnitude,

not the direction of the current.

29.60. ( ) ( )/max 1 ,ti t i e τ−= − /L Rτ =

( )3

/max max 3

3 1 1 36.94 10 H 11 ln ln ln 20.0 4 4 4 42.56 10 s

t t Li i e Rt

τ

τ

−−

−

⋅ = − − = = − = − = Ω ⋅

29.61. Use the formulas: 2B

0

1 ,2

u Bμ

= 2E 0

12

u Eε= and 2

B2

E 0 0

1 .u Bu Eμ ε

= In particular, the values of the energy

densities are:

( )2 4 3

62 2

B1 50.0 μT 9.94 10 J/m ,

m kg2 1.257 10 s A

u −

−

= = ⋅ ⋅

and

( )4 2

12 8 32

3E1 s 8.842 10 150. N/C 9.95 10 J/m .2

Am kg

u − − = ⋅ = ⋅

To compute the ratios, it is useful to remember that 20 01/ .cμ ε = This is a result from light being an

electromagnetic wave where c is the speed of light in a vacuum.

( )22 622

E

58B2

50.0 10 T3.00 10 m/s 1.00 10150. N/C

u Bcu E

− ⋅= = ⋅ = ⋅

The energy density of the magnetic field is much larger than that of the electric field.


1100

29.62. ( )B cos 2 ,NAB ftπΦ = which means that ( ) ( )Bind / 2 sin 2 .V d dt NAB f ftπ π= − Φ = The maximum occurs

when ( )sin 2 1.ftπ = N = 1, so ( ) ( )( )22 2 5 6ind 2 2 0.010 m 4.0 10 T 17 1.3 10 V.V R B fπ π π − −= = ⋅ = ⋅

29.63. The current of an RL circuit is given by: ( ) ( )/emf 1 ,tVi t e

Rτ−= − where / .L Rτ = For = 20.0 μs:t

( ) ( )/ /emf emf1 1 1 11 1 ln ln

2 2 2 2 ln 1/ 2t tV V L Rte e t t L

R R Rτ τ τ− − = − − = = − = − = −

( )( )( )

−⋅ Ω ⋅ = − =

3 63.00 10 20.0 10 s 0.0866 H.

ln 1/ 2L

29.64. The current of an RL circuit is given by ( ) ( )/max 1 ,ti t i e τ−= − where / .L Rτ =

( )( ) ( )

τ τ− −

−

= − =

⋅= − = − =Ω

/ /max max

6

0.995 1 0.00500

0.200 10 H ln 0.00500 ln 0.00500 2.12 ns500.

t ti i e e

LtR

It is interesting to note that the voltage of the battery is irrelevant to the result of the problem.

29.65. For a single loop of wire (N = 1), the induced potential difference is:

( )Bind cos .

d dV BAdt dt

θΦΔ = − = −

Since the normal vector of the loop and the magnetic field is parallel, cos 1.θ = The negative sign can be dropped and indVΔ becomes:

( ) ( ) ( )( )Δ = = + = = =2ind 3.00 T 2.00 T/s 2.00 T/s 5.00 m 2.00 T/s 10.0 V.dB dV A A t A

dt dt

Note the magnetic field, ,B

is increasing, and it is directed into the page. By Lenz’s law, the induced magnetic field, i ,B

opposes the change in magnetic flux, B .Φ In this case, iB

is directed out of the page to

oppose the increasing field, ,B

directed into the page. The induced current is therefore counterclockwise.

29.66. The following circuit has values: V = 9.00 V, = = Ω1 2 100. ,R R and L = 3.00 H.

(a) When the switch is closed at t = 0 s, the current through 1R is: = = =Ω1

1

9.00 V 0.0900 A.100.

ViR

The

current through 2R is ( ) ( ) ( )/ /2 2

2

1 0 0.t l RVi t e iR

− = − =

(b) At t = 50.0 ms = 0.0500 s, 1i is still 0.0900 A, while 2i is:

( ) ( ) ( ) Ω = − − = − = Ω 2

9.00 V 100. 0.0500 s 1 exp 0.0500 s 0.0900 1 0.189 A 0.0730 A100. 3.00 H

i

(c) At t = 500. ms = 0.500 s, 1i is still 0.0900 A, and 2i is:

( ) ( ) ( )− Ω = − − = − ⋅ = Ω

82

9.00 V 100. 0.500 s 1 exp 0.500 s 0.0900 1 5.78 10 A 0.0900 A.100. 3.00 H

i

(d) After 10.0 s, the equilibrium current of 0.0900 A has long since been reached. Before the switch is opened, the currents 1i and 2i oppose each other in the right-most loop, as shown below.


1101

When the switch is opened (after achieving an equilibrium current in the circuit), 1 2 .i i= − After opening the switch, Kirchhoff’s loop rule becomes 1 2/ 0.Ldi dt iR iR+ + = With 1 2 ,R R R= = this expression

becomes ( )/ 2 0.Ldi dt i R+ = Solving for i yields: ( ) L/0 ,ti t i e τ−= L / 2 ,L Rτ = and 0i is the achieved

equilibrium current, i = 0.0900 A. At t = 0 s, ( ) ( ) τ−− = = = =L0/1 2 0 00 0 0.0900 A.i i i e i

(e) At t = 50.0 ms = 0.0500 s, ( ) ( ) ( ) ( )( )− Ω= − = =2 100. 0.0500 s /3.00 H1 20.0500 s 0.0500 s 0.0900 A 0.00321 A.i i e

(f) At t = 500. ms = 0.500 s, ( ) ( ) ( ) ( )( )− Ω= − = ≈2 100. 0.500 s /3.00 H1 20.500 s 0.500 s 0.0900 A 0 A.i i e

29.67. THINK: A solenoid of length, l = 3.0 m, and n = 290 turns/m has a current of i = 3.0 A, and stores an energy of B 2.8 J.U = Find the cross-sectional area, A, of the solenoid. SKETCH:

RESEARCH: The energy stored in the magnetic field of an ideal solenoid is 2 2

B 0 / 2.U n lAiμ=

SIMPLIFY: Solving for A yields: B2 2

0

2.

UA

n liμ=

CALCULATE: ( )

( )( ) ( )( ) ( )( )2

2 227 1

2 2.8 J N m1.9635 J/T A 1.9635 1.9635 mV s/m J/V s4 10 T m/A 290 m 3.0 m 3.0 A

Aπ − −

= = = =⋅

ROUND: Rounding to two significant figures, 22.0 m .A = DOUBLE-CHECK: Considering the length, l, of the solenoid, this is a reasonable cross-sectional area. The units of the result are also correct.

29.68. THINK: The rectangular loop has dimensions a by b and resistance R. It is placed on the xy-plane. The magnetic field direction points out of the page and varies in time according to ( )3

0 1 .B B l c t= + Determine

the direction of the current induced in the loop, ind ,i and its value at t = 1 s (in terms of a, b, R, 0B and

1c ). SKETCH:


1102

RESEARCH: Since the magnetic field is increasing as it comes out of the page, the induced magnetic field, i ,B points into the page according to Lenz’s law. The induced current flows clockwise. The current is

found from ind ind ,V i R= where ind B / cos / .V d dt dBA dtθ= − Φ = − SIMPLIFY: With ( )cos cos 0 1,θ = ° = and A constant:

( )3 2ind 0 1 0 13 .dB dV A A B l c t AB c t

dt dt = − = − + = − Then,

2 2ind 0 1 0 1

ind3 3

.V AB c t abB c t

iR R R

= = =

CALCULATE: At t = 1 s, 0 1ind

3,

abB ci

R= clockwise

ROUND: Not applicable. DOUBLE-CHECK: By dimensional analysis, the result has units of current:

2 2 2 2 2 20 1

3 3

m T s m V s/m s A .s s V/A

abB c tR

= = = Ω

29.69. THINK: The battery with =emf 12.0 V,V is connected in series with a switch and a light-bulb. When the light-bulb draws a current of i = 0.100 A, its glow becomes visible. This bulb draws P = 2.00 W when it has been connected and when the switch has been closed for a long time. When an inductor is put in series with the bulb and the rest of the circuit, the light-bulb begins to glow t = 3.50 ms after the switch is closed. Find the size of the inductor, L. SKETCH:

RESEARCH: The resistance of the light-bulb can be determined from 2 / .P V R= When the inductor is attached, the current is given by ( ) ( ) ( )( )/ /

0 emf1 e / 1 e .tR L tR Li t i V R− −= − = −

SIMPLIFY: 22

emf .VVR

P P= = Substitute this expression into the equation for the current to get:

( ) ( ) ( ) ( ) ( )

( )( )

emf/ / / /emf emf2

emfemf

emf

emf

1 e 1 e 1 e e 1/

/ ln 1 .ln 1 /

tR L tR L tR L tR L V i tV V Pi tR V PV P

V i t tRtR L LP V i t P

− − − −= − = − = − = −

− = − = −

−

CALCULATE: ( )

= = Ω212.0 V

72.0 ,2.00 W

R ( )( )

( )( ){ }Ω

= − = −

0.00350 s 72.0 0.27502 H

ln 1 12.0 V 0.100 A / 2.00 WL

ROUND: L = 0.275 H. DOUBLE-CHECK: An inductor of this capacity in this circuit is capable of storing energy

( )( )= = =22B

1 1 0.300 H 0.100 A 1.50 mJ.2 2

U Li This is sufficient energy to power a 2.00 W light bulb for

0.750 ms. This is a reasonable value for L in this light-bulb circuit.


1103

29.70. THINK: A circular loop of cross-section A is placed perpendicular to a time-varying magnetic field of ( ) 2

0 ,B t B at bt= + + where 0 ,B a, and b are constants. Assume the field points into the plane of the page. Determine (a) the magnetic flux, B ,Φ through the loop at t = 0, (b) an equation for the induced potential difference, ind ,V in the loop as a function of time, and (c) the magnitude and direction of the induced current if the resistance of the loop is R. SKETCH:

RESEARCH: (a) Since the loop is perpendicular to the field, the magnetic flux is given by B .BAΦ = (b) From Faraday’s law, ind B / .V d dt= − Φ Since A is constant while B varies with time, this expression becomes ( )ind / .V A dB dt= − (c) The magnitude of the induced current is found from V = iR. With the applied magnetic field directed into the page and increasing in time, the induced magnetic field will point out of the page to oppose the change in magnetic flux. The induced current flows counterclockwise. SIMPLIFY: (a) ( ) ( )2

0B .t BA B at bt AΦ = = + + At t = 0, ( ) 0B 0 .B AΦ =

(b) ( ) ( ) ( )2ind 0 2dV t A B at bt A a bt

dt= − + + = − +

(c) The magnitude of indi is given by: ( )ind

ind

2,

A a btVi

R R+

= = counterclockwise.

CALCULATE: Not applicable. ROUND: Not applicable. DOUBLE-CHECK: By dimensional analysis, the units are correct:

2 Wb T m ;BAΦ = = 2/ V m T/s;V AB t= = / A V/Ω .i V R= =

29.71. THINK: A conducting rod of length, L = 0.500 m, slides over a frame of two metal bars placed in a magnetic field of strength, B = 1000. gauss = 0.1000 T. The ends of the rods are connected by two resistors,

= Ω1 100. R and = Ω2 200. .R The conducting rod moves with a constant velocity of v = 8.00 m/s. Determine (a) the current flowing through the two resistors, 1i and 2 ,i (b) the power, P, delivered to the resistors, and (c) the force, F, needed for the motion of the rod with constant velocity. SKETCH:


1104

RESEARCH: (a) The induced potential difference across the resistors is ind B / .V d dt= − Φ Since B is constant while A varies in time at a velocity of v, this expression becomes ( )ind / .V B dA dt BLv= − = − The current in each resistor can be determined from ind ind .V i R=

(b) The power delivered to the resistors is 2 21 1 2 2 .P i R i R= +

(c) The force needed to move the rod with a constant velocity is obtained by calculating the total force acting on the rod. The magnetic force on the rod, mag ,F is given by ( )mag eq/ ,F BiL B V R L= = where eqR is

the equivalent resistance. Note for n resistors in parallel, the equivalent resistance is:

eq 1 2

1 1 1 1... .nR R R R

= + + +

SIMPLIFY:

(a) ind ,V BLv= − ind1

1

,V

iR

= ind2

2

Vi

R=

(b) 2 21 1 2 2P i R i R= +

(c) 2 2mag ind

1 2 1 2

1 1 1 1F BV L B L vR R R R

= + = +

CALCULATE:

(a) ( )( )( )= − = −ind 0.100 T 0.500 m 8.00 m/s 0.400 V,V −

= =Ω1

0.400 V0.00400 A,

100. i

−= =

Ω2

0.400 V0.00200 A

200. i

(b) ( ) ( ) ( ) ( )= Ω + Ω =2 20.00400 A 100. 0.00200 A 200. 0.00240 WP

(c) ( ) ( ) ( ) = + = Ω Ω

2 2mag

1 10.100 T 0.500 m 8.00 m/s 0.000300 N100. 200.

F

ROUND: (a) =1 4.00 mAi , =2 2.00 mAi (b) P = 2.00 mW (c) =mag 0.300 mNF

DOUBLE-CHECK: The calculated values are consistent with the given values. Dimensional analysis confirms all the units are correct.

29.72. THINK: The loop on the door has dimensions h = 0.150 m, w = 0.0800 m and resistance, 5.00 .R = Ω When the door is closed, it is perpendicular ( )0θ = ° to the Earth’s uniform magnetic field,

5 E 2.6 10 T.B −= ⋅ At time, t = 0 s, the door is opened (right edge moving into the page in the figure below)

at a constant rate, with an opening angle of ( ) ,t tθ ω= where 3.5 rad/s.ω = Determine the direction and

magnitude of the induced current, ( ),i t at t = 0.200 s.


1105

SKETCH:

RESEARCH: The induced current, i, is found from ind / ,i V R= where indV is given by Bind / ,V d dt= − Φ and B cos .BA θΦ = As the door opens, the B field through the loop decreases; by Lenz’s law the induced B field points into the page, at an angle of ( )tθ from the plane of the page. The induced current flows clockwise.

SIMPLIFY: ( )( ) EB cos cos ,BA t whB tθ ωΦ = = ( )Bind E Ecos sin

d dV whB t whB tdt dt

ω ω ωΦ= − = − =

The magnitude of i is ( ) ind E sin.

V whB ti tR R

ω ω= =

CALCULATE: At t = 0.200 s:

( ) ( )( )( )( ) ( )( )57

0.0800 m 0.150 m 2.6 10 T 3.5 rad/s sin 3.5 rad/s 0.200 s0.200 s 1.407 10 A.

5.00 i

−−

⋅ = = ⋅Ω

ROUND: Rounding to two significant figures, ( )0.200 s 140 nAi = clockwise. DOUBLE-CHECK: This induced current is reasonable for a loop with such a small cross-sectional area in the Earth’s magnetic field.

29.73. THINK: The steel cylinder has radius, r = 2.5 cm = 0.025 m, and length, L = 10.0 cm = 0.100 m. The ramp is inclined at 15φ = ° and has a length, l = 3.0 m. Determine the induced potential difference, indV between the ends at the bottom of the ramp if the ramp points in the direction of magnetic North. Use

40.426 10 T−⋅ as the magnetic field of the Earth. SKETCH:

RESEARCH: The magnetic field of the Earth has a finite angle with the surface of the Earth (i.e. it is usually not parallel or perpendicular to the Earth's surface). Since the ramp’s surface points towards the magnetic north pole, it is aligned with the local magnetic field line (with exceptions at the Equator and poles). Generally, as a conductor of length L moves through a magnetic field the magnetic force acting on the electrons in the conductor is mag sin ,F q v B evB θ= × =

where θ is the angle between the velocity of

the conductor and the magnetic field. When the electric force, E ,F qE= and the magnetic force on the electrons are in equilibrium: sin .E vB θ= This means that the induced potential difference, ind ,V between the ends of the conductor is given by ind sin .V EL vBL θ= = As the cylinder rolls down the ramp (in the


1106

direction of the Earth's magnetic field, B), the angle between the cylinder's velocity vector, v, and the Earth's magnetic field, B, is zero so the induced voltage between the ends of the cylinder is zero. At the bottom of the ramp, the cylinder changes direction, and the induced potential difference between the ends is ind sin .V vBL θ= Note the magnitude of the Earth's magnetic field is B = 0.00003 T. To determine the speed, v, of the cylinder, recall that the cylinder rolls without slipping so the change in potential energy for the cylinder is equal in magnitude to the change in its kinetic energy:

2 2/ 2 / 2 .K U mv I mghωΔ = −Δ + = Here I is the moment of inertia for the cylinder: 2 ,/ 2I mr= and / .v Rω =

SIMPLIFY: Determine v: 2

2 2 2 22

2 2

1 1 1 1 12 2 2 2 2

1 1 4 4 sin2 4 3 3

vK U mv I mgh mv mrR

v v gh v gh gl

ω

ϕ

Δ = −Δ + = +

+ = = =

At the bottom of the ramp, the angle between B

and v is: 180 .θ ϕ= ° −

ind4 4sin sin sin(180 ) sin sin( )3 3

V vBL gl BL gl BLθ ϕ ϕ ϕ ϕ= = ° − =

CALCULATE: 2ind

4 64 (9.81 m/s )(3.0 m)sin15 (0.426 10 T)(0.100 m)sin(15 ) 9.435 10 V3

V − − = ° ⋅ ° = ⋅

ROUND: ind 9.4 μV.V = DOUBLE-CHECK: Considering the given values for this problem, this result is reasonable and also has

the correct units: 2ind 2 2

m m V s mm/s m T m V .

s mV

= ⋅

29.74. THINK: The battery is connected to a resistor and an inductor in series. Determine (a) the current, ( ),i t across the circuit after the switch is closed, (b) the total energy, U, provided by the battery from time, t = 0 to t = L/R, (c) the total energy, R ,U dissipated from the resistor, R, for the same time period, and (d) discuss the conservation of energy in this circuit. SKETCH:

RESEARCH: (a) In this RL circuit, current at any given time, t, is given by equation 29.29 in the text, namely

( )/0 1 ,ti i e τ−= − where 0 / ,i V R= and the time constant is / .L Rτ =

(b) The power provided by the battery is P = Vi. In the given time period, the total energy provided by the battery is .U Vidt=

(c) The power dissipated in the resistor is 2 .P i R= In the given time period, the total energy dissipated in the resistor is 2

R .U i Rdt= (d) Any discrepancy between the energy provided by the battery and the energy dissipated in the resistor is due to the fact that there is energy stored in the inductor, 2

L / 2.U Li=


1107

SIMPLIFY:

(a) ( ) ( )/1 tR LVi t eR

−= −

(b) ( ) ( )

( )

2 2 2 2 2 2 2/ / / 1

2 2 200 0 0

2

2

1

0.368

t tR L tR L R L

t

V V V L V V L V L V LU Vi t dt e dt t e e eR R R R R R R R

V LR

ττ τ τ ττ= − − − −

=

= = − = + = + − =

=

(c) ( ) ( ) ( )2 2 2 / 22 22 / / /

R 2 2 20 0 00 02 2 2

12 2

21 ln2

1 2 2 1 0 0.1682 2

tR Lt tR L tR L tR L

t

V V L V Le V LU i Rdt e dt e eR R R R

V L e V LeR R

ττ ττ τ −= − − −

=

−−

= = − = − + −

= − + + − + + =

(d) At time, / :t L Rτ= = ( ) ( )2 222 1

L 2 2

1 1 0.200 .2 2

LV V LU Li eR R

τ −= = − =

CALCULATE: Not applicable. ROUND: Not applicable. DOUBLE-CHECK: The total energy of the battery is the sum of the energy dissipated in the resistor and the energy stored in the inductor; energy is conserved.

29.75. THINK: The rectangular circuit loop has length, L = 0.600 m, and width, w = 0.150 m, with resistance, = Ω35.0 .R It is held parallel to the xy-plane with one end inside a uniform magnetic field as shown in the

figure. The magnetic field is =

Rˆ2.00 TB z along the positive z-axis to the right of the dotted line; L 0 TB =

to the left of the dotted line. Determine the magnitude of the force, app ,F required to move the loop to the

left at a constant speed of v = 0.100 m/s, while the right end of the loop is still in the magnetic field. Determine the power, P, used by an agent to pull the loop out of the magnetic field at this speed, and the power, R ,P dissipated by the resistor. SKETCH:

RESEARCH: The magnitude of the force required to move the loop will be equal to the magnitude of the force, i ,F on the current induced in the segment of the loop that lies along the y-axis in the magnetic field. That is, app 2 sin .F F iwB θ= = Since the angle, ,θ between the loop segment of length, w, and the magnetic

field is 90 :° sin 1.θ = The induced current, i, is ind / ,i V R= where indV vwB= (see equation 29.15). The


1108

power, P, used by an agent to pull the loop out of the magnetic field is given by app .P F v= The power

dissipated by the resistor is given by 2R .P i R=

SIMPLIFY: The current is ind .V vwBi

R R= =

(a) appF iwB=

(b) appP F v=

(c) 2RP i R=

CALCULATE:

(a) ( )( )( )0.100 m/s 0.150 m 2.00 T

0.85714 mA35.0

i = =Ω

( )( )( )app 0.85714 mA 0.150 m 2.00 T 0.25714 mNF = =

(b) ( )( )0.25714 m N 0.100 m/s 25.7714 μWP = =

(c) ( ) ( )= Ω =2R 0.85714 mA 35.0 25.714 μWP

ROUND: (a) app 0.257 mNF =

(b) 25.8 μWP = (c) R 25.7 μWP = DOUBLE-CHECK: All the power used to move the loop while in the magnetic field is dissipated in the resistor: R .P P=

Chapter 30: Electromagnetic Oscillations and Currents

1109


In-Class Exercises

30.1. b 30.2. b 30.3. a 30.4. a 30.5. a 30.6. d 30.7. b 30.8. d 30.9. c Multiple Choice

30.1. d 30.2. c 30.3. b 30.4. a 30.5. d 30.6. b 30.7. d 30.8. c

Questions

30.9. The impedance of an RLC circuit in series is

( )22 .L CZ R X X= + −

At resonance, .L CX X= Therefore, the impedance is at is minimum value of .Z R=

30.10. The total energy stored in a magnetic field is the magnetic energy density times the volume of the field, ,B BU u V= where ( )2

0/ 2Bu B= μ (Equation 29.35). For the = 5.00 kmd thick shell above the Earth’s

surface, the volume is, ( ) ( ) ( )3 3E E4 / 3 4 / 3 ,V R d Rπ π= + − where E 6378 kmR = is the Earth’s radius. The

total energy stored in this field above the Earth’s surface is

( )( ) ( )( ) ( ) ( )( )π π

μ π

−

−

⋅ = + − = ⋅ − ⋅ = ⋅ ⋅

242 3 33 3 6 6 12E E 7

0

0.500 10 T4 4 6.383 10 m 6.378 10 m 2.54 10 kJ.6 6 4 10 T m/AB

BU R d R

30.11. No, charges are not crossing the gap (dielectric) of the capacitor. It simply means that, because of the periodic change in the polarity of the emf source, the capacitor is being periodically charged and discharged. No charge crosses the gap of the capacitor, whether in a DC or AC circuit. (This presumes that the potential on the capacitor does not get so big that the electric field exceeds the dielectric strength.)

30.12. In an RL circuit with alternating current, the expression “the current lags behind the voltage” means that the current achieves its maximum value at a delayed time compared to the time when the applied voltage achieves its maximum value. This is due to the phenomenon of self-induction: the changing current through the coil of the inductor creates a changing magnetic flux through the coil. Faraday’s law will result in an induced emf in the coil, which will oppose the externally applied emf, in compliance with Lenz’s law. The net effect is that the current will always try to “catch up” with the applied voltage, but will “lag behind” because of the self induced emf of the inductor.

30.13. The voltages given in the problem are rms values. Of course, Kirchhoff’s rules will be obeyed at any instant of time, but not when using rms values since the voltages are out of phase with each other. This circuit does not violate Kirchhoff`s rules.

30.14. The power depends upon the voltage. For an AC circuit, the voltage oscillate about zero, so the average voltage is zero. Thus, the average power for any AC circuit would be zero regardless of amplitude, which would not be very informative as an average value. Hence, RMS power is used instead.

30.15. Each device has its own specific operating current and voltage. Each needs its own transformer with a specific ratio of primary to secondary coils to convert the normal household current and voltage into the required current and voltage in order to prevent damage to the device.


1110

30.16. When electrons arrive at one plate of the capacitor, they repel an equal number of electrons off the opposite plate. For this reason, the amount of current flowing into one plate of the capacitor is equal to the amount of current flowing out of the other plate, despite the fact that charge does not actually flow across the gap.

30.17. Any surrounding high-frequency electromagnetic waves can induce unwanted signals. In parallel pairs, noise sources may affect one wire more than the other and this can be disruptive. Twisting the pairs minimizes this effect, since for each half twist the wire nearest to the noise source is exchanged.

30.18. (a) In this circuit capacitive reactance can be neglected since there is no capacitor. The rms input voltage

rmsV and the root-mean-square current rmsI are related by ( )1/22 2 2rms rms rms/ / ,I V Z V R L= = + ω where R

is the resistance in the circuit and L is the inductance of the solenoid. The effect of the ferromagnetic core is to increase the inductance of the solenoid by the factor m .κ If the inductive reactance Lω is substantially greater than the resistance ,R inserting the core will greatly increase the total impedance. Thus, the rms current will decrease. (b) With a DC power source ( )0 ,ω = the current would have fluctuated as the magnet was being inserted due to induction. Once the insertion was complete the current would return to its original value since the resistance of the circuit is unaffected by the presence of the core.

30.19. The response of the tuner at any frequency is related to the amplitude of the input signal and how close to resonance the input signal is. By design the response should be dominated by the signal that is at resonance. However, if the signal at some other non-resonant frequency is large and somewhat close to the resonant frequency, noticeable crosstalk can occur.

30.20. A sine or cosine signal is termed “pure” or “monochromatic”; it consists of a single frequency. Any other periodic signal is a superposition of harmonics. For simplicity, consider a square wave ( )S t which takes

the value 1+ (“on”) for ( )0 / 2t T≤ < and 1− (“off”) for ( )/ 2 ,T t T≤ < repeating periodically for all time, ,t in both directions. As this is an odd function, it can be written as a sum of sine functions of period

:T ( ) ( )( )1

sin 2 / .nn

S t b n T tπ∞

=

= From Fourier analysis (see below), the coefficients nb can be determined

by multiplying both sides by ( )( )sin 2 /k T tπ and integrating from 0t = to .t T=

( ) ( )

( )( ) ( )( )( ) ( ) ( )( )( )

0/2

0 /2/2

0 /2

1 2( )sin/ 2

2 2 21 sin 1 sin

2 2 2 2cos cos2 2

1 cos cos 0 cos 2 cos

1 2 2cos

T

k

T T

Tt T t T

t t T

kb S t t dtT T

k kt dt t dtT T T

T k T kt tT k T T k T

k k kk

k

π

π π

π ππ π

π π ππ

ππ

= =

= =

=

= + − −

− − = −

= − − − + −

= −

( )( )4 if is odd;

0 if is even.

k

kk

kπ

=


1111

Hence, this square wave can be written as ( ) ( )0

2 2 14 1 sin ,2 1j

jS t t

j Tπ

π

∞

=

+= + where the harmonic

frequencies are all odd multiples of the fundamental frequency 2 / .Tπ Fourier analysis is a very important mathematical tool, named for Jean Baptist Joseph Fourier (1768-1830), with a long and curious history. If such a square wave is applied as the driving voltage of an RLC circuit, the response of the circuit will be the sum of the responses to the fundamental ( )0j = term. But if the frequency of the square wave is any odd submultiple (one-third, one-fifth, etc.) of the resonant frequency, the square wave will contain a harmonic at the resonant frequency and the circuit will resonate in response to that term. The frequency response curve of the circuit with square-wave input will contain a sequence of resonance peaks, at its resonant frequency and all odd submultiples of it.

30.21. Yes, it is possible to have the voltage amplitude across the inductor exceed the voltage amplitude of the

voltage supply. Since the voltage across each component is related by ( ) ( )2 22m ,R L CV V V V= + − the

voltage across the battery is equal to the voltage across the resistor at resonance. Therefore, at resonance, the voltage across the inductor could be anything since it is countered by the same voltage across the capacitor.

30.22. The transformer works on the principle of mutual inductance, and depends on the (back) emf that is generated in the set of two coils in the transformer due to changing magnetic flux within the coils. If the current is DC, then there is no change in flux, and therefore, no possibility of operating the two coils as a transformer.

Problems

30.23. From the inductance and capacitance, 32.0 mHL = and 45.0 μF,C = the frequency of oscillation is

( ) 1/20 .LC −=ω The total energy is constant at 2

0 / 2U q C= where 0 10.0 μC,q = and the charge varies as

( )0 0cos .q q t= ω Since energy remains constant, when the energy in both is the same, it is ( )1/ 2 .U

( ) ( )

( )( )

2 2 20 0 2 10

E 00

1 1 4

cos1 1 1 1 1 cos cos2 2 2 2 22

1 1 cos 32.0 mH 45.0 μF cos 9.42 10 s2 2

q t qU U t tC C

t LC

ωω

ω−

− − −

= = = =

= = = ⋅

30.24. (a) From conservation of energy, ( )2 2max max/ / 2,2E BU U q LiC= = where max ,q CV= with 2.00 μF,C =

0.250 HL = and 12.0 V.V = Therefore,

( ) ( )( ) ( )

22 22max max

max

2.00 μF1 12.0 V 33.9 mA.2 2 2 2 0.250 H

CVq Li CCV i VC C L

= = = = = =

(b) The angular frequency of the current is ( )ω −= 1/20 ,LC so that the frequency of the oscillation is

( )( )0 1 1 225 Hz.

2 2 2 0.250 H 2.00 μFf

LCωπ π π

= = = =

30.25. In general, the frequency of oscillation is ( )ω −= 1/20 ,LC where = 1.00 mH.L The maximum energy in the

capacitor is ( )= 2E max / 2 .U q C Since the charge varies as ( )max 0cos ,q q t= ω and at time = 2.10 mst the

energy on capacitor is half the maximum value, ( )=E 1/ 2 .U U This means


1112

( ) ( )

( )( )

( )( )

ωω ω −

− −

− −

= = = =

⋅ = = = ⋅

2 2 2max 0 2 1max

E 0 0

21 3

3 1

cos1 1 1 1 1 cos cos2 2 2 2 22

cos 1/ 2 2.10 10 s1 1 7.15 mF.1.00 10 H cos 1/ 2

q t qU U tC tC

CtLC

30.26. (a) Since the current is proportional to ( )ω0sin t where 10 1200. s ,ω −= this is at a maximum when

ω π=0 / 2;t therefore, ( ) ( )( )10/ 2 / 2 1200. s 1.31 ms.t π ω π −= = =

(b) The total energy in the circuit is = 2max / 2,U Li where max 1.00 A.i = Since the angular frequency is

( )ω −= 1/20 ,LC then ( )ω= 2

01/ ,L C where 10.0 μF.C = The total energy of the circuit is then

( )( ) ( )

max max2 210

1.00 A34.7 mJ.

2 2 2 1200. s 10.0 μF

Li iU

Cω −= = = ≈

(c) The inductance is

( ) ( )2 210

1 1 69.4 mH.1200. s 10.0 μF

LCω −

= = =

30.27. THINK: The charge on the capacitor will oscillate with time as a cosine function with a period determined by the inductance, 0.200 H,L = and capacitance, 10.0 μF.C = The potential, emf 12.0 V,V = will give the initial charge on the capacitor. Ignoring the sign of the charge, the charge on the capacitor will equal

80.0 μCQ = periodically. SKETCH:

RESEARCH: The initial (and maximum) charge on the capacitor is =max .q CV The charge will oscillate as

( )max 0cos ,q q t= ω where 0 1/ .LC=ω SIMPLIFY: The first time 1t when the charge on the capacitor is equal to Q is

( ) 1max 0 1 1

0 max

1cos cos ;Qq Q q t tq

− = = =

ω

ω

By symmetry, the second time 2t and the third time 3t are given by

0 2 0 1 0 1 2 10 0

t t t t t

= − = − = −

π πω π ω ωω ω

and

0 3 0 1 0 1 3 10 0

.t t t t t

= + = + = +

π πω π ω ωω ω

CALCULATE: ( )( )0

1 707.107 rad/s0.200 H 10.0 μF

ω = =

( )( )max 10.0 μF 12.0 V 120. μCq = =


1113

( )1

180.0 μC1 cos 0.0011895 s

707.107 rad/s 120. μCt −

= =

( )2 0.0011895 s 0.0032534 s707.107 rad/s

t π= − =

( )3 0.0011895 s 0.0056323 s707.107 rad/s

t = + =π

ROUND: Rounding the times to three significant figures gives: 1 1.19 ms,t = 2 3.25 ms,t = and

3 5.63 ms.t = DOUBLE-CHECK: Given the high frequency, small times are expected, so the answers are reasonable.

30.28. THINK: The current, max 3.00 A,i = will oscillate with time as a sine function with a period determined by the inductance, 7.00 mH,L = and capacitance, 4.00 mF.C = The charge will also vary with the same period as the current, but as a cosine function. The maximum charge on the capacitor is related to the energy of the system. SKETCH: Not required. RESEARCH: The total energy in circuit is found either when the current is at a maximum,

( )= 2max1/ 2 ,E Li or when charge on capacitor is a maximum, ( )2

max / 2 .E q C= The charge varies as

( )max 0cos ,q q t= ω where ω =0 1/ .LC SIMPLIFY:

(a) The energy in the circuit is = 2max

1 .2

U Li

(b) The maximum charge on the capacitors is found by = = =2max

max max 2 .2

qU q CU i LC

C Therefore,

the expression for charge is max cos .tq i LCLC

=

CALCULATE:

(a) ( )( )21 7.00 mH 3.00 A 0.0315 J2

U = =

(b) ( ) ( )( )( )( )

( )3.00 A 7.00 mH 4.00 mF cos 0.01587cos 189.0 C7.00 mH 4.00 mF

tq t = =

ROUND: (a) 31.5 mJU = (b) It is best not to round the coefficients of the expression. Keep the answer as ( )0.01587cos 189.0 ,q t= with the intention of rounding after particular values of t are substituted. DOUBLE-CHECK: The equation for the charge has the proper units since LC has units of s and maxi has units of C/s.

30.29. In general, the energy on the capacitor will vary as ( )ω−= / 2E max cos .Rt LU U e t The energy on the inductor

varies similarly by ( )/ 2B max sin .Rt LU U e t−= ω For max / 2,U U=

/ /max

1 ln2 ln2.2

Rt L Rt LE B

Rt LU U U U e e tL R

− −= + = = = =


1114

30.30. THINK: For an RLC circuit, the charge on the capacitor is described by equation 30.6. The capacitance is found by equating the term in the exponential to the period determined from the term in the cosine. The resistance and inductance are 50.0 R = Ω and 1.00 mH.L = SKETCH: Not required. RESEARCH: The general expression for an RLC circuit is

( )/2max cos ,Rt Lq q e t−= ω

where ( )ω ω= − 220 / 2R L and where ω =0 1/ .LC The period of the oscillation is π ω= 2 / .T The decay

rate of the exponential is 2 / .L Rτ = SIMPLIFY: Capacitance is found by letting ,T=τ which gives the result:

( ) ( )22 2222 2 20 20 2 2

02 2 2

/ 22 2 14 4 1 .2 2 4 4 4

RR LL R R RL

R L LCL L

ωωπ π ωω π π

−− = = = + = =

Therefore, ( )π=

+2 2

4 .4 1

LCR

CALCULATE: ( )

( ) ( )3

82 2

4 1.00 10 H3.9527 10 F

50.0 4 1C

π

−−

⋅= = ⋅

Ω +

ROUND: Rounding to three significant figures, 39.5 nF.C = DOUBLE-CHECK: The time constant is 52 / 4 10 s.L Rτ −= = ⋅ The angular frequency, ω, is

25 11 1.5708 10 s ,

2R

LC L− = − = ⋅

ω

and the period is 52 / 4.00 10 s.T π ω −= = ⋅ So ,T=τ as required.

30.31. THINK: The frequency of the damped oscillation is independent of the initial charge, and hence potential. It then only depends on the inductance, 0.200 H,L = the resistance, 50.0 ,R = Ω capacitance 2.00 μF,C = and =emf 12.0V.V SKETCH:

RESEARCH: In general the charge on the capacitor is

( )/2max cos ,Rt Lq q e t−= ω

where ( )ω ω= − 220 / 2R L and where ω =0 1/ .LC The frequency of oscillation is ( )ω π= / 2 .f

SIMPLIFY: ( ) ( ) ( )22 2 2

0 / 2 1/ / 4

2 2 2

R L LC R Lf

− −= = =

ωωπ π π

CALCULATE: ( )( )( ) ( ) ( )( )2 261/ 0.200 H 2.00 10 F 50.0 / 4 0.200 H

250.858 Hz2

fπ

−⋅ − Ω= =


1115

ROUND: Rounding to three significant figures, 251 Hz.f = DOUBLE-CHECK: This frequency is typical of RLC circuits.

30.32. THINK: The angular frequency of the RLC is 20% less than the angular frequency of the LC circuit where the inductance is 4.0 mHL = and the capacitance is 2.50 μF.C = Even though the current is actually a damped oscillation, the magnitude of the oscillation is describe simply by the exponential decay, similar to that of the charge. Therefore, the current is at half of its maximum when the exponential is at a half. The number of periods in a given time is the number of oscillations. SKETCH: Not required. RESEARCH: The RLC frequency is related to LC frequency by ω ω=RLC LC0.8 . The RLC frequency is

( )( )22RLC 0 / 2 ,R L= −ω ω where ω ω= =0 LC 1/ .LC The current varies as /2

max .Rt Li i e−= A wave goes

through n oscillations in time ,t nT= where T is the period. SIMPLIFY: (a) For the oscillation frequency,

2 2 22 2 2 2

RLC LC 0 0 0 0 0 2

0

0.8 0.8 0.64 0.36 2 2 4

1.2 1.2 1.2 .

R R RL L L

L LR LCLC

= − = − = =

= = =

ω ω ω ω ω ω ω

ω

(b) /2 /2max max

1 1 2 ln22 2

Rt L Rt L Li i i e e tR

− − = = = =

(c) The period is RLC

2 .T πω

= Therefore, the number of oscillations is ω

π= = RLC .

2ttn

T

CALCULATE:

(a) ( )( )

3

6

4.0 10 H1.2 48

2.50 10 FR

−

−

⋅= = Ω

⋅

(b) ( )

( ) ( )32 4.0 10 H

ln 2 0.000115525 s48

t−⋅

= =Ω

(c) ( )( )( )

( )

2

RLC 3 6 3

48 1 8000 rad/s,4.0 10 H 2.50 10 F 2 4.0 10 H

ω− − −

Ω = − = ⋅ ⋅ ⋅

and

( )( )0.000115525 s 8000 rad/s0.14709 cycles.

2n

π= =

ROUND: To two significant figures, (a) 48 R = Ω (b) 120 μst = (c) 0.15 cyclesn = DOUBLE-CHECK: Since the resistance is large compared to the inductance, the current will die off quickly, so a complete oscillations will not occur before it is at half its maximum value.

30.33. The capacitive reactance, = Ω200. ,CX is given by ( )1/ ,CX C= ω where = 10.0 μF;C therefore, the frequency is

( )( )ω = = =Ω

1 1 500. rad/s.200. 10.0 μFCX C


1116

30.34. The capacitive reactance is given by ( )ω=C 1/ ,X C where −= ⋅ 65.00 10 FC and 2 ,f=ω π where =100. Hz.f Therefore, the capacitive reactance is

( )( )π π −= = = Ω

⋅ 6

1 1 318 .2 2 100. Hz 5.00 10 FCX

fC

The maximum current through the capacitor, ,CI is given by / ,C C CI V X= where = 10.0 V.CV Therefore,

( )( )= = =

Ω10.0 V

31.4 mA.318

CC

C

VI

X

30.35. The inductive reactance is given by ,LX L= ω where = 47.0 mHL and 2 ,f=ω π where =1000. Hz.f Therefore, the inductive capacitance is

( )( )π π −= = ⋅ = Ω32 2 1000. Hz 47.0 10 H 295 .LX fL

The maximum current through the inductor, ,LI is given by / ,L L LI V X= where =12.0 V.LV Therefore,

( )( )= = =

Ω12.0 V

40.7 mA.295

LL

L

VI

X

30.36. There can be no non-zero steady state current though the circuit. The current will vary until the capacitor is fully charged. Since the battery remains in the circuit, the current is not allowed to oscillate as it would in an LC circuit. The current will rise to a maximum and fall to zero eventually, where a steady state is achieved.

30.37. (a) The resonant angular frequency of an RLC circuit is ω =0 1/ LC where 0.500 HL = and 0.400 μF.C = Therefore,

( )( )0

6

1 1 2240 rad/s0.400 10 F 0.500 HLC −

= = =⋅

ω

(b) At resonance, only the resistor contributes to the overall impedance, so = emf /I V R where =emf 40.0 VV and 100.0 .R = Ω Therefore, ( ) ( )40.0 V / 100.0 0.400 A.I = Ω =

30.38. In general, the resonant frequency is determined by ω π π= =0 0 0/ 2 1/ 2 ,f LC when =0 5.0 MHzf and

0 15 pF,C = the circuit has an inductance of L that satisfies the given relation. When 1 380 pF,C = the same L will give another resonant frequency. Therefore,

( ) ( )( )

0 2 20 00

01 0 1

11 12 2

0 0

1 1 ,42

15 pF1 1 5.0 MHz 1.0 MHz.380 pF2 2

4

f Lf CLC

Cf f f

CLC Cf C

ππ

π ππ

= =

= = = = = =

30.39. Given the frequency, 1.00 kHz,f = the angular frequency is 2 .f=ω π The phase constant for an RLC circuit is given by

( )1/tan .L C L CX X

R R−−

= =ω ω

φ

For the values 100. ,R = Ω 10.0 mHL = and 100. μF,C = the phase constant is

( )( )( ) ( )( )( )( )( )

13 6

1

3 32 1.00 10 Hz 10.0 10 H 2 1.00 10 Hz 100. 10 Ftan 0.549 rad.

100.

π πφ

−− −

−

⋅ ⋅ − ⋅ ⋅ = =

Ω


1117

The impedance for this circuit is

( )( )

( ) ( )( )( ) ( )( )( )( )

22

212 33 3 6

1/

100. 2 1.00 10 Hz 10.0 10 H 2 1.00 10 Hz 100. 10 F 117 .

Z R L Cω ω

π π−

− −

= + −

= Ω + ⋅ ⋅ − ⋅ ⋅ = Ω

30.40. For an RLC circuit with = 5.00 mHL and = 4.00 μF,C the resonant frequency, ω0 , is 0 1/ .LC=ω Therefore,

( )( )ω

− −= =

⋅ ⋅0 3 6

1 7070 rad/s.5.00 10 H 4.00 10 F

At resonance, the resistor with = Ω1.00 k ,R is the only contribution to the impedance. For a peak voltage of =m 10.0 V,V the maximum current is

( )( )= = =

Ωm

m

10.0 V10.0 mA.

1000. VIR

30.41. THINK: Given the equation, ( ) ( )ω= 12.0 V sinV t the peak voltage is clearly =m 12.0 V.V When the circuit is in resonance, the current is dictated solely by the resistance, = Ω10.0 .R This means the inductor, = 2.00 H,L and the capacitor, =10.0 μFC will not influence the current. However, the current will dictate the voltage drop across each. SKETCH: Not required. RESEARCH: At resonance, impedance is = .Z R The maximum current at resonance is m m / .I V R= The

resonant frequency is ω =0 1/ .LC The voltage drop across the inductor is ,L LV IX= where 01/ .LX C= ω SIMPLIFY: The voltage drop across the inductor is,

m m mL 0 .L

V V VL LV IX LR R R CLC

= = = =ω

CALCULATE: ( )( )

( )( )−

= =Ω ⋅ 6

12.0 V 2.00 H536.66 V

10.0 10.0 10 CLV

ROUND: To three significant figures, =L 537 V.V DOUBLE-CHECK: Even though the answer seems large compared to m ,V the voltage drop across the capacitor is the exact same. Since the voltage across each component is related by

( ) ( )2 22m ,R L CV V V V= + − the voltage across the battery is equal to the voltage across the resistor at

resonance. Therefore, at resonance, the voltage across the inductor could be anything since it is countered by the same voltage across the capacitor.

30.42. THINK: The inductive reactance and the capacitive reactance are needed to find the impedance of the RLC circuit. The AC power source oscillates with frequency 60.0 Hzf = and has an amplitude of

m 220 V.V = The resistance is 50.0 ,R = Ω the inductance is 0.200 HL = and the capacitance is = 0.040 mF.C

SKETCH: Not required. RESEARCH: The angular frequency of oscillation is ω π= 2 .f The inductive reactance is .LX L= ω The

capacitive reactance is 1/ .CX C= ω The impedance of circuit is ( )22 .L CZ R X X= + − The maximum

current through the circuit is m m / .I V Z= The maximum potential drop across each component is

m ,i iV I X= where i denotes either ,R C or .L


1118

SIMPLIFY: (a) 2LX L fL= =ω π

(b) 1 12CX

C fC= =

ω π

(c) ( )22L CZ R X X= + −

(d) mm

VI

Z=

(e) m ,RV I R= mC CV I X= and mL LV I X= CALCULATE: (a) ( )( )2 60.0 Hz 0.200 H 75.398 LX = = Ωπ

(b) ( )( )3

1 66.315 2 60.0 Hz 0.040 10 FCX

−= = Ω

⋅π

(c) ( ) ( )2 250.0 75.398 66.315 50.818 Z = Ω + Ω − Ω = Ω

(d) ( )

( )m

220 V4.329 A

50.818 I = =

Ω

(e) ( )( )4.329 A 50.0 216.457 V,RV = Ω = ( )( )4.329 A 66.315 287.085 VCV = Ω = and

( )( )4.329 A 75.398 326.409 V.LV = Ω = ROUND: (a) 75.4 LX = Ω (b) 66 CX = Ω (c) 50.8 Z = Ω (d) m 4.3 AI = (e) 220 V,RV = 290 VCV = and 330 V.LV = DOUBLE-CHECK: In order to satisfy Kirchhoff’s loop rule, the vector phasors must sum as vectors to match m :V

( )22 2

m .R L CV V V V= + − Therefore,

( ) ( )2 2m 216.457 V 326.409 V 287.085 V 220 V,V = + − =

as required.

30.43. THINK: The maximum current occurs for when the AC voltage is at its peak, m 110 V.V = The angular frequency of the oscillation is 377 rad/s.=ω The voltage acts across the total impedance of the circuit where 2.20 ,R = Ω 9.30 mHL = and 2.27 mF.C = The maximum current mI ′ occurs for a capacitance C′ that puts the RLC circuit in resonance with the supplied voltage. At resonance, the phase constant is zero and only the resistor influences the current. SKETCH: Not required. RESEARCH: The impedance of inductor and capacitor are LX L= ω and 1/ .CX C= ω The maximum

current is m m / ,I V Z= where ( )22 .L CZ R X X= + − The phase angle for the circuit is

( )( )1tan / .L CX X R−= −φ When at resonance, 0 1/ .LC= =ω ω The maximum current at resonance is

m m / .I V R=


1119

SIMPLIFY:

(a) The maximum current is ( ) ( )( )

m m mm 2 22 12

.L C

V V VI

Z R X X R L C −= = =

+ − + −ω ω

(b) The phase constant is ( ) 1

1 1tan tan .L C L CX XR R

−

− − −− = =

ω ωφ

(c) The required capacitance for resonance is

2

1 1 ,'

CLLC

′= =ωω

and maximum current at resonance is mm .

VI

R=

CALCULATE:

(a) ( )

( ) ( )( ) ( )( )( )m 212 3 3

110 V34.2675 A

2.20 377 rad/s 9.30 10 H 377 rad/s 2.27 10 F

I−

− −

= = Ω + ⋅ − ⋅

(b) ( )( ) ( )( )( )

( )

13 3

1377 rad/s 9.30 10 H 377 rad/s 2.27 10 F

tan 0.8157 rad2.20

−− −

−⋅ − ⋅

= =Ω

φ

(c) ( ) ( )

42 3

1 7.565 10 F377 rad/s 9.30 10 H

C −

−′ = = ⋅

⋅ and

( )( )m

110 V50.0 A

2.20 I ′ = =

Ω

ROUND: (a) m 34 AI = (b) 0.816 rad=φ (c) 757 μF,C′ = m 50. A,I ′ = 0 rad′ =φ DOUBLE-CHECK: The current is at a maximum when at resonance, so having m mI I′ > is reasonable.

30.44. THINK: Although not explicitly stated, since the circuit should pass a frequency of = 5.00 kHz,f it is natural to assume that it is a high-pass filter. The ratio of the voltages is out in/ 1/ 2.V V = SKETCH:

RESEARCH: The ratio of the voltages for the high-pass RC filter is given by 2 2 2out in/ 1/ 1 1/ ,V V R Cω= +

where 2 .f=ω π The phase constant for the circuit is ( )( )1tan / ,L CX X R−= −φ with 0LX = and

( )1/ .CX C= ω SIMPLIFY:

(a) 2 2 2out2 2 22 2 2

in

1 1 1 1 1 1 1 1 1 4 .2 3 23 31 1/

VR C C

V R fRR CR C= = + = = = =

+ω

ω πωω


1120

(b) The phase constant is ( )( )1 1 10 1/ 1tan tan tan .

2L C

CX XR R fRC

− − − − − = = = −

ωφ

π

CALCULATE:

(a) ( )( )π−= = ⋅

⋅ ⋅ Ω8

3 3

1 1 1.838 10 F2 5.00 10 Hz 1.00 10 3

C

(b) ( )( )( )φπ

−−

= − = − ⋅ ⋅ Ω ⋅

13 3 8

1tan 1.0472 rad2 5.00 10 Hz 1.00 10 1.838 10 F

ROUND: (a) = 18.4 nFC (b) φ = −1.05 rad DOUBLE-CHECK: The negative phase constant simply means the current lends the voltage, which makes sense since the current must first reach the capacitor before it can charge it. The values for the capacitor and phase constant seem reasonable.

30.45. THINK: For signals of frequency = 60.0 Hzf the required noise factor is =out in/ 0.00100.V V The given total impedance = 20.0 kΩ.R Since the ratio of voltages will increase as the frequency does, the lowest frequency that has 90.0% signal strength is when =out in/ 0.900.V V SKETCH:

RESEARCH: The signal ratio is ( )2 2 2out in/ 1/ 1 1/ ,V V R C= + ω where the angular frequency is 2 .fω π=

SIMPLIFY: (a) When −= = ⋅ 3

out in/ 0.001000 1.00 10 ,V V then

( )( )

( ) ( )

ωωω

π

−−= ⋅ = + = = −+

=−

13 6 2 2 2 6out2 2 22 2 2

in

6

1 11.00 10 1 10 10 1 1 1/

1 .10 1 2

VR C

V R CR C

CfR

(b) When =out

in

0.900,VV

then

( )ω

ωω

ωπ

− = = + = = − = +

= =

12 2 2out

2 2 22 2 2in

1 1 1 1 810.900 1 10.81 0.81 191 1/

81 1 81 1 .19 19 2

VR C

V R CR C

fRC RC

CALCULATE:

(a) ( ) ( )( )( )π

−= = ⋅⋅ − ⋅ Ω

9

6 3

1 1.3263 10 F1.00 10 1 2 60.0 Hz 2.00 10

C


1121

(b) ( )( )π −

= = ⋅ Ω ⋅

3 9

81 1 123884 Hz19 2 2.00 10 1.3263 10 F

f

ROUND: To three significant figures, (a) = 1.33 nFC A capacitor of capacitance 1.00 nF needs to be used in this high-pass filter. (b) =124 kHzf Frequencies of 120 kHz and higher will be passed with at least 90% of their amplitude. DOUBLE-CHECK: Since frequencies higher than 120 kHz will pass with 90% of their strength, this seems like a reasonable high-pass filter.

30.46. In general, rms m / 2;V V= therefore, m rms2 .V V=

(a) For rms 110 V,V = ( )m 2 110 V 160 V.V = =

(b) For rms 220 V,V = ( )m 2 220 V 310 V.V = =

30.47. The quality factor for an RLC circuit is defined as ( )0 Energy stored/Power lost .Q = ω For the RLC circuit,

the resonant frequency is 0 1/ .LCω = In general, the energy stored in the circuit is /0 .Rt LU U e−= The

power lost is defined as / .P dU dt= − Therefore,

( )( )

( )( )( )

/ /0 0 0

// 0

0

1 ./

Rt L Rt L

Rt LRt L

U e U e LQd R CLC R L U eU edt

− −

−−

= = =−

ω

30.48. On any product label the voltage and power displayed are the rms voltage and average power. Therefore, take rms 110 VV = and 1250 W.P = The peak value of current is related to rms current by rms m / 2.I I=

In general, rms rmscos .P I V= φ Assuming that the hair dryer acts like a resistor, cos / 1,R Z= =φ so

( )( )rms m

rms rms

1250 W 2 2 16 A.

110 VP P

I IV V

= = = =

30.49. (a) The resonant frequency of the radio tuner is related to the inductance, 3.00 mHL = and capacitance 25.0 nF,C = by 0 1/ .LCω = Keeping in mind that ω π=0 02 ,f

( )( )0

03 9

1 1 18.4 kHz.2 2 2 3.00 10 H 25.0 10 F

fLC − −

= = = =⋅ ⋅

ωπ π π

(b) At resonance, the total impedance is solely that from the resistance, 1.00 μ .R = Ω Given that the voltage drop across the resistor is rms 1.50 mV,V = the power in the circuit is

( )( )

232rms

6

1.50 10 V2.25 W.

1.00 10 V

PR

−

−

⋅= = =

⋅ Ω

30.50. THINK: The current through the circuit is driven by the AC potential, rms 50.0 VV = and 2000. Hz,f = which has a total impedance from the resistance, 100. ,R = Ω capacitance, 0.400 μFC = and inductance,

0.0500 H.L = The average power lost over the circuit is determined by the rms voltage and rms current. SKETCH: Not required. RESEARCH: The rms current is given by


1122

rmsrms 2

2

.1

VI

R LC

= + − ω

ω

The rms voltage drop across each component is rms ,i iV I X= where i denotes ,R L or .C The average power drawn from the circuit is

2rms .P I R=

SIMPLIFY: (a) Since the angular frequency is 2 ,f=ω π

rmsrms 2

2

.12

2

VI

R fLfC

=

+ −

ππ

(b) The rms voltage drop across the resistor is rms .RV I R= The rms voltage drop across the capacitor is

rmsrms .

2C CI

V I XfC

= =π

The rms voltage drop across the inductor is rms rms2 .L LV I X I fL= = π

(c) 2rmsP I R=

CALCULATE:

(a) ( )

( ) ( )( ) ( )( )

rms 2

2

6

50.0 V0.1134 A

1100. 2 2000. Hz 0.0500 H2 2000. Hz 0.400 10 F

I

ππ −

= = Ω + − ⋅

(b) ( )( )0.1134 A 100. 11.34 VRV = Ω =

( )( )( )6

0.1134 A22.56 V

2 2000. Hz 0.400 10 FCVπ −

= =⋅

( )( )( )2 0.1134 A 2000. Hz 0.0500 H 71.26 VLV π= =

(c) ( ) ( )20.1134 A 100. 1.286 WP = Ω =

ROUND: To three significant figures, (a) rms 0.113 AI = (b) 11.3 V,RV = 22.6 V,CV = 71.3 VLV = (c) 1.29 WP =

DOUBLE-CHECK: For an RLC circuit, the maximum voltage is given by ( )22 2m ;R L CV V V V= + −

therefore, as a check ( ) ( )2 2m 11.34 V 71.26 V 22.56 V ,V = + − which equals 50.003 V, which is mV

within rounding errors, so the answer is reasonable.

30.51. THINK: In order to receive the best signal, the radio should be tuned at resonance with the incoming frequency, 0 88.7 MHz.f = The inductance of the radio receiver is 8.22 μH.L = Signal strength is

m 12.9 μV.V = The similar signal with frequency 88.5 MHz,f = is not at resonance, so its total impedance is influenced by the resistor, capacitor and inductor such that its current is half that of the current for the frequency at resonance. SKETCH: Not required. RESEARCH: At resonance, 0 1/ ,LC= =ω ω where the angular frequency is 0 02 .f=ω π The impedances of the inductor and capacitor are LX L= ω and ( )1/ ,CX c= ω respectively. The impedance of the RLC


1123

circuit is ( )22 .L CZ R X X= + − At resonance, the current amplitude is m m / ,I V R= and when not at

resonance, the current amplitude is m m / .I V Z′ = SIMPLIFY:

(a) At resonance, 0 0 0 2 200

1 12 .4

f Cf LLC

= = =ω ππ

(b) When m m1 ,2

I I′ = then

( )2m m0 0

0

1 22 L C

V VR Z R X X

Z R= = = + −

( ) ( ) ( ) 122 2

0 0 0

2 24 .

3 3L C

L C

X X fL fcR R X X R

−− −= + − = =

π π

CALCULATE:

(a) ( ) ( )

1322 6 6

1 3.9167 10 F4 88.7 10 Hz 8.22 10 H

C −

−= = ⋅

⋅ ⋅π

(b) ( )( ) ( )( )( ) 1

6 6 6 13

0

2 88.5 10 Hz 8.22 10 H 2 88.5 10 Hz 3.9167 10 F11.941

3R

−− −⋅ ⋅ − ⋅ ⋅

= = Ωπ π

ROUND: To three significant figures, (a) 0.392 pFC = (b) 0 11.9 R = Ω DOUBLE-CHECK: This is an RLC circuit, so the current across the circuit decays exponentially with a time constant 2 / 1.38 μs.L R= =τ Assuming that the time constant represents the delay from when the radio picks up the signal to when it transmits it as sound, it is reasonable that the value is small.

30.52. If a power station provides power ,P and delivers it over potential difference ,V then the current out of the

station is / .I P V= The power loss over the lines is ( )22 / .P I R P V R′ = = Therefore, if 10 ,V V→ the

power loss over the lines is now ( ) ( )2 2/10 / /100 /100.P P V R P V R P′′ ′= = = Therefore, if voltage is raised by a factor of 10, the power loss is 100 times smaller.

30.53. (a) From solved problem 29.1, the coil has C 31N = turns, and the solenoid has 290 turns/cmn = and length of 12.0 cm.l = Therefore, the number of turns in the solenoid is S .N nl= If the voltage across the solenoid is S 120 V,V = and the coil and solenoid act as a transformer, then

( )( )( )( )

C S S C S CC 1

C S S

120 V 31 1.1 V.

290 cm 12.0 cmV V V N V N

VN N N nl −

= = = = =

(b) With DC current, no magnetic flux is created within the solenoid so there is no voltage in the coil.

30.54. The primary coil has P 800N = turns, and the secondary coil has S 40N = turns. (a) The voltages across each coil is given by P P S S/ / ,V N V N= so when =P 100. V,V the voltage on the secondary coil is

( )( ) ( )= = =S S P P/ 40 100. V / 800 5.00 V.V N V N (b) The current on the secondary coil when =P 5.00 AI is

( )( ) ( )= = =S P P S/ 5.00 A 100. V / 5.00 V 100. A.I I V V


1124

(c) When the voltage is DC, no magnetic flux is created within the secondary coil, so there is no voltage on the secondary coil. (d) When the voltage is DC, the voltage on the secondary coil is zero, so the secondary coil does not carry a current.

30.55. The primary coil has P 200N = turns and the secondary coil has S 120N = turns. The secondary coil drives a current I through a resistance of = Ω1.00 k .R The input voltage applied across the primary coil is

=rms 75.0 V.V The voltage across the secondary coil is S rms S P/ .V V N N= The power dissipated in the

resistor is ( ) ( )( ) ( )( )

( )= = =⋅ Ω

22rms S P

3

75.0 V 120 / 200/2.03 W.

1.00 10V N N

PR

30.56. The frequency of the source is S 60. Hz.f = The full wave rectifier is a configuration of diodes that allows all of the current in the circuit to flow in one direction. This is illustrated in Figure 30.35 of the textbook. Comparing the plot of emf as a function of the time (Figure 30.35 (a)) to the rectified current (Figure 30.35 (d)), it can be seen that the frequency is doubled, S2 120 Hz.f f= = The capacitor minimizes the size of the ripples, but the frequency stays the same.

30.57. THINK: The voltage applied to the primary side of the transformer is rms P 110 VV V= = and its frequency is 60. Hz.f = The ratio of primary coil turns to secondary coil turns in the transformer is P S/ 11.N N = The secondary coil voltage, SV is used as the source voltage for the fullwave rectifier shown in Problem 30.56. Using the equations for transformers, the rms voltage in the secondary coil can be found. To find the DC voltage DCV provided to the resistor, it is necessary to integrate over the AC voltage to obtain the time-averaged value. SKETCH:

RESEARCH: (a) The secondary rms voltage is given by S P S P/ .V V N N= For any rms voltage, the maximum voltage is

given m rms2 .V V= (b) The DC voltage DCV is the time-averaged value of the rectified emf.V . The time average is found from the equation

( )/2

DC emf0

2 ,T

V V t dtT

=

where ( ) ( )emf m sin .V t V tω= SIMPLIFY:

(a) P Sm

P

2V N

VN

=

(b) Substitute 2T = πω

into the equation:

( ) ( ) ( )/ /m mDC m 0 m0

2sin cos | 1 1 .V VV V t dt t V= = − = − − − =

π ω π ωω ω ωπ π π π


1125

CALCULATE:

(a) ( )m12 110 V 14.14 V

11V = =

(b) ( )DC2 14.14 V 9.003 VVπ

= =

ROUND: To two significant figures, (a) m 14 VV = (b) DC 9.0 VV = DOUBLE-CHECK: The number of turns in the primary is greater than the number of turns in the secondary so it is expected that the voltage in the secondary is lower.

Additional Problems

30.58. The given quantities are: the inductance, = 100. mH;L the frequency, = 60.0 Hz;f and the rms voltage,

rms 115 V.V = The average power is given by rms rms / ,P I V R Z= so to maximize the power output the

impedance must be minimized. The impedance is ( )( )22 1/ .Z R L Cω ω= + − To minimize ,Z the

expression in the brackets must equal zero:

( ) ( )( ) ( )ω

ω ω π π−− = = = = = ⋅ 5

2 2 2

1 1 1 10 7.04 10 F. 2 2 60.0 Hz 0.100 H

L CC L f L

30.59. This question deals with an LC circuit. The given quantities are the frequency, =1000. kHzf and the inductance, = 10.0 mH.L What is the capacitance, ,C of the capacitor when the station is properly tuned?

Equating the expressions 2 fω π= and 1/ LCω = and solving for :C 2 1/ ,f LCπ =

( ) ( )( )( ) ( )π π−

−= = = ⋅

⋅ ⋅

122 26 2

1 1 2.53 10 F.2 2 1.000 10 Hz 1.00 10 H

Cf L

30.60. This question deals with an RLC circuit. The circuit is driven by a generator with =rms 12.0 VV and frequency, 0 .f The inductance is = 7.00 mH,L the resistance is = Ω100. R and the capacitance is

= 0.0500 mF.C (a) The angular resonant frequency of the RLC circuit is given by 0 1/ ,LC=ω where 0 02 .f=ω π Therefore, the resonant frequency is

( )( )π π − −= = =

⋅ ⋅0 3 3

1 1 85.1 Hz.2 2 7.00 10 H 0.0500 10 F

fLC

(b) The average power dissipated in the resistor at the resonant frequency is

( )( )= = =

Ω

22rms 12.0 V

1.44 W.100.

VP

R

30.61. A “60-W light bulb” dissipates power at 60. W.P = The question gives rms 110 V.V =

(a) The maximum current is m rms2 .I I= The average power is rms rms ,P I V= so the maximum current is

( )( )m

rms

2 60. W20.77 A.

110 VP

IV

= = =

(b) The maximum voltage is ( )m rms2 2 110 V 160 V.V V= = =


1126

30.62. The given quantities are the frequency, 360 Hz,f = the inductance 25 mHL = and the resistance 0.80 .R = Ω For the current and voltage to be in phase, the circuit must be in resonance. This occurs

when the inductive reactance and the capacitive reactance are equal:

( ) ( )( ) ( )2 2 2 3

1 1 1 1 7.8 μF.2 2 360 Hz 25 10 H

L CX X L CC L f L −

= = = = = =⋅

ωω ω π π

30.63. In an RLC circuit, the inductance is = 65.0 mHL and the capacitance is =1.00 μF.C The circuit loses electromagnetic energy at a rate of Δ = −3.50% per cycle.U The energy stored in the electric field of the

capacitor is expressed by ( )2 /E max 0cos .Rt LU q e t−Δ = ω The rate of energy loss is E, final E, initialE

E E, initial

,U UU

U U−Δ

=

where time initialt is zero and finalt is the time to complete one cycle, final 02 / .t = π ω The rate of energy loss per cycle can now be written as

( ) ( )

( )

0

2 22 / 2 0 2max max

2 /E2

0 2E max

cos 2 cos 02 20.035 1.

cos 02

R L

R L

q qe eU C C eU q e

C

− −

−

−

−Δ= − = = −

π ω

π ωπ

Since 0 1/ LC=ω ,

( ) ( )

( ) ( ) ( )( )

π π

π π

−

−

−

− = = −

⋅= − = − = Ω

⋅

2 /

3

6

21 0.0350 ln 0.9650

65.0 10 Hln 0.9650 ln 0.96501.45 .

2 2 1.00 10 F

R LC L R LCeL

LRC

30.64. The transformer has P 400 turnsN = on its primary coil and S 20 turnsN = on its secondary coil. The average power output from the secondary coil is = 1200. WP and the maximum voltage output from the secondary coil is =S, m 60.0 V.V The rms current in the primary coil is given by,

SSS, rmsP, rms

P S, rms P

.P NI N

IN V N

= =

Substituting the maximum values for the rms values gives ( )( )( )( )

= = = =

P S S, mP, m

S, m S, m PP

1200. W 202 2 2.00 A.

60.0 V 40022

P N P NII

V V NN

30.65. The given quantities are the capacitance, = 5.00 μF,C the resistance, = Ω4.00 R and the battery voltage, = 9.00 V.V The capacitor is charged for a long time by closing the switch to position a. At time 0t = the

switch is closed at position b and the capacitor is discharged through an inductor with = 40.0 mH.L


1127

(a) The maximum current through the inductor is given by ω=max 0 max ,i q where maxq is the maximum

charge on the capacitor, and 0 1/ .LCω = The fully charged capacitor has charge max .q CV= Substituting for 0ω and maxq gives

( )( ) ( )

−

−

⋅= = = =

⋅

6

max 3

5.00 10 F9.00 V 0.101 A.

40.0 10 HCV Ci V

LLC

(b) The current is given by ( )max 0sin .i i tω= − For the current to be a maximum ( )0sin 1,tω = or when

0 / 2.tω π= This occurs at time

( )( )π π − − −= = ⋅ ⋅ = ⋅3 6 440.0 10 H 5.00 10 F 7.02 10 s.2 2

t LC

30.66. THINK: The resistance is = Ω10.0 kR and the capacitance is = 0.0470 μFC in the RC high-pass filter. To find the frequency f where the ratio of the output voltage to the input voltage gives

( )out in20log / 3.00,V V = − use the voltage ratio for an RC high-pass filter. SKETCH: A sketch of the circuit is provided in the problem. RESEARCH: For an RC high-pass filter, the ratio out in/V V is given by:

out

in2 2 2

1 .11

VV

R Cω

=+

The frequency is / 2 .f ω π= From the properties of logarithms, if ( )blogy x= then .yx b=

SIMPLIFY: out out

in in

20 log 3.00 log 0.150.V VV V

= − = −

This can be rewritten as −= 0.150out

in

10 .VV

Substituting for out

in

VV

gives:

0.150 0.300 0.300 2 2 22 2 2 0.300

2 2 22 2 2

1 1 1 110 10 1 10 1 10 11 11

R CR C

R CR C

ωω

ωω

− −= = + = =−++

( ) ( )0.300 0.300

1 1 1 1 .210 1 10 1

fRC RC

ωπ

= = − −

CALCULATE: ( ) ( )( )π −

= = − ⋅ Ω ⋅

0.300 3 6

1 1 339.43 Hz10 1 2 10.0 10 0.0470 10

f

ROUND: To three significant figures, = 339 Hz.f DOUBLE-CHECK: The breakpoint frequency for this RC high-pass filter is ( )B 1/ 2 338 Hz.f RC= =π Since the calculated frequency is larger than this value, the answer is reasonable.

30.67. THINK: The unknown wire-wound resistor R is initially connected to a DC power supply. When there is a voltage of =emf 10.0 VV across the resistor, the current is = 1.00 A.I Next the resistor is connected to an AC power source with =rms 10.0 V.V When the AC power source is operated at frequency

= 20.0 kHz,f a current of =rms 0.800 AI is measured. Find: (a) the resistance, R ; (b) the inductive reactance, ,LX of the resistor; (c) the inductance, ,L of the resistor; and (d) the frequency, ,f ′ of the AC power source at which .LX R=


1128

SKETCH: Not required. RESEARCH: (a) The resistance of the resistor when used with the DC source can be found using Ohm’s law, / .R V I= (b) When connected to the AC power source, the resistor can be treated as an RL series circuit. The

impedance of the RL circuit is 2 2rms rms/ .LZ R X V I= + =

(c) The inductance can be found with LX L= ω and 2 .f=ω π (d) , 2L R f ′= =ω ω π SIMPLIFY:

(a) VRI

=

(b) 2 2 2 2 2 L LZ R X X Z R= + = − Substituting rms

rms

VZ

I= gives,

22rms

rms

.LV

X RI

= −

(c) 2

L LX XL

f= =

ω π

(d) 2

RfL

′ =π

CALCULATE:

(a) ( )( )= = Ω10.0 V

10.0 1.00 A

R

(b) ( ) = − Ω = Ω

2210.0 V 10.0 7.50

0.800 ALX

(c) ( )

( )π−Ω

= = ⋅⋅

53

7.50 5.968 10 H

2 20.0 10 HzL

(d) ( )π −

Ω′ = =⋅ 5

10.0 26667 Hz2 5.968 10 H

f

ROUND: The answers should be reported to three significant figures. (a) = Ω10.0 R (b) = Ω7.50 LX

(c) −= ⋅ 55.97 10 HL (d) ′ = 26.7 kHzf DOUBLE-CHECK: Since the current decreased when the power supply was changed from DC to AC, the resistance must have increased. This additional resistance is explained by the inductive reactance of the resistor. The units for all calculated values are correct.

30.68. THINK: In this RLC circuit, ,L C and R are connected in parallel, as shown in the provided figure. The circuit is connected to an AC source providing rmsV at frequency .f A phasor diagram can be used to find an expression for rmsI in terms of rms ,V ,f , L R and .C


1129

SKETCH: Phasor diagram for current across each component in the circuit:

RI is in phase with ,V and CI leads .LI RESEARCH: Since this is a parallel circuit, the voltage rmsV is the same across all components. The current, however, has different phases and amplitudes within each component, as shown in the above phasor diagram. By a similar analysis to determining the voltage in a series RLC circuit, the phasor diagram above shows that the current is:

( )22tot rms .R L CI I I I I= = + −

Since rmsV is the same in each component, rms ,L L C C RV I X I X I R= = = where ,LX L= ω 1/CX C= ω and 2 .fω π=

SIMPLIFY: ( )2 22

22 rms rms rmsrms rms2 2

1 1 22R L C

L C

V V VI I I I V fC

X X fLR R

= + − = + − = + −

ππ

CALCULATE: Not required. ROUND: Not required. DOUBLE-CHECK: By dimensional analysis of the above expression, the units are correct:

1/2 1/22 2

rms 2 2

1 s F 1 1 1 VV V A.H s Ω Ω ΩΩ Ω

I = + − = + − = =

30.69. THINK: (a) The loop of wire has a diameter of = =5.00 cm 0.0500 md and it carries current = 2.00 A.i Find the magnetic energy density, BU at the loop’s center. (b) Find the current, 'i in a straight wire that gives the same value of BU at a point,

= =4.00 cm 0.0400 mr from the wire. SKETCH: (a) (b)

RESEARCH: (a) The magnetic energy density is ( )( ) 2

B 01/ 2 .U Bμ= The magnetic field, B for the wire loop is given by

Equation 28.8 in the textbook, ( )μ

= 0loop .

2 / 2i

Bd

(b) The magnetic field due to current traveling in a straight wire is given by Equation 28.4, ( )μ π=wire 0 '/ 2 .B i r


1130

SIMPLIFY:

(a) 2B loop

0

1 ,2

U Bμ

=

substituting for loopB gives: ( )

2 20 0

B 20

1 1 .2 2 / 2 2

i iU

d dμ μ

μ

= =

(b) 2B wire

0

1 ,2

U Bμ

=

substituting for wireB gives:

μ π πμ μ μπ

= = =

2 2 2 2 2 220 B B

B 2 20 0 0

' 8 81 ' ' .2 4

i r U r UU i i

r

CALCULATE:

(a) ( )( )

( )π −

−⋅

= = ⋅27

3 3B 2

4 10 T m/A 2.00 A1 1.005 10 J/m2 0.0500 m

U

(b) ( ) ( )π

π

−

−

⋅= =

⋅

22 3 3

7

8 0.0400 m 1.005 10 J/m' 10.053 A

4 10 T m/Ai

ROUND: The answers should be reported to three significant figures. (a) −= ⋅ 3 3

B 1.01 10 J/mU (b) =' 10.1 Ai DOUBLE-CHECK: The calculated values have the proper units. It is expected that the current required to generate a given magnetic field would be much larger for a straight wire than for a loop.

30.70. THINK: The bulb of average power 75000 WP = operates at a current of rms 200. A,I = a voltage of

rms 440. V,V = and a frequency of 60.0 Hz.f = The inductive reactance of the bulb is not negligible so its impedance needs to be considered. The inductive reactance can be neglected, so 0.CX = SKETCH:

RESEARCH: The average power of the bulb is 2rms .P I R= The rms voltage is rms rmsZ,V I= where

2 2LZ R X= + and .LX L= ω

SIMPLIFY: The resistance of the bulb is

2rms

.P

RI

=

The inductance of the bulb is

( )( )( )( ) ( )

22 22 2 2 2rms rms

rms rms rms22 rms

2 22 2rms rms

rms rms

1 1 .2

V VI R L I V L R

IR L

V VL R R

I f I

ω ωω

ω π

= + = = −

+

= − = −


1131

CALCULATE: ( )( )2

75000 W1.875

200. AR = = Ω

( )( )( ) ( )

22440. V1 1.875 0.003053 H

2 60.0 Hz 200. AL

π

= − Ω =

ROUND: To three significant figures, = Ω1.88 R and = 3.05 mH.L DOUBLE-CHECK: These are reasonable values. If this was a DC source Ohm’s Law would give,

= = =/ 440. V / 200. A 2.20 Ω,R V I which is comparable to the total calculated impedance. All values have the correct units.

30.71. THINK: A resistor R is connected across an AC source which oscillates at angular frequency .ω Show that the power dissipated in R oscillates with frequency 2 .ω SKETCH:

RESEARCH: The power is 2 .RP i R= For an AC power supply, ( )sin .R Ri I t= ω Also useful is the

trigonometric identity ( ) 2cos 2 1 2sin .θ θ= −

SIMPLIFY: ( ) ( )( ) ( )( )2 2 2 2 21 cos 2 1sin 1 cos 2 .2 2R R R R

tP i R I t R I R I R t

−= = = = −

ωω ω It can be seen from the

above equation that the power oscillates with a frequency twice that of the voltage. CALCULATE: Not required. ROUND: Not required. DOUBLE-CHECK: Power, P, is proportional to 2.i Since i varies proportionately with V, it must be the case that 2i varies proportionately with 2 .V Since V varies proportionately with ,ω it must be the case that 2V varies proportionately with 2 .ω Therefore by transitivity, P is proportional to 2 .ω Therefore, there exists a constant, c, such that 2 .P cω= So the change in P with respect to time, / ,dP dt will be proportional to 2 / ,d dtω or 2 .ω

30.72. THINK: The resistor has a resistance of = Ω300. R and is connected in series with a capacitor with = 4.00 μF.C The AC power supply has rms 40.0 V.V = Find:

(a) the frequency f at which ;C RV V= (b) the current rmsI at which this occurs. SKETCH:


1132

RESEARCH: (a) ,C CV iX= and 1/ ,CX C= ω while .RV iR= Recall 2 .fω π=

(b) In this circuit, rms rms ,V I Z= where 2 2CZ R X= + (as there is no inductor).

SIMPLIFY:

(a) 1 1 1 2 2C R CV V iX iR R R f

C fC RC= = = = =

ω π π

(b) rms rms rms rmsrms 2 2 2 2

12C

V V V VI

Z RR X R R= = = =

+ +

CALCULATE:

(a) ( )( )π= =

Ω1 132.6 Hz

2 300. 4.00 μFf

(b) ( )( )= =

Ωrms

40.0 V1 0.09428 A300. 2

I

ROUND: To three significant figures, = 133 Hzf and =rms 0.0943 A.I

DOUBLE-CHECK: These values are reasonable given the initial conditions. Dimensional analysis confirms the units are correct.

30.73. THINK: The electromagnet has 200N = loops, a length = 0.100 m,l and a cross-sectional area = 25.00 cm .A Find its resonant frequency 0f when it is attached to the Earth.

SKETCH:

RESEARCH: The resonant frequency is 0 1/ .LCω = Recall 2 .fω π= The radius of the Earth is 66.38 10 mr = ⋅ and for a spherical capacitor, 04 .C r= πε From Chapter 29, 2

0L n lAμ= for a solenoid, where / .n N l≡

SIMPLIFY: ( ) ( )

0 320 0

0 0

1 142 / 4

lfN ArN l lA r π μ επ μ πε

= =

CALCULATE:

( )( )

( )( )( )( )π π − − −= =

⋅ ⋅ ⋅ ⋅0 3 7 12 4 3 6

0.100 m1 376.8 Hz4 200 4 10 T m/A 8.854 10 F/m 5.00 10 m 6.38 10 m

f

ROUND: To three significant figures, =0 377 Hz.f DOUBLE-CHECK: This is a reasonable frequency for an electromagnet. Dimensional analysis confirms the units are correct.

30.74. THINK: The inductance of the inductor is 1.00 H.L = The resonance of the series RLC circuit is to occur at frequency 0 60.0 Hz.f = The voltage across the capacitor (or inductor), CV (or LV ) is to be 20.0 times that across the resistor, .RV Find the capacitance C and the resistance .R


1133

SKETCH:

RESEARCH: At resonance, the angular frequency is 0 1/ ,LC=ω where 0 02 .f=ω π At resonance,

emfRV V= and ,L CV V= − where .L LV IX= Recall V IR= and 0 .LX L= ω

SIMPLIFY: 2 2 20 0

1 1 ,4

CL f Lω π

= = emf 0 00

L

2 2 R R

L LV V f L V f L

V IX L RR R V

= = = =π πω

Since 20.0 ,L RV V= ( )0 02

.20.0 10.0

R

R

V f L f LR

V= =

π π

CALCULATE: The capacitance of the capacitor must be ( ) ( )

622

1 7.0362 10 F.4 60.0 Hz 1.00 H

Cπ

−= = ⋅

The resistance of the resistor must be ( )( )60.0 Hz 1.00 H

18.8496 .10.0

Rπ

= = Ω

ROUND: Rounding to three significant figures, 7.04 μFC = and 18.8 .R = Ω DOUBLE-CHECK: At resonance, 0.L CX X− = To check this, plug in the value found for the capacitance:

( )( ) ( )( )0 0 60 0

1 1 12 2 60.0 Hz 1.00 H 0,2 2 60.0 Hz 7.0362 10 FL CX X L f L

C f C −− = − = − = − =

⋅ω π π

ω π π

as required.

30.75. THINK: The RC low-pass filter has a breakpoint frequency of =B 200. Hz.f Find the frequency at which the output voltage divided by the input voltage is =out in/ 0.100.V V SKETCH:

RESEARCH: For a RC low-pass filter, the breakpoint frequency is: ( )B 1/ ,RCω = where B B2 .f=ω π The ratio of the input voltage to output voltage is

out

2 2 2in

1 .1

VV R C

=+ ω

SIMPLIFY: B BB

1 12 2

f RCRC f

= = =ω ππ


1134

( )( )

2 22 2 222 2in in in

2out B out B out B

11 1 1 12 2 2

V V VRC

V f V f V f

= + = + = + − =

ω ωω ωπ π π

2 2

in inB B

out out

2 1 1V V

f f fV V

= − = −

ω π

CALCULATE: ( ) = − =

21200. Hz 1 1989.97 Hz0.100

f

ROUND: To three significant figures, =1990 Hz.f DOUBLE-CHECK: Since out in/V V is less than 1/ 2 (the value associated with the breakpoint frequency), by the above sketch, the frequency f must be greater than the breakpoint frequency B .f

30.76. THINK: The components of the RLC circuit are connected in series, and their values are = Ω20.0 ,R = 10.0 mHL and −= ⋅ 65.00 10 F.C They are connected to an AC source of peak voltage = 10.0 VV and

frequency =100. Hz.f Find: (a) The amplitude of the current I. (b) The phase difference φ between the current and the voltage. (c) The maximum voltage across ,R ,L and .C SKETCH:

RESEARCH: (a) The amplitude of the current in an RLC series circuit is

( )m

m 22,

L C

VI

R X X=

+ −

where LX L= ω and 1/ .CX C= ω Recall that 2 f=ω π and m rms2 .V V=

(b) The phase difference between the current and the voltage is ( )1tan .L CX X

R− −

=

φ

(c) For the maximum voltage across each circuit component, m ,L LV I X= mC CV I X= and m .RV I R= SIMPLIFY:

(a) ( )( )

rmsm 22

2

2 1/ 2

VI

R fL fC=

+ −π π

(b) ( )1 2 1/ 2

tanfL fC

R− −

=

π πφ

(c) m2 ,LV fI L= π m

2CI

VfC

=π

and m .RV I R=


1135

CALCULATE:

(a) ( )

( ) ( )( ) ( )( )ππ −

= = Ω + − ⋅

m 2

2

6

2 10.0 V0.04523 A

120.0 200 rad/s 0.0100 H200 rad/s 5.00 10 F

I

(b) ( )( ) ( )( )( )

( )π π

φ−

− − ⋅ = = − Ω

6

1200 rad/s 0.0100 H 1/ 200 rad/s 5.00 10 F

tan 1.507 rad20.0

(c) ( )( )( )π= =0.04523 A 200 rad/s 0.0100 H 0.2842 VLV

( )( )( )π −

= =⋅ 6

0.04523 A14.40 V

200 rad/s 5.00 10 FCV

( )( )= Ω =0.04523 A 20.0 0.9046 VRV ROUND: To three significant figures, (a) =m 45.2 mAI (b) φ = −1.51 rad (c) = 0.284 V,LV =14.4 V,CV = 0.905 VRV

DOUBLE-CHECK: It must be true that ( )22 2m .R L CV V V V= + − To check this, plug in the calculated

values for each side of the equation:

( )( ) ( )( ) ( ) ( ) ( )( )

= = =

+ − = + − =

2 22m rms

2222

2 2 10.0 V 200. V

0.9046 V 0.2842 V 14.40 V 200. V,R L C

V V

V V V

as required.


1136

Chapter 31: Electromagnetic Waves

In-Class Exercises

31.1. e 31.2. c 31.3. e 31.4. c 31.5. b 31.6. a 31.7. d 31.8. c 31.9. b Multiple Choice

31.1. c 31.2. b, c, e 31.3. b 31.4. d 31.5. a 31.6 a 31.7. c 31.8. b

Questions

31.9. (a) The intensity of the light passing through the first polarizer is 1 0 / 2.I I= The angle between the transmission axis of the first and second polarizer as a function of time is ( )1,2 45 ,t tθ ω= ° + where t is in seconds and ω is in rad/s. The intensity of the light passing the second polarizer is then

( ) ( )2 22 1 1,2 0 2

1cos cos 45 .2

I I I t I tθ ω= = ° + = The angle between the transmission axis of the second and

third polarizer as a function of time is ( )2,3 45 .t tθ ω= ° − The intensity of the light passing the third polarizer is:

( ) ( ) ( ) ( ) ( )22 2 23 2 2,3 0 0 3

1 1cos cos 45 cos 45 cos 45 cos 45 .2 2

I I I t t I t t I tθ ω ω ω ω = = ° + ° − = ° + ° − =

Now, use the trigonometric identity, ( ) ( )1cos cos cos cos :2

u v u v u v = + + −

( ) ( ) ( ) ( )

( )

1cos 45 cos 45 cos 45 45 cos 45 4521 1cos90 cos2 cos2 .2 2

t t t t t t

t t

ω ω ω ω ω ω

ω ω

° + ° − = ° + + ° − + ° + − ° +

= ° + =

Therefore, ( ) ( )2

2 23 0 0 0

1 1 1 1 1cos2 cos 2 cos 2 .2 2 2 4 8

I I t I t I tω ω ω = = =

Then, use the identity,

2 1 cos2cos ,2

uu += which in this case means: ( ) ( )2 1 cos 4cos 2 .

2t

tω

ω+

= So:

( ) ( ) ( )3 0 0 0

1 cos 41 1 1 cos 4 .8 2 16 16

tI t I I I t

ωω

+= = +

So, the intensity of the light making it past the third polarizer will oscillate about the value of 0 /16I as a cosine function with amplitude of 0 /16I and an angular frequency that is four times the angular frequency at which the polarizer is rotating. Thus, the intensity oscillates between a minimum value of zero (when polarizer 2 is parallel to either polarizer 1 or polarizer 3) and a maximum value of 0 / 8I when polarizer 2 is at 45° with polarizer 1 and 3. The result is thus consistent with the 0 / 8I result of Example 31.4, where polarizer 2 was at a fixed angle of 45°. (b) The transmission axis of a polarizer is a direction in the plane of the polarizer, not a single specific line in the plane of the polarizer. Therefore, moving the second polarizer parallel to itself in any direction will not change anything for the light passing through the second polarizer. Light that is incident on the first polarizer but that does not pass through the second polarizer will not pass through the third polarizer at all. In other words, if the light is initially incident on the total surface area of the first polarizer, the total amount of light (i.e. number of photons) that passes through the third polarizer after the second polarizer


1137

is displaced by a distance d < R, will be proportional to the fraction of surface area of overlap between all three polarizers.

31.10. Charge moving up and down along the antenna creates an electric dipole on the antenna. This produces an electric field along z±

at point A (parallel to the antenna). From Ampere’s law, the current produces a magnetic field along x±

at point A. Since radiation is moving along the y± direction away from the

antenna, the possible directions for E

and B

are: (a) E

in positive directionz − and B

in the positive direction.x −

(b) E

in the negative directionz − and B

in the negative direction.x −

See Figure 31.16 as a visual aid.

31.11. Assuming that the randomly polarized light source (the sun) was replaced by a polarized source, the results are still correct since for randomly polarized light, the average of 2E is the same as the average of

2E for a polarized light.

31.12. A magnetic monopole is a magnet with only one pole. The magnetic field produced by the monopole is similar to the electric field produced by an electric charge. The magnetic field vector is directed radially outward, that is, .B r∝

If a charged particle is moving parallel to ,r its motion is not affected by the magnetic field. However, if the particle is moving perpendicular to ,r then its motion is affected and there is a perpendicular force to its velocity producing a helical motion.

31.13. Both signals would arrive at the same time, since the speed of both signals is the speed of light. This is correct provided there is no medium between the Earth and the Moon. It is known that the speed of light in a medium depends on its refractive index. The refractive index of the medium depends also on the frequency of light.

31.14.

Ampere’s law is defined as:

0S

,B dS J dAμ=

where J

is a charge density within the surface. Since the surface is closed, the Amperian loop can be made very small, such that 0.B dS =

The integral J dA

represents the net rate of charge transport out of

the region bounded by the surface, S. In a static situation, the charge in the region is constant and the integral is 0.J dA =

For a dynamical situation, the integral must be equal to the negative rate of change

of charge in the region, that is, in 0.

dQJ dA

dt= − ≠

Therefore, there is inconsistency in Ampere’s law. The Maxwell-Ampere law, however, take the form:

0 0 0S S

.dB dS J dA E dAdt

μ μ ε= +


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Applying Gauss’s law, it is found that: in in0 0 0.

dQ dQdt dt

μ μ− + = Therefore, the Ampere-Maxwell law is

satisfied. For Faraday’s law: S

.dE dS B dAdt

= −

Both sides give zero since Gauss’ law states:

S

0.B dA =

That is, there is no magnetic charge in the region enclosed by S. Therefore, there is no

inconsistency.

31.15. According to Maxwell’s equations, the velocity of light always has a fixed value regardless of the observer’s speed. This is in direct contradiction to Newton’s laws of motion based on the Galilean addition law for velocities. According to the Galilean addition law, the velocity of light should not be the same in all inertial frames. Therefore, Maxwell’s equations and Newton’s laws of motion are mutually inconsistent.

31.16. Our vision begins with chemical reactions in the rod and cone cells of the retinas of our eyes, which release neurotransmitters in response to electromagnetic radiation in the visible range. These are resonant processes –the transfer of energy from the electromagnetic field to the nervous system is enhanced by a matching of frequencies. As in all resonances, high amplification can only be achieved in a narrow resonance peak. In order to achieve the sensitivity necessary for seeing, a narrow bandwidth is needed. Therefore, the narrow frequency band is necessary for our eyes to have sufficient sensitivity. This sensitivity allows seeing in high and low intensity situations. This is the reason that it is impossible to have a wide range of frequencies that can be seen.

31.17. (a) From energy conservation, the power per unit area or intensity of radiation from a point source must be inversely proportional to 2 .r (b)The radiation field falls off with distance at the same rate as the electrostatic field of a point charge which falls off according to 2/ .E kQ r=

31.18. As discussed in Section 31.10, LCD displays use polarizing filters in one of their display components. Therefore, the light emitted by the LCD is a polarized light. Since some sunglasses also have polarizing filters for their lenses, the intensity of LCD light passing through the glass varies as the sunglasses are rotated. It can be concluded that the sunglasses must be polarizing the light.

31.19.

Since the light reaching 2P is polarized, the transmitted intensity is given by 22 1 cos .I I θ= Similarly, the

intensity passing through 3P is given by ( ) ( )2 2 2 23 2 1cos 90 cos sin sin 2 / 4.I I Iθ θ θ θ= ° − = = Therefore, as

the intermediate filter is rotated, the intensity of light passing through the polarizers will increase to a maximum at 45θ = ° and then it will decrease to zero at 90θ = ° and it will continue in this pattern every 45° increase in the rotation of the second polarizer.


1139

Problems

31.20.

Applying Maxwell’s law of induction along a closed loop of radius, r, gives: E

0 0 ,d

B dSdt

μ ε Φ• =

where EΦ is given by 2E .EA E RπΦ = =

Substituting the expression for EΦ into the equation for Maxwell’s law gives:

( )2 20 0 0 0 2 .dE dEB dS R B r R

dt dtμ ε π π μ ε π= =

Thus, the magnetic field is: 2

0 01 .2

R dEBr dt

μ ε =

Substituting R = 0.0600 m, r = 0.100 m and

/ 10.0 V/m sdE dt = yields:

( )( ) ( ) ( )2

7 12 180.0600 m1 4 10 H/m 8.85 10 F/m 10.0 V/m s 2.00 10 T.2 0.100 m

B π − − −= ⋅ ⋅ = ⋅

Because /dE dt is positive, the direction of B

is counterclockwise, as shown in the figure above.

31.21. THINK: A magnetic field can be produced by a current and by induction due to a change in an electric flux. To solve this problem, use the Maxwell-Ampere law. There is no current between the plates, but there is a change in the electric flux. The wire carries a current 20.0 A.i = The parallel plate capacitor has radius 4.0 cm,R = and separation 2.0 mm.s = The radius of interest is 1.0 mmr = from the center of the parallel plates. SKETCH:

RESEARCH: Since there is no current between the capacitor plates, the Maxwell-Ampere law becomes: E

0 0 .d

B dSdt

μ ε Φ=

SIMPLIFY: Applying this law along a circular Amperian loop with a radius, ,r R≤ as shown above. Since B

is parallel to ,dS

the left-hand side of the above equation is 2 .B dS B dS B rπ= =

Assuming the

electric field, ,E

is uniform between the capacitor plates and directed perpendicular to the plates, the electric flux through the loop is 2

E r .EA E rπΦ = = Thus, the Ampere-Maxwell law becomes:

( ) 20 02 .dEB r r

dtπ μ ε π=


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Therefore, the magnetic field is: 0 0 .2

r dEBdt

μ ε =

Since the electric field of the capacitor is 0/ ,E σ ε= the

rate of change of the electric field is given by:

( ) ( )0 R0 0 0 R

1 1 1/ / .dqdE d d d q Adt dt dt dt A dt

σσ εε ε ε

= = = =

Since i = dq/dt, 20

.dE idt Rε π

= Using this result, the magnetic field is: 0 0 02 2

0

.2 2

r iiB rR R

μ ε με π π

= =

CALCULATE: ( )( )

( )( )

75

2

4 10 H/m 20.0 A0.010 m 2.5 10 T

2 0.040 mB

π

π

−−

⋅= = ⋅

ROUND: Two significant figures are required: 52.5 10 TB −= ⋅ . DOUBLE-CHECK: This is the same as calculating a magnetic field inside a wire with radius, R. Applying Ampere’s law gives:

20 0

2 2 .2 2

irB i rr R R

μ μππ π π

= =

This is the same result as above.

31.22. THINK: To determine the electric field, apply Faraday’s law of induction. The solenoid is 20.0 cm long, 2.00 cm in radius, and has 500 turns. The current varies from 3.00 A to 1.00 A in 0.100 s. SKETCH:

RESEARCH: The magnetic field inside a solenoid is given by 0 0 / .B ni Ni Lμ μ= = Applying Faraday’s law along a loop with radius, r, gives:

B1 .2

dE

r dtπΦ = −

SIMPLIFY: Substituting 2B 0 /BA Ni r Lμ πΦ = = into the above equation yields:

20 0 0 2 11 .

2 2 2N r Nr Nr i idi diE

r L dt L dt L tμ π μ μ

π − = − = − = − Δ


( )7

44 10 H/m 500 0.0100 m 1.00 A 3.00 A 3.14 10 V/m

0.100 s2 0.200 mE

π −−

⋅ − = − = ⋅

ROUND: Keeping three significant figures: 43.14 10 V/m.E −= ⋅ DOUBLE-CHECK: The direction of the induced electric field must be such that the magnetic field induced by the current opposes the change in magnetic flux. Because the magnetic flux is decreasing, the induced magnetic field will be in the same direction as the original magnetic field. The fact that the calculated electric field is positive confirms that this requirement is satisfied.


1141

31.23.

The displacement current, d ,i produced by a rate of change in the electric field of a parallel plate capacitor is d 0 E / .i d dtε= Φ The flux, E ,Φ is given by E 0 0/ / .A qσ ε εΦ = = Therefore, the displacement current is:

d 00

1 10.0 μA.dqidt

εε

= =

31.24. THINK: This problem is similar to problem 31.21 except that here, the rate of change of the potential difference across the capacitor is given. In order to get the induced magnetic field, apply Maxwell’s law of induction. The parallel plate capacitor has radius 10.0 cmR = and separation 5.0 mm.d = The potential in increasing at a rate of 1200 V/s. The radius of interest is 4.0 cmr = from the center of the capacitor. SKETCH:

RESEARCH: Applying Maxwell’s law of induction along a circular loop with a radius, ,r R≤ and assuming a uniform electric field yields 0 0 E / .B dS d dtμ ε= Φ

SIMPLIFY: ( ) 0 0E

0 0 0 0 0 0

/2

d V d Ad dEA dVB dS B r Adt dt dt d dt

μ επ μ ε μ ε μ εΦ= = = = =

The magnetic field is: ( ) ( )2 3

0 0 0 02 2 .

2 2d V d Vr r

B rdt dtR d R d

μ ε π μ επ

Δ Δ = ⋅ =


( ) ( )37 12

152 3

4 10 H/m 8.85 10 F/m 0.04 m1200 V/s 8.5640 10 T

2 0.10 m 5.0 10 mB

π − −−

−

⋅ ⋅ = = ⋅ ⋅

ROUND: Rounding to two significant figure gives 158.6 10 T.B −= ⋅ DOUBLE-CHECK: Treating the area between the parallel plates as a solid conductor carrying a current of magnitude equal to the displacement current, d 0 / ,i AdE dtε= the problem becomes one of finding the magnetic field inside a current carrying wire. Applying Ampere’s Law,

30 d 0 0 0 0

2 2 2

/ /,

2 2 2i AdE dt r dV dt

B r rR R R d

μ μ ε μ επ π

= = =

which is the same result as that obtained by applying

Maxwell’s law of induction.

31.25. THINK: To determine the displacement current, the electric field inside the conductor is needed.


1142

SKETCH:

RESEARCH: The displacement current is defined as: d 0 E / .i d dtε= Φ The electric flux inside the conductor is: ( )E / .EA V L AΦ = = SIMPLIFY: Since V = iR, the electric flux becomes / .iRA LΦ = Therefore, the displacement current is:

d 0 .A dii RL dt

ε =

Using /R L Aρ= or / ,RA Lρ = the displacement current simplifies to: d 0 .diidt

ε ρ=

CALCULATE: Not required. ROUND: Not required. DOUBLE-CHECK: Since the current depends in part on the resistance of the current carrying conductor, and the resistance depends on the geometry and resistivity of the material, it makes sense that the current is some function of the resistivity.

31.26. The amplitude of the B field of an electromagnetic field is related to the electric field by / .B E c= Therefore,

78

250 V/m 8.3 10 T.3.00 10 m/s

B −= = ⋅⋅

31.27. The distance traveled by light is given by:

( )8 9 ft3.00 10 m/s 1.00 10 s 0.300 m 0.300 m 3.28 0.984 ft.m

x c t − = Δ = ⋅ ⋅ = = =

31.28. The interval of time taken by light to travel:

(a) from the Moon to the Earth is: 8

8

3.8 10 m 1.3 s,3.00 10 m/s

dtc

⋅Δ = = =⋅

(b) from the Sun to the Earth is: 11

8

1.5 10 m 500 s 8.3 min,3.00 10 m/s

t ⋅Δ = = =⋅

(c) from Jupiter to the Earth is: 11

8

6.2 10 m 2066 s 34 min.3.00 10 m/s

t ⋅Δ = = =⋅

31.29. (a) The time delay from New York to Baghdad by cable is 7

8

1 10 m 0.03 s.3.00 10 m/s

dtc

⋅Δ = = =⋅

(b) The time delay via satellite is given by / .t d cΔ = The distance, d, is given by twice the distance from

New York to the satellite, that is, ( ) ( )2 22 36000 km 5000 km/2 2 36345 km 72691 km.d = + = ⋅ = The

time delay is: 7

8

7.269 10 m 0.24 s.3.00 10 m/s

t ⋅Δ = =⋅

When the signal travels by the cable, the time delay is very short, so it is not noticeable. However, the time delay for the signal traveling via satellite is about a quarter of a second. This means in a conversation, Alice will find that she receives a response from her fiancé after 0.5 s, which is quite noticeable.


1143

31.30. THINK: The speed of electromagnetic waves in a vacuum is different from the speed of such waves in different media. The difference depends on the dielectric constant, κ , and the relative permeability, mκ , of the material. SKETCH: Not required. RESEARCH: The speed of electromagnetic waves in a material is 1/v με= and the speed in a vacuum

is 0 01/ .c μ ε= The permittivity is 0 ,ε κε= and the permeability is m 0 .μ κ μ= SIMPLIFY: The ratio of the speed of electromagnetic waves in a vacuum to the speed in a material is:

0 0m

0 0

1/.

1/cv

μ ε με κκμ εμε

= = = This ratio is the index of refraction.

CALCULATE: Not applicable. ROUND: Not applicable. DOUBLE CHECK: The calculated ratio is the index of refraction, which is a measure of how much the speed of light, or other electromagnetic waves, is reduced in a medium compared to the speed in a vacuum.

31.31. The relation between the wavelength of light and the frequency is .f cλ = Therefore, the frequencies for the wavelengths of 400 nm and 700 nm are:

814

1 9

3.00 10 m/s 7.5 10 Hz400 10 m

f −

⋅= = ⋅⋅

and 8

142 9

3.00 10 m/s 4.3 10 Hz.700 10 m

f −

⋅= = ⋅⋅

The range of frequencies is 144 10 Hz⋅ to 148 10 Hz.⋅

31.32. Using the relation between frequency and wavelength, the operating frequency of the signal of a cell phone is / .f c λ= Since / 4,L λ= the frequency is:

( )8

83.00 10 m/s 9.4 10 Hz 940 MHz.4 4 0.080 mcfL

⋅= = = ⋅ =

31.33. THINK: To solve this problem, the frequency of oscillation of an RLC circuit must be determined. The circuit has a capacitor 122.0 10 F,C −= ⋅ and must have a resonance frequency such that it will generate a radio wave with wavelength 150 m.λ = SKETCH: A sketch is not required. RESEARCH: The angular frequency of the RLC circuit in resonance is 0 1/ .LCω =

SIMPLIFY: Using 0 2 fω π= and / ,f c λ= the above equation becomes: 2 1 .cLC

πλ

= The inductance

required in the RLC circuit is: ( )

2

2 .2

Lc C

λπ

=

CALCULATE: ( )

( ) ( )

2

28 12

150 m0.00317 H

2 3.00 10 m/s 2.0 10 FL

π −= = ⋅ ⋅

ROUND: Rounding to two significant figures yields 3.2 mH.L = DOUBLE-CHECK: A wavelength of 150 m corresponds to a frequency of 62 10 Hz.⋅ Such a large frequency necessarily requires a fairly small inductance.

31.34. THINK: The radio frequencies given are: 1 91.1 MHz,f = 2 91.3 MHz,f = and 3 91.5 MHz.f = To determine the wavelength width of the band-pass filter used in a radio receiver, the wavelengths of the three radio frequencies are required.


1144

SKETCH: A sketch is not required. RESEARCH: Wavelength is related to frequency by / .c fλ = The maximum bandwidth required to distinguish between two adjacent frequencies is given by ( )12 1 22λ λ λ= − for 1f and 2 ,f and

( )23 2 32λ λ λ= − for 2f and 3 .f Thus, the maximum allowable bandwidth to distinguish all three

frequencies is ( )12 23min , .λ λ λΔ = Δ Δ SIMPLIFY: Simplification is not necessary. CALCULATE: The corresponding wavelengths of the three radio frequencies are given by:

8

1 61

3.00 10 m/s 3.293 m,91.1 10 Hz

cf

λ ⋅= = =⋅

8

2 62

3.00 10 m/s 3.286 m,91.3 10 Hz

cf

λ ⋅= = =⋅

and finally,

8

3 63

3.00 10 m/s 3.279 m.91.5 10 Hz

cf

λ ⋅= = =⋅

The differences of two adjacent wavelengths are:

( )12 2 3293 mm 3286 mm 14 mmλΔ = − = and ( )23 2 3286 mm 3279 mm 14 mm.λΔ = − = Therefore, the maximum allowable wavelength bandwidth is 14 mm.λΔ = ROUND: Rounding is not necessary. DOUBLE-CHECK: A larger wavelength width in the band pass filter would allow overlap between two signals, resulting in interference. This result is reasonable.

31.35. The magnitude of a Poynting vector is given by: 2

Power Power .Spherical Area 4

SRπ

= = Therefore, the magnitudes

of the Poynting vectors are:

(a) ( )

22

1.5 W 1.3 W/m ,4 0.30 m

Sπ

= =

(b) ( )

22

1.5 W 1.2 W/m ,4 0.32 m

Sπ

= =

(c) ( )

22

1.5 W 0.12 W/m .4 1.00 m

Sπ

= =

31.36. (a) The electric field experienced by an electron is:

( )( )( )

9 2 2 1911

2 29

8.99 10 N m / C 1.602 10 C5.8 10 V/m.

0.050 10 m

kqEr

−

−

⋅ ⋅= = = ⋅

⋅

(b) The intensity of a laser beam is related to the rms electric field by:

( )( )( )

2112 20 2rms 7 8

0

5.8 10 V/m1 8.9 10 W/m .4 10 H/m 3.00 10 m/s

I Ecμ π −

⋅= = = ⋅

⋅ ⋅

31.37. The intensity of the laser beam is I = P/A. This intensity is related to the amplitude of the electric field by ( )2

0/ 2 .I E cμ= Therefore, the amplitude of the electric field in the beam is:

( )( )( )( )

7 8 36 60

23

2 4 10 H/m 3.00 10 m/s 3.00 10 W21.697 10 V/m 1.70 10 V/m.

0.500 10 m

cPE

A

πμ

π

−

−

⋅ ⋅ ⋅= = = ⋅ ≈ ⋅

⋅


1145

31.38. The electric field of an electromagnetic radiation is related to its magnetic field by .E cB= Therefore, the maximum E in the region is ( )( )8 5

m m 3.00 10 m/s 0.00100 T 3.00 10 V/m.E cB= = ⋅ = ⋅ The period of

oscillation is: 1 1 1 s.1 Hz

Tf

= = = The magnitude of the Poynting vector is:

( )( )58 2 8 2

m m m 70

3.00 10 V/m 0.001 T1 2.3873 10 W/m 2.39 10 W/m .4 10 H/m

S E Bμ π −

⋅= = = ⋅ = ⋅

⋅

31.39. The average value of the Poynting vector, ave ,S is:

( )( )( )

2

2 2ave m 7 8

0

100. V/m1 13.3 W/m .2μ 2 4 10 H/m 3.00 10 m/s

S Ec π −

= = =⋅ ⋅

(a) The average energy density is: ( )( )22 12 2 8 30

1 1 8.85 10 C / N m 100. V/m 4.43 10 J/m .2 2

u Eε − −= = ⋅ = ⋅

(b) The amplitude of the magnetic field is: 78

100. V/m 3.33 10 T.3.00 10 m/s

EBc

−= = = ⋅⋅

31.40. THINK: The maximum electric field of a beam of light is given as 6m 3.0 10 V/m.E = ⋅

SKETCH: A sketch is not necessary. RESEARCH: (a) The magnitude of a magnetic field is related to the magnitude of an electric field by / .B E c= (b) The intensity of the wave is given by ( )2

m 0/ 2 .I E cμ= (c) If the electric field is above this maximum value, the air will be ionized by the presence of the electric field. The energy of the wave will be dissipated in the ionized air. SIMPLIFY: Simplification is not necessary. CALCULATE:

(a) 6

28

3.0 10 V/m 1.0 10 T3.00 10 m/s

B −⋅= = ⋅⋅

(b) ( )

( )( )

2610 2

7 8

3.0 10 V/m1.19 10 W/m

2 4 10 H/m 3.00 10 m/sI

π −

⋅= = ⋅

⋅ ⋅

ROUND: Round the results to two significant figures. (a) 21.0 10 TB −= ⋅ (b) 10 21.2 10 W/mI = ⋅ DOUBLE-CHECK: This is a very large magnetic field, and a very high intensity, as expected for a field at the breakdown threshold. The results are reasonable.

31.41. THINK: A laser beam has a power of 10.0 W and a beam diameter of 1.00 mm. Assume the intensity of the beam is the same throughout the cross section of the beam. SKETCH: A sketch is not required. RESEARCH: (a) The intensity of the laser beam is given by I = P/A. Area 2 .A rπ= (b) The intensity is related to the rms electric field by ( )2

rms 0 rms 0/ .I E c E cIμ μ= = (c) The time-averaged Poynting vector is equal the intensity of the beam, ave .S I=

(d) ( ) ( ) 2

0

,,

E x tS x t

cμ

= and ( ) ( )m, sin .E x t E kx tω φ= − +


1146

(e) The rms magnetic field is rms rms / .B E c= SIMPLIFY:

(d) Substituting the expression for ( ),E x t gives: ( ) ( )2 2m

0

1, sin .S x t E kx tc

ω φμ

= − + Because ( )0,0 0,S =

take 0.φ = Therefore, ( ) ( )2, 2 sin .S x t I kx tω= − Note that 2 2 /f cω π π λ= = and 2 / .k π λ= CALCULATE:

(a) ( )

7 223

10.0 W 1.2732 10 W/m0.500 10 m

Iπ −

= = ⋅⋅

(b) ( )( )( )7 8 7 2 4rms 4 10 H/m 3.00 10 m/s 1.2732 10 W/m 6.932809 10 V/mE π −= ⋅ ⋅ ⋅ = ⋅

(c) 7 2ave 1.2732 10 W/mS = ⋅

(d) ( ) ( ) ( )

( ) ( )

87 2 2

9 9

7 2 2 7 1 15

2 3.00 10 m/s2, 2 1.2732 10 W/m sin514.5 10 m 514.5 10 m

2.5464 10 W/m sin 1.22122 10 m 3.66366 10 Hz

S x t x t

x t

ππ− −

−

⋅ = ⋅ − ⋅ ⋅ = ⋅ ⋅ − ⋅

(e) 4

4rms 8

6.932809 10 V/m 2.30936 10 T3.00 10 m/s

B −⋅= = ⋅⋅

ROUND: (a) 7 21.27 10 W/m .I = ⋅ This intensity is much larger than the intensity of sunlight on Earth

( )21400 W/m .

(b) 4rms 6.93 10 V/mE = ⋅

(c) 7 2ave 1.27 10 W/mS = ⋅

(d) ( ) ( ) ( )( )7 2 2 7 1 15, 2.5464 10 W/m sin 1.22122 10 m 3.66366 10 HzS x t x t−= ⋅ ⋅ − ⋅ ; Don’t round coefficients

here, instead round the value of S after substituting values for x and t. (e) 4

rms 2.31 10 TB −= ⋅ DOUBLE-CHECK: The laser has a very high power output in a very narrow beam. This is a desirable property in a laser. The results make sense.

31.42. THINK: The Poynting vector is proportional to .E B×

Assume that the electric field in a conductor is uniform. The conductor is placed along the y-axis and the current is flowing along the positive y-direction. This means the electric field is in the positive y-direction. SKETCH: (a) (b)


1147

RESEARCH: The Poynting vector is defined as 0/ .S E B μ= ×

(a) The electric field on the surface of the conductor is ( ) ˆ/ .E V L y=

Assume a long cylindrical conductor and the magnetic field on the surface is:

( )0 0ˆ ˆ ˆcos sin .2 2

I IB x z

R Rμ μ

θ θ θπ π

= = −

(b) The integral of S dA is: ,S dA S dA= −

since .dA dAr=

SIMPLIFY: (a) The Poynting vector is the cross product:

( ) ( ) ( )0

0

1 ˆ ˆ ˆ ˆ ˆ ˆcos sin cos sin .2 2 2

IV VI VIS y x z z x rL R RL RL

μ θ θ θ θμ π π π

= × − = − − = −

This means that the Poynting vector is directed toward the cylindrical conductor with a magnitude of ( )/ 2 .S VI RLπ=

(b) Taking the integral over the surface of the cylinder:

( ) ( ) 2R R2 2 2 .

2VIdA RL S dA S RL RL VI IR I I R

RLπ π π

π= = − = − = − = − = −

Note that the subscript, R, is to distinguish that RR means resistance. CALCULATE: Not necessary. ROUND: Not necessary. DOUBLE-CHECK: Dimensional analysis shows that the units of the calculated result are 2W/m , as required for the Poynting vector.

31.43. (a) The Poynting vector is 21.40 kW/m .S =

( )( )( )2 3 2 8 70

0

1 1.40 10 W/m 3.00 10 m/s 4 10 T m/A 726.49 V/m 726 V/mS E E Scc

μ πμ

−= = = ⋅ ⋅ ⋅ = ≈

68

726.49 V/m 2.422 10 T 2.42 μT.3.00 10 m/s

EBc

−= = = ⋅ ≈⋅

(b) 3 2

r 8

1.00 10 W/m 3.33333 μPa 3.33 μPa,3.00 10 m/s

IPc

⋅= = = ≈⋅

( )( )6 2r r 3.33333 10 Pa 0.750 m 2.50 μN.FP F P A

A−= = = ⋅ =

31.44. (a) First, determine the force needed to accelerate a 10.0 ton spaceship by 21 m/s . Newton’s second law, ,F ma= gives: ( )( )3 2 410.0 10 kg 1.00 m/s 1.00 10 N.F = ⋅ = ⋅ Now, the radiation pressure is given by

r / .P F A= The radiation pressure is related to the intensity of the radiation by r /P I c= (total absorption). Comparing the two equations for the radiation pressure gives:

( )( )4 89 2

3 2

1.00 10 N 3.00 10 m/s 2.14 10 m .

1.40 10 W/mF I FcAA c I

⋅ ⋅= = = = ⋅

⋅

This area is large. Moreover, even if the scientists are able to get the astronauts to another planet, how do they get them back to Earth?

(b) For perfect reflection: ( )9 2 9 2r

2 1 2.14 10 m 1.07 10 m .2 2

I F FcP Ac A I

= = = = ⋅ = ⋅

31.45. The net force on the sail is .F A P= Δ The area, A, is ( )22 3 8 210.0 10 m 3.142 10 m .A Rπ π= = ⋅ = ⋅ The

differential pressure, ,PΔ is:


1148

2 .I I IPc c c

Δ = − =

The intensity, I, is given by the Stefan-Boltzman law as:

( )( )44 8 2 4 6 25.67 10 W/m K 2.725 K 3.126 10 W/m .I Tσ − −= = ⋅ = ⋅ 6 2

148

3.126 10 W/m 1.042 10 Pa3.00 10 m/s

P−

−⋅ Δ = = ⋅

⋅

( )( )8 2 14 6 3.142 10 m 1.042 10 Pa 3.27 10 NF − − = ⋅ ⋅ = ⋅

31.46. THINK: There will be a constant force on the astronaut due to the radiation pressure from the laser. This force can be determined from the laser power, and then the time required to reach the shuttle can be determined. d = 20.0 m, m = 100.0 kg and P = 100.0 W. SKETCH:

RESEARCH: r2IPc

= (totally reflecting), ,PIA

= r ,F maPA A

= = 212

x at= (constant acceleration)

SIMPLIFY: rr ,

P AmaP aA m

= = ( )

r

2 /2 2 2 P AI A P PP a

c c m Ac mc = = = =

21 2 22 2

d mc dmcx d at t da P P

= = = = =

CALCULATE: ( )( )( )8

420.0 m 100.0 kg 3.00 10 m/s

7.746 10 s100.0 W

t⋅

= = ⋅

ROUND: 47.75 10 s 21.5 ht = ⋅ = DOUBLE-CHECK: The time decreases as the laser power increases or the mass decreases. This is what would be expected.

31.47. THINK: The applied force can be determined from the given data. From the force, the radiation pressure, and then the laser power used in the demonstration can be determined. d = 2.00 mm, t = 63.0 s, m = 0.100 g and 2r = 1.00 mm. SKETCH:

RESEARCH: / ,I P A= 2 ,A rπ= r2 ,I maPc A

= = 212

d at=

SIMPLIFY: 2 ,P IA I rπ= = 2 ,2 2

mca mcaIA rπ

= = 2

2dat

= 22 2 2

2 2 22

mca mca mc d mcdP rr t t

ππ

= = = =


( )

3 8 32

2

0.100 10 kg 3.00 10 m/s 2.00 10 m1.512 10 W

63.0 sP

− −−

⋅ ⋅ ⋅= = ⋅


1149

ROUND: 21.51 10 W 15.1 mWP −= ⋅ = DOUBLE-CHECK: Note that the result does not depend on the spot size of the laser. The power is all that matters. This result is reasonable.

31.48. THINK: The radius of the particle can be determined if the required mass of the particle is known because the density is given. The mass of the particle can be determined by comparing the radiation force and the gravitational force. 32000. kg/m ,ρ = 111.50 10 m,d = ⋅ 302.00 10 kg,M = ⋅

rad grav/ 1.00% 0.0100,F F = = 21400. W/m .I =

SKETCH:

RESEARCH: grav 2

G ,MmFd

= 34 ,3

m rπ ρ= rrad ,F P A= 2 ,A rπ= r 2 / .P I c=

SIMPLIFY: 3grav 2 2

G G 4 ,3

Mm MF rd d

π ρ = =

( )2rad r

2IF P A rc

π= =

( )

32 2 2

rad grav 2

4 G32 2 30.01 0.01

0.01 2 G40.01 G3

r Mr I I d IdF F rc c c Md M

π ρπ π

ρπρ

= = = =

CALCULATE:

( )( )( ) ( )( )( )( )

22 115

8 3 11 3 1 2 30

3 1400. W/m 1.500 10 m5.901 10 m

0.0100 2 3.00 10 m/s 2000. kg/m 6.673 10 m kg s 2.00 10 kgr −

− − −

⋅= = ⋅

⋅ ⋅ ⋅

ROUND: 59.0 μmr = DOUBLE-CHECK: If the radiation pressure on the particle is to be equal to the gravitational force on it, its radius would have to be smaller by a factor of 100. This indicates that very small particles would be pushed away from the sun, while more massive objects would be pulled toward the sun. This is consistent with observation. The result makes sense.

31.49. THINK: Given the density and volume, the mass can be determined. Given the power and spot size of the laser, the intensity and the radiation pressure of the laser can be determined. To determine how many lasers are needed, calculate the total force required and divide this by the force per laser applied.

31.00 mg/cm ,ρ = D = 2.00 mm, t = 0.100 mm, P = 5.00 mW, d = 2.00 mm. SKETCH:

RESEARCH:

(a) The weight is given by w = mg, where ( )2/ 2 .m D tπ ρ= Note that 3 31 mg/cm 1 kg/m .=


1150

(b) rIPc

= (absorbing material), ,PIA

= 2

.2dA π =

(c) las r .F P A= The number of lasers needed is given by las/ .N w F= SIMPLIFY:

(a) 2

4D t gw π ρ=

(b) 2 2

4 4 ,PI Pd dπ π

= = r 2

4I PPc d cπ

= =

(c) las r ,IA P PF P A Ac cA c

= = = = las

w wcNF P

= =

CALCULATE:

(a) ( ) ( )( )( )23 3 3 2

92.00 10 m 0.100 10 m 1.00 kg/m 9.81 m/s

3.082 10 N4

wπ − −

−⋅ ⋅

= = ⋅

(b) ( )( )

33 2

23

4 5.00 10 W1.592 10 W/m ,

2.00 10 mI

π

−

−

⋅= = ⋅

⋅

3 26 2

r 8

1.592 10 W/m 5.30510 N/m3.00 10 m/s

IPc

−⋅= = =⋅

(c) ( )( )

( )9 8

3

3.082 10 N 3.00 10 m/s184.9

5.00 10 WN

−

−

⋅ ⋅= =

⋅

ROUND: (a) 93.08 10 N 3.08 nNw −= ⋅ = (b) 21.59 kW/m ,I = 2

r 5.31 μN/mP = (c) N = 185 lasers DOUBLE-CHECK: Even though the object is very light, it would still require a large power output to produce enough radiation pressure to overcome the force of gravity.

31.50. The first filter is out of alignment by 15.0° with the incident light. The second filter is out of alignment by 30.0° with the incident light. The intensity of the transmitted light is

( ) ( ) ( )2 2 2 2 2 20 1 2cos cos 1.00 cos 15.0 cos 30.0 0.69976 W/m 0.700 W/m .I I θ θ= = ° ° = ≈

31.51.

( ) ( ) ( )2 3 2 2 20 cos 10.0 10 W cos 90.0 30.0 10 W cos 60.0 2.50 mWI I θ − −= = ⋅ ° − ° = ° =

31.52. THINK: Only half the intensity gets through the first polarizer since the incident light is un-polarized. After this, multiply the transmission for each polarizer to obtain the final intensity. 10 .θ = °


1151

SKETCH:

RESEARCH: 1 0 / 2,I I= 21 cosn nI I θ−= (n = 2, 3, 4, 5)

SIMPLIFY: ( )2 22 1 0cos cos / 2,I I Iθ θ= = ( )2 4

3 2 0cos cos / 2,I I Iθ θ= = ( )2 64 3 0cos cos / 2,I I Iθ θ= =

( )2 85 4 0cos cos / 2I I Iθ θ= =

CALCULATE: ( )85 0 0cos 10 / 2 0.4424I I I= ° =

ROUND: 5 00.442I I= DOUBLE-CHECK: Only 44.2% of the original intensity passes through the polarizers. The first polarizer decreases the intensity by 50%, but the subsequent polarizers allow the majority of the light to pass through due to the smaller angles. This is a reasonable result.

31.53. THINK: First calculate the intensity of the light after it first passes through the two polarizers. Once the intensity is calculated, the magnitude of the electric and magnetic fields can be determined. The angles of the first and second polarizers are 1 35θ = ° and 2 55θ = °, respectively. The laser spot size diameter is

1.00 mmd = and the laser power is P = 15 mW. SKETCH:

RESEARCH: 21 0 1cos ,I I θ= ( )2

2 2 11 cos ,I I θ θ= − 0 ,PIA

= 2

,2dA π =

2

0

1 ,2

EIcμ

=

E c

B=

SIMPLIFY: ( ) ( )2 2 22 1 2 1 0 1 2 1cos cos cos ,I I Iθ θ θ θ θ= − = − ( )2 2

0 2 1 2 12 2

4 4 cos cosP P PI IA d d

θ θ θπ π

= = = −

02 ,E Icμ= B = E/c

CALCULATE: ( )

( )( ) ( )

32 2 4 2

2 23

4 15 10 Wcos 35 cos 55 35 1.132 10 W/m

1.00 10 mI

π

−

−

⋅= ° ° − ° = ⋅

⋅

( )( )( )4 2 8 7 1 32 1.132 10 W/m 3.00 10 m/s 4 10 T m A 2.921 10 V/mE π − −= ⋅ ⋅ ⋅ = ⋅3

68

2.921 10 V/m 9.737 10 T3.00 10 m/s

B −⋅= = ⋅⋅

ROUND: 4 22 1.13 10 W/m ,I = ⋅ 32.92 10 V/m,E = ⋅ 69.74 10 TB −= ⋅

DOUBLE-CHECK: The initial intensity of the laser light is about 4 21.9 10 W/m .⋅ The initial electric and magnetic fields are also significantly larger. It is expected that some of the intensity of the laser beam would be blocked by the polarizers.


1152

Additional Problems

31.54.

The total distance traveled by the beam is 2d.

( )( )8 36

3.00 10 m/s 50.0 10 s2 7.50 10 m2 2

d ctv c dt

−⋅ ⋅= = = = = ⋅

31.55. The total average power on all of the photovoltaic panels is equal to the average power times the area:

( )( )( )2300. W/m 3.00 m 8.00 m 7.20 kW.=

The total electrical energy for 30 days is the product of the total power times 30 days times the efficiency for converting the solar power into electricity: ( )( )( )total 7.20 kW 720 h 0.100 518 kW h.E = = This is enough for a small energy-efficient home.

31.56. Use the fact that radiation pressure scales with intensity, and intensity goes linearly with the intrinsic power of the star, and inversely as the square of the distance away. Thus, if it is known that at the Earth’s orbit, the intensity is 21.35 kW/m , this intensity can be related to the radiation pressure at Earth’s orbit, and then to the hypothetical distance of Uranus’ orbit away from Betelgeuse. Note that the radius of the earth's orbit is 111.50 10 m Er ≈ ⋅ and Uranus's orbit is 122.88 10 m.Ur ≈ ⋅

SrE 2

E

P,P

r∝ rB 2

PB

U

Pr

∝

( )( )2 22

4 2rU E ErU rE2

rE U4 2

P P 110 27.1 I 27.1 I 27.1 1350 W/mP 19.2P

3.66 10 W/m

B B

S US

P r rRP rr

= = = = = = ⋅ =

= ⋅

For a perfect absorber we have: 2

4 28

36600 W/m 1.22 10 N/m , 3.00 10 m/s

rUrU

IP

c−= = = ⋅

⋅and for a perfect reflector we

have: 2

4 28

2 2 36600 W/m 2.44 10 N/m .3.00 10 m/s

rUrU

IP

c−⋅= = = ⋅

⋅Since no information is provided on the reflectivity

of the surface being acted on by the radiation, the final solution is: 4 2 4 21 10 N/m 2 10 N/m , rUP− −⋅ < < ⋅

31.57. 2

8 2 506 2

0

power 200. W 2.00 10 W/m 2 3.88 10 V/marea 21.00 10 m

ES E c Sc

μμ

= = = ⋅ = = = ⋅ ⋅

Note the 2 is in the denominator from: 22 2

2

0 0

. Therefore, and S = = . 2 22

rmsrms rms

EE E EE Ec cμ μ

= = The

wavelength has nothing to do with the solution.

31.58. 8

6

3.00 10 m/s 35 cm850 10 Hz

cf cf

λ λ ⋅= = = =⋅

31.59. ,P IAε= 2

0

1 ,2

EIcμ

=

A = LW, and 0.18.ε =


1153

( ) ( )( )( )

( )( )22

8 70

673 V/m 1.40 m 0.900 m 0.180 136 W

2 2 3.00 10 m/s 4 10 T m/AE LWP

cε

μ π − = = =

⋅ ⋅

31.60. The displacement current between the capacitor plates is the same as the conventional current in the rest of the circuit.

( ) ( )( )/ 3emf6

25 V 0.3621 sexp 1.0288 10 0.36785 0.37845 mA24300 24300 14.9 10 F

t RCVi e

R− −

−

− = = = ⋅ = Ω Ω ⋅

( )( )( ) ( )11dd 0 12 4 2

02 2

0.37845 mA 4.3 10 V/ m s8.85 10 C / N m 1.00 10 m

idE dEi Adt dt A

εε − −

= = = = ⋅⋅ ⋅

31.61. ( )( )( )8 7 2rms 0 3.00 10 m/s 4 10 T m/A 38.2 W/m 120. V/mE c Iμ π −= = ⋅ ⋅ =

31.62. ( )( )8 3 3.00 10 m/s 5 10 T 1.50 MV/mE c E cBB

−= = = ⋅ ⋅ =

31.63. The antinodes are spaced half a wavelength apart, / 2,d λ= where / .c fλ =

( )8

29

3.00 10 m/s 6.25 10 m 6.3 cm2 2 2.4 10 Hzcdf

−⋅ = = = ⋅ =

⋅

31.64. 21400 W/mI =

(a) ( )( )( )2

2 8 7rmsrms 0

0

1400 W/m 3.00 10 m/s 4 10 T m/A 726.5 V/mE

I E Icc

μ πμ

−= = = ⋅ ⋅ =

( ) 3max rms2 2 726.5 V/m 1027 V/m=1.0 10 V/mE E= = = ⋅

(b) 6 6maxmax 8

1027 V/m 3.425 10 T 3.4 10 T3.00 10 m/s

EB

c− −= = = ⋅ = ⋅

⋅

31.65. THINK: The peak magnetic field can be determined from the speed of light and the peak electric field. The power of the bulb can be determined from its intensity, which can be determined from the electric and magnetic fields. Use the values r = 2.25 m and E = 21.2 V/m. SKETCH:

RESEARCH: ,E cB

= ,PIA

= 2

02EIcμ

=

SIMPLIFY: (a) B = E/c

(b) 2

0

.2EP IA Acμ

= = Light from a light-bulb is emitted isotropically, that is equally in all directions. To

determine the power a distance, d, away from the light-bulb, the intensity at all points a distance, d, from the light-bulb must be summed. Hence, A should be the surface area of a sphere of radius, r:

2 22

0

44 .2

r EA r Pc

ππμ

= =


1154

CALCULATE:

(a) 88

21.2 V/m 7.067 10 T3 10 m/s

B −= = ⋅⋅

(b) ( ) ( )

( )( )2 2

8 7

4 2.25 m 21.2 V/m37.92 W

2 3.00 10 m/s 4 10 T m/AP

ππ −

= =⋅ ⋅

ROUND: (a) 87.07 10 TB −= ⋅ (b) P = 38.0 W DOUBLE-CHECK: These values are consistent with the power output for a regular household light bulb.

31.66. THINK: To determine the temperature of the star, the power radiated by the star must be known. To determine the power, the intensity is needed. The intensity can be determined from the electric field and the distance. D = 15 AU, E = 0.015 V/m, S2 ,r r= 5

S 6.955 10 km.r = ⋅ SKETCH:

RESEARCH: 4 2 44 ,P A T r Tσ π σ= = 2

2024

P EIcd μπ

= =

SIMPLIFY: ( )2 2 2 2

2 4S

0 0

4 2 4 2 ,2

d E d EP r Tc c

π π π σμ μ

= = = 1/4 1/42 2 2 2

2 2S 0 S 0

28 4

d E d ETr c r c

ππ σμ σμ

= =

CALCULATE:

( )( ) ( )( ) ( )( )( )

1/42 211

25 3 8 2 4 7 2 8

15 1.49598 10 m 0.015 V/m72.37 K

4 6.955 10 10 m 5.670 10 W m K 4 10 N/A 3.00 10 m/sT

π− − − − −

⋅ = = ⋅ ⋅ ⋅ ⋅ ⋅

ROUND: T = 72 K DOUBLE-CHECK: Although numerically correct, this temperature is far too low for a star.

31.67. THINK: From the power and the spot size, the intensity of the beam can be determined. From the intensity, the electric field can be determined. For the total energy, multiply the energy density by the volume of the beam. P = 5.00 mW, d = 2.00 mm, l = 1.00 m. SKETCH:

RESEARCH: 2rms

0

,E PI

c Aμ= = 2 ,A dπ= 2

0 rms12Eu Eε=

SIMPLIFY:

(a) ( )

0rms 2/ 2

cPE

dμ

π=


1155

(b) 2 2 2

2 0 rmstot 0 rms

12 2 8E

E d ldE u V E lε πε π = = =

CALCULATE:

(a) ( )( )( )

( )7 2 8 3

rms 23

4 10 N/A 3.00 10 m/s 5.00 10 W774.6 V/m

1.00 10 mE

π

π

− −

−

⋅ ⋅ ⋅= =

⋅

(b) ( )( ) ( )( )2212 3 1 4 2 3

12tot

8.8542 10 m kg s A 774.6 V/m 1.00 m 2.00 10 m8.34495 10 J

8E

π− − − −−

⋅ ⋅= = ⋅

ROUND: (a) rms 775 V/mE =

(b) 12tot 8.34 10 JE −= ⋅

DOUBLE-CHECK: It is expected that a laser pointer would have small electric field and generate a small amount of energy, considering it’s intended use. A laser pointer with more power would be dangerous.

31.68. THINK: The total power incident on the roof is the intensity of the light times the area of the roof. From the intensity, the radiation pressure can be determined, and from this the force can be determined.

21.00 kW/m ,I = l = 30.0 m, w = 10.0 m. SKETCH:

RESEARCH: P = IA, A = lw, r / .p I c= SIMPLIFY: (a) P = Ilw (b) r /p I c= CALCULATE: (a) ( )( )( )3 2 51.00 10 W/m 30.0 m 10.0 m 3.00 10 WP = ⋅ = ⋅

(b) 3 2

6 2r 8

1.00 10 W/m 3.33 10 N/m3.00 10 m/s

p − ⋅= = ⋅ ⋅

ROUND: (a) 53.00 10 WP = ⋅ (b) 6 2

r 3.33 10 N/mp −= ⋅ DOUBLE-CHECK: The radiation force is small, as expected, while the amount of power incident on the roof is fairly large. This large source of energy can be harnessed by the use of solar panels.

31.69. THINK: The laser will apply a force to the particle. Assume the particle starts from rest and 1% of the laser intensity is reflected from it. The laser applies a force to a known mass, but only the acceleration can be calculated. Without a time there is no way to calculate the speed. Acceleration does provide a sense of "how fast" something moves. It just has units of m/s per second. Use the values: 500./192 TW,P =

2.00 mm,d = 3 3 32.00 g/cm 2.00 10 kg/m ,ρ = = ⋅ and 0.0100.ε =


1156

SKETCH:

RESEARCH: ,rF ma p A= = 2 ,rIp

cε= ,PI

A=

3 343 2 6

d dm π ρπ ρ = =

SIMPLIFY: 3 3

2 2 2 2 6 12rp A A I A P P P Pam m c mc A mc c d d c

ε ε ε ε επ ρ π ρ

= = = = = =

CALCULATE: ( )( )

( ) ( )( )12

7 233 3 3 8

12 0.0100 500. 10 W/1922.072 10 m/s

2.00 10 m 2.00 10 kg/m 3.00 10 m/sa

π −

⋅= = ⋅

⋅ ⋅ ⋅

ROUND: To three significant figures, 7 22.07 10 m/s .a = ⋅ DOUBLE-CHECK: This is a reasonable result only for very short times. When the particle approaches the speed of light, relativistic effects will need to be taken into account. For example, the length of time it takes to reach 75% the speed of light is:

( )8

7 2

0.75 3.00 10 m/s0.750.75 20 s.1 10 m/s

cv c at ta

⋅= = = = =

⋅

The answer for this problem is only valid well before this time.

31.70. THINK: Power will be dissipated out of the curved surface of the resistor. The result should be 2 .P i R= This can be derived by determining the expressions for the electric and magnetic fields at the surface of the resistor and using the Poynting vector definition of power/area. SKETCH:

RESEARCH: 0

,EBSμ

= ,P SA= 2 ,A rLπ= ,V iR= ,VEL

= 0

2i

Br

μπ

=

SIMPLIFY: ( ) ( ) 20

0 0

222

iEB rL VP SA rL Vi iR i i RL r

μππμ μ π

= = = = = =

CALCULATE: Not applicable. ROUND: Not applicable. DOUBLE-CHECK: This is consistent with previous results for the power dissipated by a resistor.

31.71. THINK: The direction of the Poynting vector is the direction in which energy is transported, in this case radially away from the antenna. From the definition of the Poynting vector as power/area, the magnitude can be calculated given the power and the radial distance. Once the Poynting vector is known, the electric field can be determined. 43 10 W,P = ⋅ d = 12 km.


1157

SKETCH:

RESEARCH: 0

1 ,S E Bμ

= ×

2

0 0

,P EB ES SA cμ μ

= = = =

rms ,2

EE = 24 ,A dπ= rms rmsF qE=

SIMPLIFY: 24PSdπ

=

(a) Not applicable. (b) 0 rms 0 / 2,E cS E cSμ μ= = 0 / 2rms eF q cSμ=

CALCULATE: ( )

45 2

23

3 10 W 1.658 10 W/m4 12 10 m

Sπ

−⋅= = ⋅⋅

5 2ˆ1.658 10 W/mS r−= ⋅

(radially away from the antenna). (a) The radio waves will in general be polarized. The detailed nature of the polarization will depend on the method in which electrons are accelerated up and down the antenna.

(b) ( ) ( )( )( )19 7 8 5 2 21rms 1.602 10 C 4 10 T m/A 3.00 10 m/s 1.658 10 W/m / 2 8.9558 10 NF π− − − −= ⋅ ⋅ ⋅ ⋅ = ⋅

ROUND: 5 21.66 10 W/mS −= ⋅ (a) Not applicable. (b) 21

rms 8.96 10 NF −= ⋅ DOUBLE-CHECK: Dimensional analysis shows the results all have the correct units.

31.72. THINK: To answer these questions, use the classical equation for the momentum and angular momentum, and use the quantum equation for the energy. SKETCH:

RESEARCH: ,U EPc c

= = U ELω ω

= =

SIMPLIFY:

(a) EPc c

ω= =

From the dispersion relation of light: / .k cω=

Then the vector momentum can be written:

.P k P k= =

(b) EL Lωω ω

= = = =

The angular momentum of a photon is constant! For the direction, note that the angular momentum, ,L

is perpendicular to the linear momentum, :P

.L r P= × Therefore, / .L k k= ±

(c) The spin quantum number is given by: / 1.s L s= = ±


1158

CALCULATE: Not applicable. ROUND: Not applicable. DOUBLE-CHECK: These results are classically unreasonable. However, they are correct nonetheless. They all stem from the purely quantum fact that light is quantized in units of . For light of frequency, ,ω

ω is the smallest amount of energy the electromagnetic wave can be measured to have.

31.73. THINK: To determine how long it takes the ice to melt, first determine how much total energy is required to melt the ice cube, and then determine the intensity of the microwaves at the location of the ice cube. To determine the number of photons hitting the ice per second, the energy of one photon must be calculated and compared to the total radiation power incident on the ice. 0 250. W,P = l = 2.00 cm, d = 10.0 cm,

30.960 g/cm ,ρ = 10.0 cm.λ = The fraction of incident light absorbed by ice is 0.100.ε = SKETCH:

RESEARCH: The energy required to melt the ice is f 333.55 J/g.c = The intensity of light at the cube is

20 / 4 .I P dπ= The radiation power incident on the cube is 2 .Il The power absorbed by the cube is

2 / .P Il E tε= = The mass of the ice is given by 3 .m lρ= The energy of one photon is given by

ph / .E hf hc λ= =

SIMPLIFY: The energy required to melt the ice is given by 3m f f .E mc l cρ= = The power absorbed by the

cube is given by:

2 2 02 .

4P EP Il l

tdε ε

π

= = =

The time required to melt the cube can be determined as follows: 2 2

02 2

0

4 ,4

l PE Edtt d l P

ε ππ ε

= = 3m fE E l cρ= =

3 2 2f f

200

4 4.

l c d ld ctPl P

πρ πρεε

= =

The total power incident on the cube is given by: 2 2 20 / 4 J/s.Il P l d xπ= = x J/s is supplied by N photons of

energy, ph ,E every second:

( )ph/ J/s J J ,

sNhc Nhc NhcNE x x N x

hcλ λ

λ λ= = = =

2 20 0

2 2J s s.4 4P l P l

x Nd hd c

λπ π

= =

CALCULATE: ( )( )( ) ( )

( )

234

4 0.960 g/cm 2.00 cm 10.0 cm 333.55 J/3.219 10 s 8.942 h

0.100 250. J/s

gt

π= = ⋅ =

( )( ) ( )( )( ) ( )

2

23234 10

250. J/s 2.00 cm 10.0 cm s4.003 10

4 6.626 10 J s 10.0 cm 3.00 10 cm/sN

π −= = ⋅

⋅ ⋅

ROUND: t = 8.94 h (or 8 hours 57 minutes), 234.00 10N = ⋅ DOUBLE-CHECK: The number of photons per second, N, is reasonable. The time is correct, although a real microwave will work much faster. This is because a real microwave is not a single point source. Also, a microwave has shielding which serves to reflect all waves hitting the walls, which keeps the intensity of the radiation high.


1159

31.74. THINK: The Poynting vector is directed along the propagation axis, ˆ.z Its magnitude is given by the laser power and the beam’s cross-sectional area. The electric field is along the polarization direction, and the magnetic field is directed in the y-direction, perpendicular to both E

and .S

P = 6.00 kW, 10.6 μmλ =

and 100.0 μm.d = Note that due to the properties of electromagnetic waves the relative phase between the magnetic field and electric field is 0 degrees. SKETCH:

RESEARCH: ( )0ˆsin ,E E kz t xω= −

2 / ,k π λ= ( )0

ˆsin ,B B kz t yω= −

/ ,k cω = 0 0

0

2 ,E B PS

Aμ= = where P

is the average power, 2

,2dA π =

and 0

0

.E

cB

=

SIMPLIFY: 2

2 8P PSA dπ

= = and ( ) 2

0 0 0

0 0

/.

E E c ES

cμ μ= = Set these expressions equal to each other to get:

20 0

02 20

88 ,E P cP E

cd dμ

μπ π= = 0 0

0 2

8.

E PB

c d cμ

π= =

Then, with 2 ,k πλ

= 2 :cck πωλ

= = 02

8 2 2 ˆsinP c z ctE x

dμ π π

λ λπ = −

and 0

2

8 2 2 ˆsin .P z ctB yd cμ π π

λ λπ = −


( )3 7 8

70 24

8 6.00 10 W 4 10 T m/A 3.00 10 m/s2.400 10 V/m

1.000 10 mE

π

π

−

−

⋅ ⋅ ⋅= = ⋅

⋅

72

0 8

2.400 10 V/m 8 10 T,3 10 m/s

B −⋅= = ⋅⋅

5 16

2 2 5.928 10 m10.6 10 m

π πλ

−−= = ⋅

⋅

( )5 1 8 14 12 5.928 10 m 3 10 m/s 1.778 10 scπλ

− −= ⋅ ⋅ = ⋅

ROUND: ( ) ( )7 5 1 14 1 ˆ2.40 10 V/m sin 5.93 10 m 1.78 10 sE z t x− −= ⋅ ⋅ − ⋅

( ) ( )2 5 1 14 1 ˆ8.00 10 T sin 5.93 10 m 1.78 10 sB z t y− − −= ⋅ ⋅ − ⋅

DOUBLE-CHECK: Check that 0

0

:E

cB

= 7

82

2.40 10 V/m 3.00 10 m/s.8.00 10 T−

⋅ = ⋅⋅

Also, it is necessary that :ckω =

14 18

5 1

1.78 10 s 3.00 10 m/s.5.93 10 m

−

−

⋅ = ⋅⋅

These results are reasonable.


1160

Chapter 32: Geometric Optics

In-Class Exercises

32.1. c 32.2. b 32.3. c 32.4. c 32.5. d Multiple Choice

32.1. c 32.2. b 32.3. a 32.4. c 32.5. b 32.6. a

Questions

32.7. In a step-index fiber, there is a discontinuity of the index of refraction at the core-cladding interface. Consequently, light will undergo total internal reflection at the core-cladding interface, and will propagate through the fiber in a zigzag path. By contrast, in a graded-index fiber, the refractive index changes gradually in moving from core to cladding. A light ray entering the core of a graded-index fiber will continuously change its direction as the refractive index continuously changes. Consider light incident at the core at an angle of i .θ The incident ray moves from the core, of refractive index core ,n to a refractive index of core ,n n− Δ where nΔ is a small change in the index of refraction. For subsequent refractions, assume that the index of refraction changes by the same amount and the new incident angle is the previous final angle. The first three refractions are given by:

( ) (1) (1) corecore i core f f i

core

sin sin sin sinn

n n nn n

= − Δ = − Δ

θ θ θ θ

( ) ( )(1) (2) (2) (1)corecore f core f f f

core

sin 2 sin sin sin2

n nn n n n

n n − Δ

− Δ = − Δ = − Δ θ θ θ θ

( ) ( )(2) (3) (3) (2)corecore f core f f f

core

22 sin 3 sin sin sin .

3n n

n n n nn n

− Δ− Δ = − Δ = − Δ

θ θ θ θ

Combining these equations gives:

(3) (2) (1)core core coref f f

core core core

core core core corei i

core core core core

2 2sin sin sin

3 3 22

sin sin .3 2 3

n n n n n nn n n n n nn n n n n nn n n n n n n n

− Δ − Δ − Δ= = − Δ − Δ − Δ − Δ − Δ

= = − Δ − Δ − Δ − Δ

θ θ θ

θ θ

For N refractions,

( ) coref i

core

sin sin .N nn N n

= − Δ

θ θ

As the light approaches the cladding, core cladding ,n N n n− Δ → so

coref i

cladding

sin sin .n

n

=

θ θ

In the limit, Snell’s Law is recovered. If coren and claddingn are chosen correctly, then total internal reflection

will occur. The difference between the two fibers is that for a step-index fiber, the reflection occurs instantly (zigzag path), but for a graded-index fiber, the reflection occurs gradually (sinusoidal path) since the light gets refracted along the way.


1161

32.8. Ray A will leave the plexiglass at some angle, .φ This ray extrapolated back will reconnect with ray B, which does not get refracted since it is normal to the interface. The location where the rays reconnect is where the image will appear to form.

From the diagram, two equations are apparent: tand D= θ and tan .d D′= φ

Equating these gives, tantan tan .tan

D D D D′ ′= = θθ φφ

Therefore, the apparent height of the text is: tan1 .tan

y D D D ′= − = −

θφ

The angle, ,φ is given by Snell’s Law:

( )( ) ( )1 11

1 22

1.51sin sin sin sin sin sin 25.0 39.65 .

1.00nn nn

− − = = = ° = °

θ φ φ θ

Therefore, the height is:

( ) ( )( )

tan 25.02.00 cm 1 0.875 cm.

tan 39.65y

°= − = °

32.9. Due to spherical aberration, light rays a distance d from the center axis of a mirror of curvature R will cross the optical axis a distance y from the center of the mirror, where y is approximated as:

2

21 .2 2R dy

R

≈ −

When there is no spherical aberration, the image is formed at the focal point; that is, / 2.y f R= =

Therefore, to reduce the spherical aberration produced by the mirror, ( )2 2/ 2 0.d R → Since the mirror

cannot be changed (R cannot change), the only way to reduce the spherical aberration is to make the height of the object, d, small.

32.10. (a) If one were to look directly down to the bottom of the pool, all the light rays will be perpendicular to the surface and so will not be refracted. Therefore, the pool will still appear to be 4 feet deep. (b) If one were to look at an angle, the cone of rays coming from a point at the bottom of the pool would refract away from the normal when they cross the water/air interface. If the rays were then extrapolated back, they would intersect at a point above the bottom of the pool. Therefore, the pool would appear to be less deep than it actually is. Figure 32.38 illustrates this for a similar situation.

32.11. When the light enters a medium, it will interact with the atoms that make up the material. Electrons will vibrate and then re-emit the light, which will then go along and interact with another electron. This effect will continue all the way through the material. The interactions with the electrons take a finite amount of time, causing delays along the way through the material. This gives the appearance that the light is traveling at a different speed.


1162

32.12. If the fiber is bent too much, then the incident angle of light on the edge of the fiber exceeds the critical angle, and light is lost, reducing signal transmission.

32.13. The drum acts like a concave or converging mirror. From Table 32.1, it can be seen that when the object moves from just inside the focal point of a converging mirror to just outside the focal point, the image changes from an upright image to an inverted image. Hence, if she starts with her finger close enough to the drum that she sees an upright image and slowly moves it away, the point at which the image flips is the focal point, and the distance from her finger to the mirror at that point is the focal length. Twice the focal length is the radius of curvature.

32.14. True. The speed of the light wave in vacuum is given by .v c f= = λ In a medium, the speed is given by / .v c n f ′= = λ Since the frequency f is constant, these equations can be solved in terms of wavelength:

/ .n′ =λ λ Therefore, the wavelength of any type of light in water 1.33n = is less than its wavelength in air 1.00n =

32.15. A ray of light incident at any angle on a corner cube of mirrors will always be reflected back at the same angle as the incident ray due to the law of reflection. The array of cubes is then necessary to provide a larger surface to strike. The smaller the corner cube, the closer to the original path the reflected path will be, minimizing any possible change in path length between the two rays. Therefore, when light is incident on the array, the reflected beam will be sent back virtually along the same path as the incident beam. The change in path length of the two beams will be insignificant compared to the large distance between the Earth and the Moon, thus still providing excellent precision in the measurement.

32.16. (a) Inside the prism, the reflected beam would hit each leg of the prism at 45°. For the prism-water interface, the critical angle is:

( )( )

2 2H O H O1 1c c

glass glass

1.333sin sin sin 61.28 .

1.520n nn n

− −

= = = = ° θ θ

Therefore, when the light strikes the legs at 45° when in water, total internal reflection will not occur and most of the light will exit the prism. Some of the light will get reflected, but its intensity will be low, rendering the prism ineffective. (b) Prisms are used for several reasons: (1) Cost; quality mirrors require expensive reflecting coatings while prisms do not. (2) Quality; total internal reflection reflects all of the light while all mirrors reflect with some loss in intensity. (3) Durability; there is no reflective coating on the prisms to corrode or degrade over time. It is only necessary to keep it clean and dry.

32.17. It the upper half of the mirror is covered then only the bottom portion of the mirror will focus the rays. The image will be at the same location, but will appear dimmer since fewer light rays will be brought to a focus.

32.18. The light entering the water will bend towards the vertical, so you will observe the light at a steeper angle than you would if you were not under water. Therefore, the Sun will appear to be higher in the sky than it actually is.

32.19. The spoon can be treated as a spherical convex mirror. Decent estimates for the radius of curvature of the spoon and object distance are 5.0 cmR = and o 15.0 cm,d = respectively. Since the spoon is a convex mirror, the radius of curvature is negative. The basic mirror equation then gives the image distance:

( ) ( )

11

io i o

1 1 1 2 2 1 2 1 2.14 cm 2.1 cm.5.0 cm 15.0 cm

dd d f R R d

−− + = = = − = − = − ≈ − −


1163

The magnification is: ( )( )

i

o

2.14 cm0.14.

15.0 cmdmd

−= − = − =

32.20. No, it is not possible to attain temperatures exceeding the temperature of the photosphere of the Sun. The second law of thermodynamics prevents this from happening. For a perfectly efficient process, the maximum attainable temperature is 6000 K.

Problems

32.21. For plane mirrors, the object distance, o 1.0 m,d = is always equal to the image distance, but the image is located behind the mirror. Therefore, the location of the image is i o 1.0 m.d d= − = −

32.22.

The distance D of the final image of the yellow hat from the lower mirror is

o 3.00 m 0.400 m 3.40 m.D d L= + = + =

32.23.

The object has an image in each mirror equidistant from the mirror and the object, so o i .d d= This means the distance between the two images is the hypotenuse of a triangle, where each leg is a length,

o i o2 .d d d+ = So, the distance between the images is

( ) ( ) ( )2 2 22o o o2 2 8 8 2 m 6 m.D d d d= + = = =

32.24. THINK: If a photon bounces once, it has 99.997% of its energy, so after n bounces it has ( )0.99997 n of its original energy left. The longest length in the cube room, sides l = 3.00 m, is the diagonal along the cube. The average distance a photon goes in the room is half the longest distance. SKETCH:


1164

RESEARCH: The maximum distance of the room is 3 .d l= The average distance is then / 2.d d =

The energy after n bounces is ( ) 00.99997 .nE E′ = The time for one bounce is / .t d cΔ = The total time is .t n t= Δ

SIMPLIFY: The number of bounces to reduce to 0.0100E′ = is:

( ) ( )( )0 0

ln 0.01000.99997 0.0100 .

ln 0.99997nE E E n′ = = =

The total time of light dissipation is: 3 .

2 2d nd lt n t n nc c c = Δ = = =

CALCULATE: ( )( )

( )( )8

3 3.00 m ln 0.01000.001329 s

ln 0.999972 3.00 10 m/st

= = ⋅

ROUND: To three significant figures, 1.33 ms.t = DOUBLE-CHECK: Such a small time would be undetectable for an average human. This is expected, considering the speed of light.

32.25. The focal length is given by .2Rf = For 25 cm,R = − the focal length is:

( )25 cm13 cm.

2 2Rf

−= = ≈ −

32.26.

For a radius of curvature of 10.0 cmR = and an object distance of o 30.0 cm,d = the image distance is:

( ) ( )

11

io i o

1 1 1 2 2 1 2 1 6.00 cm.10.0 cm 30.0 cm

dd d f R R d

−− + = = = − = − =

Therefore, the position of the image is ( ) ( )i o i 30.0 cm 6.00 cm 24.0 cm.x d d= − = − =

The magnification is i o i o/ / .m h h d d= = − For an object height of o 5.00 cm,h = the image height is:

( )( )( )

i oi o

o

6.00 cm 5.00 cm1.00 cm.

30.0 cmd h

h mhd

= = − = − = −

Since i 0,h < the image is inverted. Since i 0,d > the image is real.

32.27. For a radius of curvature of 14.0 mR = − and an object distance of o 11.0 m,d = the image distance is:

( ) ( )

11

io i o

1 1 1 2 2 1 2 1 4.28 m.14.0 m 11.0 m

dd d f R R d

−− + = = = − = − = − −

The magnification of the mirror is: ( )( )

i

o

4.28 m0.389.

11.0 mdmd

−= − = − =


1165

32.28. For a focal length of 10.0 cmf = − and an object distance of o 30.0 cm,d = the image distance is:

( ) ( )

11

io i o

1 1 1 1 1 1 1 7.50 cm.10.0 cm 30.0 cm

dd d f f d

−− + = = − = − = − −

For an object height of o 5.00 cm,h = the image height is:

( )( )( )

i oi

o

7.50 cm 5.00 cm1.25 cm.

30.0 cmd h

hd

−= − = − =

Since i 0,d < the image is virtual. Since i 0,h > the image is upright.

32.29. For an object a distance of o 2.0 md = in front of a convex mirror with magnification 0.60,m = the image distance is

ii o

o

.dm d mdd

= − = −

The focal length is: ( )( )

( )o

o i o o o

0.60 2.0 m1 1 1 1 1 1 1 3.0 m.1 0.60 1

mdm ff d d f d md md m

−= + = − = = = = −− −

32.30. THINK: The object, at o 100. cm,d = is behind the second mirror (which is located at the focal point of the first mirror of focal length 1 20.0 cm)f = of focal length 2 5.00 cm.f = Assume the second mirror is a two-way mirror so that the light rays from the object go through it, reflect from the first mirror, and then reflect from the second mirror. The reflections will continue until the final image is formed outside of both mirrors. SKETCH:

RESEARCH: The relevant equations are the mirror equation and the magnification equation:

o i

1 1 1 ,d d f

+ = i

o

.dmd

= −

SIMPLIFY: When the object is first reflected by mirror 1, the image distance and magnification are: 1

1o 1 1 1 o

1 1 1 1 1 dd d f f d

−

+ = = −

and 11

o

.dmd

= −

This image must form behind mirror 2 since mirror 2 is at the focal point of mirror 1. Therefore, the image is a virtual object for mirror 2 at a distance of ( )2 1 1 .d d f= − − The image distance and magnification are:


1166

1

32 3 2 2 2

1 1 1 1 1 dd d f f d

−

+ = = −

and 32

2

.d

md

= −

The total magnification of the system is then 1 31 2

o 2

.d d

m m md d

= =

CALCULATE: ( ) ( )

1

11 1 25.0 cm,

20.0 cm 100. cmd

−

= − =

( ) ( )( )2 25.0 cm 20.0 cm 5.00 cmd = − − = −

The final image distance and total magnification is:

( ) ( )

1

31 1 2.50 cm

5.00 cm 5.00 cmd

−

= − = − and

( )( )( )( )

25.0 cm 2.50 cm0.125

100. cm 5.00 cmm = − =

−

ROUND: Remaining at 3 significant figures, the final image location is between the two mirrors, a distance of 3 2.50 cmd = from mirror 2. The total magnification is 0.125.m = DOUBLE-CHECK: For two plane mirrors facing each other, an infinite number of virtual images are formed. For two concave mirrors, it is expected that a real image must form after some number of reflections.

32.31. THINK: An arbitrary point on the elliptical mirror can be chosen: ( ), .p x y+ − Two ray vectors exist that

point from p to ( ), 0 ,c± where 2 2 .c a b= − The normal line, which is perpendicular to the surface of the elliptical mirror, can be determined. The dot product can be used to determine the angle between the two ray vectors and the normal vector. If perfect reflection occurs, the angles between the normal vector and the two ray vectors should be the same. SKETCH:

RESEARCH: The two ray vectors are defined as ( ) ˆ ˆu x c x yy= − − + and ( ) ˆ ˆ,v x c x yy= − + + and they

make an angles of uθ and vθ with .N

The normal vector to a surface is defined as:

( ) ( ), ,ˆ ˆ.f x y f x y

N x yx y

∂ ∂= +

∂ ∂

SIMPLIFY: First, determine the normal vector: ( ) ( ) 2 22 2

2 2 2 2 2 2

, , 22ˆ ˆ ˆ ˆ ˆ ˆ.f x y f x y y y yx x xN x y x y x y

x y x ya b a b a b∂ ∂ ∂ ∂= + = + + + = + ∂ ∂ ∂ ∂

For point ( ),x y− :

2 2

22 ˆ ˆ.yxN x ya b

= −

The unit vectors of u and v are given by:


1167

( )( )2 2

ˆ ˆˆ x c x yyuu

u x c y

− − += =

− +

and

( )( )2 2

ˆ ˆˆ .

x c x yyvvv x c y

− + += =

+ +

The dot product of N

with the two unit vectors gives ˆ ˆ cos cosu uN u N u Nθ θ= =

and ˆ ˆ cos cos .v vN v N v Nθ θ= =

It is known from the law of reflection that the two angles must be equal. Therefore, if ˆ ˆN u N v= is

shown, then the proof will be complete.

( )( )( )

( )

22 22

2 2 22 22

2 2 2 22

22 ˆ ˆ ˆ ˆ 4ˆ

2

yx xcyx x y x c x yya a ba b

N ux xc c yx c y

− + −− − − + = =

− + +− +

Recall that 2 22 2

2 2 2 21 1 .y yx xa b b a

+ = = − Also, ( )2 2

2 2 2 22 21 1 .x xy b a c

a a

= − = − −

Substitution gives:

( )( )

2 2 2

2 422

2 222 2 2 2 22 2 2 2

22

2 22

2 2

2 2 22

2

24 14 1ˆ

22 1

4 24 .

2

xc x cxca aaN u

x cx x xc c a c xx xc c a caa

x ca xca a

x c aa xca

− +− + = =

− + + − − +− + + − −

− +

= =− +

( )( )( )

( )

22 22

2 2 22 22

2 2 2 22

22 ˆ ˆ ˆ ˆ 4ˆ

2

yx xcyx x y x c x yya a ba b

N vx xc c yx c y

− − −− − + + = =

+ + ++ +

Making the same substitutions as above gives:

( )( )

2 2 2

2 422

2 222 2 2 2 22 2 2 2

22

2 22

2 2

2 2 22

2

24 14 1ˆ

22 1

4 24 .

2

xc x cxca aaN v

x cx x xc c a c xx xc c a caa

x ca xca a

x c aa xca

+ +− − = =

+ + + − − ++ + + − −

+ +

= =+ +

2

4ˆ ˆ ,N u N va

= = u v u v cos cos θ θ θ θ = = , since the angles are both in the same quadrant.

CALCULATE: Not applicable. ROUND: Not applicable. DOUBLE-CHECK: The above derivation proves that, given the law of reflection, the vector that goes through a focal point will be reflected through the other focal point of an elliptical mirror.

32.32. The velocity of light in a medium with refractive index n is / .v c n= For crown glass with index of refraction of 1.52,n = the velocity is:

( )( )

88

3.00 10 m/s1.97 10 m/s.

1.52cvn

⋅= = = ⋅


1168

32.33. The critical angle is given by c 2 1sin / .n nθ = The critical angles of the optical fiber in air, water and oil are:

1c,air

1.000sin 421.5

θ − = = °,

1c,water

1.333sin 631.5

θ − = = °

and 1c,oil

1.5sin 90 .1.5

θ − = = °

32.34. The helium-neon laser light of wavelength vac 632.8 nmλ = is in water with an index of refraction of 1.333.n =

(a) The velocity is:

( )( )

88

2.998 10 m/s2.249 10 m/s.

1.333cvn

⋅= = = ⋅

(b) The frequency remains unchanged (it is independent of the medium), so using values in a vacuum gives:

( )( )

814

vac 9vac

2.998 10 m/s 4.738 10 Hz.

632.8 10 mcc f fλ

λ −

⋅= = = = ⋅

⋅

(c) The wavelength is: ( )

( )vac 632.8 nm

474.7 nm.1.333n

λλ = = =

(d) Technically the color does not change because it is the frequency of light that our eyes receive and interpret. Therefore, the color is still that of 144.738 10 Hz⋅ on the spectrum (red-like).

32.35. To get fully polarized light, the incident light must strike the water-plate glass interface at the Brewster angle:

1 12B

1

1.73tan tan 52.4 .1.33

nn

θ − − = = = °

32.36. THINK: Regardless of the angle of incidence, light rays from the air will enter the water. However, some light rays coming from underwater will hit the surface at or above the critical angle and will undergo total internal reflection, creating a virtual mirror. For light incident from water to air, the indices of refraction are 1 1.333n = and 2 1.000.n = This means that the only clear window is the circle created by the cone, with the angle from the vertical equal to the critical angle of the water/air interface. The tip of the cone is

2.00 mh = below the surface of the water. SKETCH:

RESEARCH: The critical angle is defined as c 2 1sin / .n nθ = The radius of the circle is ctan .r h θ= The diameter of the window is 2 .d r= SIMPLIFY: The diameter of the window is:

1 2c

1

2 2 tan 2 tan sin .nd r h hn

θ − = = =


1169

CALCULATE: ( ) 1 1.0002 2.00 m tan sin 4.538 m1.333

d − = =

ROUND: To 3 significant figures, 4.54 m.d = DOUBLE-CHECK: This value is reasonable considering the depth of the observer.

32.37. THINK: Since the normal line of the first surface bisects the opposite angle, the refracted ray must hit the other angled surface. Simple geometry must be utilized to determine all the angles involved. The index of refraction of air and the prism are a 1.00n = and p 1.23,n = respectively.

SKETCH:

RESEARCH: Since the incident beam is parallel to the base, the incident angle is i 30θ = °. Snell’s Law is used to determine refracted angles: i i j jsin sin .n n=θ θ

SIMPLIFY: At the first interface:

1 aa i p 1 1 i

p

sin sin sin sin .n

n nn

θ θ θ θ−

= =

Based on the geometry shown in the figure above, 1 190= ° −φ θ and ( )2 1180 60 .= ° − ° +φ φ Therefore,

( )2 1 1 1120 120 90 30 .= ° − = ° − ° − = ° +φ φ θ θ Also, 2 290 .= ° −θ φ Therefore,

( ) 1 a2 1 i

p

90 30 60 sin sin .nn

θ θ θ−

= ° − ° + = ° −

At the second interface:

p p1 1 1 ap 2 a 3 3 2 i

a a p

sin sin sin sin sin sin 60 sin sin .n n n

n nn n n

θ θ θ θ θ− − − = = = ° −

The change in direction is equal to the sum of the changes in angle at each interface: ( ) ( ) ( )f i 1 3 2 i 1 3 1 i 3

p1 1 af i i

a p

60 60 ,

60 sin sin 60 sin sin .n nn n

θ θ θ θ θ θ θ θ θ θ θ

θ θ θ− −

= − + − = − + − ° − = − ° +

= − ° + ° −


( )( ) ( )1 1

f

1.23 1.0030 60 sin sin 60 sin sin 30 16.322

1.00 1.23θ − −

= ° − ° + ° − ° = °

ROUND: Rounding to three significant figures, f 16.3= °.θ DOUBLE-CHECK: The change in direction depends on the initial incident angle, the refractive index of air and the refractive index of the prism, as expected. This is a reasonable angle for the ray of light to be deflected after going through a prism.


1170

32.38. THINK: The light is refracted as it crosses the air-glass interface and the glass-air interface. The air and the glass have a refractive index of a 1.00n = and g 1.55,n = respectively.

SKETCH:

RESEARCH: The angle of refraction at each interface can be determined using Snell’s Law: 1 1 2 2sin sin .n nθ θ=

SIMPLIFY: At the first interface,

1 aa i g 1 1 i

g

sin sin sin sin .n

n nn

θ θ θ θ−

= =

Based on the geometry of the glass block,

1 a2 1 i

g

30.0 30.0 sin sin .nn

θ θ θ−

= ° − = ° −

At the second interface,

g g1 1 1 ag 2 a 3 3 2 i

a a g

sin sin sin sin sin sin 30.0 sin sin .n n n

n nn n n

θ θ θ θ θ− − − = = = ° −

The angle from the horizontal is

g1 1 aBT 3 i

a g

30.0 30.0 sin sin 30.0 sin sin .n nn n

θ θ θ− − = ° − = ° − ° −

CALCULATE: ( )( )

( )( ) ( )1 1

BT

1.55 1.0030.0 sin sin 30.0 sin sin 20.0 2.632

1.00 1.55θ − −

= ° − ° − ° = °

ROUND: Rounding to three significant figures, BT 2.63θ = °. DOUBLE-CHECK: This result is reasonable.

32.39. THINK: The maximum incident angle max 14.033α = ° corresponds to the light ray that reaches the core-cladding interface at an angle equal to the critical angle. Knowing the critical angle and the refractive index of core 1.48,n = the index of refraction of the cladding can be determined using Snell’s Law. SKETCH:


1171

RESEARCH: Snell’s Law is given by 1 1 2 2sin sin ,n nθ θ= and the critical angle is given by c 2 1sin / .n nθ =

SIMPLIFY: By Snell’s Law, 1 aa max core max max max

core

sin sin sin sinn

n nn

α β β α− = =

. The critical angle is

1 ac max max

core

90 90 sin sin .n

nθ β α−

= ° − = ° −

At the core-cladding interface:

1 acladding core c core max

core

sin sin 90 sin sin .n

n n nn

θ α− = = ° −

The percent difference between the index of refraction of the core and the index of refraction of the cladding is:

( ) ( )cladding 1 amax

core core

% difference 1 100% 1 sin 90 sin sin 100% .n n

n nα−

= − = − ° −

CALCULATE: ( )( ) ( ) ( )1 1.00

% difference 1 sin 90 sin sin 14.033 100% 1.3513%1.48

− = − ° − ° =

ROUND: To three significant figures, the percent difference between the index of refraction of the core and the index of refraction of the cladding is 1.35%. DOUBLE-CHECK: This result is reasonable. One would expect the difference to be small.

32.40. THINK: The colors of a rainbow occur because white light from the sun is refracted into its component colors by water droplets in the atmosphere. A rainbow is observed at an angle of 42° from the direction of the sunlight, because at this angle, the intensity of the various colors is greatest. This occurs because, for angles less than 42°, the separation of the colors is less pronounced and rays merge to form white light. The angle of 42° represents the maximum angle at which light rays exit a spherical water droplet. SKETCH:

RESEARCH: The path of the ray inside the water droplet can be determined using Snell’s Law, the law of reflection, i r ,θ θ= and the geometry of circles and triangles.

SIMPLIFY: The angle, 0 ,θ is given by Snell’s Law: 1 aa i w 0 0 i

w

sin sin sin sin .n

n nn

θ θ θ θ− = =

Due to

the geometry of a circle, the incident angle at point B is equal to 0 .θ This is true of the incident angle at C as well. Therefore, the refracted ray leaving at C is iθ by Snell’s Law. The angle θ ′ is the change in direction of the light ray. For the two refractions and the one reflection, the total change in direction is:

( ) ( ) 1 a0 i 0 0 i 0 i i i

w

2 4 2 4sin sin 2 .nn

θ θ θ θ θ θ θ θ θ θ− ′ = − + + − = − = −


1172

The maximum value of θ ′ occurs when i/ 0.d dθ θ′ = The following derivatives can be found in a table of derivatives:

( )1

2

1sin1

d duudx dxu

− =−

and sin cos .d x xdx

=

The value of iθ for the maximum value of θ ′ is given by: 2

2a a ai i i2

i w w w2a

iw

2 2 22 2 2 2a a w

i i i iw w a

4 cos 2 0 2 cos 1 sin

1 sin

4 cos 1 sin 4cos sin .

n n ndd n n nn

n

n n nn n n

θ θ θ θθ

θ

θ θ θ θ

′= − = = −

−

= − + =

Using the trigonometric identity 2 2sin 1 cosx x= − gives:

( )2 2 2 2

2 2 2w w w ai i i 2

a a a

2 21 w a

i 2a

4cos 1 cos 3cos 1

cos3

n n n nn n n

n nn

θ θ θ

θ −

−+ − = = − =

− =


2 21

i 2

1.333 1.000cos 59.4105

3 1.000θ −

− = = °

( )( ) ( ) ( )1

max

1.0004sin sin 59.4105 2 59.4105 42.078

1.333θ −

′ = ° − ° = °

ROUND: Rounding to four significant figures, the maximum angle is 42.08 .° Therefore, the observed angle for a rainbow is 42.08° from the direction of the sunlight. DOUBLE-CHECK: This is the angle that the question asked to derive.

32.41. THINK: Fermat’s Principle states that the path taken by a ray between two points in space is the path that takes the least amount of time. The law of reflection can be found by using this principle. To accomplish this, determine the time it takes for a ray to travel from one point to another by hitting the mirror. Using calculus, this time can be minimized and the law of reflection is recovered. SKETCH:

RESEARCH: The time it takes the ray to reach the mirror is / .t d v= To minimize the time, set / 0.dt dx =

SIMPLIFY: ( ) ( )22 2 21 21 2

1d d nt d d h x h l xv v v c

= + = + = + + + −

The path of least time is determined from:


1173

( ) ( ) ( )( )

( )( )

( )i r2 2 2 2 2 22 2

i r i r

1/ 2 2 1/ 2 20 sin sin

sin sin 0

x l x l xdt n n x ndx c c ch x h xh l x h l x

θ θ

θ θ θ θ

− − = = − = − = − + ++ − + − − = =

CALCULATE: Not applicable. ROUND: Not applicable. DOUBLE-CHECK: The law of reflection was recovered using Fermat’s Principle.

32.42. THINK: Fermat’s Principle states that the path taken by a ray between two points in space is the path that takes the least amount of time. Snell’s Law can be determined by using this principle. To accomplish this, determine the time it takes for a ray to travel from one point to another by traveling through both materials. Using calculus, this time can be minimized and the Snell’s Law is recovered. SKETCH:

RESEARCH: The time it takes the ray to reach the point is / .t d v= To minimize the time, set / 0.dt dx =

SIMPLIFY: ( )22 2 21 2 1 2 1 21 2

1 2

/ 4 / 4d d n n n nt d d x D h x Dv v c c c c

= + = + = + + − +

The path of least time is determined from:

( ) ( ) ( )( )

( )( )

1 2 1 2

2 2 2 2 2 22 2

1 21 2 1 1 2 2

1/ 2 2 1/ 2 20

/ 4 / 4/ 4 / 4

sin sin 0 sin sin

x h x h xn n n ndt xdx c c c cx D x Dh x D h x D

n n n nc c

θ θ θ θ

− − = = − = − + + − + − +

− = =

CALCULATE: Not applicable. ROUND: Not applicable. DOUBLE-CHECK: Fermat’s Principle was used to derive an equation involving the indices of refraction and angles from the horizontal, as desired. The resulting equation is Snell’s Law.

Additional Problems

32.43. (a) The image distance is 50 cm behind the mirror. (b) The image has the same height, h = 2.0 m. (c) The image is upright. (d) The image is virtual.

32.44. (a) The frequency of the ray does not change in the medium, so:

( )( )

814

7air

3.00 10 m/s4.29 10 Hz.

7.00 10 mcf

λ −

⋅= = = ⋅

⋅


1174

(b) The speed inside the liquid is: ( )

( )8

8

2

3.00 10 m/s1.84 10 m/s.

1.63cv

n⋅

= = = ⋅

(c) The wavelength of the refracted ray is: ( )

( )air

2 air 2

700. nm1 429 nm./ 1.63

v cf n c n

λλλ

= = = = =

32.45. For the image to be twice the size of the object, the magnification is:

ii o

o

2 2 .dm d dd

= = = ±

The spherical mirror equation is:

( ) ( )o i o o

oo

1 1 1 2 1 1 2 2

2 1 2 122 4

d d f R d d RR

dd R

+ = = ± =

± ±= =

The object can be placed at:

( )o3 3 20.0 cm 15.0 cm4 4

d R= = = or ( )

o

20.0 cm5.00 cm,

4 4Rd = = =

to produce an image that is twice the size of the object. If the object is placed at 15.0 cm, the image distance will be ( )i 2 15.0 cm 30.0 cm.d = = Since i 0,d > this image will be real. If the object is placed at

5.00 cm, the image distance will be ( )i 2 5.00 cm 10.0 cm.d = − = − Since i 0,d < this image will be virtual.

32.46. The critical angle is given by c 2 1sin / .n nθ = Thus, the critical angle for a water-air interface is:

1 12c

1

1.000sin sin 48.611.333

nn

θ − − = = = °.

32.47.

The reflected ray has the same angle to the normal as the incident ray. The refracted ray has an angle given by Snell’s Law:

( )( )

1 1ii i 0 0 0 i

0

1.000sin sin sin sin sin sin30.0 22.0

1.333nn nn

θ θ θ θ− − = = = ° = °.

The angle between the reflected and refracted rays is v r 0180.0 180.0 30.0 22.0 128.0 .θ θ θ= ° − − = ° − ° − ° = °

32.48. The focal point of the ornament is / 2 / 4.f R d= = By convention, a convex mirror has a negative value for R, so d is negative. Using the mirror equation, the image distance is:


1175

1

ii o o o

1 1 1 4 1 4 1 .dd f d d d d d

−

= − = − = −

Saint Nicholas will see his reflection at:

( ) ( )

1

i4 1 1.97 cm.

8.00 cm 156 cmd

−

= − = − −

The image is virtual since i 0.d <

32.49. The critical angle is given by:

12 2c c

1 1

sin sin .n nn n

θ θ − = =

The critical angle for the diamond-air interface is:

1c, a

1.000sin 24.44 .2.417

θ − = = °

The critical angle for the diamond-water interface is:

1c, w

1.333sin 33.47 .2.417

θ − = = °

Therefore, the critical angle in water is 9.03° greater than the critical angle in air.

32.50. Since / 2,f R= Table 32.1 shows that

(a) for o ,d R> the image is real,

(b) for o/ 2 ,R d R< < the image is real,

(c) and for o / 2,d R< the image is virtual.

32.51. Since the incident angle is equal to the reflected angle, 1 40.0 .θ = ° The refracted angle θ is given by Snell’s Law:

( )( ) ( )1 11

1 1 2 12

1.000sin sin sin sin sin sin 40.0 28.9 .

1.333nn nn

θ θ θ θ− − = = = ° = °

32.52. THINK: The object is moved around from one point to another. Using the magnification of the two points and the change in the image distance, the focal point of the mirror and the change in the distance of object can be determined. The magnification of the image at the first position is 2m = and the magnification of the image at the second position is 3.m′ = The difference between the image distances is

i i i 75 cm.d d d′Δ = − = SKETCH:

RESEARCH: The magnification is given by i o/ .m d d= The mirror equation is: o i

1 1 1 .d d f

+ =


1176

SIMPLIFY: The object distances are o i /d d m= and o i / .d d m′ ′ ′= The mirror equation gives:

( ) ( )io i i i i

11 1 1 1 1 ,mm d m f

f d d d d d+

= + = + = = +

( ) ( )io i i i i

11 1 1 1 1 .mm d m f

f d d d d d′ +′ ′ ′= + = + = = +

′ ′ ′ ′ ′

The focal length is given by

( ) ( ) ( ) ( )i

i i i 1 1 .d

d d d m f m f m m f fm m

Δ′ ′ ′Δ = − = + − + = − =′ −

The change in the object distance is: ( ) ( ) ( ) ( )

( )i i

o o o i

1 1.

m m f m m f m m m md dd d d f dm m mm mm m m mm

′ ′ ′ ′+ − + − −′′Δ = − = − = = = Δ

′ ′ ′ ′ ′−

CALCULATE: ( )( )75 cm

75 cm3 2

f = =−

( )( )( )( ) ( )o

2 375 cm 12.5 cm

3 2 2 3d

−Δ = = −

−

ROUND: To two significant figures, the focal length of the mirror is 75 cmf = and the object was moved

o 13 cm.dΔ = DOUBLE-CHECK: Since the image is larger after it is moved, the object should be moved towards the mirror. This is indicated by the negative value for the change in the object distance.

32.53. THINK: The rays of light from the point are refracted before they reach the person, according to Snell’s Law. Because the index of refraction of air is less than that of water, the image appears shallower. The point is 3.00 md = from the surface and 2.00 mw = from the edge of the pool. SKETCH:

RESEARCH: The angle of the ray is given by Snell’s Law: w w a asin sin .n n=θ θ The triangles also relate the angles to the lengths:

w 2 2sin w

w d=

+θ and a 2 2

sin .ww h

=+

θ

SIMPLIFY: Combining the above equations gives: w a

w w a a2 2 2 2sin sin .

n w n wn n

w d w h= = =

+ +θ θ

Solving for the apparent depth gives:


1177

( ) ( )( ) ( )

2 2 2 2 2 2 2 2 2 2w a w a

2 2 2 2 2 2 2 2 2 2 2 2w a w a a w a

w

1 .

n w h n w d n w h n w d

n h n n w n d h n n w n dn

+ = + + = +

= − + = − +

CALCULATE: ( ) ( ) ( )( )( ) ( ) ( )2 2 2 2 21 1 1.3 2.00 m 1 3.00 m 1.92 m1.3

h = − + =

ROUND: Remaining at 3 significant figures, the apparent depth of the pool is 1.92 m.h = DOUBLE-CHECK: The apparent depth is less than the true depth of the pool, as expected.

32.54. THINK: For the smallest incident angle, total internal reflection at the surface occurs at the critical angle. Snell’s Law and the geometry of the prism can be used to find this incident angle. SKETCH:

RESEARCH: Snell’s Law at the air-glass interface is a i g 2sin sinn nθ θ= and the critical angle at the glass-

air interface is given by c a gsin / .n nθ = The angles are related by c 2 2 c90 70 20 .θ θ θ θ° = ° + − = − °

SIMPLIFY: The critical angle is given by ( )1c a gsin / .n nθ −= Therefore, ( )1

2 a gsin / 20 .n nθ −= − ° The

incident angle is given by:

g g g1 1 1a ai 2 i

a a g a g

sin sin sin sin 20 sin sin sin 20 .n n nn nn n n n n

θ θ θ− − − = = − ° = − °

CALCULATE: ( )( )

( )( )

1 1i

1.5 1.0sin sin sin 20 33.87

1.0 1.5θ − −

= − ° = °

ROUND: The angle 70° was given in a geometric figure, so treat it as having two significant figures. To two significant figures, the smallest incident angle is i 34 .θ = ° DOUBLE-CHECK: This result is reasonable.

32.55. (a) Time reversal leaves the charge and electric field the same, but reverses the current and magnetic field. The time reversal solution is obtained with:

( ) ( ),t tρ ρ→ − ( ) ( ),j t j t→ − −

( ) ( ),E t E t→ −

and ( , ) ( , ).B x t B x t→ − −

By plugging these transformations in Maxwell’s equations in Table 31.1, it is seen that the negative signs cancel out in each of the equations, and the original equations are recovered. (b) One-way mirrors do not violate Maxwell’s equations since light can go both ways through a one-way mirror. A one-way mirror is a partially silvered mirror mounted between a brightly lit room and a darkened room. The mirror is partially reflective and partially transparent. The key to its operation is the difference in lighting between the two rooms. In the brightly lit room, reflected light overwhelms light


1178

transmitted from the dark room and the one-way mirror looks like a mirror. Seen from the dark room, the light transmitted from the bright room overwhelms reflected light and the one-way mirror looks like a window into the bright room. Ordinary windows demonstrate the same effect. A window in a brightly lit room looks like a mirror from the room side at night, but a window to the outdoors in daylight.

32.56. THINK: The path a ray of light takes between two points minimizes the time required for the trip. This problem asks for the time it takes light to travel between two points via various paths. The situation is depicted below. SKETCH:

RESEARCH: The time it takes the light ray to travel its path is / .t d v= The speed of the ray in a medium is / .v c n= SIMPLIFY: (a) The time of travel for light along path a is:

w 1 a 21 2a

1 2 1 2

1 .cos cosn d n dl lt

v v c θ θ

= + = +

(b) The time of travel for light along path b is:

w 1 a 2 w 1 a 21 2b

1 2 3 3 3

1 ,cos cos cosn d n d n d n dl lt

v v c cθ θ θ +

= + = + =

where the angle 3θ is given by:

1 2 1 1 2 23 3

1 2 1 2

tan tantan tan .

w w d dd d d d

θ θθ θ+ += =

+ +

Thus w 1 a 2

b1 1 1 2 2

1 2

.tan tancos tan

n d n dt

d d cd dθ θ−

+=

+ +

(c) The time of travel for light along path c is:

( ) ( )2 22 2w 1 ac 2 1 2 w 1 a 2 1 1 2 2

1 tan tan .n d n

t d w w n d n d d dc c c

θ θ = + + + = + + +

(d) The time of travel for light along path d is:

w a a 2 w a a 21 2d

1 2 2 2 2

1 .cos cos cosn d n d n d n dl lt

v v c cθ θ θ +

= + = + =


1179

CALCULATE:

(a) ( )( )( )

( )( )( )

( )8

a 8

1.333 1.50 m 1.000 1.70 m1 1.5873 10 scos 32.0 cos 45.03.00 10 m/s

t − = + = ⋅ ° °⋅

(b) ( )( ) ( )( )

( ) ( ) ( ) ( )( ) ( ) ( )

8b

1 8

1.333 1.50 m 1.000 1.70 m1.5980 10 s

1.50 m tan 32.0 1.70 m tan 45.0cos tan 3.00 10 m/s

1.50 m 1.70 m

t −

−

+= = ⋅

° + °⋅ +

(c) ( )( ) ( ) ( ) ( ) ( ) ( ) ( )( )

( )

22

c 8

1.333 1.50 m 1.000 1.70 m 1.50 m tan 32.0 1.70 m tan 45.0

3.00 10 m/st

+ + ° + °=

⋅

81.7109 10 s−= ⋅

(d) ( )( ) ( )( )

( ) ( )8

d 8

1.333 0.937 m 1.000 1.70 m1.3902 10 s

3.00 10 m/s cos 45.0t −+

= = ⋅⋅ °

ROUND: Round the results to three significant figures. (a) 8

a 1.59 10 st −= ⋅

(b) 8b 1.60 10 st −= ⋅

(c) 8c 1.71 10 st −= ⋅

(d) 8d 1.39 10 st −= ⋅

(e) Path d has the shortest travel time, but the rays are not actually starting at the location where the fish appears to be. Path a, given by Fermat’s Principle (using Snell’s Law), has the smallest travel time for light from the fish to the observer. Therefore, Fermat’s Principle holds since path a is the actual path taken by the light. DOUBLE-CHECK: The path given by Fermat’s principle (i.e. Snell’s Law) does take the least amount of time, as expected.

32.57. The focal length of a liquid mirror is 2/ 2 ,f g ω= where ω is the angular velocity of the rotating mirror. The angular velocity is:

( )( )

29.81 m/s1.40 rad/s.

2 2 2.50 mgf

ω = = =

32.58. THINK: A proposal for a space telescope is to place a rotating liquid mirror, of focal length 347.5 mf =

and diameter 100.0 m,d = on the Moon, where the gravitational acceleration is 2M 1.62 m/s .g =

SKETCH:

RESEARCH: The focal length of a rotating mirror is given by 2/ 2 .f g= ω The linear speed of a rotating point a distance r from the axis of rotation is .v rω= The height of the liquid can be determined by considering the conservation of energy. The kinetic energy of the liquid is ( ) 21/ 2K mv= and the potential energy is M .U mg h= SIMPLIFY: (a) The angular velocity of the liquid is given by


1180

M .2g

fω =

(b) The linear speed of a point on the perimeter is

M .2 2

gdv rf

ω= =

(c) The height of the liquid at any point on the mirror is given when the potential energy and kinetic energy are equal:

( )22 2 2M2

MM

2

M

2 / 21 1 .2 2 2 2 2 16

dgr dK U mv m r mg h hg f g f

ωω

= = = = = =

CALCULATE:

(a) ( )

( )2

21.62 m/s

4.82798 10 rad/s2 347.5 m

ω −= = ⋅

(b) ( ) ( )

( )21.62 m/s100.0 m

2.41399 m/s2 2 347.5 m

v = =

(c) ( )

( )

2100.0 m1.79856 m

16 347.5 mh = =

ROUND: Round the answers for parts (a) and (b) to three significant figures, and the answer for part (c) to four significant figures. (a) The angular velocity of the mirror is 24.83 10 rad/s.−= ⋅ω (b) The linear speed of a point on the perimeter of the mirror is 2.41 m/s.v = (c) The perimeter is at a height of 1.799 mh = above the center of the mirror. DOUBLE-CHECK: Each result has the appropriate units.

Chapter 33: Lenses and Optical Instruments

1181


In-Class Exercises

33.1. a 33.2. e 33.3. d 33.4. e 33.5. d 33.6. a 33.7. a Multiple Choice

33.1. b 33.2. b 33.3. c 33.4. d 33.5. a 33.6. d 33.7. b 33.8. d 33.9. b 33.10. b 33.11. b Questions

33.12. The dots are on the lens of the glasses, so they are too close to be brought into focus by the eyes of the painter. Since they are so small, they will not appear in what the painter sees. However, the dots will block the light coming from other objects, reducing the brightness of other objects.

33.13. When the diver is wearing the mask, light rays enter the eye from the air (index of refraction is 1) so the diver’s vision is normal. When the mask is removed light rays enter the eye from the water (having an index of refraction of 1.33). As a result, the strength of the lens of the eye decreases and objects that are near will not be able to be brought into focus and the diver becomes farsighted. As the index of refraction of the medium approaches that of the lens (in this case, 1.40),n = the focal length of the lens approaches infinity and even distant objects will appear blurred.

33.14. In order to focus light properly, the index of refraction of the lenses of his eyes must be greater than the index of refraction of the surrounding medium. Since his lens has the same index of refraction as that of air, the focal length of the lens will be at infinity. This means that everything will be totally unfocused and he will only be able to detect changes in brightness and color.

33.15. A lens cannot focus all colors to the same point, due to chromatic aberration. The index of refraction of a lens varies with the wavelength of light. By allowing only one wavelength to pass through their telescopes, astronomers eliminate chromatic aberration. The disadvantage is that the intensity of the light is reduced and images appear fainter.

33.16. In order to start a fire the image of the Sun must be focused to a small area. Focusing the light concentrates the energy of the Sun’s rays, creating a large amount of heat at that point and making it possible to start a fire. If the glasses are for myopia (nearsightedness) then they are diverging lenses. Since diverging lenses only produce virtual images, light cannot be focused on a point. If the glasses are for hyperopia (farsightedness) then they are converging lenses. Since converging lenses can create real images, light can be focused to a point. Therefore, it is possible to start a fire with eye glasses, but only if they are for correcting hyperopic vision.

33.17. The magnification produced by the lens is due to its ability to refract light. Since the difference between the index of refraction of water and glass is less than the difference between that of air and glass, light will refract less at a water/glass boundary. Hence, when the lens is submerged in water, the magnification will decrease.

33.18. Light is reflected in all directions from each point of an object. In order to create an image of an object, the light arriving at one point of an image must be originating from one point on the object. Imagine what is involved in seeing, the light from each object in the field of view enters the eye and is projected onto a particular spot on the retina by using a lens to focus the light. Without the lens in our eye, all of the rays diverging from a particular point on any object would not be focused and would be projected over the


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entire retina. Without using optical elements, an image can be made by allowing light to pass through a very small hole. Such a device is called a “pinhole camera” where light passes in a straight line from a point on an object through the hole and then onto one point on the image. Essentially, a pinhole camera eliminates the angular spread of light reaching the image from a point on an object. The drawback is that only the light along a straight path enters so the image will be faint since only a small amount of light can enter through the hole.

33.19. (a) A ray diagram through the system is presented below:

From the ray diagram it can be seen that in a telecentric system, due to the stop aperture being at the common focal point, only the rays that are parallel (or near parallel) to the axis of the system will contribute to the image formed. The image magnification does not depend on the distance from the system.

(b) Based on the geometry of the system, the magnification is given by 2

1

.f

mf

=

(c) To achieve the maximum resolution, the image of the circular object must cover the entire short dimension (5.00 mm) of the CCD detector. Therefore,

= = = = =i 22 1

o 1

5.00 mm 0.100 0.100 .50.0 mm

h fm f fh f

No specific values for 1f and 2f can be determined, but the first lens will have to have a focal length ten times longer than the second lens. In addition, to accept only parallel rays, the first lens must have a diameter larger than the diameter of the disk to be imaged, and the second lens a diameter larger than the diagonal dimension of the CCD detector.

33.20. (a) The “speed” of a lens is directly connected to the speed with which a photographic exposure can be made in any given lighting situation. The amount of light through the lens per unit time is proportional to the area of the lens aperture, i.e. the square of the aperture diameter or the inverse square of the f- number. Hence the exposure time or shutter speed required is inversely proportional to the aperture area, or proportional to the square of the f-number. A “fast” lens (low f-number) requires a faster shutter speed or shorter exposure than a “slower” lens of larger f-number. The traditional values for f-numbers correspond to factors of two in aperture area or inverse factors of two in shutter speed. (b) The Keck Observatory document Interfacing Visitor Instruments to the Keck Telescopes gives the maximum diameter of the Keck II primary mirror as 10.95 m; it has an area equal to a circular aperture 9.96 m. The focal length of the primary is 17.5 m. So the f-number of the primary mirror is:

( ) ( )-number / 17.5 m / 9.96 m 1.76,f f D= = = which is fairly fast in comparison with ordinary camera lenses. The text gives 2.40 mD = and 57.6 mf = for the primary mirror of the Hubble Space Telescope. These imply

( ) ( )-number / 57.6 m / 2.40 m 24.0,f f D= = =


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which is slow compared to an ordinary camera lens. The National Astronomy and Ionosphere Center/Arecibo Observatory document The 305 meter Radio Telescope gives the diameter of the Arecibo radio telescope as 305 m, and its focal length, the height of the receiving platform above the dish as 450 feet or 137 m. The f-number of this primary mirror is

( ) ( )-number / 137 m / 305 m 0.449,f f D= = = which is very fast compared to an ordinary camera lens.

33.21. In an image, the portion of a scene within the depth of field of a lens appears in focus. Of course, a lens can only focus at one distance, but the decrease in sharpness away from this point on the image may be gradual enough so that it appears in focus to the human eye. If a large aperture is placed in front of a lens, rays reaching the lens far from the optical axis (small f-number) will be bent through large angles. Therefore, rays exiting the lens will intercept the optical axis at large angles and the range of distances over which an image will be in focus will be small. That is, the depth of field is small for large apertures. Conversely, a lens with a small aperture (high f-number) excludes highly diverging rays so that the rays exiting the lens approach the optical axis at shallow angles. Thus, the range of distances over which an image will be in focus will be larger. In this case, the depth of field is large. The limiting case of a very small aperture or high f-number approximates the pinhole camera, which forms images by excluding all rays except those passing through the pinhole. It has no focal length, and can form images of objects at any distance in any plane beyond the pinhole.

33.22. For astronomical mirrors the accuracy and precision of the reflection properties of the mirror are paramount. First-surface mirrors are used for astronomical instruments to avoid refraction through the glass before and after reflection off of the coating, and the accompanying distortion and dispersion (“chromatic aberration,” as the refraction would be different for different wavelengths of light). For household mirrors such precision is not required. Second-surface mirrors are used because of their greater durability since the reflective coating is protected by the glass covering.

33.23. When your friend adjusts the binoculars to his vision, the light rays exiting them are focused by his eyes onto his retina. However, since your friend wears glasses and you do not, the lenses in your eyes are different from his. Therefore, when you use the binoculars on his setting, the light rays are not properly focused onto your retina, so a re-adjustment is required to suit your eyes.

33.24. The ray tracing diagram is shown below:

Since the light rays diverge from the lens and it is the extrapolated rays that actually produce the image, the image is virtual. It is seen from the diagram that the image height is less than the object height.

33.25. To determine what image will be produced when the eyepiece is closer to the objective lens than the image formed by the objective lens, 1 ,i ray diagrams can be used. There are two possible final image types depending on whether 1i is located outside of the focal point of the eyepiece, 2 ,f or between 2f and the eyepiece.


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In the first case, the final image, 2 ,i is virtual, upright and enlarged and in the second case the final image,

2 ,i is real, inverted and enlarged.

33.26. (a) A nearsighted person can only focus on objects that are near. Without corrective lenses, light rays come to a focus at a point before the retina. Diverging lenses are required to diverge the rays before entering the eye so that the focal point advances onto the retina. (b) A farsighted person can only focus on objects that are far away. Without corrective lenses, light rays come to a focus at a point after the retina. Converging lenses are required converge the rays more before entering the eye so that the focal point recedes onto the retina.

33.27. The make-shift microscope has converging lenses, one with focal length 1 6.0 cmf = and the other with

2 3.0 cm.f = The lenses are separated by a distance = 20. cm.L The magnification of a microscope is given by: ( ) ( )= − 0 e0.25 m / .m L f f It therefore does not matter which lens is used as the eyepiece and which is used as the objective. However, it is more practical to use the lens with the smaller focal length as the objective in order to bring the object closer to the microscope.

Problems

33.28. The set up is as shown:

(a) Assume the lens is a thin lens. For the image to form at a distance i 3d f= on the right side of the lens,

od must be: ( )( )i

o i o i i

31 1 1 1 1 1 1.5 .3o

f ffdd f

d d f d f d d f f f+ = = − = = =

− −

(b) The magnification m must be i

o

3 2,1.5

d fmd f

= − = − = − where the negative sign denotes that the image

is inverted.

33.29. The distance to the image id is:

oi

o i i o o

1 1 1 1 1 1 fd

dd d f d f d d f

+ = = − =−

Therefore, the magnification is


1185

o

o

o o

9.0 cm3.0.

6.0 cm 9.0 cm

fdd f fm

d d f

− = − = − = − =

− −

33.30. The radius of curvature of the front surface of the ice lens is =1 15.0 cmR and that for the back is = −2 20.0 cm.R To start a fire with the ice lens the twigs would need to be placed at the location where the

light rays are focused. Presume the source of the light rays (the Sun) is at infinity; o .d = ∞ Use the Lens-Maker’s formula in the following form: ( )( )o i 1 21/ 1/ 1 1/ 1/ .d d n R R+ = − − Since ice has an index of refraction of 1.31,n = this becomes:

( ) ( )−

+ = − − = + = ∞ −

1

ii

1 1 1 1 1 11.31 1 0.31 27.6 cm15.0 cm 20.0 cm 15.0 cm 20.0 cm

dd

It would be best to put the twigs about 27.6 cm from the ice lens in order to create a fire.

33.31. For the purposes of this question, the laser can be treated as an object at a distance od with height −= ⋅ 3

o 1.06 10 m.h The image height is 6i 10.0 10 mh −= ⋅ and the distance to the image is

= =i 20.0 cm 0.200 m.d Since the image is to be formed behind the lens and reduced in size, the lens must be a converging lens, and the object should be greater than 2 f away from the lens, where f is the focal length. This means that both the object distance od and the image distance id are positive. From the magnification equation i o i o/ / ,m d d h h= = the object distance must be

( )( ) ( )3 6o i o i/ 0.200 m 1.06 10 m / 10.0 10 m 21.2 m.d d h h − −= = ⋅ ⋅ =

This large value for od is consistent with how a laser beam is highly collimated, that is the rays are almost parallel. From the thin equation o i1/ 1/ 1/ ,d d f+ = the focal length is

( ) ( ) 11/ 1/ 21.2 m 1/ 0.200 m 5.0472 m 0.198 m.f f−= + = = For incoming rays that are parallel with the optical axis, the focal point is the focus. Since these rays are highly collimated it is reasonable that image location is near the focal point.

33.32. The plastic cylinder, shown below, has length 30. cm,L = and the radius of curvature of each end is 10. cm.R = The index of refraction of the plastic is 1.5.n = The object distance is o 10. cmd = from the

left end.

Assume the object is in a medium with an index of refraction of 1n = (like air, or a vacuum). The image distance from the left end of the plastic cylinder is:

( )( )

( )( )( )( )( ) ( )

o,1i,1

o,1 i,1 o,1

1.5 10. cm 10. cm11 30. cm.1 10. cm 1.5 1 10. cm

n nRdn dd d R d n R

− + = = = = − − − − −

The negative sign indicates that the image is to the left of the cylinder. Therefore, the object for the right end is at a distance o,2 i,1 60. cm.d d L= + = The image distance from the right end of the cylinder is:

( )( )

( )( )( )( ) ( )( )

o,2i,2

o,2 i,2 o,2

10. cm 60. cm11 40. cm.1 60. cm 1 1.5 1.5 10. cm

n Rdn dd d R d n nR

− − + = = = = − − − − −

Therefore, a real image is formed 40. cm to the right of the right end of the cylinder.


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33.33. THINK: The object height is o 2.5 cm,h = and is o 5.0 cmd = from a converging lens of focal length 3.0 cm.f = The thin lens equation can be used to find the image distance and the magnification can be

found from this. SKETCH:

RESEARCH: The magnification m is: i o/ .m d d= − The thin lens equation is: o i1/ 1/ 1/ .d d f+ =

SIMPLIFY: The image distance is: oi

o i i o o

1 1 1 1 1 1 .fd

dd d f d f d d f

+ = = − =−

Therefore, the

magnification is: = −−o

.fmd f

CALCULATE: The magnification is: 3.0 cm 1.500.5.0 cm 3.0 cm

m = − = −−

Since the magnification is negative,

the image is inverted and since 1,m > the image is enlarged. ROUND: To two significant figures, the magnification of the image is 1.5.m = − DOUBLE-CHECK: The ray tracing shown above confirms that the image is inverted and enlarged. As seen in Table 33.1, this is what it is expected for < <o 2 .f d f

33.34. THINK: There are three different locations for placing a real object in front of a thin convex lens which results in a real image. Consider each case separately: Case 1: The object distance is >o 2 .d f Case 2: The object distance is =o 2 .d f Case 3: The object distance is > >o2 .f d f (Note that when =od f no image is formed and when <o ,d f the image is virtual.) The thin lens equation can be used to find the minimum distance between a real object and a real image. SKETCH:

RESEARCH: For a thin lens, = +o i1/ 1/ 1/ .f d d When the image of a real object is on the opposite side of the lens, the image is real and both od and id are positive by convention.


1187

SIMPLIFY: For separation distance L between the object and the image, write + =o i .d d L Then = −i o .d L d The thin lens equation becomes:

( ) ( )− −= + = =

− − +o o o o

o o o o

1 1 1 .d L d d L d

ff d L d L d d L

Solving for L gives: = − =−

22 o

o oo

.d

Lf d L d Ld f

Case 1: If ∞ > >o 2 ,d f the distance L lies between ( )

( )( )→∞

> > ∞ > >− −o

22o

o

2lim 4 .

2d

fdL L f

d f f f

Case 2: If =o 2 ,d f the distance L is ( )

( )

224 .

2f

L ff f

= =−

Case 3: If > >o2 ,f d f the distance L lies between ( )

( )( )

( )

2 22 4 .

2f f

L f Lf f f f

> > > > ∞− −

So the

minimum separation distance between a real object and a real image for a thin convex lens is 4 .L f= CALCULATE: Not required. ROUND: Not required. DOUBLE-CHECK: This is consistent with the ray diagrams.

33.35. THINK: (a) An air-filled cavity bound by two spherical surfaces is created inside a glass block. The two spherical surfaces have radii of curvatures of 1 30.0 cmR = − and 2 20.0 cm.R = − Both values are negative because each surface is concave. The LED is a distance o,1 60.0 cmd = from the cavity. The thickness of the cavity is 40.0 cm.d = The index of refraction for glass and air is g 1.50n = and a 1.00,n = respectively. The

equations for thick lenses can be used to find the final position of the image of the LED through the cavity. The image formed by the first (left) surface will act as the object for the second (right) surface. SKETCH:

RESEARCH: In the paraxial approximation, 1 2 2 1

o i

.n n n nd d R

−+ =

SIMPLIFY: For the first surface the interface is glass to air, so the image distance is:

( )g a g a o,1 1a

,1o,1 i,1 1 o,1 a g 1 g

.i

n n n n d Rnd

d d R d n n R n

−+ = =

− −

For the second surface the interface is air to glass, so the image distance is:


1188

( )−

+ = =− −

g g a g o,2 2ai,2

o,2 i,2 2 o,2 g a 2 a

.n n n n d Rn

dd d R d n n R n

CALCULATE: The image for the first lens is: ( )( )( )

( )( ) ( )( )i,1

1.00 60.0 cm 30.0 cm120. cm.

60.0 cm 1.00 1.50 30.0 cm 1.50d

−= = −

− − −

The negative sign indicates that the image is to the left of the first surface, so object distance for the second surface is given by o,2 i,1 40.0 cm 120. cm 160. cm.d d d= + = + − = Therefore, the image formed is at a distance of

( )( )( )( )( ) ( )( )i,2

1.50 160. cm 20.0 cm48.0 cm.

160. cm 1.50 1.00 20.0 cm 1.00d

−= = −

− − −

The negative sign indicates that the image is to the left of the second surface and that it is virtual. ROUND: All values are given to two significant figures. The final position of the virtual image of the LED through the cavity is 48.0 cm to the left of the second surface (or 8.00 cm to the left of the first surface). DOUBLE-CHECK: The ray diagram below is consistent with the calculated position of the final image:

The dotted lines drawn from the center of each curve to the respective curve’s surface makes a normal line to that surface. The dashed lines represent rays being extrapolated back to find the location of the virtual image. Note the light rays refract at each interface, bending away from the normal when going from glass to air, and bending toward the normal when going from air to glass.

33.36. The magnification of a lens is given by i o/ .m d d= − The object distance is given as =o 3.00 cm.d From the thin lens equation:

oi

o i o

1 1 1 .fd

df d d d f

= + =−

Therefore, the magnification is: 5.00 cm

2.50.3.00 cm 5.00 cmo

fmd f

= − = − =− −

33.37. The angular magnification of a magnifying glass is approximately near / .m d fθ ≈ With a given focal length of = 5.0 cm,f and assuming a near point of =near 25 cm,d the magnifying power of this lens with the

object placed at the near point is ( ) ( )θ = =25 cm / 5.0 cm 5.0.m

33.38. Generally, magnification is defined as the ratio of image height to object height, i o/ .m h h= With an object height of =o 1.0 mmh and an image height of =i 10. mm,h the magnification is

= =10. mm /1.0 mm 10.m The angular magnification of a magnifying glass is approximately

( )θ ≈ 25 cm / ,m f where a near point of 25 cm is assumed. The focal length of the magnifying glass is:

≈ = =25 cm 25 cm 2.5 cm.10.

fm


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33.39. THINK: The person’s near-point distance is near 24.0 cm.d = The magnifying glass gives a magnification

nearm that is 1.25 times larger when the image of the magnifier is at the near point that when the image is at infinity, that is near 1.25 .m m∞= Find the focal length of the magnifying glass, .f SKETCH:

RESEARCH: When the image is place at infinity, the angular magnification of a magnifying glass is: near / .m d f∞ = When the image is placed at the near-point, the text shows that the equation becomes ( )near near / 1.m d f= +

SIMPLIFY: near near nearnear near

1.25 0.251.25 1 1 0.25

d d dm m f d

f f f∞= + = = =

CALCULATE: The focal length is ( )= =0.25 24.0 cm 6.00 cm.f ROUND: To three significant figures, the focal length of the magnifying glass is 6.00 cm.f = DOUBLE-CHECK: The focal length should be less than neard if the image is to form at near .d

33.40. A beam of parallel light has a diameter 1 1.00 mm.d = It passes through the first lens of focal length

1 10.0 cm,f = then a second lens of focal length 2 20.0 cm.f = The emerging light is again parallel. (a) Light from the first lens is focused at its focal point 1.f Since the light exiting the second lens is parallel, the object location for the second lens must be located a distance in front of that lens equal to its focal length, 2 .f The total separation between the two lenses is; therefore, the sum of their focal lengths:

10.0 cm 20.0 cm 30.0 cm.L = + = (b) A triangle can be formed for the original lens with a height of 0.50 mm (the radius of the beam) and length of 10.0 cm ( )1 .f For the second lens a triangle can be drawn whose height is the outgoing beam’s

radius and whose length is 20.0 cm ( )2 .f See the diagram below:

Since these triangles are similar triangles (same angles), the ratio of length to height must be the same for both: ( ) ( )1 1 2 2 2 1 2 1/ / / 0.50 mm 20.0 cm /10.0 cm 1.00 mm.r f r f r r f f= = = = And the width of the

outgoing beam is ( )2 2 1.00 mm 2.00 mm.d = =

33.41. The total magnification is the product of the magnification after passing through the first lens, 1 ,m and the magnification of the second lens, 2 .m Magnification is = − =i o i o/ / .m d d h h The focal length of each lens is 5.0 cm,f = and the distance that the insect is from the first lens is =o,1 10.0 cm.d Using the thin lens equation the image distance from the first lens is:


1190

( )( )o,1i,1

o,1

5.0 cm 10.0 cm10. cm.

(10.0 cm 5.0 cm)fd

dd f

= = =− −

Then ( ) ( )1 i,1 o,1/ 10.0 cm / 10.0 cm 1.00m d d= − = − = − This image is inverted, but the size does not change. This image acts as an object for the second lens, and is a distance o,2 i,1d L d= − from the second lens, where L is the separation distance of the two lenses, 12 cm.L = Using the thin lens equation, the image distance from the second lens is:

( )( )( )( )

o,2i,2

o,2

5.0 cm 12 cm 10.0 cm3.3 cm.

12 cm 10.0 cm 5.0 cmfd

dd f

−= = = −

− − −

Then ( ) ( )2 i,2 o,2/ 3.3 cm / 12 cm 10.0 cm 1.7.m d d= − = − − − = This image is oriented the same way as the

object (inverted). The final magnification of the insect is 1 2 ( 1.0)(1.7) 1.7.m m m= = − = − Therefore, the final image of the insect has a size of ( )i o 1.7 5.0 mm 8.5 mm.h mh= = − = − With respect to the original insect, the final image is enlarged, inverted (since magnification is negative) and virtual (since i,2d is negative).

33.42. THINK: Three converging lenses of focal length 5.0 cmf = are arranged with a spacing of 20. cmL = between them. They are used to image an insect o,1 20. cmd = away. In each case, the image formed by the preceding lens will act as the object for the next lens. (a) To find the location and orientation of the image, the thin lens equation can be applied consecutively for the different lenses. (b) If the final image is to the right of the third lens then the image is real, and if it is to the left then it is virtual. (c) Every time a real image is formed by a convex lens, the image is inverted. SKETCH:

RESEARCH: In each case, the image formed by the preceding lens will act as the object for the next lens. The thin lens equation is o i1/ 1/ 1/ .f d d= + SIMPLIFY: (a) Find the first image location, ,1 :id

o,1i,1

o,1 i,1 i,1 o,1 o,1

1 1 1 1 1 1 .d f

df d d d f d d f

= + = − =−

This image acts as the object for the second lens. The second image location, i,2 ,d is:

o,2i,2

o,2 i,2 i,2 o,2 o,2

1 1 1 1 1 1 .d f

df d d d f d d f

= + = − =−

The final image location, ,3 ,id is:

o,3i,3

o,3 i,3 i,3 o,3 o,3

1 1 1 1 1 1 .d f

df d d d f d d f

= + = − =−


1191

CALCULATE:

(a) The first image is at location: ( )( )

( ) ( )i,1

20. cm 5.0 cm6.667 cm.

20. cm 5.0 cmd = =

− This image acts as the object for

the second lens. Since i,1d is positive, o,2 20. cm 6.667 cm 13.33 cm.d = − = The second image is at

location: ( )( )

( ) ( )i,2

13.33 cm 5.0 cm8.00 cm.

13.33 cm 5.0 cmd = =

− This image acts as the object for the third lens. Since i,2d

is positive o,3 20. cm 8.00 cm 12 cm.d = − = The final image is at location:

( )( )( ) ( )i,3

12 cm 5.0 cm8.57 cm.

12 cm 5.0 cmd = =

−

(b) Since i,3d is positive, the final image is on the right side of the third lens, so the image is real. (c) Since the image of each object is inverted, and there are an odd number of lenses, the final image is inverted. ROUND: (a) To two significant figures, the image is located i,3 8.6 cmd = to the right of the third lens. DOUBLE-CHECK: It is reasonable that the image due to each lens is real since each object is outside of the focal length of each lens. Using the equation for magnification the orientation of the final image is verified (recall od and id are positive in each case):

i,1 i,2 i,31 2 3

o,1 o,2 o,3

0.d d d

m m m md d d

= = − − − <

Therefore, the final image is inverted.

33.43. THINK: Two identical thin convex lenses, each of focal length, ,f are separated by a distance 2.5 .d f= An object is placed in front of the first lens at a distance o,1 2 .d f= The thin lens equation can be used to find the location, orientation, and size of the final image. The image formed by the first lens will act as the object of the second lens. SKETCH:

RESEARCH: The thin lens equation, o i1/ 1/ 1/ ,f d d= + can be used in succession to determine the final image location. The magnification is given by: i o i o/ / .m d d h h= − = The total magnification is the product of the magnification of the first lens, 1 ,m and the magnification of the second lens, 2 ;m that is,

1 2 .m m m=

SIMPLIFY: For the first image: o,1i,1

i,1 o,1 o,1

1 1 1 .d f

dd f d d f

= − =−

Since o,1 ,d f> i,1d is positive so

the object distance for the second lens is o,2 i,1.d d d= − For the final image:

o,2i,2

i,2 o,2 o,2

1 1 1 .d f

dd f d d f

= − =−


1192

The magnification is i,1 i,21 2

o,1 o,2

.d d

m m md d

= = − −

CALCULATE:

(a) ( )

( )i,1

22 ,

2f f

d ff f

= =−

o,2 2.5 2 0.5 .d f f f= − = Therefore, ( )

( )i,2

0.5.

0.5f f

d ff f

= = −−

The final image is at

the focal point on the left side of the second lens. It must be a virtual image.

(b) The total transverse magnification of the system is: 2 2.2 0.5

f fmf f

−= − − = −

(c)

(d) Since i,2d is negative, the final image is virtual. Since 0,m < the final image is inverted. Since 1,m > the final image is enlarged. ROUND: Not required. DOUBLE-CHECK: The ray tracing diagram of part (c) is consistent with the calculations of part (a) and part (b).

33.44. THINK: Two converging lenses with focal lengths =1 5.00 cmf and =2 10.0 cmf are placed = 30.0 cmL apart. An object of height =o,1 5.00 cmh is placed =o,1 10.0 cmd to the left of the first lens. The thin lens equation can be used to. The image formed by the first lens will act as the object for the second lens. The thin lens equation can be used consecutively to find the position i,2d of the final image produced by this lens system. The magnification equation can be used to find the final image height i,2 .h SKETCH:

RESEARCH: The thin lens equation is given by: = +o i1/ 1/ 1/ .f d d The total magnification is the product of the magnification after passing through the first lens 1,m and the magnification of the second lens, 2 ;m that is 1 2 .m m m= The magnification is given by: = − =i o i o/ / .m d d h h

SIMPLIFY: The first image is at location: = − =−

o,1 1i,1

i,1 1 o,1 o,1 1

1 1 1 .d f

dd f d d f

Since >o,1 1 ,d f i,1d is positive

so the object distance for the second lens is = −o,2 i,1 .d d d For the final image:

= − =−

o,2 2i,2

i,2 2 o,2 o,2 2

1 1 1 .d f

dd f d d f

The magnification of the final image is

= = − −

i ,1 i,21 2

o,1 o,2

.d d

m m md d

Therefore, the final image height is

= − −

i ,1 i,2i,2 o,1

o,1 o,2

.d d

h hd d


1193

CALCULATE: The image distance of the first lens is ( )( )

( ) ( )= =−i,1

10.0 cm 5.00 cm10.0 cm.

10.0 cm 5.00 cmd The object

distance for the second lens is then: = − =o,2 30.0 cm 10.0 cm 20.0 cm.d The final image distance is

( )( )( ) ( )= =

−i,2

20.0 cm 10.0 cm20.0 cm.

20.0 cm 10.0 cmd The final image is 20.0 cm to the right of the second lens. This

image is real since >i ,2 0.d The final height is = − − =

i,210.0 cm 20.0 cm5.00 cm 5.00 cm.10.0 cm 20.0 cm

h The

image is the same size as the object and since the height is positive, the image is upright. ROUND: To three significant figures, the location of the final image is =i,2 20.0 cmd to the right of the second lens and the final image height is =i,2 5.00 cm.h DOUBLE-CHECK: Since for each converging lens o ,d f> the image produced must be real. Upon each pass through a lens, the image is inverted. Thus after two lenses, the final image is upright.

33.45. The object is =o,1 10.0 cmh tall and is located =o,1 30.0 cmd to the left of the first lens. Lens 1L is a biconcave lens with index of refraction 1.55n = and has a radius of curvature of 20.0 cm for both surfaces. The first surface has negative radius of curvature as its surfaces is concave with respect to the object: 1 20.0 cm.R = − The second surface is convex with respect to the object, so its radius of curvature is positive: 2 20.0 cm.R = Lens 2L is 40.0 cmd = to the right of the first lens 1.L Lens 2L is a converging lens with a focal length of 2 30.0 cm.f = The image formed from the first lens acts as the object for the second lens. The position of the image formed by lens 1L is found from the Lens Maker’s Formula with the thin

lens approximation: ( ) + = − −

o i 1 2

1 1 1 11 .nd d R R

Then the image distance is:

( )−

= − − − = − −

1

i,11 1 11.55 1 11.32 cm.

20.0 cm 20.0 cm 30.0 cmd

This image is on the left side of lens 1L and it acts as the object for lens 2 .L The object distance for lens

2L is = + = + =o,2 i,1 40.0 cm 11.32 cm 51.32 cmd d d from lens 2 .L From the thin lens equation, the

image distance of lens 2L is: −

= − = − =

1

i,2i,2 2 o,2

1 1 1 1 1 72.2 cm.30.0 cm 51.32 cm

dd f d

Since this

distance is positive, the final image is real and is 72.2 cm to the right of lens 2 .L The focal length of lens

1L is required for a ray diagram. The Lens Maker’s Formula gives:

( ) ( )1 1

1 11 2

1 1 1 11 = 1.55 1 18.2 cm.20.0 cm 20.0 cm

f n fR R

− − = − − − − = − −


1194

33.46. THINK: Light rays are described at any point along the axis of the system by a two-component column

vector containing ,y the distance of the ray from the optic axis, and ',y the slope of the ray. Components of the system are described by 2 2× matrices which incorporate their effects on the ray; combinations of components are described by products of these matrices. (a) Construct the matrix for a thin lens of focal length .f (b) Write the matrix for a space of length .x (c) Write the matrix for the two-lens “zoom lens” system described in the text. SKETCH: Not required. RESEARCH: (a) As stated, a thin lens does not alter the position of a ray, but increases (diverging) or decreases (converging) its slope an amount proportional to the distance of the ray from the axis. The constant of proportionality between the distance of the ray from the optic axis and the change in its slope must be

1/ ,f− so that a ray initially parallel to the axis (zero slope) will descend to the axis after traveling a distance f from the lens. So the matrix corresponding to a thin lens of focal length f is

( ) 1 0,

1/ 1L f

f

= −

where it is assumed that the displacement of a ray from the axis is the first component of the column vector describing the ray, and its slope is the second. (b) A space between components does not alter the slope of a ray; the distance of the ray from the axis changes by the slope of the ray times the length of the space. As described, the matrix for a space of length x along the optic axis

( ) 1.

0 1x

S x

=

(c) The zoom lens described consists of a lens of focal length 1f followed by a space of length ,x then a second lens of focal length 2 .f The corresponding matrix is ( ) ( ) ( )2 1 .Z L f S x L f= SIMPLIFY: For part (c),

( ) ( ) ( ) ( ) ( )1

2 11 2 1 2 22 1

1 /1 0 1 1 0.

/ 1/ 1/ 1 /1/ 1 0 1 1/ 1x f xx

Z L f S x L fx f f f f x ff f

− = = = − + −− −

CALCULATE: Not required. ROUND: Not required. DOUBLE-CHECK: A ray parallel to the optical axis will descend to the optical axis a distance i,2d from the second lens. Using the matrix above, the distance i,2d is given by the negative of the original distance from the axis divided by the effective slope of the two-lens system:


1195

( )( )

−−

= =− +

− +

2 11i,2

2 1

1 2 1 2

1.

1 1

xf x ffd

x f fxf f f f

In the text, the effective focal length efff of the combination is measured from the first lens. This result

implies ( )( )

−= + = +

− +2 1

eff i,22 1

,f x f

f x d xx f f

in exact agreement with the analysis in the text.

33.47. The typical length of a human eyeball is 2.50 cm. Use the thin lens equation: + =o i1/ 1/ 1/ .d d f

(a) When od is large, ≈ i .f d The effective focal length for viewing objects at large distances is 2.50 cm.f =

(b) When the object is at a typical near point, = ≈o near 25 cmd d and the image forms at the back of the eye at =i 2.50 cm,d the effective focal length is:

11 1 2.273 cm 2.3 cm.25 cm 2.50 cm

f−

= + ≈ =

33.48. The effective focal length of two thin lenses placed close together is: eff 1 2

1 1 1 .f f f

= + The cornea in a typical

human eye has a fixed focal length of 1 2.33 cm.f = For very distant objects the effective focal length of the lens-cornea system was found to be eff 2.50 cmf = in the previous problem. In this case the focal length

2f of the lens of the eye would have to be: 1 1

2eff 1

1 1 1 1 34.3 cm.2.50 cm 2.33 cm

ff f

− − = − = − = −

For objects at the near point the effective focal length of the lens-cornea system was found to be eff 2.273 cmf = in the previous problem. In this case the focal length 2f of the lens of the eye would have

to be: 1 1

2eff 1

1 1 1 1 92.9 cm.2.273 cm 2.33 cm

ff f

− − = − = − =

Therefore, the lens in the human eye must have a range of focal lengths between 34.3 cm− and 92.5 cm.

33.49. Jane’s near point is near 125 cmd = and the computer screen is =o 40. cmd from her eye. Use the thin lens equation: + =o i1/ 1/ 1/ .d d f Also, the power of a lens (in diopters) is 1/D f= where f is in meters. (a) The object distance is just the distance to the computer screen: =o 40. cm.d (b) The image distance is Jane’s near point: = − = −i near 125 cm.d d It is negative because the image appears on the same side of the eye as the object (the image is virtual).

(c) The focal length is 11 1 59 cm.

40. m 125 cmf

− = + = −

(d) Jane’s near point is 1.25 m; to read the computer screen at =o 0.40 m,d the image must be located at the near point, = −i near .d d The power of this corrective lens would be:

= = + = + = +−o i

1 1 1 1 1 1.7 diopter.0.40 m 1.25 m

Df d d

(e) Since the focal length is positive, the corrective lens is converging.


1196

33.50. Bill’s far point is far 125 cm.d = Use the thin lens equation: + =o i1/ 1/ 1/ .d d f Also, the power of a lens (in diopters) is 1/D f= where f is measured in meters.

(a) The objects he wishes to see are far away, so the object distance is .od = ∞ (b) The image distance is = − = −i far 125 cm.d d It is negative because the image appears on the same side of the eye as the object (the image is virtual).

(c) The focal length is 11 1 125 cm.

125 cmf f

− = + = − ∞ −

(d) Bill’s far point is 1.25 m, so images of distant objects ( )= ∞od must be located at the far point, = −i far .d d The power of this corrective lens would be:

≡ = + = + = −∞ −o i

1 1 1 1 1 0.800 diopter.1.25 m

Df d d

(e) Since the focal length is negative, the corrective lens is diverging.

33.51. The newspaper is located at =o 25 cm.d The converging part of the lens has a focal length of c 70. cm.f = The diverging part of the lens has a focal length of d 50. cmf = − (it is negative because the lens is a diverging lens). The converging lens places the image at the near point. Since the image is on the same side of the lens as the object, i near .d d= − From the thin lens equation

− + = − = = − = − =

1

nearo i o near c o c

1 1 1 1 1 1 1 1 1 1 39 cm;25 cm 70. cm

dd d f d d f d f

The diverging lens places the image at the far point. Since the image is on the same side of the lens as the object, far i .d d= − The objects are at .od = ∞ From the thin lens equation,

− + = − = = − = − = ∞ −

1

faro i o far d o d

1 1 1 1 1 1 1 1 1 1 50. cm;50. cm

dd d f d d f d f

33.52. The radius of curvature for the outer part of the cornea is 31 8.0 10 m,R −= ⋅ while the inner portion is

relatively flat, so 2 .R = ∞ The radius of curvature 1R is positive because the surface facing the object is convex. The index of refraction of the cornea and the aqueous humor is 1.34.n = (a) The power of the cornea is 1 1/ .D f= From the Lens Maker’s Formula, the power of the cornea is:

( ) ( )1 31 2

1 1 1 1 11 1.34 1 42.5 diopter 43 diopter.8.0 10 m

D nf R R −

= = − − = − − = ≈ ∞⋅

(b) The combination of the lens and the cornea has a power of =eff 50. diopter.D For two adjacent lenses, the effective focal length is eff 1 21/ 1/ 1/ .f f f= + Rewriting in terms of power yields eff 1 2 .D D D= + The power 2D of the lens is 2 eff 1 50. diopter 42.5 diopter 7.5 diopter.D D D= − = − =

33.53. THINK: As objects are moved closer to the human eye the focal length of the lens decreases. The shortest focal length is min 2.3 cm.f = The thin lens equation can be used to determine the closest one can bring an object to a normal human eye, o,norm ,d and still have the image of the object projected sharply onto the retina, which is =i,norm 2.5 cm.d behind the lens. A near sighted human eye has the same minf but has a retina that is 3.0 cm behind the lens. The thin lens equation can be used to determine the closest one can bring an object to this nearsighted human eye, o,near ,d and still have the image of the object projected sharply on the retina at =i,near 3.0 cm.d SKETCH: Provided with the problem.


1197

RESEARCH: In each case the object is in front of the lens, and the image is formed behind the lens, so both od and id are positive. The thin lens equation is: + =o i1/ 1/ 1/ .d d f The angular magnification is given by near / .m d fθ ≈

SIMPLIFY: −

= − = −

1

oo i i

1 1 1 1 1 .dd f d f d

The ratio of angular magnifications is:

near, normnorm near

near norm near, near

.dm f

m f d

=

Since the object is placed at the near point for the image to form on the retina and near norm minf f f= = , this becomes

= =

o, norm o, normnorm near

near o, near norm o, near

.d dm f

m d f d

CALCULATE: For the normal eye the minimum distance is: −

= − =

1

o, norm1 1 28.75 cm.

2.3 cm 2.5 cmd

For the elongated eye the minimum distance is: −

= − =

1

o, near1 1 9.86 cm.

2.3 cm 3.0 cmd The ratio of

angular magnifications is norm

near

28.75 cm 2.916.9.86 cm

mm

= =

ROUND: To two significant figures, =o, norm 29 cm,d =o, near 10. cm,d and norm near 2.9 .m m= DOUBLE-CHECK: The nearsighted eye should have a closer near point than the normal eye.

33.54. THINK: The power of the eyeglasses lens is 5.75 diopters.D = − The negative power implies that the lenses are diverging lenses, and that the person is indeed nearsighted. Objects at a far distance must have an image formed at the person’s near point to be resolved. The lenses are 1.00 cmL = in front of his corneas. The thin lens equation can be used to find the prescribed power of his contact lenses. SKETCH:

RESEARCH: The power of a lens is 1/D f= where f is measured in meters. The thin lens equation is

+ =o i1/ 1/ 1/ .d d f Since the person is nearsighted it can be assumed that the objects are distant: = ∞o .d The near point neard is = +near id d L (see sketch above). The image distance id is negative because the image forms on the same side of the lens as the object. The image formed from the contact lenses must be at the near point as well. In this case = −i neard d since there is no space between the contacts and the cornea.

SIMPLIFY: With the glasses, − −

= + = − + = − + = + = + ∞

1 1

near io glasses

1 1 1 1 1 .d d L L L f L Lf d f D

With ,od = ∞ the power of the contacts is contactsnear

glasses

1 1 .1

Dd

LD

= − =

+


1198

CALCULATE: ( )contacts 1 2

1 5.437 diopter.1/ 5.75 m 1.00 10 m

D− −

= − = −− + ⋅

ROUND: To three significant figures, the prescribed power of the contact lenses is contacts 5.44 diopter.D = −

DOUBLE-CHECK: The power of the contacts should be slightly less than the power of the glasses, since the contacts are on the eye.

33.55. Equation 33.7 gives the effective focal length of a two lens system as: ( )( )

2 1eff

2 1

.f x f

f xx f f

−= +

− +

For a separation of 50. mmx = between the lenses, with focal lengths of = ⋅ 21 2.0 10 mmf and

= − ⋅ 22 3.0 10 mmf for the first and second lens, respectively, the effective focal length is:

( )( )( )

2

eff

2

2 2

3.0 10 mm 50. mm 2.0 10 mm50. mm 350 mm.

50. mm 3.0 10 mm 2.0 10 mmf

− ⋅ − ⋅= + =

− − ⋅ + ⋅

For a separation of = ⋅ 21.0 10 mmx between the lenses, the effective focal length is:

( )( )( )

2 2 22

2 2 2eff

3.0 10 mm 1.0 10 mm 2.0 10 mm1.0 10 mm 250 mm.

1.0 10 mm 3.0 10 mm 2.0 10 mmf

− ⋅ ⋅ − ⋅= ⋅ + =

⋅ − − ⋅ + ⋅

33.56. The distance between the lens and the film is 10.0 cm. Initially, objects that are very far away appear properly focused on the film, so the distance from the lens to an object can be taken as = ∞o .d Since the images form on the film, the image distance is =10.0 cm.id Approximating the lens as a thin lens, the

focal length of the lens is ( ) ( )− −+ = = + = ∞ + =1 1o i o i1/ 1/ 1/ 1/ 1/ 1/ 1/10.0 cm 10.0 cm.d d f f d d To

properly focus an object =o 100. cmd away, the film must lie at the location where the image forms, at

i :d ( )−= − = − =1i o i1/ 1/ 1/ 1/10.0 cm 1/100. cm 11.1 cm.d f d d Therefore, you would have to move

the lens about 1.1 cm in order for it to focus an object 1.00 m away.

33.57. The focal length of the original lens is fixed at 60. mmf = and the zoom lens has a variable focal length. The object is a distance = ∞od from the lens. Using the thin lens equation for the original lens, the image appears at = =i 60. mm.d f With the zoom lens set to a focal length of ' 240. mm,f = the image appears at i ' ' 240. mm.d f= = The ratio of magnifications of each lens is:

original i o i

zoom i o i

/ 60. mm 1 .'/ ' 240. mm 4.0

m d d dm d d d

−= = = =

−

The zoom lens (at ' 240. mmf = ) produces an image that is 4.0 times the size of the image produced by the original 60. mmf = lens.

33.58. THINK: The first lens is the diverging lens of focal length = −1 10.0 cm;f the second lens is the converging lens of focal length =2 5.00 cm.f The two lenses are held = 7.00 cmL apart. A flower of length =o,1 10.0 cmh is held upright at a distance =o,1 50.0 cmd in front of the diverging lens. The thin lens equation can be used to find the location i,2d of the final image, and the magnification equation can be used to find the orientation, size i,2 ,h and the magnification m of the final image.


1199

SKETCH:

RESEARCH: The thin lens equation is: + =o i1/ 1/ 1/ .d d f The magnification equation for each lens is:

= = −i o i o/ / .m h h d d For multiple lenses, the total magnification is the product of the magnification of each lens: 1 2 .m m m=

SIMPLIFY: The image produced by the diverging lens is formed at position: =−

o,1 1i,1

o,1 1

.d f

dd f

For a

diverging lens, 0.f < Since >o,1 0,d from the above equation <i,1 0;d that is, the image is on the left side

of the lens. This image acts as the object for the converging lens at a distance of = +o,2 i,1 .d L d The

position of the image produced by the converging lens is: =−

o,2 2i,2

o,2 2

.d f

dd f

The final magnification is

= − −

i,1 i,2

o,1 o,2

.d d

md d

The size of the final image is =i,2 o,1 .h mh

CALCULATE: The image formed by the first lens is at location: ( )( )

( ) ( )−

= = −− −i,1

50.0 cm 10.0 cm8.3333 cm.

50.0 cm 10.0 cmd

The object distance for the second lens is: = + − =o,2 7.00 cm 8.3333 cm 15.33 cm.d The final image formed by the second lens is at location:

( )( )= =

−i,2

15.33 cm 5.00 cm7.4201 cm.

15.33 cm 5.00 cmd

The total magnification is −

= − − = −

8.3333 cm 7.4201 cm0.08067.

50.0 cm 15.33 cmm

The size of the final image is ( )( )= − = −i,2 0.08067 10.0 cm 0.8067 cm.h Since 0,m < the final image is inverted. ROUND: To three significant figures: the final image is =i,2 7.42 cmd to the right of the convex lens, the magnification of the final image is = −0.0807m and the size of the final image is = −i,2 0.807 cm.h DOUBLE-CHECK: These results are consistent with the ray diagram:


1200

33.59. The magnification of a microscope is given by the equation: ( ) o e25 cm / .m L f f= The magnitude of the

desired magnification is 23.0 10 .m = ⋅ Treating the lens attached to the tube as the objective lens with focal length, o 0.70 cm,f = the focal length, e ,f of the eyepiece required, should be

( ) ( )( )( )( )e 2

o

25 cm 25 cm 20. cm2.4 cm.

0.70 cm 3.0 10L

ff m

= = =⋅

(Note that the designation of eyepiece and objective to

the two lenses is independent of the magnification.)

33.60. The objective lens in a laboratory microscope has a focal length of o 3.00 cmf = and provides an overall

magnification of = ⋅ 21.0 10 .m The distance between the two lenses is 30.0 cm.L = The focal length of the eyepiece, e ,f is given by:

( ) ( ) ( )( )( )

= = = =⋅e 2

o e o

25 cm 25 cm 25 cm 30.0 cm 2.5 cm.

3.00 cm 1.0 10L L

m ff f f m

33.61. The focal length of the objective lens is =o 7.00 mm.f The distance between the objective lens and the eyepiece lens is = 20.0 cm.L The magnitude of the magnification is 200.m = The viewing distance to the image is =i, 2 25.0 cm.d The focal length of the eyepiece, e ,f can be found from the equation for the

magnification of a microscope: ( )

= =i, 1 i, 2

o, 1 o, 2 o e

25.0 cm.

Ld dm

d d f f The focal length of the eyepiece is:

( ) ( )( )( )( )= = =e

o

0.250 m 25.0 cm 20.0 cm3.57 cm.

0.700 cm 200.L

ff m

The best choice is the lens marked with a 4.00 cm focal length.

33.62. The focal length of the eyepiece is e 2.0 cm.f = The focal length of the objective lens is o 0.80 cm.f = The relaxed viewing distance is typically =i,2 25 cm.d The distance between the lenses is 16.2 cm.L = In a microscope, the image of the objective lens forms approximately at the focal length of the eyepiece (see Figure 33.32 in the text) so that = + ≈ +i,1 o,2 i,2 e .L d d d f Then ≈ − = − =i,1 e 16.2 cm 2.0 cm 14.2 cm.d L f The object distance from the objective lens, o,1d is given by the thin lens equation, = +o o,1 i,11/ 1/ 1/ .f d d

Then ( ) ( ) ( )( )−−= − = − =

11o,1 o i,11/ 1/ 1/ 0.80 cm 1/ 14.2 cm 0.85 cm.d f d

33.63. THINK: The distance between the two lenses of the microscope, ,L is fixed. The magnitude of the magnification is to vary from 1 150m = to 2 450m = for substituted eyepieces of various focal lengths. The equation for the magnification of a microscope can be used to determine the focal length. The longest focal length eyepiece corresponds to the smallest magnification.


1201

SKETCH:

RESEARCH: The equation for the magnification of a microscope is: ( )

= =i, 1 i, 2

o, 1 o, 2 o e

25 cm.

Ld dm

d d f f

SIMPLIFY:

(a) ( ) ( )

1o e o e, 1

25 cm 25 cm

L Lm m

f f f f= = and

( )2

o e, 2

25 cm Lm

f f=

( ) ( )2 1e, 1e, 2 e, 1

1 o e, 2 o e, 1 e, 2 2

25 cm 25 cm

m mL L ff f

m f f f f f m

= ÷ = =

(b) ( ) ( )

oo e 1 e, 1

25 cm 25 cm .

L Lm f

f f m f= =

CALCULATE:

(a) ( )e, 215060. mm 20.0 mm450

f = =

(b) ( )( )( )( )o

25 cm 35 cm9.72 mm

150 6.0 cmf = =

ROUND: To two significant figures, (a) the shortest focal length of the eyepiece is e, 2 20. mm,f = and (b) the focal length of the objective lens should be o 9.7 mm.f = DOUBLE-CHECK: Using the calculated value of e,2 20 mmf = and its corresponding magnification

2 450m = yields a focal length for the objective lens of:

( ) ( )( )( )o

2 e, 2

25 cm 25 cm 35 cm9.7 mm.

450 2.0 cmL

fm f

= = =

33.64. The angular magnification of a refracting telescope is o e/ .m f fθ = − With an objective lens of focal length

o 10.0 m,f = and an eyepiece of focal length 2e 2.00 10 m,f −= ⋅ the magnification of this telescope is:

210.0 m / 2.00 10 m 500.,mθ−= − ⋅ = − where the negative sign indicates that the image is inverted.

33.65. The angular magnification of a refracting telescope is o e/ .m f fθ = − With an objective lens of focal length

o 100. cm,f = and an eyepiece of focal length e 5.00 cm,f = the magnification of this telescope is: 100. cm / 5.00 cm 20.0,mθ = − = − where the negative sign indicates that the image is inverted.

33.66. The angular magnification of a refracting telescope is e o o e/ / .m f fθ θ θ= − = − The telescope has an eyepiece focal length of e 25.0 mmf = and an objective focal length of o 80.0 mm.f = The magnification of this telescope is, therefore: o e/ 80.0 mm / 25.0 mm 3.20.m f f= = − = − Using the small angle approximation of tan ,θ θ≈ the angle subtended by the moon (the object) when viewed by the unaided

eye is (in radians) ( )6

3moon8

moon

2 1.737 10 m29.037 10 rad.

3.844 10 moRd

θ −⋅

= = = ⋅⋅

Thus, the angle subtended by the

image of the moon through the eyepiece is: ( )3 2e 3.20 9.037 10 rad 2.89 10 rad.omθ θ − −= = ⋅ = ⋅


1202

33.67. Galileo’s telescope had an objective lens with a focal length of =o 40.0 inchesf and an eyepiece lens with a focal length of =e 2.00 inches.f The angular magnification of the refracting telescope is o e/ .m f fθ = −

Therefore, θ = − = −40.0 inches 20.0,2.00 inches

m where the negative sign indicates that the image is inverted.

33.68. The two distant stars are separated by an angle of o 35 arcseconds.θ = The stars are observed to be separated by e 35 arcminutes 2100 arcsecondsθ = = through a refracting telescope. (There are 60 arcseconds in one arcminute.) This telescope has an objective lens of focal length o 3.5 m.f = The focal length of the eyepiece, e ,f is found from the equation for the angular magnification of a refracting telescope, e o o e/ / .m f fθ θ θ= − = − Then ef is

( )( )o oe

e

3.5 m 35 arcseconds5.8 cm.

2100 arcsecondsf

fθ

θ= = =

33.69. THINK: The telescope is a refracting telescope with a magnification of 180.m = It is adjusted for a relaxed eye when the two lenses are 1.30 mL = apart. The telescope is designed such that the image formed by the objective lens (which appears at its focal length o )f lies at the focal length of the eyepiece. Then the distance L between the two lenses is the sum of the two focal lengths: o e .L f f= + The magnification equation for a telescope can be used to find the focal length of each the objective lens, o ,f and the eyepiece lens, e .f SKETCH:

RESEARCH: The angular magnification of a refracting telescope is o e/ .m f fθ = − With two equations and two unknowns, the two focal lengths of and ef can be determined.

SIMPLIFY: o e o e/ ,m f f f f mθ θ= = ( )o e e e e / 1 .L f f f m f f L mθ θ= + = + = +

CALCULATE: ( )e1.30 m 7.182 mm,

1 180f = =

+ ( )( )o 7.182 mm 180 1.293 m.f = =

ROUND: The focal length of the eyepiece is 7.2 mm. The focal length of the objective lens is 1.3 m. DOUBLE-CHECK: The focal point of the objective should be much greater the focal length of the eyepiece for a refracting telescope.

33.70. THINK: The objective focal length of both telescopes is o 95.0 cmf = and the eyepiece focal length of both telescopes is e 3.80 cm.f = Telescope A has an objective diameter of oA 10.0 cmD = while telescope B has an objective diameter of oB 20.0 cm,D = and for the eyepiece diameter, eB eA2 .D D= (a) The angular magnifications of telescopes A and B can be found by using the magnification equation for telescopes. Both telescopes have the same angular magnification since both of their lenses have the same focal lengths. (b) The brightness of an image is proportional to the area of the lenses.


1203

SKETCH:

RESEARCH: (a) The angular magnification of a refracting telescope is o e/ .m f fθ = −

(b) The area of a lens is 2 / 4,A Dπ= where D is the diameter of the lens. SIMPLIFY: Not required. CALCULATE:

(a) 95.0 cm 25.0,3.80 cm

mθ = − = − where the negative sign indicates that the image is inverted.

(b) 2 2

A A A

B B A

1 .2 4

A D DA D D

= = =

Therefore, the images of telescope B are four times brighter than the

images of telescope A. ROUND: (a) To three significant figures, the magnification of telescopes A and B is 25.0.mθ = − DOUBLE-CHECK: (a) The magnification of the telescopes should have a magnitude greater than 1. (b) It is reasonable that the images of telescope B are brighter since more light enters through its larger lens.

33.71. THINK: Some reflecting telescope mirrors utilize a rotating tub of mercury to produce a large parabolic surface. The tub is rotating on its axis with an angular frequency .ω Conservation of energy can be used to show that the focal length of the resulting mirror is: 2/ 2 .f g ω= SKETCH:

RESEARCH: Consider a single drop of mercury in the rotating tub. The kinetic energy of this drop of mercury is given by:

( )221 12 2

K mv m rω= =

The gravitational potential energy relative to the bottom of the lowest point of the surface is given by: gU mgh=

SIMPLIFY: By conservation of energy, g :K U=

( )2 2

212 2

rm r mgh hg

ωω = =


1204

Now, the equation of a parabola with its vertex at the origin is given by 2

2 24 4 ,4rx fy r fh h

f= = =

where f is the focal length. Substitution gives: 2 2 2

2 .4 2 2

gr r ff g

ωω

= =

CALCULATE: Not required. ROUND: Not required. DOUBLE-CHECK: It is reasonable that the focal length is inversely proportional to the angular frequency of the tub since a faster rotation results in a steeper parabola.

Additional Problems

33.72. The object height is =o 4.0 cm.h It is projected onto a screen using a converging lens with a focal length of 35 cm.f = The image on the screen is = −i 56 cmh in size. (It is negative to represent the fact that the image has been inverted; the image must be real to be projected onto a screen, and for a converging lens a real image is always inverted). The distance from the lens to the screen is id and the distance from the object to the screen is +i o .d d The magnification is: = = −i o i o/ / .m h h d d Then

( ) ( )= − = − − =o o i i i i/ 4.0 cm / 56 cm /14.d h d h d d From the thin lens equation, ( )= −i o o/ .d d f d f Substitution for od gives the distance from the lens to the screen:

( )( ) ( ) = − = = + = = = = ≈ −

i ii i

i

/14 1 15 14 1 14 15 15 35 cm 5.25 m 5.3 m./14 14 14 14 14

d f d fd f d f f fd f

Therefore, the distance from the object to the screen is ( )+ = + =i o 5.25 m 5.25 m /14 5.6 m.d d

33.73. The eyeglasses of a near sighted person use diverging lenses and create virtual images of objects for the near sighted wearer. When a normal person wears these eyeglasses, the person with normal vision will only be able to focus on these virtual images if they fall within the focusable distances of a normal eye, which is from 25 cm out to infinity. Since only the most distant objects can be focused on, the objects at infinity must be making virtual images at the normal near point of 25 cm. This will happen when:

−= + = ∞ + − = − 1o i1/ 1/ 1/ 1/ 1/ 0.25 m 4.0 m .f d d Note that id is negative because the image is virtual.

The prescription strength of the eyeglasses is about 4.0 diopter.−

33.74. The focal length of the spectacles is the reciprocal of the power, so the focal length is 1/ 0.20 diopter 5.0 m.f = − = − Therefore, light from a distant object will form a virtual image 5 m in

front of the spectacles. Since this is a distance at which your eye can bring objects to a focus, you will still be able to focus on distant objects. The problem comes from near objects. This is a diverging lens (negative focal length), so light from nearby objects will be even more divergent, and therefore, more difficult for your eye to focus. Since the near point of your eye is 20. cm , virtual images formed by the spectacles cannot be closer than 20. cm. Otherwise, your eye will not be able to focus. From the thin lens equation

( )

−− + = = − = − = −− ⋅

11

o 2o i i

1 1 1 1 1 1 1 21 cm.20. cm5.0 10 cm

dd d f f d

Thus, the range over which you will be able to see is from 21 cm to infinity. The spectacles have hardly changed your range because they are low in power.

33.75. In the derivation of the Lens Maker’s Formula, the following relation can be inferred (in the text it was assumed that 1 1) :n = ( )+ = −1 o 2 i 2 1/ / / .n d n d n n R In this case, the refracting surface is flat so R is


1205

infinite ( ).R = ∞ The equation can be rearranged as ( )= −o 1 2 i/ .d n n d With the fish (the object) in water,

1 1.33n = and with you in air, 2 1.n = The apparent depth of the fish is the virtual image distance, = −i 10. cm.d (it is negative because it is on the same side of the surface as the object and therefore, a

virtual image.) Then ( )( )= − − =o 1.33 /1 10. cm 13 cm.d The fish is actually 13 cm under the surface of the water, and must be grabbed at this position.

33.76. The mirror has a focal length of = 40.0 cm.f To project the image onto a screen, the image must be real, and therefore, the mirror must be a concave mirror with 0.f > The bird has a height =o 10.0 cmh and is

=o 100. md away from the mirror. From the mirror equation, ( )= −i o o/ .d d f d f From the equation for magnification, = = −i o i o/ / ,m h h d d the image height is:

( ) ( )( )( ) ( )

−= − = − = − = − = −

− ⋅ −o o oi

i o o 4o o o

40.0 cm 10.0 cm/0.402 mm.

1.00 10 cm 40.0 cmd f d f fhdh h h

d d d f

The image of the bird is inverted, but it is much smaller than one centimeter in size. Therefore, he will not make good on his claim.

33.77. The object is =o 6.0 cmd away from a thin lens of focal length 9.0 cm.f = The image distance id is determined from the thin lens equation: + =o i1/ 1/ 1/ .d d f Therefore,

−− = − = − = −

11

io

1 1 1 1 18 cm.9.0 cm 6.0 cm

df d

The image is 18 cm from the lens, and on the same side of the lens as the object (the negative sign indicates that it is a virtual image).

33.78. The spherical lens bulges outwards in the middle on both sides so it is a convex lens. The surfaces are ground to radii of 0.25 m and 0.30 m. The radii will have opposite signs, and since there will be an absolute value it does not matter which is taken to be negative. Take 1 0.25 mR = and 2 0.30 m.R = − Using the Lens Maker’s Formula, the power of the lens is:

( ) ( )glass1 2

1 1 1 1 11 1.5 1 3.7 diopter.0.25 m 0.30 m

D nf R R

= = − − = − − =−

33.79. The convex surface is part of a sphere with radius 0.45 m.r = The concave surface is part of a sphere with radius 0.20 m,R = and r and R have the same sign. Using the Lens Makers Formula, the power of the lens is:

( ) ( )glass1 1 1 1 11 1.5 1 1.4 diopter.

0.45 m 0.20 mD n

f r R= = − − = − − =

Since the lens is a diverging lens, the answer should be taken to be negative. The answer is 1.4 diopter.D = −

33.80. The farsighted person can clearly see an object if it is at least 2.5 m away; therefore, for this person the image distance is = −i 2.5 m.d Using the thin lens equation, the power of the lenses required to read a book a distance =o 0.20 md away is:

= = + = + =−o i

1 1 1 1 1 4.6 diopter.0.20 m 2.5 m

Df d d

Since the power is positive, the will require glasses with converging lenses.


1206

33.81. The magnifying glass is a converging lens. If you hold the magnifying glass at =i 9.20 cmd above your desk you can form a real image on the desk of a light directly overhead. The distance from the light to the table is 235 cm.h =

Using the thin lens equation, where = −o i ,d h d the focal length of the magnifying glass is: − − − = + = + = + = − −

1 1 1

i o i i

1 1 1 1 1 1 8.84 cm.9.20 cm 235 cm 9.20 cm

fd d d h d

33.82. The girl needs to hold the book at a distance 15 cm from her eyes to clearly see the print. This is her near point. (a) The girl is nearsighted since she can see objects close to her eye. Therefore, she requires diverging lenses in order to see the book 25 cm away. (b) The thin lens equation can be used to find the focal length of the lens: = +o i1/ 1/ 1/ .f d d Substituting

=o 25 cmd and = −i 15 cmd gives a focal length of ( ) 11/ 25 cm 1/15 cm 38 cm.f −= − = −

33.83. The focal length of the camera lens is 38.0 mm.f = The lens must be moved a distance dΔ to change

focus from a person at = ⋅ 4o 3.00 10 mmd to a person that is at = ⋅ 3

o ' 5.00 10 mm,d where Δ = −i i ' .d d d

Using the thin lens equation, =−

oi

o

fdd

d f and =

−o

io

'' .

'fd

dd f

Therefore, the lens must be moved a distance

( )( )( )

( )( )( )

⋅ ⋅Δ = − = − = − =

− − ⋅ − ⋅ −

4 3o o

i i 4 3o o

38.0 mm 3.00 10 mm 38.0 mm 5.00 10 mm'' 0.243 mm.

' 3.00 10 mm 38.0 mm 5.00 10 mm 38.0 mmfd fd

d d dd f d f

33.84. The magnitude of a telescope’s magnification is 41.m = The focal length of the eyepiece is e 0.040 m.f = The magnitude of the magnification is given by: o e/ .m f f= Solving for of gives:

( )( )o e 41 0.040 m 1.6 m.f m f= = =

33.85. THINK: The object is =o,1 2.0 cmh high and is located at 0 0 m.x = A converging lens with focal length 50. cmf = is located at = =L o,1 30. cm.x d A plane mirror is located at m 70. cm,x = so the distance

between the lens and the mirror is m L 40. cm.L x x= − = The image formed by the lens will act as the object for the plane mirror. The thin lens equation can be used to determine the position i,2x and the size

i,2h of the final image. SKETCH:


1207

RESEARCH: The thin lens equation is + =o i1/ 1/ 1/ .d d f The magnification of a lens is = = −i o i o/ / .m h h d d For plane mirrors, =i od d and =i o .h h

SIMPLIFY: When the thin lens equation is rearranged to solve for the image distance, it becomes

=−

oi

o

.fd

dd f

The image produced by the lens is located a distance of =−o,1

i,1o,1

fdd

d f from the lens. Since

> o,1 ,f d i,1d will be negative, and therefore, on the same side of the lens as the object. This image acts as

the object for the mirror, and is a distance = +o,2 i,1d L d from the plane mirror. The final image is the

image created by the plane mirror, and will appear =i,2 o,2d d to the right of the mirror. The final image position is given by = +i,2 m i,2 .x x d Since the mirror does not change the height of the image, the

magnification is due to the lens, and the final height of the image is = − i,1i,2 o,1

o,1

.d

h hd

CALCULATE: The image distance for the lens is ( )( )

( ) ( )= = −−i,1

30. cm 50. cm75 cm.

30. cm 50. cmd The object distance

for the plane mirror is = + − =o,2 40. cm 75 cm 115 cm.d Therefore, the position of the final image is

= + =i,2 70. cm 115 cm 185 cm.x The size of the final image is ( )( )

( )−

= − =i,2

75 cm 2.0 cm5.0 cm.

30. cmh

ROUND: To two significant figures, the final image is =i,2 190 cmx to the right of the object and the size of the final image is =i,2 5.0 cm.h DOUBLE-CHECK: Since <od f for the converging lens, the image of the lens must be virtual, enlarged and upright. The plane mirror cannot change these attributes, so the calculated results agree with these expectations ( )> >i,2 o,1 0 .h h

33.86. The distance from the lens to the retina at the back of the eye is 2.0 cm. The focal length can be found with the thin lens equation: ( )−= + 1

o i1/ 1/ .f d d (a) The focal length of the lens when viewing a distant object

( )= ∞od is ( ) 11/ 1/ 2.0 cm 2.0 cm.f

−= ∞ + = (b) The focal length of the lens when viewing an object

=o 25 cmd from the front of the eye is ( ) 11/ 25 cm 1/ 2.0 cm 1.9 cm.f

−= + =

33.87. You require lenses of power 8.4 diopter.D = − A negative power infers that the focal length is negative, so diverging lenses are being used. In a nearsighted eye, light comes to a focus before it reaches the retina and diverging lenses are required to correct this. Therefore, you are nearsighted. For nearsighted eyes, corrective lenses focus distant objects ( )= ∞od at the near point, so = −i near .d d Solving the thin lens equation for neard gives:

( )−= = + = ∞ − − = = − − =1o i near near1/ 1/ 1/ 1/ 1/ 1/ 1/ 8.4 m 0.12 m.D f d d d d D

Without glasses the book must be held 12 cm from your eye in order to read clearly.

33.88. Jack has a near point of near 32 cm 0.32 md = = and the power of the magnifier is 25 diopter.D = (a) The focal length is given by 1 /f D= and the angular magnification of a magnifier for an image

formed at infinity is near .d

mf

= Therefore, ( )( )1near 0.32 m 25 m 8.0.m d D −= = =


1208

(b) If the final image is at the near point then −−

= = − =

near neari

o o o

.d ddm

d d d Using the thin lens equation:

−−

= + = − = + = +

11

nearo

i o i near near

1 1 1 1 1 1 1 .fd

df d d f d f d f d

Therefore the magnification is:

( ) ( )( )near nearnear near

near

near

1 1 1 1 25 m 0.32 m 9.0.d d

m f d Ddfd f f

f d

= = + = + = + = + =

+

33.89. The diameter of the glass marble ( )=g 1.5n is = =2.0 in 5.1 cm.d The radius of curvature of the marble is

then / 2.R d= Holding the marble a distance of = =o,1 1.0 ft 30. cmd from your face, the distance of the image formed by the first side of the marble is:

( )( )

( )( )( )( )( ) ( )

−+ = = = =

− −− −gg g o,1

i,1o,1 i,1 o,1 g

2 1 1.5 5.1 cm 30. cm1 9.217 cm.2 30. cm 1.5 1 5.1 cm2 1

nn n ddd

d d d d n d

This image acts as the object for the second surface, for which the radius of curvature is negative (concave), = −5.1 cm.d Since >i ,1 ,d d the image for the second surface appears past it, so = −o,2 i,1 .d d d Therefore, the final image distance can be computed as follows.

( )( )

( )( )( )

gg i,1o,2i,2

o,2 i,2 o,2 g g i,1 g g

2 11 ,2 1 2 1

nn d d dddd

d d d d n dn d d n dn

− −+ = = =

− + − − +

( )( )( )( ) ( )( )

− += = =

+ − + −i,2

5.1 cm 9.217 cm 5.1 cm3.324 cm 1.3 in.

2 9.217 cm 5.1 cm 1 1.5 5.1 cm 1.5d

The magnification is ( )( )( )

( )( )= − = − = − = −− +

i ,1 i,2 i,1 i,2

o,1 o,2 o,1 i,1

9.217 cm 3.324 cm0.070,

30.48 cm 9.217 cm 5.1 cmd d d d

md d d d d

where

the negative sign indicates that the image is inverted.

33.90. THINK: The diverging lens has a focal length of 2 30.0 cm.f = − It is placed a distance 15.0 cmx = behind a converging lens with focal length, 1 20.0 cm.f = The thin lens equation can be used to find the image location for an object that is located at infinity in front of the converging lens. The image formed by the converging lens will act as the object for the diverging lens. SKETCH:

RESEARCH: The thin lens equation is: = +i o

1 1 1 .f d d


1209

SIMPLIFY: For the converging lens: = + = + =∞ i, 1 1

1 o, 1 i, 1 i, 1

1 1 1 1 1 .d ff d d d

The object distance of the

diverging lens can now be written as: = − = −o, 2 i, 1 1.d x d x f Substituting this into the thin lens equation for the diverging lens gives:

( ) ( )

−

= + = + = − − −

1

i, 22 o, 2 i, 2 1 i, 2 2 1

1 1 1 1 1 1 1 .df d d x f d f x f

CALCULATE: ( )

− = − = − −

1

i, 21 1 6.00 cm

30.0 cm 15.0 cm 20.0 cmd

ROUND: To three significant figures, the object at infinity will be focused =i, 2 6.00 cmd to the right of the diverging lens. DOUBLE-CHECK: This result agrees with the diagram shown above. It is expected that the diverging lens causes the focal point to be beyond the focus of the converging lens.

33.91. THINK: The instructor wants the lens to project a real image of a light bulb onto a screen a distance 1.71 mD = from the bulb. The thin lens equation can be used to find the focal length that is required to

achieve a magnification of 2.m = SKETCH:

RESEARCH: The image is real and enlarged; therefore, the focal length f must be smaller than the object distance o .d The distance from the bulb to the screen is = +o i ,D d d where id is the image distance. The

magnitude of the magnification is = i o/ .m d d The thin lens equation is: = +i o

1 1 1 .f d d

SIMPLIFY: Also, = = =i o i o2 / 2 .m d d d d Therefore, = + = + =i o o o o2 3D d d d d d or =o / 3.d D

From the thin lens equation, ( ) ( )

11

i o

1 1 1 1 2 .2 / 3 / 3 9

f Dd d D D

−− = + = + =

CALCULATE: ( )2 1.71 m 0.380 m9

f = =

ROUND: To three significant figures, the focal length required is 38.0 cm.f = DOUBLE-CHECK: The calculated focal length has the correct units. The answer seems reasonable considering the values provided in the question.

33.92. THINK: The length of the refracting telescope is 55 cmL = and it has a magnification of 45.m = The equation for the magnification of a telescope can be used to find the focal length of its objective, of and the focal length of its eye lens, e .f The length of a refracting telescope is just the sum of the focal lengths,

o e .L f f= +


1210

SKETCH:

RESEARCH: The magnification of a refracting telescope is: o e/ .m f f=

SIMPLIFY: The focal length of the eye lens is: ( ) ( )o e e e e e1 .1

Lf f m L f m f f m fm

= = + = + =+

The focal length of the objective lens is: ( )o .1L m

fm

=+

CALCULATE: ( )e

55 cm1.196 cm,

1 45f = =

+

( )( )( )o

55 cm 4553.80 cm

1 45f = =

+

ROUND: To two significant figures, the focal length of the objective lens is o 54 cmf = and the focal length of the eye lens is e 1.2 cm.f = DOUBLE-CHECK: As shown in the diagram, it is expected that o e .f f>

33.93. THINK: The converging lens has a focal length =L 50.0 cm.f It is 175 cmL = to the left of a metallic sphere. This metallic sphere acts as a convex mirror of radius = −100. cmR (the radius of curvature of a diverging mirror is negative) and focal length = = −m / 2 50.0 cm.f R The object of height, = 20.0 cm,h is a distance =o,1 30.0 cmd to the left of the lens. The thin lens equation, the mirror equation, and the magnification for a system of optical elements can be used to find the height of the image formed by the metallic sphere, i,2 .h The image formed by the lens acts as the object for the mirror. SKETCH:

RESEARCH: The thin lens equation is i o

1 1 1 ,f d d

= + The magnification (for lenses and mirrors) is

i o i o/ / .m h h d d= = − The total magnification m is the product of the magnification of the lens and the mirror: L m .m m m=

SIMPLIFY: For the lens, the thin lens equation can be rearranged as: o,1 Li,1

o,1 L

.d f

dd f

=−

Since L o,1 ,f d> i,1d

is negative, so the image is on the same side as the object (the image is virtual). This image acts as the object for the mirror at a distance of o,2 i,1d L d= + from the metallic sphere. The location of the image

produced from the sphere is o,2 mi,2

o,2 m

.d f

dd f

=−

The final image height is


1211

( )( )( )

− − = = = = = − + − −

o,1 L o,2 m

i,1 i,2 o,1 L o,2 m L mi,2 L m

o,1 o,2 o,1 o,2 o,1 Lo,1 L m

o,1 L

.

d f d fd d d f d f f f hh mh m m h h hd d d d d f

d f L fd f


( ) ( )( ) ( )

−= =

− + − − −

i,2

50.0 cm 50.0 cm 20.0 cm8.3333 cm

30.0 cm 50.0 cm30.0 cm 50.0 cm 175 cm 50.0 cm

30.0 cm 50.0 cm

h

ROUND: To three significant figures, the height of the image formed by the metallic sphere is =i,2 8.33 cm.h

DOUBLE-CHECK: It is expected that <i,2 .h h For a converging lens, an image produced by an object placed within the focal length of the lens is enlarged, virtual and upright. For a diverging mirror, the image is always virtual, upright and reduced. Therefore, the height of the final image should be less than the height of the object since both the lens and the mirror act to reduce it.

33.94. THINK: The lens has a focal distance, 10.0 cm.f = The laser beam exits a pupil of diameter,

o 0.200 cmD = that is located a distance o 150. cmd = from the focusing lens. Consider the case when the image of the exit pupil forms on the sample. (a) The thin lens equation can be used to find the distance,

i ,d from the sample to the lens and (b) the magnification equation can be used to find the diameter, i ,D of the laser spot on the sample (this is the image of the exit pupil). SKETCH:

RESEARCH:

(a) The thin lens equation is: i o

1 1 1 .f d d

= +

(b) The magnification is i i

o o

.D dmD d

= = −

SIMPLIFY:

(a) 1

io

1 1 df d

−

= −

(b) i oi ii

o o o

d DD d D

D d d= − = −

CALCULATE:

(a) 1

i1 1 10.714 cm

10.0 cm 150. cmd

−

= − =

(b) ( )( )

i

10.714 cm 0.200 cm0.1429 mm,

150. cmD = − = − where the negative sign indicates that the image is

inverted. ROUND: Round to three significant figures.


1212

(a) The sample is located i 10.7 cmd = past the lens. (b) The image of the exit pupil has a diameter of i 0.143 mm.D = DOUBLE-CHECK: The laser beam is being focused on the sample so it is reasonable that the diameter of the laser sport on the sample is smaller than the exit pupil.

33.95. THINK: The computer monitor is at a distance of 0.55 mL = from his eyes. The image of the monitor must be located at his near point, near 1.15 m.d = Since the image is located in front of the lens (the image is virtual), the image distance is ( )= − −i near e .d d d Since the lens-eye distance for his glasses is known to be

e 0.020 m,d = the object distance from the lens to the computer monitor is = −o e .d L d The thin lens equation can be used to find the lens power required. SKETCH:

RESEARCH: The thin lens equation is given by = +o i

1 1 1 .f d d

SIMPLIFY: The power of the lens is defined as 1/D f= where f is in meters. Therefore,

e near e

1 1 1 .Df L d d d

= = −− −

CALCULATE: 1 1 1.0018 diopter0.55 m 0.020 m 1.15 m 0.020 m

D = − =− −

ROUND: To two significant figures, his optician should prescribe a power of 1.0 diopter.D = DOUBLE-CHECK: Since the power is positive, a converging lens must be used. Since the object is inside his near point a converging lens is expected in order to correct his vision.

33.96. THINK: An image of a far away object produced by an objective lens of a telescope is located at the focal point of the objective lens. This image becomes the object for the eyepiece. The focal length of the eyepiece is e 8.0 cmf = and the image is to be projected on a screen that is a distance of 150 cmL = past the original location of the eyepiece. The thin lens equation can be used to determine how far the eyepiece must be moved.


1213

SKETCH:

RESEARCH: The thin lens equation is given by: = +

o i

1 1 1 .f d d

SIMPLIFY: The distance from the object to the eyepiece is = +o ed f d and the distance from the image to the eyepiece is = −i .d L d Therefore, the thin lens equation becomes:

( )( ) ( )( ) ( )( ) ( )e ee e e

e e e e

1 1 1 L d f d L ff d L d f L f

f f d L d f d L d f d L d− + + +

= + = = + − = ++ − + − + −

( ) ( )2 2 2 2e e e e e e 0.d L f d f L f L f d f L d f − + − + = + + − + =

Solving this quadratic equation for d yields: ( ) ( )2 2

e e e4.

2f L f L f

d− − ± − −

=

CALCULATE: Substituting the numerical values gives:

( ) ( ) ( )2 28.0 cm 150 cm 8.0 cm 150 cm 4 8.0 cm0.452 cm or 142 cm.

2d

− − ± − −= =

The most realistic distance is 0.452 cm.d = ROUND: To two significant figures, the eyepiece should be moved a distance of 4.5 mmd = towards the screen.

DOUBLE-CHECK: Since ,L d>> the thin lens equation can be approximated by e e

1 1 1 .f f d L

≈ ++

Solving

this equation for d gives 1 1 1

e ee e

1 1 1 1 1 1 8.0 cm 0.45 cm.8.0 cm 150 cm

f d d ff L f L

− − − + = − = − − = − − =

This approximation is the same as what was obtained above.


1214

Chapter 34: Wave Optics

In-Class Exercises

34.1. c 34.2. d 34.3. c 34.4. b 34.5. b 34.6. a Multiple Choice

34.1. c 34.2. c 34.3. d 34.4. a 34.5. a 34.6. a 34.7. b 34.8. c Questions

34.9. The fringe width, defined as the distance between two bright or dark fringes, is given by / .y L dλΔ = (a) If the wavelength is increased, the fringe width will increase or the pattern will expand. (b) If the separation distance between the slits is increased, the fringe width will decrease or the pattern will shrink. (c) If the apparatus is placed in water or the wavelength is decreased, the fringe width will decrease or the pattern will shrink.

34.10. Diffraction effects depend on the ratio between the size of an obstacle and the wavelength of light. If the diffraction effect for a sound wave is similar to that of light, the wavelength of a sound wave should be similar to light. Let us assume the wavelength of sound is about 500 nm.λ = Since the speed of sound is about 340 m/s, the frequency corresponding to this wavelength is

9

340 m/s 680 MHz.500 10 m

vfλ −= = =

⋅

34.11. A radio telescope is so much larger than an optical telescope because the wavelength of a radio wave is much larger than the wavelength of visible light. Since the resolution of a telescope is proportional to the ratio / Dλ ( D is the diameter of the telescope), in order to get similar resolution as the visible light, the diameter of the telescope must be larger. With similar reasoning, since the wavelength of x-ray’s are much less than visible light, the diameter of an x-ray telescope can be smaller than a visible light telescope.

34.12. Yes, light can pass through such a slit. Using Huygen’s principle, where each point on the wave front of light acts as a source of a spherical wave, the diffraction pattern of a very narrow slit is produced by a single spherical wave. The intensity as a function of angle from the direct beam is

( ) ( )( )( )( )

π θ λθ

π θ λ=

2

o 2

sin sin /,

sin /

aI I

a where oI is the intensity at 0.θ = Since d is less than ,λ the ratio λ/a is

less than 1. As a consequence, the intensity falls off with the angle ;θ but it never reaches the minimum.

34.13. (a) A hologram is an interference pattern produced by the interference of two light sources (object and reference sources). The recorded pattern acts as a diffraction grating for the light shining on it. The scale of a diffraction pattern is set by the wavelength of the light. The size of the image produced by the hologram is proportional to the wavelength of the light that produced the hologram. Therefore, if white light is used, it will produce a set of nested images of different colors, the size of each image is proportional to its wavelength. (b) The size of each image is proportional to the wavelength. The longest wavelengths of the visible light, is those of red light, produce the largest images. Conversely, the violet light, the smallest wavelength, produces the smallest image.


1215

34.14. No, it will not. No interference pattern will be produced since the light source is not a coherent light source.

34.15. There are two advantages: (a) The intensity of the collected radio wave is increased. (b) The effective diameter of the telescopes is increased producing a better resolution.

34.16. The maximum of a diffraction pattern located at angles determined by the equation θ λ=sin / .m d For a maximum to be visible on screen, the angle must be less than 90° or λ /m d than 1. This means there is an upper limit on the value of m that satisfies the above equation. Therefore, the number of maxima is finite.

34.17. For a circular aperture telescope, the minimum angle resolvable or the limiting angle is given by Rayleigh’s criterion, θ λ=R 1.22 / ,D where λ is the wavelength and D is the diameter of the aperture. Since the blue light has smaller wavelength than the red light, the minimum angle for the blue light is also smaller than for the red light. Therefore, two blue stars are more resolvable than two red stars.

34.18. Bright spots on the screen behind a diffraction grating are produced when there is a constructive interference. The condition for the constructive interference is sind mθ λ= or ( )θ λ−= 1sin / .m d Since green light has smaller wavelength than red light, it will produce bright spots at smaller angles. Therefore, the green bright spots will be closer together.

Problems

34.19. The wavelength of EM radiation in a medium with a refractive index n is o / nλ λ= where oλ is the wavelength of light in a vacuum. Similarly the speed of light in the medium is = / .v c n

(a) The wavelength of a helium-neon laser in Lucite is o 632.8 nm 421.9 nm1.500n

λλ = = =

(b) The speed of light in the Lucite is 8

82.998 10 m/s 1.999 10 m/s.1.500

cvn

⋅= = = ⋅

34.20. The wavelength of light in a medium is o / .nλ λ= Thus, the wavelength of the light from a HeNe laser in

water is 632.8 nm 474.7 nm.1.333

λ = = The color of the light in water is the same as the color in the air, since

the color of a light is determined from its frequency, not its wavelength. The frequency of light does not change as it passes different medium.

34.21. One wavelength corresponds to a phase difference of 2 .π Therefore, the minimum path difference which

causes a phase shift by / 4π is ( )/ 4 1 1 700. nm 87.5 nm.2 2 8 8

x θ πλ λ λπ π

ΔΔ = = = = =

34.22. A constructive interference occurs when the path difference between two coherent light sources is a multiple of wavelength. A destructive interference occurs when the path difference is ( )1/ 2 .x m λΔ = + By dividing the path difference by the wavelength, the properties of the interference can be determined.

The ratio of the path difference and wavelength is 2

59

20.25 10 mratio 4.500 10 .450.0 10 m

xλ

−

−

Δ ⋅= = = ⋅⋅

The ratio is a

multiple of the wavelength. Therefore, the interference is constructive.


1216

34.23. For a Young’s interference experiment, the maxima of the interference pattern is located at / .y m L dλ= Substituting 1m = for the first maximum intensity yields / .y L dλ= Therefore, the distance between the

slits and the screen is ( )( )3 3

9

5.40 10 m 0.100 10 m1.0 m.

540 10 mydLλ

− −

−

⋅ ⋅= = =

⋅

34.24. The maxima of the fringe pattern is located at / .y m L dλ= The separation between the central maximum intensity ( )0m = to the next maximum intensity ( )1m = is / .y L dλΔ = Note that d is the distance between the centers of the two slits, that is, 1.00 mm 1.50 mm 2.50 mm.d = + = Thus, the separation

between the maxima is ( )( )9

3

633 10 m 5.00 m0.001266 m 1.27 mm.

2.5 10 my

−

−

⋅Δ = = ≈

⋅

34.25. THINK: The intensity of light is proportional to the square of the electric field. The light has wavelength 514 nmλ = and the slits are separated by a distance of 0.500 mm.d = The intensity of the radiation at

the screen 2.50 m away from each slit is 180.0 W/cm2 (not the maximum intensity, max ).I However, this intensity is not needed to find the position where max / 3.I I= SKETCH:

RESEARCH: The intensity of the light produced by the interference from two narrow slits on a distant screen is given by:

2max4 cos .dyI I

Lπλ

=

SIMPLIFY: For max 1/3/ 3, :I I y y= →

2 11/3 1/3maxmax 1/3

1 14 cos cos cos .3 12 12

dy dyI LI yL L d

π π λλ λ π

− = = =

CALCULATE: Substituting the numerical values gives

( )( )( )

91

1/3 4

514 10 m 2.50 m 1cos 0.001045 m.5.00 10 m 12

yπ

−−

−

⋅ = = ⋅

ROUND: To three significant figures, 1/3 1.05 mm.y = DOUBLE-CHECK: As a comparison the first minimum intensity is located at

( ) ( )( )( )( )

9

4

1/ 2 514 10 m 2.50 m1/ 21.29 mm.

5.00 10 mL

yd

λ −

−

⋅= = =

⋅

The result for 1/3y is less than 1.29 mm, as expected.

34.26. THINK: The new wavelength as light passes through a medium of refractive index of n is given by

o / .nλ λ= The 10th dark fringe corresponds to a path difference of ( )1/ 2 ,x m λΔ = + with 9.m =


1217

SKETCH:

RESEARCH: The path difference between two paths 1(P and 2 )P is given by ( )sin 1 ,x d n tθΔ = − − where t is the thickness of a glass slide. The central fringe is when 0,xΔ = that is ( )sin 1 .d n tθ = − This central fringe corresponds to the 10th dark fringe for the interference without the glass slide. The condition for the 10th dark fringe is ( )θ λ= +sin 9 1/ 2 .d

SIMPLIFY: From the equations in the Research step, it can be concluded that ( ) ( )1 9 1/ 2 .n t λ− = +

Therefore, the refractive index is λ = + +

19 1.2

nt

CALCULATE: Putting in the numerical values gives = + + =

1 633 nm9 1 1.5011.2 12000 nm

n

ROUND: Keeping three significant figures yields =1.50.n DOUBLE-CHECK: This value is within the expected range for glass.

34.27. The minima of the interference pattern produced by a thin film is related to its thickness by 2 / .t m nλ= The first dark band which corresponds to the thinnest and is when m D= or when the thickness is much less than .λ The next dark bands are for 1m = and 2.m = Therefore, the thicknesses that produces the

dark bands are ( )λ= = = ≈1

1 550 nm 208 nm 210 nm2 2 1.32

tn

and λ λ= = = = ≈22 550 nm 417 nm 420 nm.2 1.32

tn n

34.28.

Since airn is less than oil ,n there will be a phase change of ( )1/ 2 λ and 180° in the light reflected by the air-oil interface. However, for the oil-water interface, there will be no phase change since oil water .n n> Therefore, in order to get a constructive interference, the path difference between two reflected light

waves must be oil

1 .2

x mnλ Δ = +

Using 2 ,x tΔ = it becomes

oil

12 .2

t mnλ = +

The wavelength that

satisfies this requirement is ( )( )oil 2 100.0 nm 1.472 294 nm .

1/ 2 1/ 2 1/ 2tn

m m m= = =

+ + +λ Since 0,1,2...,m = the only

possible white light that is reflected is for 0.m = Thus ( )2 294 nm 588 nm.λ = =


1218

34.29.

The first interface (air-hafina) causes a phase change of 180° in the first reflected light wave ( )1 .r The second interface does not cause a phase change since 1 2 .n n> Therefore, to get a constructive interference

in the reflected light, the path difference must be 1

12 .2

x t mnλ Δ = = +

The minimum thickness of the

thin film is when 0,m = that is, ( )1

1.06 μm1 0.139 μm 139 nm.4 4 1.90

tnλ= = = =

34.30.

A constructive interference is needed in the reflected light. There are two possible answers to this problem depending on the value of lens .n If 1 lens ,n n> using similar reasoning as in problem 34.29, the minimum

thickness is ( )1

1 800.0 nm 145 nm.4 4 1.38

tn

= = =λ If 1 lens ,n n< there will be a phase change of 180° in the light

reflected by MgF2-lens interface. Therefore, the condition for constructive interference is the path

difference 12 / .x t m nλΔ = = The minimum thickness ( )1m = is ( )1

1 800.0 nm 290. nm.2 2 1.38

tn

= = =λ While

both answers are technically correct, it makes sense to assume that the refractive index of the film is greater than that of the lens in order to achieve the absolute minimum thickness. Hence, choose 145 nm as the final answer.

34.31. THINK: It is assumed that the refractive index of mica is independent of wavelength. In order to solve the problem, the condition for destructive interference of the reflected light is required. The film has thickness 1.30 μm.t = The wavelengths of interest are 433.3 nm, 487.5 nm, 557.1 nm, 650.0 nm, and 780.0 nm. SKETCH:


1219

RESEARCH: Since air mica ,n n< the light reflected by the first interface 1I has a phase change of 180 .° The light reflected by the second interface ( )2I has no phase change. The condition for destructive interference in the reflected light is

air 2 ( 0,1, 2,...).m t mn

= =λ

For two adjacent wavelengths with 2 1 ,λ λ> 2 1 1.m m= − Therefore,

1 11 2

2 2and 1 .nt ntm mλ λ

= − =

SIMPLIFY: Solving these two equations for the refractive index n gives:

( )1 2

1 2 1 2 2 1

2 2 1 11 2 1 .2

nt nt nt nt

λ λλ λ λ λ λ λ

= + − = = −

CALCULATE: Choosing two adjacent wavelengths, 1 432 nmλ = and 2 483 nmλ = and substituting into

the above equation yields ( )( )

( ) ( )( )9 9

6 9

433.3 10 m 487.5 10 m1.499.

2 1.30 10 m 487.5 433.3 10 mn

− −

− −

⋅ ⋅= =

⋅ − ⋅

ROUND: To three significant figures, the refractive index of the mica is 1.50.n = DOUBLE-CHECK: Choosing another two adjacent wavelengths, 1 650.0 nm=λ and 2 780.0 nm,=λ the refractive index is found to be

( )( )( ) ( )( )

9 9

6 9

650.0 10 m 780.0 10 m1.50.

2 1.30 10 m 780.0 650.0 10 mn

− −

− −

⋅ ⋅= =

⋅ − ⋅

This is in agreement with the previous result.

34.32. THINK: To determine the condition for a bright band (constructive interference), the phase shift at the interfaces and the path difference between the two exiting beams of light need to be determined. Since both Beam 1 and Beam 2 pass through the same thickness of glass, the refractive index of glass is not needed to solve the problem. This means that the location of the bright bands will be the same for any material. SKETCH:

RESEARCH: Since gn is larger than air ,n there is no phase change in the reflected light 1.r But for the

reflected light 2 ,r there is a phase change of 180 .° Therefore, the condition for constructive interference is,

( ) airair

2 1122 4

mt m t

λλ

+ = + =

( 0, 1, 2,...).m =

This can be related to the location x of the bright fringes from the geometry of the set up. The air wedge has length 2

max 8.00 10 mx −= ⋅ and at this location it has thickness 5max 2.00 10 m.t −= ⋅

SIMPLIFY: If θ is the angle of the wedge:


1220

max

maxtan .

tx

=θ

In general, the location of bright fringes is: ( ) air maxmax

brightmax max

2 1.

4m xtx

xt t

λ+= =

The number of bright bands is found by setting bright maxx x= and solving for :m

( ) air max

max air

2 1 41 2 1 ,

4m t

mt

λλ

+= + =

max

air

2 1 .2

tm

λ= −

CALCULATE: The location of the bright bands as a function of m is:

( )( )( ) ( ) ( )

9 24

bright 5

633 10 m 8.00 10 m2 1 2 1 6.33 10 m.

4 2.00 10 mx m m

− −−

−

⋅ ⋅= + = + ⋅

⋅

The number of bright bands is:

( )( )

5

9

2 2.00 10 m 1 62.69.2633 10 m

m−

−

⋅= − =

⋅

ROUND: To three significant figures, ( ) ( )4bright 2 1 6.33 10 m, 0, 1, 2,... .x m m−= + ⋅ = The number of full

bright bands is 62.m = DOUBLE-CHECK: Setting 62.69m = should give 8.00 cm:x =

( )( ) 4bright 2 62.69 1 6.33 10 m 0.0800 m 8.00 cm,x −= + ⋅ = = as required.

34.33. THINK: The path length difference between the two beams and phase shifts at the interfaces need to be considered. For a plano-convex lens with focal length 0.8000 mf = and index of refraction l 1.500,n = the Lens-Maker’s Formula can be used to determine the radius of curvature of the lens. The third bright circle is observed to have a radius of 30.8487 10 m.r −= ⋅ SKETCH:

RESEARCH: Since l air,n n> there is no phase change in the reflected beam 1.r However, there is a phase change of 180° for the beam reflected by the mirror. Because the path length difference between the two beams is 2x dΔ = and there is a phase change of 180° in one of the beams, the condition for constructive interference is ( )2 1/ 2x d m λΔ = = + with 0,1,2... .m = The Lens-Maker’s Formula is given by:

( )l1 2

1 1 11 .nf R R

= − −

SIMPLIFY: Using 2 2 ,d R R r= − − the wavelength is given by ( )2 22

.1/ 2

R R r

mλ

− −=

+ If ,R r>>

2 2R r− can be approximated by,


1221

1/22 22 2

2

11 .2

r rR r R RRR

− = − ≈ −

Therefore, the wavelength simplifies to

( )2

.1/ 2r

m Rλ =

+

An expression for R can be found by using the Lens-Maker’s Formula using 1R R= for the radius of curvature of the bottom surface of the lens and 2R → ∞ for the plane surface:

( ) ( )ll l

11 1 11 1 .nn R f n

f R R− = − − = = − ∞

The wavelength of light is therefore:

( )( )2

l

.1/ 2 1

rf m n

λ =+ −

CALCULATE: Substituting 2m = for the third bright circle ( 0m = corresponds to the first) yields:

( )( )( )( )

230.8487 10 m720.29 nm.

0.8000 m 2 1/ 2 1.500 1λ

−⋅= =

+ −

ROUND: Rounding the answer to four significant figures gives 720.3 nm.λ = DOUBLE-CHECK: This is within the range of wavelengths of visible light.

34.34. In a wavelength meter, the number of counted fringes corresponds to the number of wavelengths in the path difference. Since the path difference is 2 ,x dΔ = Δ the number of fringes is / 2 / .N x dλ λΔ = Δ = Δ Therefore, the number of fringes for two wavelengths are 1 12 /N d λΔ = Δ and 2 22 / .N d λΔ = Δ (a) Taking a ratio of 1NΔ and 2NΔ gives 1 2 2 1/ / .N N λ λΔ Δ = If 1λ is a known wavelength, then the

unknown wavelength is ( )4

12 1 4

2

6.000 10 632.8 nm 488.0 nm.7.780 10

NN

λ λΔ ⋅= = =Δ ⋅

(b) The displacement, ,dΔ is ( )41 1 6.000 10 632.8 nm

0.01898 m 18.98 mm.2 2

Nd λ ⋅ΔΔ = = = ≈

34.35. The number of fringes is given by the ratio of the path difference and the wavelength, that is,

( )3 29/ 2 / 2 0.381 10 m / 449 10 m 1697 17.0 10 .N x dλ λ − −= Δ = = ⋅ ⋅ = ≈ ⋅

34.36. THINK: The phase difference of two light beams is given by 2 /xθ π λ= Δ where xΔ is the path difference between the two beams and 9550.0 10 mλ −= ⋅ is the wavelength of each beam. SKETCH:

RESEARCH: If the number of round trips is 100N = and the length of the interferometer arm is denoted by 4000. mL = then the total distance traveled by each beam is total 2 .L NL=


1222

SIMPLIFY: If there is a decrease in the length of one path and an increase in the length of the other path due to gravitational waves, each by a fractional change of 211.000 10 ,δ −= ⋅ then the net fractional change is 2 .δ Therefore, the difference in path length between the two beams is 4 .x NLδΔ = The phase difference is

( )2 4 8 .NL NLπ δ πδθ

λ λ= =

CALCULATE: Substituting in the numerical values yields ( )( )( )21

89

8 1.000 10 100 4000. m 1.8278 10 rad.

550.0 10 m

πθ

−−

−

⋅= = ⋅

⋅

ROUND: The value of 100N = can be taken as an exact number. Rounding the answer to four significant figures gives 81.828 10 rad.θ −= ⋅ DOUBLE-CHECK: A very small phase change is expected since the effect that gravitational waves have on the path length of light is always neglected.

34.37. The minima of a single slit are given by: sin ,a mθ λ=( )( )1 653 nm

1230 nm.sin32.0

a = =°

34.38. The width of the central maximum is given by: 2 /w L aλ= from problem 34.1. ( )( )

( )( )λ

−

−

⋅= = =

⋅

3

9

0.0500 m 0.135 10 m5.33 m

2 2 633 10 mwaL

34.39. The minima of a single slit width are given by: sin .a mθ λ= The first minimum corresponds to 1,m = sin .a θ λ= Minima do not appear for 90θ = ° or larger angles. Solving for a gives:

/ sin 600. nm.a aλ θ λ= = = If a is any larger θ would be less than °90 , since θ λ=sin / .a

34.40. The dark fringes of a single slit are given by: sin .a mxθ = The second dark fringe corresponds to 2,m = ( )θθ λ λ −°

= = = = ⋅ =20.0200 m sin43.0sinsin 2 0.682 10 m 0.682 cm.2 2

aa

34.41. Using Rayleigh’s Criterion, the minimum angular resolution for green light is:

( )91 1 6

R

1.22 550 10 m1.22sin sin 2.7 10 degrees.14.4 md

λθ−

− − − ⋅ = = = ⋅

34.42. The first diffraction minimum is given by: sin 1.22 / .dθ λ=

The angle θ is then given by 1 2.0 mm 1tan 0.02862 2 m

− = °

and ( )

( )91.22 570 10 m1.22

sin sin 0.0286d λ

θ

−⋅= =

° where λ

is taken to be 570 mm, the average wavelength of sunlight. 1.39 mm 1.4 mm.d = ≈


1223

34.43. The angular resolution is given by Rayleigh’s Criterion ( )1R sin 1.22 / .dθ λ−= For the Hubble Space

Telescope the value is: ( )( )θ − − −= ⋅ = ⋅ °1 9 5R sin 1.22 450 10 m / 2.40 m 1.31 10 . For the Keck Telescope the

value is: ( )( )θ − − −= ⋅ = ⋅ °1 9 6R sin 1.22 450 10 m /10.0 m 3.15 10 . For the Arecibo radio telescope, the value is:

( )( )θ −= = °1R sin 1.22 0.210 m / 305 m 0.0481 . The radio telescope is clearly worse than the other telescope

in terms of angular resolution. The Keck Telescope is better than the Hubble Space Telescope due to its larger diameter.

34.44. Angular resolution is given by the Rayleigh Criterion Rsin 1.22 / ,dθ λ= R 1.22 / dθ λ= in radians from

very small angles ( )( )

7

1.22 0.100 m436 km.

2.80 10 radiansd −= ≈

⋅

34.45. (a) Rayleigh’s Criterion is given by: ( )9

1 1R

1.22 550 10 m1.22sin sin 0.0077 .0.00500 md

λθ−

− − ⋅ = = = °

(b)

From the diagram the distance is given by ( )−= =

⋅ °3

1.5 m 1 11 km.2 tan 7.7 10 / 2

L

34.46. For the first dark fringe due to double slit interference:

( ) ( ) ( )sin 1/ 2 / 1/ 2 .d m d y L m= + = +θ λ λ

The width of the central maximum is twice ,y so 2 .w y= Using 0,m =

( )( )( )

95

1.60 m 635 10 m2.42 10 m.

2 0.0420 mL Ld

y w

−−

⋅= = = = ⋅λ λ

The missing bright fringe is due to single slit destructive interference, with 1.m = The size of the individual slits is

, whereL m La yy d

′= =′λ λ

with 4m = for the forth bright spot due to double slit interference. Therefore,

( )56

2.42 10 m6.05 10 m

4 4da

−−

⋅= = = ⋅

The slit separation is 4 times the slit width causing the fourth double slit maximum to be missing due to single slit interference.

34.47. THINK: Light of wavelength 600 nmλ = illuminates two slits. The slits are separated by a distance 24 μmd = and the width of each slit is 7.2 μm.a = A screen 1.8 mw = wide is 2.0 mL = from the slits.


1224

The problem can be approached by determining the number of fringes that appear due to the double slit and eliminate those removed by the minima due to single-slit diffraction. SKETCH:

RESEARCH: The maximum angle maxθ is given by maxtan / 2 .w Lθ = The bright fringes occur when sin / .m dθ λ= The disallowed fringes occur when sin / .n aθ λ= SIMPLIFY: The maximum number of bright fringes that can appear on the screen is

1

maxmax

sin tan2sin

.

wdLd

mθ

λ λ

−

= =

The disallowed fringes occur when

.m n m dd a n aλ λ= =

CALCULATE: The number of bright fringes is:

( ) ( )( )

1

maxmax

1.8 m24 μm sin tan

2 2.0 msin16.4.

600. nmd

mθ

λ

− = = =

The disallowed fringes occur when:

= = =24 μm 10 .7.2 μm 3

m dn a

The only scenario this can occur for (since max 16)m = is 10m = and 3.n = Therefore, the only disallowed value of m is 10, so there are 15 bright fringes on either side of the central maximum. ROUND: To the nearest integer, there are 31 fringes on the screen. DOUBLE-CHECK: Without the effects from single-slit diffraction there would be 33. It is expected that there would be fewer fringes due to the effects of single-slit diffraction.

34.48. THINK: Equations for the angular positions of the dark fringes due to single-slit and double-slit diffraction can be used to determine a relation between the slit width a and the slit separation .d Then this can be used to find the number of fringes present with the blue filter. The equation for the width of the central diffraction maximum is required to find the new width using blue light.


1225

SKETCH:

RESEARCH: For the red light, the nine interference maxima correspond to four bright fringes (and five dark fringes) on either side of the central diffraction maximum. The angular positions of the dark fringes due to single-slit diffraction are given by:

( )sin 1, 2, 3,... .m maλθ = =

The angular positions of the dark fringes due to double-slit diffraction are given by:

( )1sin 0,1, 2,... .2

m mdλθ = + =

The width of the central diffraction maximum is given by:

( )2 1, 2, 3,... .m Lw maλ= =

SIMPLIFY: The angular position of the first ( )1m = dark fringe due to single-slit diffraction is equal to

the angular position of the fifth ( )= 4m dark fringe due to double-slit diffraction, so


1226

r r1 94 .2 2

ada dλ λ = + =

Since the slit width stays constant: b br

b rr b r

22.

m Lm La w ww w

λ λλλ

= = =

CALCULATE: For blue light, the angular position of the first ( )1m = dark fringe due to single-slit

diffraction is equal to the angular position of the fifth ( )4m = dark fringe due to double-slit diffraction, so

b b b b1 1 1 9 42 2 9 / 2 2 2

m m m ma d a aλ λ λ λ = + = + + = =

Therefore, the number of fringes is independent of wavelength. There will still be nine bright fringes. The width of the central diffraction maximum for blue light is:

( )( ) ( )b

450 nm4.50 cm 3.02 cm

670 nmw = =

ROUND: To two significant figures, b 3.0 cm.w = DOUBLE-CHECK: It is reasonable that the width of the central diffraction maximum will decrease slightly for the blue light.

34.49. (a) The first minimum on either side of the central maximum is given by: sin ,a mθ λ= 1,m = sinθ θ≈ for small angles .aθ λ= From the graph 0.1,θ ≈ / 0.1 10 .a λ λ= = (b) Note that the mth interference maxima for a double slit setup is given by: sin ,d mθ λ= sinθ θ≈ for small angles .d mxθ = From 0 to 0.1 radians there are 10 interference maxima

/ 10 / 0.1 100 .d mλ θ λ λ= = = (c) / 10 /100 1/10a d λ λ= = so the ratio is 1 : 10. (d) Without ,λ there is insufficient information to find a or .d

34.50. Constructive interference of a grating is given by sinm dλ θ= we have ( ) ( )unknown unknown3 sin 2 600. nm 2 / 3 600. nm 400. nm.dλ θ λ= = = =

34.51.

From the above diagram, 0.332 mtan .1.00 m

θ = For a diffraction grating with 1,m = the wavelength of light is

( )1 7

5

0.332 m1sin sin tan 4.49 10 m 449 nm.1.00 m7.02 10 / m

dλ θ λ − −

= = = ⋅ = ⋅

34.52. THINK: A diffraction grating with width 25.000 10 ma −= ⋅ and 200N = grooves is used to resolve two beams of wavelength a 629.8 nmλ = and b 630.2 nm.λ = The condition for constructive interference of the grating is required to determine the angular position of the beams.


1227

SKETCH:

RESEARCH: The expression for the angle of constructive interference from a diffraction grating is

sin .d mθ λ= For first-order diffracted beams use 1.m = Resolving power is given by: ,R Nmλλ

= =Δ

where λ is the average wavelength. SIMPLIFY: The spacing of the gratings is / .d a N= The angle of the first-order diffraction peak is

1 1sin sin .Nd aλ λθ − − = =

The order of diffraction required to resolve the two lines is b a

1 1 .mN N

λ λλ λ λ

= =Δ −

CALCULATE: ( )( )

( )9

11a 2

200 629.8 10 msin 0.144340

5.000 10 mθ

−−

−

⋅ = = ° ⋅

( )( )( )

91

1b 2

200 630.2 10 msin 0.144431

5.000 10 mθ

−−

−

⋅ = = ° ⋅

( )( )

630.0 nm 1 7.875630.2 nm 629.8 nm 200

m = = −

ROUND: Taking 200N = to be an exact number, the angles should be rounded to four significant figures: 1a 0.1443θ = ° and 1b 0.1444 .θ = ° Since the order of diffraction must be an integer, rounding up is appropriate: 8,m = or the eighth-order diffracted beams. DOUBLE-CHECK: It is reasonable that the angles of the first-order diffracted beams are very close since their wavelengths are very close. The high order of m is necessary due to the closeness of two spectral lines and is to be expected.

34.53. THINK: The condition for constructive interference for the grating is required. For each order of diffraction ,m compute the wavelengths λ that fall into the range of visible light. The question gives the range for white light as the interval (400. nm - 700. nm). Wavelength is inversely proportional to m, and hence, an interval of allowable values for wavelength must correspond to an interval of allowable values for m. It is sufficient to find the least value of m for which the wavelength is in the interval, and then to increment m until the wavelength falls outside the given interval. Use the known values of 45.0θ = ° and

( ) 15 14.00 10 m .d−−= ⋅


1228

SKETCH:

RESEARCH: For constructive interference, sin .d mθ λ= SIMPLIFY: The wavelength is given by

sin .dm

θλ =

CALCULATE: Set m = 1, 2, 3, …,:

( ) ( )15 14.00 10 m sin 45.0

mλ

−−⋅ °=

1 1767.8 nm2 883.9 nm3 589.3 nm

visible4 441.9 nm5 353.6 nm

mmmmm

λλλλλ

= == == =

= = = =

ROUND: To three significant figures, the wavelengths that will be visible are 589 nm and 442 nm. DOUBLE-CHECK: Constructive interference occurs at integral multiples of the wavelength. As m increases, λ decreases, so there will be no values of λ more than 400 when m is greater than 5.

34.54. Bragg’s Law is given by 2 sin .a mθ λ= First order implies: 1,m = ( ) ( )2 sin / 2 sin 0.256 nm 2sin 23.0 0.200 nm.a m aλ θ θ= = = ° =

Additional Problems

34.55. The number of lines per centimeter is related to the slit separation :d θ =sin .d mx No second order spectrum occurs if for the smallest wavelength 90 ,θ = °

( ) 6 4sin90 2 400. nm 800. nm 1/ 1.25 10 lines/m 1.25 10 lines/cm.d d d° = = = ⋅ = ⋅

34.56. This is similar to two slit interference where destructive interference is desired along the 45° line. ( )sin 1/ 2d mθ λ= + for destructive interference. It is important to note that θ here is the angle to the

bisector of the line joining the antennas. θ in this case is θ = ° − ° = °90.0 45.0 45.0 . Also / ,c fλ =

θ θ ⋅ = + = = = °⋅

8

6

1 1 1 1 1 3.00 10 m/s 1 2.41 m.2 sin 2 sin 2 sin45.088.1 10 / sec

C Cd mf f

34.57. The width of the central maximum is given by twice the distance of the first minima. / 1.22 /y L dλ= first diffraction minimum, d is the diameter of the aperture ( )1.22 / ,y L dλ=


1229

( )22 1.22 width of central maximum,Lydλ= =

( ) ( )( )( )6 9

3

2 1.22 384 10 m 633 10 m2 1.220.593 m.

1.00 10 mL

dw

λ −⋅ ⋅= = =

⋅

34.58. (a) The maximum occurs for 90 ,θ = ° sin sin90 ,d m d mθ λ λ= ° = 1 1 15.79.

1000 / cm 633 nmd mλ

= = = The maximum is 15.m =

(b) For 10000 / cm, 1 1 1.579.10000 / cm 633 nm

m = = The maximum is 1.m =

34.59. The distance moved in an interferometer is given by water2 ,d Nλ=

air air

water water water water water

1.33 ,fC Cn

V f fλ

λ λ= = = =

since air water ,f f= water air / .nλ λ=

( )( )37air

air

2 1.33 0.200 10 m22 6.65 10 m 665 nm.800

N nddn Nλ λ

−−

⋅= = = = ⋅ =

34.60. Destructive interference is given by ( ) air

coating

1/ 22

mt

nλ+

= for 0m = corresponding to minimum thickness

air

coating

1 1 405 nm 64.1 nm.4 4 1.58

tn

λ= = =

For CD illuminated with infrared light of wavelength 750 nm, λ

= = =air

coating

1 1 750 nm 119 nm,4 4 1.58

tn

almost

double the thickness.

34.61. It is assumed that the refractive index of the material that the body of the airplane is made from is greater than that of the polymer coating. For this case, there will be a phase change at both interfaces of the coating, so the condition for destructive interference is given by

1air

air1 12 2 .2 2

m t m tnn

λ λ−

+ = = +

The maximum wavelength for which the plane is invisible occurs for 0,m =

( )( )air, max 4 4 5.00 mm 1.50 30.0 mm.tnλ = = = It makes sense to consider the maximum wavelength.

34.62. The bright spot from a double slit source is given by: / .y m L dλ= So the distance between two consecutive bright spots is given by:

52 7

mt m2.00 10 m6.00 cm 6.00 cm 6.00 10 m 5.00 10 m 500. nm.

2.40 mL dy y

d Lλ λ

−− −⋅− = = = = ⋅ = ⋅ =

34.63. Constructive interference for a thin film is given by ( ) air1/ 2

2 .m

tn

λ+= For the minimum thickness,

0 :m = air

coating

1 1 550 nm 104 nm.4 4 1.32

tn

λ= = =

34.64. The angle of deflection is given by: sin ,m dλ θ= sin /y Lθ = with the small angle approximation for 1,m = / .dy Lλ = The wavelengths to be resolved are 588.995 nm and 589.5924 nm.


1230

( )λλ Δ Δ −Δ = = = =Δ

589.5924 nm 588.995 nm 80 cm 238.96 nm2 mm

d y L dL y

So the number of lines is given by 1.5 cm 62,800.238.96 nm

N = =

34.65. The distance moved is related to the wavelength by:

( )6

9

2 200. 10 m22 666.7 667 fringes.600. 10 m

dN d Nλλ

−

−

⋅= = = = ≈

⋅

34.66. The Rayleigh criterion is given by: ( )( )λθ − − − − = = = ⋅ = ⋅

1 1 3 4R

1.22 400 nm1.22sin sin 7.99 10 degrees 1.39 10 rad3.5 mmd

For small angles R tan /y Lθ θ≈ = Δ where yΔ is the smallest object separation able to be resolved. Since yΔ is to be as small as possible, L is chosen to be the near point:

( )( )− − −Δ = ⋅ ⋅ = ⋅4 2 51.39 10 rad 25 10 m 3.5 10 m.y

34.67. The Rayleigh criterion is given by:

R1.22 1.22sin ,y

d L dλ λθ Δ

= =

where 384000 kmL = is the distance to the Moon.

( ) ( )λ−

−

⋅Δ = = ⋅ = ≈

⋅

96

2

1.22 550 10 m1.22 384 10 m 2147.2 m 2.1 km12 10 m

y Ld

34.68.

The angles are exaggerated. The first wave has a phase change of .π The second has a path difference of

2t and a phase change of 2 .nt πλ

The factor of n accounts for the difference of wavelength in air and in the

soap bubble. 1 12 ,2 2

tn mλ λ − = +

/ 2t m nλ= for 1,m = ( )500 nm 176 nm.

2 2 1.42t

nλ= = =

34.69. The Rayleigh Criterion is given by: R1.22 1.22sin y

d L dλ λθ Δ

= = with the small angle approximation

where yΔ is the minimum separation distance.

( )100. mm1.22 1.22 1.00 nm 122 nm1.00 mm

Lyd

λΔ = = =

34.70. THINK: The Michelson interferometer uses a light source with a wavelength of air 600. nmλ = to measure the thickness t of a piece of glass with refractive index 1.50.n = Upon insertion of the glass, the fringe pattern shifts by 1000NΔ = fringes. The presence of the glass causes a change in number of wavelengths travelled by the light, which is equal to the number of fringes that the pattern is shifted by.


1231

SKETCH:

RESEARCH: The number of wavelengths travelled by the light in a distance L is given by / .N L λ= The index of refraction of the glass can be expressed in terms of the speed of the light in air and glass:

air

glass

,fcn

v fλλ

= =

The wavelength of the light in the glass is glass air / .nλ λ=

SIMPLIFY: A factor two is needed to account for the light going through the section of air and glass twice:

( ) ( ) ( )glass airglass air air air air

22 2 2 1 ./

L L L L LN N N nnλ λ λ λ λ

Δ = − = − = − = −

( )air

2 1N

Ln

λ Δ=

−

CALCULATE: ( )( )

( )4600. nm 1000.

6.00 10 m2 1.50 1

L −= = ⋅−

ROUND: To three significant figures, 600. μm.L = DOUBLE-CHECK: The final expression indicates that the width of the glass is proportional to the increase in the number of fringes which is reasonable, since as the glass gets thicker we expect the phase change to be larger.

34.71. THINK: Upon reflection, light undergoes a phase change of half a wavelength at the first interface, but not at the second interface. Since maxima are seen for two adjacent wavelengths, the layer thickness can be found by using the conditions for constructive interference. SKETCH:

RESEARCH: Since air mica ,n n< the light reflected by the first interface has a phase change of 180 .° The light reflected by the second interface has no phase change. The condition for constructive interference in the reflected light is


1232

air1 2 ( 0, 1, 2,...).2

m t mn

λ + = =

For two adjacent wavelengths with 2 1 ,λ λ> 2 1 1.m m= − Therefore,

1 2 11 2

2 1 2 1and 1 .2 2

nt ntm m mλ λ

= − = − = −

SIMPLIFY: Solving these two equations for the thickness t gives:

( )1 2

1 2 1 2 2 1

2 1 2 1 1 12 1 .2 2 2

nt nt nt tn

λ λλ λ λ λ λ λ

− = + − = = −

CALCULATE: ( )( )

( )( )480 nm 560 nm

1070 nm 1.07 μm2 1.57 560 nm 480 nm

t = = =−

ROUND: To two significant figures, the thickness of the mica layer is 1.1 μm.t = DOUBLE-CHECK: As expected, the layer thickness is much larger than the observed wavelengths.

34.72. THINK: For a slit separation of 51.00 10 m,d −= ⋅ the distance between the 1m = and 3m = maxima on a screen 1.00 mL = away can be found by using the expression for constructive interference of a double-slit arrangement. The wavelength of light used is 500. nm.λ = SKETCH:

RESEARCH: The positions of the bright fringes for a double-slit arrangement are given by:

( )1, 2, 3,... .m Ly mdλ= =

SIMPLIFY: 3 13 2L L Ly y y

d d dλ λ λΔ = − = − =

CALCULATE: ( )( )

( )9

5

2 500. 10 m 1.00 m0.100 m

1.00 10 my

−

−

⋅Δ = =

⋅

ROUND: To three significant figures, 0.100 m.yΔ = DOUBLE-CHECK: The distance between minima must be the same as the distance between maxima. For the corresponding minima,

( ) ( )3 1

3 1/ 2 1 1/ 2 2 ,L L Ly y y

d d dλ λ λ+ +

Δ = − = − =

as required.

34.73. THINK: Show (a) =2 2x Rd and (b) λ = +

1/21 2nx n R for Newton’s ring apparatus. (c) For

10.0 m,R = and a plane glass disk of diameter = 5.00 cm,D with light of wavelength 700. nm,λ = find the number of bright fringes observed. Note that maximum radial distance = =max / 2 0.0250 m.x D


1233

SKETCH: (a)

RESEARCH: (a) x, R and d are related by the Pythagorean Theorem, ( )22 2 .x R d R+ − = (b) Since there is a phase shift from a reflected light from the plane glass disk, it needs and additional phase shift by an angle of π (half wavelength) due to path difference. The condition for constructive interference (bright fringes) is ( ) ( )2 1/ 2 , 0,1,2... .d n nλ= + = (c) Use the result from (b) to find the number of fringes. SIMPLIFY: (a) ( )22 2 2 2 2 2 2 ,x R d R x R Rd d R+ − = + − + = neglecting the 2d term, which is very small, gives the

equation, 2 2 .x Rd=

(b) λ = +

12 ,2

d n but since = =2

2 2 :2xx Rd dR

λ λ λ = + = + = +

1/2221 1 12 .

2 2 2 2x n x n R x n RR

(c) Solving for n from the result of (b): ( ) ( )λλ

= + = −

22 max

max1 1 .2 2

xx n R n

R

CALCULATE:

(c) ( )

( )( )

2

9

0.0250 m 1 88.8.2700. 10 m 10.0 m

n−

= − =⋅

So, in addition to the central maximum there are approximately

89 fringes. ROUND: (c) 89 bright fringesn = DOUBLE-CHECK: In parts (a) and (b), the appropriate equations have been derived. In part (c), the quantity found is unit-less, as would be expected.


1234

Chapter 35: Relativity

In-Class Exercises

35.1. c, d, e 35.2. a, c, d, e 35.3. e 35.4. a) True b) False c) True 35.5. a 35.6. b, c, d 35.7. a 35.8. d Multiple Choice

35.1. a 35.2. d 35.3. c 35.4. d 35.5. a 35.6. d 35.7. c 35.8. c Questions

35.9. A direct corollary of Einstein’s special theory of relativity postulates that no entity or interaction in the universe can propagate with a speed greater than the speed of light in vacuum. Therefore, instantaneous effects of events originating at one point in space on another point in space are impossible. The translational motion of a perfectly rigid object would imply that, by moving one end of the object, the other end of the object would also move instantaneously, without any time delay. This contradicts Einstein’s theory.

35.10. The y-axis is the time given in μs and the x-axis is the ‘distance’ ( )/x c given in units of μs also, since the

speed of light can be written = ⋅ = ⋅ =8 23.00 10 m/s 3.00 10 m/μs 0.300 km/μs.c To hit the target, the world line from 13 μst = − of the person (Eddie and/or Martin) must lie inside the past light cone of the target at 0x = and 0.t = As seen in the diagram, Eddie’s world line is inside the past light cone of the target from 13 μst = − to 2 km / 0.3 km/μs 20 / 3 μst = − = − and so Eddie could hit the target. However, Martin’s world line lies outside of the light cone for all time after 13 μst = − and so he could not have hit the target. Eddie and Martin find out the target has been hit at the point where their individual world lines intersect the light cone from the target at the origin at some time after the target is hit at t = 0. As shown in the diagram, Eddie finds out the target has been hit at 20 / 3 μst = and Martin finds out it has been hit at

5 km / 0.3 km/μs 50 / 3 μst = − = −


1235

35.11. If the lens was situated perfectly, there would be indeed be a halo, since the alignment is typically not

exact, we see arcs instead. Likewise, the curvature is a result of the mass, so if the object does not have a uniform mass distribution, different rays would be affected non-uniformly.

35.12. In the relativistic limit, velocities must be added relativistically (using the Lorentz transformation), not classically (using the Galilean transformation), as your friend is suggesting. Let F ′ be the frame of the rocket and F be the frame of the Earth. The torpedo has a speed of ' 2 / 3u c= with respect to the rocket (frame F ′ ) and the rocket travels at a speed of 2 / 3v c= with respect to Earth (frame F ). According to the Lorentz transformation the velocity, u, of the torpedo in the Earth’s frame is

( ) ( )( )2 2 2

2 / 3 2 / 3' 12 .131 '/ 1 4 / 9 /

c cu vu cvu c c c

++= = =+ +

This is less than the speed of light, so no violation of the theory of relativity occurs.

35.13. Yes, the observer still sees the positive charge attracted to the wire. If the positive charge is moving, with velocity v in the lab frame, parallel to the current, then it is actually moving anti-parallel to electrons, which have velocity u−

in the lab frame. Since the positive charge sees only a magnetic field, this must mean that the wire is electrically neutral, i.e. there are equal positive charges (ion cores) per unit length as there are negative charges per unit length. When the wire is seen in the reference frame of the positive charge, the positive charge is stationary while the ion cores are moving away from the positive charge with velocity .v− The electrons are also moving away from the positive charge with a velocity

2' .1 /

u vu vvu c

− −= < −+

Both the electrons and ion cores have their separation contracted due to their velocities. Since the electrons are; however, moving faster than the ion cores, their separation is smaller than the separation of


1236

the ion cores, meaning the positive charge now sees a net electric charge in any given length of wire and is therefore, attracted to the wire via an electric force instead of the magnetic force in the lab frame.

35.14. The pilot of the rocket sees the garage length contracted. At the speed of the rocket the value of γ is:

( )1/221/22

2 2

0.8661 1 2.

cvc c

−− = − − = γ

The rocket pilot therefore thinks that the garage has a length that is reduced by the γ factor of 2; that is, ( )/ 2 / / 4L Lγ = .

35.15. Since the rod makes an angle with the x-axis, it has a projected length on both the x and y axes. Since the velocity is in the x-direction, only the projection of the length on the x-axis will be contracted, meaning the y-projection length remains unchanged. Since the angle is given by ( )1tan / ,y xθ −= as x decreases, the angle increases as viewed by an observer on the ground.

35.16. The primary reason that this presents no contradiction is that the two observations are made in reference frames that are not equivalent. As such, the measurements cannot be directly compared simply by making comparison of observed dimensions. The Earth’s shape is distorted from the usual spherical shape due to the fact that length contraction that occurs in the direction of the observers motion only – perpendicular to the axis of rotation for the first astronaut and along the axis of rotation for the second astronaut. If the two observers really want to compare what they’ve seen, they must exchange information that includes their own relative speed and direction with respect to the Earth.

35.17. The Lorentz transformation for the positions relating the coordinates in the moving frame (primed coordinates) to our reference frame (unprimed coordinates) takes the form

( )' ,x x vt= −γ with γ and z-coordinates unchanged, and γ given by

1/22

21 .vc

−

= −

γ

Hence, the moving clock at ' 0x = has coordinate x vt= and the clock at 'x l= has coordinate / .x vt l γ= + The time readings are then related by the Lorentz transformation,

2' .vxt tc

= −

γ

For the clock at ' 0x = the reading is ( ) 2

0 2 2 2 2

1' 1 .v vt v t tt t t t t t t t

c c

= − = − = − − = − + = γ γ γ γ

γγ γ

For the clock at 'x l= the reading is ( ) 2

1 2 2 2 2 2 2

/ 1' 1 .v vt l v vl vl t lvt t t t t t

c c c c c +

= − = − − = − − − = −

γγ γ γ

γγ γ γ

These results display two important effects. First, time dilation is apparent, as the advance of the 't values is slowed compared to the advance of t by factor 1/ .γ Second, relativity of simultaneity is also manifest, as the readings on the moving clocks – which are synchronized in their own reference frame – differ by

2/lv c at fixed time t in our reference frame. The clock behind in position is “ahead” in time reading. That is, “the same time” at different positions is a reference-frame-dependent notion. This effect is often overlooked, but most purported relativistic kinematics are resolved unambiguously once it is take into account.


1237

35.18. Velocities are added using the relativistic velocity transformation. Assume that the velocities are along the x-axis. Then the transformation equation is

2 ,11 /x yu vu c

xyuv c−−′ = =

−−

where x and y represent the fractions of the speed of light of the two sub-light velocities being added. Now, since 1,x < it follows that 2 1.x < Multiply both sides of this inequality by 21 y− (which is positive since

1y < ), to obtain ( )2 2 21 1 .x y y<− − Expand, and add the negative terms to the opposite sides to get 2 2 2 21 .yx y x+ < + Subtract 2xy from both sides, to yield: 2 2 2 22 1 2 .x xy y xy x y− + < − + Factoring both

sides as squares gives the inequality: ( ) ( )2 21x y xy− < − . Divide both sides by the right-hand side (which

is positive since 1xy < ) which results in the inequality ( )( )

2

2 1.1

x y

xy

−<

− Taking square roots of both sides

preserves the inequality (with absolute values), so 1.1x y

xy−

<−

It follows that the velocity added

relativistically is still less than c, since .1 1x y x yu c c c

xy xy− −′ = = <

− −

35.19. Classically, conservation of kinetic energy in an elastic collision for identical particles of mass m means that

2 2 21i 1f 2f

1 1 10 ,2 2 2

mv mv mv+ = +

Where 1iv is the velocity before the collision and 2iv and 2fv are the velocities after the collision. If the

particles have the same mass this reduces to 2 2 21i 1f 2f ,v v v= + which can only be true if the velocities are

perpendicular (since conservation of momentum requires also that 1i 1f 2fv v v= + ). Let the energy and

momentum of the originally moving particle be E and .p Let the two particles have total energies after the collision of 1E and 2 ,E and momenta after the collision of 1p and 2 ,p respectively. Energy-momentum conservation implies the relationships:

21 2

1 2 .E mc E E

p p p+ = +

= +

The term 2 2 2E p c− is a scalar invariant so it is the same before and after the collision, implying:

( ) ( ) ( )

( ) ( ) ( )

2 2 22 2 2 2 21 2 1 2

2 2 2 4 2 2 2 2 2 2 2 2 21 1 2 2 1 2 1 2

2 2 2 2 2 4 2 2 2 2 2 2 21 1 2 2 1 2 1 2

2 2 2

2 2 2

E mc p c E E c p p c

E Emc m c p c E E E E p c p c p p c

E p c Emc m c E p c E p c E E p p c

+ − = + − +

+ + − = + + − − −

− + + = − + − + −

Using the term 2 2 2 2 4 ,E p c m c− = this reduces to 2 4 2 2 4 2

1 2 1 22 2

1 2 1 2

2 2 2 2 2m c Emc m c E E p p cEmc E E p p c

+ = + −= −

Hence, the dot product of the momenta 1p and 2p is given by

( )2 2

1 2 1 2

2 21 1 .

p p c E E Emc

E E mc E Emc

= −

= + − −

Energy 1E can take values from 2mc to E (as can 2E ). Therefore, the function on the right-hand side of this equation increases monotonically from zero to the value


1238

( )2214

E mc− for ( )2 21

1 ,2

mc E E mc≤ ≤ +

and decreases monotonically back to zero for ( )21

1 .2

E mc E E+ ≤ ≤ It is never negative over the allowed

range of 1.E This implies 1 2 0,p p ≥ with equality only for 21E mc= or 1 ,E E= i.e., only if one of the

particles remains at rest after the collision. Otherwise the dot product is positive, meaning the two particles emerge from the collision on trajectories forming an acute angle. Therefore, it is not necessary for the velocities of the two particles to be perpendicular.

35.20. The spaceship is accelerating, and since special relativity deals only with objects moving with constant velocity, one might think that general relativity is required to solve this problem. However, the fact that the spaceship is accelerating is irrelevant since at any point in the trajectory, its velocity is constant. Since the direction of the speed is constantly changing, the length will also appear to be warped along the curvature of the orbit. The observed length of the spaceship is

( ) ( )2 200 0 01 / 1 0.800 0.600 .

LL L v c L L

γ= = − = − =

So, the length would look to be 60.0% of the original length.

Problems

35.21. The speed of light converted from SI to ft/ns is:

8 89

1 s 3.2808 ft2.9979 10 m/s 2.9979 10 m/s 0.984 ft/ns.1 m10 ns

c = ⋅ = ⋅ =

You can see that our result is quite close to 1 foot per nanosecond, which makes this a great way to visualize the speed of light: light moves about a foot in a time interval of a billionth of a second!

35.22. Convert the acceleration due to gravity from SI units into units of 2ly/year .

7365.25 days 24 hours 3600 s1 year 1 year 3.1556 10 s1 year 1 day 1 hour

= = ⋅

272 2

15

1 ly 3.1556 10 s9.81 m/s 1.03 ly/year1 year9.461 10 m

g ⋅ = = ⋅

Just like in problem 35.21, the numerical coefficient comes out to be very close to 1. However, unlike the answer in 35.21, the answer to the present problem is more of a curiosity than a useful number for any practical purposes.

35.23. The boat has a velocity of v with respect to the water. The velocity of the water is u downstream. So in order for the boat to directly cross the river, the boat must be headed upstream at an angle such that the

velocity of the boat with respect to the ground is 2 2 .v u− The cross-stream time across the river of width D with this velocity is

cs 2 2

2 .Dtv u

=−

Going upstream, the boat has velocity ,v u− and going downstream it is .v u+ Over a distance ,D the upstream-downstream time is:

( ) ( )( )( ) 2 2ud

2 .D v u D v uD D Dvt

v u v u v v uu v u+ + −

= + = =− + − −+


1239

The ratio of times is then: ( )2 2 2 2

cs2 2

ud

2 / .2 /

t D v u v ut vDv v u

− −= =−

35.24. For v = 0.8c = 45 c,

γ = 1

1− v / c( )2= 1

1− 4 / 5( )2= 1

1−16 / 25= 1

9 / 25= 5

3≈1.6667.

35.25. (a) Another astronaut on the ship sees the meter stick in the same (rest) frame as the astronaut holding the stick and so its length remains unchanged at one meter. (b) For a ship moving at 0.50 ,v c= the length of the meter stick as measured by an observer on Earth is

( ) ( ) ( )2 200 1 / 1.00 m 1 0.50 / 0.87 m.

LL L v c c c= = − = − =

γ

35.26. (a) According to a clock on Earth the trip takes

( )( )

80

8

3.84 10 m2.6 s.

0.50 3.00 10 m/sL

tv

⋅Δ = = =

⋅

(b) According to a clock on the spaceship the trip takes,

( ) ( ) ( )2 20 1 / 2.56 s 1 0.50 / 2.2 s.tt t v c c cΔΔ = = Δ − = − =

γ

(c) On the ship, the distance to the Moon is contracted to :L

( ) ( ) ( )2 28 80 1 / 3.84 10 m 1 0.50 / 3.3 10 m.L

L D v c c c= = − = ⋅ − = ⋅γ

35.27. The time that passes in the rest frame of the Earth is 30. yr.tΔ = The time that passes in the mother’s frame is 0 10. yr.tΔ = Therefore,

( ) ( )2

2 20 00

1/2 1/22 20

1 1 / 1 /

10. 1 1 0.94 .30.

t tt t v c v c

t t

tv c c c

t

γγ

Δ Δ Δ = Δ = = − = − Δ Δ

Δ = − = − = Δ

35.28. The muon’s lifetime tΔ when it is moving at 0.90v c= will be longer than 2.2 μstΔ = when it is at rest in the laboratory frame due to time dilation:

( )( )

( )

660

0 2 2

2.2 10 s5.0 10 s.

1 / 1 0.90 /

tt t

v c c cγ

−−

⋅ΔΔ = Δ = = = ⋅

− −

35.29. The fire truck of length 0 10.0 mL = is traveling fast enough so a stationary observer sees its length contracted to 8.00 m.L = Therefore,

( ) ( )1/2 1/22 2 2

2 200

0 0

8.00 m1 / 1 / 1 1 0.600 .10.0 m

L L LL L v c v c v c c cL Lγ

= = − = − = − = − =

(a) The time taken from the garage’s point of view is ( )( )

8g 8

8.00 m 4.44 10 s.

0.600 3.00 10 m/sLtv

−= = = ⋅⋅

(b) From the fire truck’s perspective the length of the garage will be contracted to


1240

( ) ( ) ( )2 200 1 / 8.00 m 1 0.600 / 6.40 m.

LL L v c c c

γ= = − = − =

Therefore, the truck will not fit inside the garage from the fire truck’s point of view since the length of the truck from its rest frame is 10.0 m.

35.30. The rest frame time taken by Phileas Fogg is 0 80 days,tΔ = while time dilation makes the time seem like 81 days.tΔ = Therefore

( )( ) ( )21/22 20 0

01 1 / 1 / .

t tt t v c v c

t tΔ Δ

Δ = Δ = = − = − Δ Δ γ

γ

Therefore,

Δ = − = − = Δ

1/21/2 220 80 days1 1 0.16 .

81 dayst

v c c ct

35.31. THINK: The planet is 0 35 lyL = away, but the astronauts cannot travel as fast as c and hence will take longer than 35 years in the NASA (Earth) reference frame while it will take only 0 25 yearstΔ = in the astronauts’ reference frame. The astronauts will see the distance as being contracted. SKETCH:

RESEARCH: The time it takes to reach the planet as observed from Earth is 0 / .t L vΔ = The relationship

between tΔ and 0tΔ is 0 ,t tΔ = Δγ where ( )( ) 1/221 / .v c−

= −γ

SIMPLIFY:

(a) ( )

2 20 0 0

0 2 1 /

L L tt t

v v v cΔ

Δ = = Δ = −

γ

( )( ) ( )( )2 2 2

1/22 22 2 20 0 00 0

0 0 0

1 / 1 /L L L

v v c v v v t c L ct t c t

− = − + = = + Δ Δ Δ Δ

(b) 0 .L

L =γ

CALCULATE:

(a) Since 0 / 35 years,L c =

1/2225 years1 0.81373 .35 years

v c c

− = + =

(b) ( ) ( )235 ly 1 0.81373 = 20.343 lyL = −

ROUND: The answers should be given to two significant figures. (a) 0.81v c= (b) 20. lyL = DOUBLE-CHECK: The velocity found for the astronauts is less than the speed of light and the distance of the planet from the perspective of the astronauts does contract; so these values are reasonable. Also, the astronauts believe that 25 years pass during their trip. Their length contracted distance to the planet is


1241

20.343 ly. This means their speed in terms of c during the trip is ( ) ( )20.343 ly / 25 yr 0.81c= which agrees with the value found.

35.32. THINK: Since the velocity of frame F is in the x-direction, the projection of the length of the rod on the x-axis will experience a contraction, while the projection on the y-axis will remain unchanged. The angle that the meter stick makes with the x-axis changes from 0 37= °θ to 1 45= °θ in frame .F ′ Trigonometry can give equations relating the angles to the speed and length. SKETCH:

RESEARCH: In both frames, 0 1 sin sin .y yL L L Lθ θ′ ′= = In frame ,F 0cos ,xL L= θ and in frame ,F ′

1cos .xL L′ ′= θ The x-axis contraction is given by / .x xL L γ′ = SIMPLIFY:

(a) 00 1

1

sinsin sin .

sinL L L L

θθ θθ

′ ′= = In frame ,F ′ the x-axis projection is

220 0 0 0

11 1 1

sin sin cos tan1cos tan tan tan

xx

L L L LL L

′ ′= = = =

θ θ θ θθθ γ θ γ γ θ

( )1/22 2

2 0 0

1 1

tan tan1 / 1 .

tan tanv c v c

− = = −

θ θθ θ

(b) The length of the rod in frame F ′ is 0

1

sin.

sinL L

θθ

′ =

CALCULATE:

(a) ( )( )

1/22tan 37

1 0.6574tan 45

v c c ° = − = °

(b) ( ) ( )( )

sin 371.00 m 0.8511 m

sin 45L

°′ = =

°

ROUND: The answers should be rounded to two significant figures. (a) 0.66v c= (b) 0.85 mL′ = DOUBLE-CHECK: The velocity does not exceed the speed of light and the length does contract; therefore, the answers are reasonable.

35.33. THINK: The tip of the triangle is the direction of the speed, 0.400 ,v c= so that only the length, 50.0 m,L = will be contracted and the width, 20.0 m,w = is not affected. The length of the ship L is not

the same as the length of a side of the ship .l Relate the observed angle θ ′ to the speed of the ship.


1242

SKETCH:

RESEARCH: The lengths are related to the angles, in both frames, by cos / 2,l wθ = cos / 2,l wθ′ ′ = sin ,L l θ= sin ,L l θ′ ′ ′= and tan 2 / .L wθ = The length of the ship contracts by / .L L γ′ =

SIMPLIFY: Determine l′ in terms of :l coscos cos .

2 cosw l l l lθθ θ

θ′ ′ ′= = =

′

The contracted length is then

( )2sin tan 2sin cos tan tan 1 / .L l LL l l v cw

′ ′ ′ ′ ′= = = = = = −θ θθ θ θ θγ γ γ

Therefore, ( )21 2( ) tan 1 / .Lv v cw

− ′ = −

θ

The plot of the angle between the base and side of the ship as a function of the speed of the ship as measured by a stationary observer is shown below.

CALCULATE: ( ) ( )( ) ( )θ −

′ = = − = °

21 2 50. m0.40 tan 1 0.40 / 77.69

20. mv c c c

ROUND: To three significant figures, ( )0.400 77.7 .v cθ ′ = = ° DOUBLE-CHECK: As v approaches c, the expression under the square root approaches zero and hence the angle will also approach zero. This agrees with the graph where the angle is smaller at higher velocities. When ,v c= the side of the ship would effectively contract to zero, thus making an angle of zero with the width.


1243

35.34. Since the light whose rest wavelength, 0 480 nm,λ = appears as 660 nm,λ = it is red-shifted, so you must be travelling away from the light.

( ) ( )2 2 2 20 0 0 c v c v c v c v

c v c v+ += = − = +− −

λ λ λ λ λ λ

( ) ( )( ) ( )

2 22 20

2 2 2 20

660 nm 480 nm0.31

660 nm 480 nmv c c c

− − = = = + +

λ λλ λ

35.35. The light with wavelength 0 650 nmλ = is blue-shifted and appears as 520 nm,λ = as expected since the driver is travelling towards the light. Therefore,

( ) ( )2 2 2 20 0 0 c v c v c v c v

c v c v− −= = + = −+ +

λ λ λ λ λ λ

( ) ( )( ) ( )

2 22 202 2 2 20

650 nm 520 nm0.22 .

650 nm 520 nmv c v c c

− − = = = + +

λ λλ λ

You would have been traveling 0.22c, or 22% of the speed of light. This explanation would likely result in a speeding ticket!

35.36. Since the light has a rest wavelength of 0 532 nm=λ and must appear to have 560 nm,λ = it must be red-shifted so it must travel away from the meteor.

( ) ( )2 2 2 20 0 0 c v c v c v c v

c v c v+ += = − = +− −

λ λ λ λ λ λ

( ) ( )( ) ( )

2 22 20

2 2 2 20

560 nm 532 nm0.051

560 nm 532 nmv c c c

− − = = = + +

λ λλ λ

35.37. Since the car, moving with a speed 1000 m 1 h32.0 km/h 8.889 m/s,1 km 3600 s

v = =

is moving away from

the radar of frequency 0 10.6 GHz,f = the shift in frequency is,

( ) ( ) ( )( ) ( )

89

0 0 8

3.00 10 m/s 8.889 m/s1 10.6 10 Hz 1 314.078 Hz.

3.00 10 m/s 8.889 m/sc vf f f fc v

⋅ − − Δ = − = − = ⋅ − = − + ⋅ +

Therefore, the frequency is red-shifted by 314 Hz.

35.38. THINK: Since the spaceship is moving towards the station, the wavelength will be blue-shifted, resulting in the original wavelength of 0 632.8 nm=λ being reduced to 514.5 nm.λ = Using the relativistic formula for wavelength shift the speed of the ship can be deduced. SKETCH:

RESEARCH: Since the ship is moving towards the station, the relevant formula for wavelength shift is

0 .c vc v

λ λ −=+

The shift parameter is by definition: 0

0

.zλ λ

λ−

=


1244

SIMPLIFY: ( ) ( )2 2

2 2 2 2 00 0 0 2 2

0

c v c v c v c v v cc v c v

−− −= = + = − = + + +

λ λλ λ λ λ λ λλ λ

CALCULATE: ( ) ( )( ) ( )

2 2

2 2

632.8 nm 514.5 nm0.20405 ,

632.8 nm 514.5 nmv c c

− = = +

( ) ( )( )

514.5 nm 632.8 nm0.186946

632.8 nmz

−= = −

ROUND: To four significant figures, 0.2041v c= and 0.1869.z = − DOUBLE-CHECK: The velocity is less than the speed of light and the shift parameter is negative, which is what it should be for blue shift, so it makes sense.

35.39. In Sam’s reference frame, each event occurs at the following points: 0 m,Ax = 0 s,At = = 500. mBx and 0 s.Bt = To find the timing of the events in Tim’s reference frame, use the Lorentz transformation

( )2/ .t t vx c′ = −γ Therefore, 0 sAt ′ = and

( )( )( ) ( )

γ −−−′ = = = − ⋅⋅ −

52 28

0.999 500. m3.73 10 s.

2.9979 10 m/s 1 0.999B

vxtc

(a) Therefore, Tim experiences event B before event A. (a) For Tim, event A occurs 53.73 10 s−⋅ after event B.

35.40. Let an inertial reference frame F be at rest and let another inertial reference frame F ′ move at a constant speed v along a common x-axis with respect to reference frame F. According to the relativistic velocity addition formula,

( )2

2

1 /

,1 1

u vuvu c

c c vc v c vu c u cvc v c v

cc

−′ =−

−− −′= = = = =−− −

as required. Thus, the result is independent of the specific value of v.

35.41. Let all speeds be in a common x-direction. Let frame F be the ground and frame F ′ be the frame of your car. The speed of your car with respect to the ground is 50.0 m/sv = and the speed of the oncoming car is

50.0 m/su = − in frame F. Using the relativistic velocity transformation, the relative speed of the oncoming car is

( ) ( )( )( ) ( )

2 28

50.0 m/s 50.0 m/s99.99999999999862 m/s 100. m/s.

1 / 1 50.0 m/s 50.0 m/s / 2.9979 10 m/s

u vuuv c

− −−′ = = = − ≈ −+ + − ⋅

The relative velocity is about the same as a Galilean velocity transformation 2 100 m/s,u u v u′ = − = = − since the speed of the cars is so small compared to the speed of light. In order to detect a difference, fourteen significant figures would need to be kept. This shows how close the values are.

35.42. Assuming all speeds are measured along the same direction, let 0.90v c= be the speed of the ship (frame F ′ ) relative to Earth (frame F) and let 0.50u c′ = be the speed of the missile relative to the ship. The speed of the missile as seen from the Earth is given by

( ) ( )( )( )

+′ += = =′+ +2 2

0.50 0.900.97 .

1 / 1 0.90 0.50 /c cu vu c

vu c c c c

35.43. (a) The total distance travelled, as measured by Alice is


1245

( ) ( )20 2 3.25 ly 1 0.65 / 4.940 ly 4.9 ly.L

L c cγ

= = − = ≈

(b) The total time duration for the trip as measured by Alice is ( )

( )4.940 ly

7.6 years.0.65

Ltv c

= = =

35.44. THINK: The spaceship that Alice boards travels at a speed of 0.650u c= to a station 0 3.25 lyL = away. The question asks for the speed v Alice must travel so that she measures a relative speed of 0.650u c= on the return journey. In Alice’s frame, the distance of the return flight will be length contracted. The relativistic velocity transformation and length contraction formulae can be used to solve the problem. SKETCH:

RESEARCH: (a) The relativistic velocity transforms as

2 .1 /

u vuvu c−′ =

−

(b) The time of the return flight as measured by Alice is / ,t L v= where 0 /L L= γ is the length contracted distance in her frame. SIMPLIFY: (a) The speed of the spaceship is given by

2 2 2 2 .1 / 1 /

u v uvu uvu u uu u u v u u v vuv c c c uu c

′ ′ ′− −′ ′ ′= − = − − = − =′− −

(b) The time for Alice’s return flight is ( )20 0 1 / .L L

t v cv v

= = −γ

CALCULATE: (a) To Alice, the Earth is moving toward her with a speed of 0.650 ,u c′ = − so

( ) ( )( )( )

2

0.650 0.6500.91388 .

0.650 0.6501

c cv c

c cc

− −= =

−−

(b) The time duration of the flight as measured by Alice is ( )

( ) ( )23.25 ly1 0.91388 / 1.4438 years.

0.91388t c c

c= − =

ROUND: The answers should be given to three significant figures. (a) As required, the velocity of the ship relative to the Earth is 0.914 .v c= (b) The duration of Alice’s return flight as measured by her is 1.44 years.t = DOUBLE-CHECK: The speed 0.914v c= gives

( ) ( )( )( ) 2

0.650 0.9140.650 .

1 0.650 0.914 /c c

u cc c c

−′ = = −

−


1246

35.45. THINK: The arrow has a velocity of 0.300u c′ = in Robert’s reference frame. The railroad car has a length of =0 100. mL and travels at a speed of 0.750 .v c= The velocity transformation equations and the equation for length contraction can be used to determine the values observed by Jenny. SKETCH:

RESEARCH: As observed by Jenny,

(a) the railroad car is length contracted: 0 ,L

Lγ

=

(b) the velocity of the arrow is given by the inverse relativistic velocity transformation: 2 ,1 /

u vuvu c′ +=

′+

(c) the time of the arrow’s flight is given by the inverse Lorentz transformation: 2 ,vxt tc

γ′ ′= +

and

(d) the distance traveled by the arrow is given by the inverse Lorentz transformation: ( ).x x vtγ ′ ′= + SIMPLIFY: Here 0x L′ = is the length of the railroad car and 0 /t L u′ ′= is the time of the arrow`s flight in Robert`s frame of reference. As observed by Jenny,

(a) ( )20 1 / ,L L v c= −

(c) the time taken by the arrow to cover the length of the car is ( )

022

1 ,1 /

L vtu cv c

= + ′ − and

(d) the distance covered by the arrow is ( )

0

21 .

1 /

L vxuv c

= + ′ −

CALCULATE:

(a) ( ) ( )2100. m 1 0.750 / 66.14 mL c c= − =

(b) ( ) ( )( )( ) 2

0.300 0.7500.85714

1 0.750 0.300 /c c

u cc c c

+= =

+

(c) ( )

( ) ( ) ( ) ( ) 6

28

100. m 1 0.750 2.059 10 s0.3002.9979 10 m/s 1 0.750 /

tc c

−

= + = ⋅ ⋅ −

(d) ( )( )

( )( )2

100. m 0.7501 529.2 m

0.3001 0.750 /

cx

cc c

= + =

−

ROUND: The answers should be given to three significant figures. As observed by Jenny, (a) the railroad car is 66.1 mL = long, (b) the velocity of the arrow is 0.857 ,u c= (c) the time it takes the arrow to cover the length of the railroad car is 2.06 μs,t = and (d) the arrow covers a distance of 529 m.x =


1247

DOUBLE-CHECK: The railroad car length is contracted from Jenny’s viewpoint, as expected. Multiplying the answer to part (b) by the answer to part (c):

( )( )( )8 60.8571 2.9979 10 m/s 2.059 10 s 529 m,x −= ⋅ ⋅ =

as found in part (d). So, the answers are consistent.

35.46. THINK: The speed of an object can be described by the relation tanhv c θ= where θ is known as the rapidity. The question asks to prove that two velocities adding via the Lorentzian rule, corresponds to adding the rapidity of the two velocities. The question also asks for the Lorentz transformation of two coordinate systems using the rapidity. The Lorentz transformation equations can be used to solve this problem. SKETCH:

RESEARCH:

(a) The Lorentzian rule for adding two velocities is 1 22

1 2

.1 /

u uvu u c

+=

+ The Lorentz transformation between

two frames with relative velocity v in the x direction is given by the equations ( ),x x ct′ = −γ β ,y y′ = ,z z′ = and ( )/ .t t x c′ = −γ β

Velocities that add according to the Lorentzian rule correspond to adding the rapidity of each:

( ) 1 2 1 21 2 2

1 21 2

tanh tanhtanh

1 tanh tanh1 /u u c cv cu u c

+ += + = =

++θ θθ θ

θ θ

(b) For the derivation it is useful to know that the hyperbolic tangent is related to exponentials by

tanh .x x

x x

e exe e

−

−

−=+

The following relations are also useful: 2 21 sech tanh= +θ θ and sinhtanh .

coshθθθ

=

SIMPLIFY: (a) According to the Lorentzian rule,

( )

( )( ) ( )( )( )( )

1 1 2 2

1 1 2 2

1 1 2 2

1 1 2 2

1 1 2 2 1 1 2 2

1 1 2 2

1 21 22

1 21 2

tanh tanhtanh tanh1 tanh tanh1 tanh tanh /

1

e e e ec c e e e ev c cc c c e e e e

e e e e

e e e e e e e ec

e e e e e

− −

− −

− −

− −

− − − −

− −

− −+++ + += = =++ − −+ + +

− + + + −=

+ + +

θ θ θ θ

θ θ θ θ

θ θ θ θ

θ θ θ θ

θ θ θ θ θ θ θ θ

θ θ θ θ θ

θ θθ θθ θθ θ

( )( )( )

( ) ( )1 21 2

1 21 1 2 2 1 21 2

2 2 tanh ,2 2

e ec ce e e e e

− ++

− +− − +

−= = +− − +

θ θθ θ

θ θθ θ θ θ θθ θ

as required. (b) If tanhv c θ= then,

( )( ) ( )( ) ( ) ( )1 1 1 1

2 22 2 2 22 21 / 1 tanh / 1 tanh sech coshv c c c− − − −

= − = − = − = =γ θ θ θ θ and

tanh tanh .v cc c

θβ θ= = =

The Lorentz transformation becomes ( ) cosh cosh tanh cosh sinh ,x x ct x ct x ct′ = − = − = −γ β θ θ θ θ θ ,y y′ =


1248

,z z′ = and

( )/ cosh cosh tanh cosh sinhx xt t x c t tc c

′ = − = − = −γ β θ θ θ θ θ

CALCULATE: Not required. ROUND: Not required. DOUBLE-CHECK: Note that the transformation is similar to a transformation from one coordinate system to another where they differ by the angle :θ

cos sinx x y′ = +θ θ and sin cos .y x y′ = − +θ θ

35.47. The relativistic momentum is .p mvγ= If the momentum is equal to p mc= then mv mcγ = or

( ) ( )

2 2 2

22

1 1 1 1 2 .21 /1 /

c c c c cvv v v vv cv c

= = = = − = = −−

γ

This can be left in exact form, or written as 0.707 .v c≈

35.48. (a) The energy of the electron is 2 .E mcγ= For the energy to be 10 times greater than its rest energy of 2

0 ,E mc=

( )( )2

2

1 1 99= 10 1 / 0.995100 1001 /

v c v c cv c

= − = = =−

γ

(b) The momentum is ( )( )210 0.511 MeV/ 99 /100 5.08 MeV/ .p mv c c c= = =γ

35.49. The kinetic energy of the colliding beams in the center-of-mass reference frame is related to the fixed-target equivalent, or lab reference frame by

( ) ( )( ) ( )( )( )( )( )= + = + = ⋅

2 2cmlab cm 6

2

2 2 197 100. GeV4 4 197 100. GeV 4.02 10 GeV.

197 1.00 GeVp

KK K

m c

This is an incredibly large energy.

35.50. The work done on the proton is equal to the change in kinetic energy of the proton.

( )( )

( )( )( )

2 2 2 2p p p p2

227 8

2

11 11 /

1 1 1.672 10 kg 2.9979 10 m/s1 0.997 /

11.14477 GeV 11.14 GeV.

W K m c m c m c m cv c

c c

γ γ

−

= Δ = − = − = − −

= − ⋅ ⋅ −

= ≈

35.51. The energy of the proton is ( )

( )2p 2

1 938 MeV 1200 MeV.1 0.61 /

E m cc c

= = =−

γ


1249

35.52. THINK: Two protons in an accelerator are on a head-on collision course. In the lab reference frame (frame F) the protons reach a speed of 0.9972 .v c= The relativistic velocity transformation and the relativistic formula for kinetic energy can be used to solve the problem. SKETCH:

RESEARCH: The speed of the proton in the other proton’s rest frame (frame F ′ ) is given by

2 .1 /

u vuvu c−′ =

− The kinetic energy of a relativistic particle is ( ) 21 .K mcγ= − The mass of a proton is

2p 938.27 MeV/ .m c=

SIMPLIFY: Let u denote the speed of the proton in the lab frame. In the proton reference frame, the speed of the other proton is

( )( ) ( )2 2

2 .1 / 1 /

v v vuv v c v c

− −′ = =

− − +

The kinetic energy K of the protons in the lab reference frame is the sum of the kinetic energy of each proton:

( ) ( ) ( )( )

2 2 2 21 2 p p p p2

11 1 2 1 2 1 .1 /

K K K m c m c m c m cv c

= + = − + − = − = − −

γ γ γ

The kinetic energy K ′ in the proton reference frame is ( )( )

2 2p p2

11 1 .1 /

K m c m cu c

′ = − = − −

γ

CALCULATE:

(a) ( )

( )2

2 0.99720.999996

1 0.9972 /

cu c

c c′ = =

+

(b) ( )

( )2 2

2

12 1 938.27 MeV/ 23217.35 MeV1 0.9972 /

K c cc c

= − = −

(c) ( )

( )2 2

2

1 1 938.27 MeV/ 333689.6 MeV1 0.999996 /

K c cc c

′ = − = −

ROUND: (a) To six significant figures, the speed of one proton with respect to another is 0.999996 .u c′ = (b) To four significant figures, in the lab reference frame, the particles have a kinetic energy of

23220 MeV.K = (c) To four significant figures, in the proton’s reference frame, the other proton has a kinetic energy of 333700 MeV. DOUBLE-CHECK: These are typical speeds and energies for protons to have in proton accelerators.

35.53. THINK: Electrons acquire kinetic energy as they accelerate through the potential difference. The speed acquired by the electron after moving through this potential can be found and then the appropriate


1250

classical and relativistic formulae can be used to find the total energy and momentum. Many of the answers only make sense if they are given to three significant figures, so rounding will be nonstandard. SKETCH:

RESEARCH: (a) The kinetic energy gained by the electron in moving through the potential difference V is equal to the work done by the potential difference: .W K qV= = (b) The kinetic energy of a relativistic particle is ( ) 01 .K E= −γ (c) The relativistic values for the total energy and momentum are R 0E E= γ and R .p mv= γ Classically,

these values are given by 2C C

12

E K mv= = and C C .p mv=

The rest mass energy of an electron is 0 511 keV.E = SIMPLIFY: (a) K eV= (b) The speed of the particle is found using the relativistic formula ( ) 01 :K E= −γ

( )( )

( )

22 2 20 0 00 0

220 0 00

21 1 .1 /

K E E K KEK E Ev c c c

E K E K EK Ev c

+ − + += = = − = = + ++ −

γ

(c) The relativistic values for the total energy and momentum are R 0 0 ,E E K E= = +γ and

20 20 0

R 020 0

22 / .

K KEK E Ep mv c K KE c

E K Ec+ +

= = = + + γ

Classically, the total energy and momentum are C ,E K= and

C C 02 / 2 2 / .p mv m K m Km KE c= = = = CALCULATE: (a) ( )= =5.00 kV 5.00 keVK e

(b) ( ) ( )( )

( ) ( )+

= =+

25.00 keV 2 5.00 keV 511 keV0.1389

5.00 keV 511 keVv c c

(c) ( ) ( )= + =R 5.00 keV 511 keV 516 keVE

( ) ( )( )= + =2R 5.00 keV 2 5.00 keV 511 keV / 71.659 keV/p c c

=C 5.00 keVE

( )( )C 2 5.00 keV 511 keV / 71.484 keV/p c c= =

ROUND: (a) The kinetic energy that the electron acquires is = 5.00 keV.K


1251

(b) The electron has a speed of 0.139 ,v c= thus the electron will have only a small difference between its classical and relativistic values, but this can still be considered a relativistic speed. (c) The relativistic and classical energies are R 516 keVE = and 5.00 keV, respectively. (The difference is due to the fact that the relativistic energy includes the rest energy). The relativistic and classical momenta are R 71.7 keV/p c= and C 71.5 keV/ ,p c= respectively. DOUBLE-CHECK: The classical and relativistic momenta are similar, as expected for such a low speed.

35.54. The momentum before the collision must equal the momentum after the collision.

1 2 1 2

1 1 1 1 1 1 2 2 2 .p p p p

m v m v m v′ ′+ = +′ ′ ′ ′= +γ γ γ

The ratio is ( )( )

( )( )

( )( )

2 2

2 1 1 1 1

1 2 2

2

0.700 0.500

1 0.700 / 1 0.500 /7.63.

0.200

1 0.200 /

c c

c c c cm v vm v c

c c

γ γγ

−−

− −′ ′−= = =

′ ′

−

35.55. THINK: Two particles collide inelastically. One particle has a mass of 1m m= and momentum 1 .p mc=

The second particle has a mass of 2 2 2 .m m= Conservation of energy and momentum can be used with the relativistic energy equation to determine the speed and mass of the new particle. SKETCH:

RESEARCH: The relativistic momentum is .p mvγ= The energy of the particles is 2 2 2 2 4E p c m c= + after the collision. SIMPLIFY: (a) The speed of the projectile with momentum 1p mc= before the collision is given by

( ) ( ) ( )2 2 2 2 21 1 1 1 1 11 / / 1 / 2 .

2mc cv c v c v c v c v c vmγ

= = − = − = =

(b) The total energy is conserved before and after the collision. Therefore,

( ) ( )

f i

2 2 2 4 2 2 2 4 2 2 2 41 1 2 2

222 2 2 4 2 2 4 4

2 2 2 4 2 2

2 2 2 4 2

2 2 2 2 2

0 2 2

2 2 2

3 218

E E

p c M c p c m c p c m c

p c M c mc c m c m c

p c M c mc mc

p c M c mcp M c m c

=

+ = + + +

+ = + + +

+ = +

+ =+ =


1252

From conservation of momentum, 1 .p p mc= = Therefore, the above equation becomes:

( )2 2 2 2 2

2 2 2 2

1817

17

mc M c m cM c m c

M m

+ =

=

=

Note that there is more mass than there was before the collision. Some kinetic energy has become mass energy. (c) Using the conservation of momentum

( )

( )

2 2 2 2 2 2 21

2 2 22

1 /

1817

p p Mv mc Mv mc v c M v m c m vmc mc cv

M m m m

γ= = = − = −

= = =+ +

CALCULATE: Not required. ROUND: Not required. DOUBLE-CHECK: It is reasonable that the speed of the new particle is smaller than the speed of the projectile. The momentum of the new particle is given by

( )( )( )( ) ( )2 2

17 / 18 17 17 .18 118 1 1/181 ( / ) 1 / 18 /

m cMvp Mv mc mc mcv c c c

γ= = = = = =−−− −

This is the initial momentum of the projectile, as expected by conservation of momentum.

35.56. THINK: To derive the Lorentz transformation for momentum, follow Derivation 35.3. In this case, the momentum is similar to the position coordinates and the energy is analogous to the time. SKETCH:

RESEARCH: The energy is given by 2E mcγ= and the momentum is given by .p mvγ= In order to use

the energy as a momentum, it must be of the form 2 .v Ep Ecc

= = β

SIMPLIFY: In frame ,F the vectors are

,xOA p=

xpO A

′′ =

γ and 2 .EvOO

c′ =

Using the equation OA O A OO′ ′= +

gives

2 2 .xx x x

p Ev Evp p pc c

′ ′= + = −

γγ

(1)

For frame ,F ′ the vectors are

,xpOA =

γ ,xO A p′ ′=

and 2 .E vOOc′′ =

Using the equation OA O A OO′ ′= +

gives


1253

2 .xx

p E vpc′′= +

γ (2)

Substituting from equation (1) for xp′ into equation (2) gives 2 2 .xx

p Ev E vpc c

′ = − +

γγ

Solving for :E′

2 2 2

2 2

1 1 11 .xx x x

pc c cE p E p E E vpv v v

′ = − + = − + = − −

γ γ γ γ γ γγ γ β γ

From 2

1 ,1

=−

γβ

it is easy to show that 2 2

1 11 1.β γ

− =

Therefore, ( ).xE E vp′ = −γ Of course, for

motion in one dimension (the x-direction), y yp p′ = and .z zp p′ = Thus the Lorentz transformation for momentum and energy is established. CALCULATE: Not required. ROUND: Not required. DOUBLE-CHECK: This result matches with the required expressions.

35.57. THINK: The Lorentz transformations for energy and momentum in the frame F ′ can be used to write the quantity 2 2 2E p c′ ′− in terms of the values in the unprimed frame F. SKETCH: Not required. RESEARCH: The Lorentz transformations are

( ),xE E vp′ = −γ ( )2/ ,x xp p vE c′ = −γ y yp p′ = and .z zp p′ =

SIMPLIFY: Apply the transformations:

( ) ( )

( )( ) ( )( )( ) ( )( )

222 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2

2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2

22 2 2 2 2 2 2 2 2 2

2 22 2 2 2 2 2 2 2 2

/

2 2 /

1 /

1 / 1 /

x y z x x y z

x x x x y z

x y z

x y z

E p c E p c p c p c E vp p vE c c p c p c

E Evp v p p c p vE v E c p c p c

v c E v c p p c p c

v c E v c p c p c p c

′ ′ ′ ′ ′ ′− = − − − = − − − − −

= − + − + − − −

= − + − − −

= − − − − − =

γ γ

γ γ γ γ γ γ

γ γ

γ γ 2 2 2 .E p c−

CALCULATE: Not required. ROUND: Not required. DOUBLE-CHECK: The statement in the problem has been proved using only the Lorentz transformation equations. One could check the result for special and limiting cases. For example, if v = 0 then γ = 1 and the Lorentz transformations reduce to E E′ = and ,x xp p′ = so the result holds. When p = 0 in the frame F,

( )( )( )

22 2 2 2 2 2 2 2 2 2 2 2 2 2 2

22 2 2 2 2 2 2 2 2

/

/ 1 / .

x y zE p c E p c p c p c E vE c c

E v E c v c E E

′ ′ ′ ′ ′ ′− = − − − = − −

= − = − =

γ γ

γ γ γ

35.58. The gravitational potential at the surface of the Earth – taking the potential to be zero at infinity – is the same as would be produced by a point mass m⊕ at the center of the Earth. Hence, the desired ratio is:

( )( )( )( ) ( )

11 3 2 24

102 2 28 6

6.674 10 m / kg s 5.9736 10 kg6.962 10 ,

2.998 10 m/s 6.371 10 m

Gmc c r

−

−⊕

⊕

⋅ ⋅Φ = − = − = − ⋅⋅ ⋅

a dimensionless quantity. The deviation from flat space-time geometry produced by the Earth’s gravitation is rather small.

35.59. (a) Using the formula for the Schwarzschild radius, the Schwarzschild radius corresponding to the mass of the Sun is


1254

( )( )( )( )

11 3 2 30

SS 2 28

2 6.674 10 m / kg s 1.989 10 kg22.954 km,

2.998 10 m/s

GMr

c

−⋅ ⋅= = =

⋅

a characteristic size scale for stellar-mass black holes. (b) The Schwarzschild radius corresponding to a proton mass is

( )( )( )( )

11 3 2 27

54PS 2 28

2 6.674 10 m / kg s 1.673 10 kg22.485 10 m.

2.998 10 m/s

Gmrc

− −

−⋅ ⋅

= = = ⋅⋅

This is much smaller than the femtometer size scale usually associated with protons: it is orders of magnitude smaller than the Planck scale (see Chapter 39), generally considered the smallest scale on which our basic notions of length make sense. Hence, it is unlikely that a proton could usefully be described via a classical black-hole geometry.

35.60. The time dilation between the Earth and the satellite is 20 0

11 .2

t t t Δ = Δ ≈ + Δ

γ β The difference is

( )( ) ( ) ( )β −

⋅ = = ⋅ = ⋅

23

2 118

4.00 10 m/s1 1 8.90 10 s/ Earth second 89.0 ps/ Earth second .2 2 2.9979 10 m/s

This corresponds to a difference of 67.69 10 s/day 8 μs/day.−⋅ ≈

35.61. The Schwarzschild radius of a black hole is S 2

2 .GMRc

= The black hole at the center of the Milky Way in

Example 12.4 was found to be 63.72 10⋅ solar masses. The mass of the Sun is 301.989 10 kg.⋅ The Schwarzschild radius of this black hole is

( )( )( )( )

11 2 2 6 306

S 2 98

2 6.674 10 N m / kg 3.72 10 1.989 10 kg AU10.99 10 m149.60 10 m2.9979 10 m/s

0.0735 AU.

R−⋅ ⋅ ⋅ = = ⋅ ⋅ ⋅

=

Additional Problems

35.62. In the garage’s reference frame, the limousine is length contracted. The speed required for it to fit into the garage is

( ) ( ) ( )

( ) ( ) ( )( )γ

= = − = −

= − = − =

2 2 200 0

220

1 / / 1 /

1 / 1 35.0 ft / 50.0 ft 0.71 .

LL L v c L L v c

v L L c c c

In the limousine’s reference frame, the length of the garage is length contracted by a factor of

γ = 50.0/35.0 = 1.43.

35.63. The relativistic momentum of an electron is given by ( )2R / 1 /e ep m v m v v cγ= = − where me is the mass

of the electron. The classical momentum is C .p mv= Therefore, the percentage difference between the classical and relativistic momenta is

( ) ( ) ( ) ( )( )R C

C

100 % 100 % 1 100 %e e

e

p p m v m vp

p m v− −

Δ = = = −γ

γ

For an electron moving at ( )82.00 10 m/s 2 / 3 ,v c= ⋅ =


1255

( ) ( )2 21/ 1 / 1/ 1 2 / 3 1.342.v c= − = − =γ Its relativistic momentum is

( ) ( )( )31 8 22R 1.342 (2 / 3) 0.8944 9.109 10 kg 3.00 10 m/s 2.44 10 kg m/s,e ep m v m cγ − −= = = ⋅ ⋅ = ⋅

which differs from its classical value by ( )( )1 100 % 34 %.p γΔ = − = For an electron moving at

( )3 52.00 10 m/s 2.00 10 / 3.00 ,v c−= ⋅ = ⋅

( ) ( )22 51/ 1 / 1/ 1 2.00 10 / 3.00 1.000v cγ −= − = − ⋅ =

Its relativistic momentum is

( ) ( ) ( )( )( )35 5R

1 8 271.000 2.00 10 / 3 9.109 10 2.00 10 / 3 3.00 10 m/s 1.82 10 kg m/s.e ep m v m cγ −− −−= = ⋅ = ⋅ ⋅ ⋅ = ⋅

This does not differ appreciably from its classical value since ( )( )1 100 % 0− =γ to many decimal places. For small velocities, the classical momentum of the electron is a good approximation.

35.64. Let the Earth be frame F and rocket A be the moving frame .F ′ The speed of rocket B in frame F is then 0.95 .u c= The speed of frame F ′ with respect to frame F is 0.75 .v c= The speed of rocket B relative to

rocket A is then ( ) ( )( )( )2 2

0.95 0.750.70 .

1 / 1 0.95 0.75 /c cu vu c

uv c c c c−−′ = = =

− −

35.65. The Newtonian and relativistic kinetic energies of a particle are ( ) 2N 1/ 2K mv= and ( ) 2

R 1 ,K mcγ= − respectively. In Newtonian mechanics, the difference in their kinetic energy is

( ) ( )( )( )

Δ = − = − = −

= − =

2 2 2 2 2 2 2N 1 2 1 2

2 2 2 2

1 1 1 1 0.9999 0.99002 2 2 21 0.9999 0.9900 0.511 MeV/ 5.03 keV.2

K mv mv m v v mc

c c

The difference using special relativity is

( ) ( ) ( )( ) ( )

( )

γ γ γ γ Δ = − − − = − = − − −

= − = − −

2 2 2 2R 1 2 1 2 2 2

1 2

2 2

1 11 11 / 1 /

1 1 0.511 MeV 32.5 MeV.1 0.9999 1 0.9900

K mc mc mc mcv c v c

35.66. (a) The clock of the friend waiting in B will show a longer time interval due to time dilation. The person traveling experiences time “slowing down” relative to a stationary observer. (b) The time dilation is given by 0 .t tγΔ = Δ Since the velocity of the airplane is small compared to the

speed of light, γ can be approximated as 211 .2

≈ +γ β The difference in time between the two clocks is

( )( )2

20 0 0 0 8

1 1 240 m/s 3.00 h 3600 s/h 3.5 ns.2 2 3.00 10 m/s

t t t t tγ β Δ − Δ = Δ − Δ = Δ = = ⋅

35.67. The mass can be found from the energy:

( )( )( )

122 4

2 28

15.0 4.00 10 J6.78 10 kg 0.678 g.

3.00 10 m/s

EE mc mc

−⋅

= = = = ⋅ =⋅


1256

35.68. The speed can be found using the equation for length contraction:

( ) ( ) ( ) ( )2

2 2 2 200 0 0

90.0 cm1 / / 1 / 1 / 1 0.436 .100. cm

LL L v c v c L L v L L c c c

γ = = − = − = − = − =

35.69. Using the relativistic velocity transformation, the speed of object A relative to object B as measured by an observer on object B is

( ) ( )( )( )

A B2 2

A B

0.600 0.6000.882 .

1 / 1 0.600 0.600 /c cv vu c

v v c c c c− −−′ = = =

− − −

35.70. The length contraction factor is one-third so 3.=γ Therefore, the relative velocity is

( )( ) ( )2 2

2

1 1 1 83 1 / 1 /9 9 31 /

v c v c v cv c

= − = − = =−

35.71. The average speed on the trip, which took 40.0 hours to travel 2200.0 miles, was 55.0 mph. Since the

velocity of the vehicle is small compared to the speed of light, γ can be approximated as 211 .2

≈ +γ β

Therefore, the difference in time between your watch and your professor’s watch (your watch runs slow) is

( )( )

2

20 0 0 0 8

1 h 1609.3 m55.0 mph3600 s 1 mi1 1 40.0 h 3600. s/h 0.484 ns

2 2 3.00 10 m/st t t t tγ β

Δ − Δ = Δ − Δ = Δ = = ⋅

This amount of time is very tiny and could not be a reason for being late.

35.72. Because of the second postulate of relativity, both observers measure the speed of light to be the same. (a) The speed of light measured on the spaceship is c. (b) The speed of light measured on the asteroid is also c.

35.73. The distance of 100. ly was measured by someone on one of the space stations. Someone on the spaceship will measure a different distance, one that is shorter according to the formula for length contraction,

0 / .L L= γ The time it takes to travel from one space station to the next as measured by someone on the spaceship is

( ) ( )( ) ( )2 20 0

1

100. ly1 / 1 0.950 / 32.8684 years 32.9 years.

0.950L LLt v c c c

v v v cγ= = = − = − = ≈

As seen by someone on the space station, the time will be ( )( )2

100. ly105 years.

0.950Ltv c

= = =

35.74. The electron gains kinetic energy from the potential: ( ) 21 .K mc qVγ= − = Solving for the velocity :v

( )( ) ( )

( )

22

2 2 2 22

22

2 2

2

1 11 1 1 1 1 /1 /

1 1 / 1 1 .1

qV qV qVmc qVmc mc mcv cv c

qVv c v cmcqV

mc

−

− = − = = + = + −−

− = = − +

+

γ γ


1257

The rest mass energy of the electron is 2 0.511 MeVmc = and the potential energy is

( )61.0 10 V 1.0 MeV.qV e= ⋅ = Thus the electron attains a speed of

( )( )

21.0 MeV

1 1 0.94 .0.511 MeV

v c c−

= − + =

35.75. As seen by those on the ship, the round trip distance is length contracted to 0 ,L

L =γ

where

0 / 4000.0 yr.L c = If the speed of the ship is v and the journey must take only 40.000 yrt = then the required speed is

( ) ( )( )

( ) ( )( )

2 2 2 2 20 00 2

0

2

1 / /1 /

0.99995 .1 40.000 yr / 4000.0 yr

L L cv v c ct L v c v vt t ct L

cv c

γ= = − = − =

+

= =+

35.76. THINK: The particle is moving at a speed of 0.800 .v c= The mass of the particle is unknown, but the momentum of the particle is 201.00 10 N s.p −= ⋅ This is all that is required to find the energy of the particle. SKETCH:

RESEARCH: The energy and momentum of a relativistic particle are 2E mcγ= and p mvγ= respectively.

SIMPLIFY: 2

2 2p pcE mc mcmv v

γ= = =

CALCULATE: ( )( )

( )20 82

121.00 10 N s 2.9979 10 m/s

3.747 10 J 23.392 MeV0.800

cpcEv c

−−

⋅ ⋅ ⋅= = = ⋅ =

ROUND: To three significant figures, the energy of the particle is 123.75 10 J−⋅ or 23.4 MeV. DOUBLE-CHECK: This is a typical energy for a high energy particle. For 0.800 ,v c= the value of γ is found to be 5/3. Hence the mass of the particle is 28/ 0.25 10 kgm p vγ −= = ⋅ , which is a reasonable mass for an atomic particle. Using this mass, the energy of the particle is

2 28 8 2 12(5 / 3)(0.25 10 kg)(2.9979 10 m/s) 3.747 10 J,E mc − −= = ⋅ ⋅ = ⋅γ which agrees with the calculated value.

35.77. THINK: The running back is travelling at 55.0% the speed of light relative to the field. He throws the ball to a receiver running at 65.0% the speed of light relative to the field in the same direction. The speed of the ball relative to the running back is 80.0% the speed of light. The relativistic velocity transformation can be used to find the speed that the receiver perceives the ball to be travelling at. Recall that the speed of light is the same in all reference frames.


1258

SKETCH:

RESEARCH: The velocity of the ball with respect to the running back is 0.800 .xu c′ = The velocity of the running back with respect to the field is rb 0.550 .v c= The inverse Lorentz transformation can be used to find the velocity xu of the ball in the field frame:

,rb rb2

,rb rb

.1 /

xx

x

u vu

u v c′ +

=′+

Using a Lorentz transform gives the speed of the ball relative to the receiver: rec

,rec 2rec

,1 /

xx

x

u vu

u v c−′ =

−

where rec 0.650v c= is the velocity of the receiver relative to the field. SIMPLIFY: Not required. CALCULATE:

(a) ( ) ( )( )( )

( ) ( )( )( ),rec 22

0.800 0.550 0.9375 0.6500.9375 0.7360

1 0.9375 0.650 /1 0.800 0.550 /x x

c c c cu c u c

c c cc c c

+ −′= = = =

−+

(b) Photons travel at the speed of light and the speed of light is the same in any reference frame; therefore, the photons would appear to be travelling at the speed of light to the receiver. ROUND: (a) To three significant figures, the speed of the ball perceived by the receiver is

e8

,r c 0.736 2.21 10 m/s.xu c′ = = ⋅ DOUBLE-CHECK: The calculated value of the football’s relative speed was less than the speed of light as it must be, since no massive object can travel at the speed of light.

35.78. THINK: The 14 C electrons have kinetic energy 00.305 ,K E= where 0E is the rest energy. The baseline between the detectors is 2.0 m.xΔ = Find the necessary timing accuracy needed by the detectors to show that the expression for the relativistic momentum, and not the expression for the non-relativistic momentum, is correct. SKETCH: Not required. RESEARCH: The rest energy is 2

0 .E mc= The non-relativistic momentum is nr e nr ,p m v= and the non-

relativistic kinetic energy is ( ) 2nr e nr1/ 2 ,K m v= where em is the electron’s mass. The non-relativistic

velocity nrv can be determined from these equations. The relativistic momentum is r e r ,p m v= γ where

( )2r1/ 1 / .v cγ = − The relativistic kinetic energy is ( ) 01 .K E= −γ The relativistic velocity rv can be

determined from these equations. Finally, the time needed to travel a distance xΔ is / .t x v= Δ SIMPLIFY: Non-relativistic case:

( )2nr e nr 0 nr 0 e

1 0.305 2 0.305 /2

K m v E v E m= = =

Substituting 20 eE m c= into the equation gives


1259

( )2nr e e nr

nr

2 0.305 / 0.6100.610

x xv m c m c tv cΔ Δ= = = =

Relativistic case: ( ) 0 01 0.305 1.305K E E= − = =γ γ

( )( ) ( ) ( )

( ) ( )( ) ( )( )

( )( )

22 2 2 2 2

2 22r

2 22 22 2

r r r2 2r

1 1.305 1.305 1.3051 /

1.305 1 1.305 1.305 1.305 1 0.6425

1.305 1.305 1

rr

c c v cc vv c

x xv c v c c tv c

= = = − =−−

− Δ Δ = − = = = =

−

γ

CALCULATE: ( )( )nr 8

2.0 m8.5358 ns,

0.610 3.00 10 m/st = =

⋅

( )( )

( )( )

= =⋅ −

2

r 28

2.0 m 1.30510.376 ns

3.00 10 m/s 1.305 1t

ROUND: To two significant figures, nr 8.5 nst = and =r 10. ns.t By comparison of the calculated values

nrt and r ,t the necessary timing accuracy is on the order of 1 ns. DOUBLE-CHECK: The calculated values for nrt and rt had the correct units.

35.79. THINK: The spacecraft travels a distance of 31.00 10 lyd −= ⋅ in a time of 20.0 hrstΔ = as measured by an observer stationed on Earth. The length of the journey, 0 ,tΔ as measured by the captain of the spacecraft will be shorter due to time dilation. SKETCH: Not required. RESEARCH: The speed of the spacecraft is given by / .v d t= Δ The expression for time dilation is given by 0 .t tγΔ = Δ

SIMPLIFY: ( )

( ) ( )2200 02

1 / 1 /1 /

tt t t v c t t d c t

v c

ΔΔ = Δ = Δ − Δ = Δ − Δ

−

CALCULATE: Since 3/ 1.00 10 yr,d c −= ⋅

( ) ( )( )( )

23 3

0

1.00 10 yr 8.766 10 hr/yr20.0 hr 1 17.977 hr.

20.0 hrt

− ⋅ ⋅ Δ = − =

ROUND: To three significant figures, 0 18.0 hr.tΔ = DOUBLE-CHECK: The time measured by the captain is shorter than the time measured by the observer on the Earth. This makes sense because the captain is traveling at the same speed as the spacecraft (e.g. the captain is at rest with respect to the spacecraft). According to the time dilation theory, a moving clock runs slower than a clock at rest.

35.80. THINK: A hypothetical particle with rest mass 21.000 GeV/m c= and kinetic energy 1.000 GeVK = collides with an identical particle at rest. The two particles fuse to form a single new particle. Total energy and momentum are both conserved in the collision. Find (a) the momentum p and speed v of the first particle and (b) the rest mass newm and speed newv of the new particle. SKETCH:


1260

RESEARCH: The total energy is 20 ,E mc E Kγ= = + where 2

0 1.000 GeVE mc= = and ( ) 01 .K E= −γ

The relationship between energy and momentum is given by 2 2 2 2 4 .E p c m c= + SIMPLIFY: (a) The momentum of the first particle is given by

( )

2 22 2 2 2 4 2 2 2 2 0

0 2

2 2 20 0 0

/ 2 / .

E EE p c m c p c E p

cp E K E c E K K c

−= + = + =

= + − = +

The speed of the first particle is given by

( ) ( ) ( )2

0 00 0 22 2

0

1 1 1 .1 / 1 /

E EK E K E v c

K Ev c v c

= − + = = − + − −

(b) The rest mass newm of the new particle can be found by using the relationship between energy and momentum:

2 2 22 2 2 2 4 2 2 2 2 2new newnew new new new new new new4 / .

E p cE p c m c m m E p c c

c−

= + = = −

By energy and momentum conservation, the newly formed particle has the same total energy and momentum as the two original particles did prior to the collision, so

new 02 3.000 GeV,E E K= + = and new ,p p= which was found in part (a). The speed of the new particle is given by:

( ) ( )2 2 2 2 2

2new new new new new newnew new new new 2 2 22

newnewnew

1 /1 /

m v m v m c vp m v p

c vv cv c= = = =

−−−γ

2 2 2 2 2 2 2 newnew new new new new new 2 2 2

new new

.p c

p c p v m c v vm c p

− = =+

CALCULATE:

(a) ( )( ) ( )22 1.000 GeV 1.000 GeV 1.000 GeV / 1.73205 GeV/p c c= + =

( )( ) ( )( )

2

2

1.000 GeV1 0.86603

1.000 GeV 1.000 GeVv c c

= − = +

(b) ( ) ( )2 2 2 2 2new 3.000 GeV 1.73205 GeV/ / 2.44949 GeV/m c c c c= − =

( )( ) ( )

new 2 22 2

1.73205 GeV/0.57735

2.44949 GeV/ 1.73205 GeV/

c cv c

c c c= =

+

ROUND: To four significant figures, (a) 1.732 GeV/ ,p c= 0.8660v c= (b) 2

new 2.449 GeV/ ,m c= new 0.5774v c= DOUBLE-CHECK: The mass of the new particle is on the same order as the mass of a proton,

2p 0.938 GeV/ ,m c= so it is reasonable. The calculated speeds are large, but are realistic for small masses.

35.81. THINK: In considering accelerating bodies with special relativity, the acceleration experienced by the moving body is constant; that is, in each increment of the body’s own proper time, ,dτ the body acquires


1261

velocity increment dv gdτ= as measured in the body’s frame (the inertial frame in which the body is momentarily at rest). Given this interpretation, (a) Write a differential equation for the velocity v of the body, moving in one spatial dimension, as measured in the inertial frame in which the body was initially at rest (the “ground frame”). (b) Solve this equation for ( ),v t where both v and t are measured in the ground frame. (c) Verify that the solution behaves appropriately for small and large values of .t (d) Calculate the position of the body ( ),x t as measured in the ground frame. (e) Identify the trajectory of the body on a Minkowski diagram with coordinates x and ,ct as measured in the ground frame. (f) For 29.81 m/s ,g = calculate how much time t it takes the body to accelerate from rest to 70.7% of ,c as measured in the ground frame, and how much ground-frame distance, ,xΔ the body covers in this time. SKETCH: Not required. RESEARCH: In moving from the ground frame to the next frame, the body’s velocity was incremented by

.dv Since we are interested in a differential equation for the velocity as measured in the ground frame, an inverse Lorentz transformation from the next frame to the ground frame is necessary:

nextground 2 2

next

.1 / 1 /

u v v dvu v dvu v c vdv c

+ += + =+ +

The increment of the body’s proper time dτ is related to the increment of ground-frame time dt by time

dilation, ( )( )1/221 / .d v c dt= −τ The trajectory of the body in a space-time diagram will be determined by

examining the position as a function of time, which is determined in part (d). SIMPLIFY: (a) Ignoring squares and higher powers of differentials,

( ) ( )( )2

2 21 ... 1 / ...,1 /

v gd vgdv dv v gd v g v c dvgd c c

τ ττ ττ

+ + = = + − + = + − + + or ( )( )21 / .dv g v c d= − τ

But the increment of proper time dτ is related to the increment of ground-frame time dt by time dilation so the differential equation, in terms of ground frame quantities, becomes

( )( ) ( )( ) ( )( )( )( )

1/22 2 2

3/22

1 / 1 / 1 /

1 /

dv g v c d g v c v c dt

dv g v cdt

= − = − −

= −

τ

(b) The above differential equation separates, yielding

( )( )( ) ( )

( )( )( )3/2 1/20 0 22.

1 / 1 /

t v t v tdvg dt gtv c v t c

′′ = =′− −

This is readily solved, giving ( )( )( )1/221 /

gtv tgt c

=+

for the ground-frame velocity of the accelerating body

as a function of ground-frame time. (c) For ,gt c<< i.e., the Newtonian limit, the above result takes the form ( ) ,v t gt≅ exactly as expected. The relativistic limit, as time approaches infinity is:

( )( ) ( )2 2

lim lim .1 / /t t

gt gtv t cgt c gt c→∞ →∞

= = =+


1262

That is, the velocity of the accelerating body asymptotically approaches ,c as expected. (d) The position follows from the velocity through integration:

( ) ( )( )

1/2 1/22 22 2 2 2 2

1/20 0 2

0

1 11 /

t

t t gt dt gt gtc c c c cx t v t dtg g g g c g cgt c

′ ′ ′ ′ ′= + = + = + + = + ′ +

(e) The above result implies the simple relation ( )22 4 2/ .x ct c g− = The right-hand side is constant. Hence, the trajectory is a branch of a hyperbola on a Minkowski diagram. (f) Consider the ground-frame speed as a function of ground-frame time from part (b),

( )( )2

.1 /

gtv tgt c

=+

The time t required for the body to accelerate from rest to 0.707v c= is given by:

( )( )( ) ( )

( )2 22

2 21 / .

1 / 1 /

gt vv v gt c gt tgt c g v c

= + = =+ −

The ground-frame distance travelled in this time is ( ) ( )0 .x x t xΔ = − As stated in the problem, the

ground-frame position at ground-frame time 0t = is ( ) 20 / .x c g= Then

( )( )

1/2 222 2 2

2

/1 1 1 .

1 /

v cgtc c cxg c g g v c

Δ = + − = + − −

CALCULATE:

(f) ( )( )

( ) ( )( )

87

22

0.707 2.998 10 m/s3.055 10 s 353.6 days,

9.81 m/s 1 0.707 /t

c c

⋅= = ⋅ =

−

( )( )

( )( )( )( )

2 2815

22

2.998 10 m/s 0.707 /1 1 3.793 10 m 0.4009 ly

9.81 m/s 1 0.707 /

c cx

c c

⋅ Δ = + − = ⋅ = −

ROUND: The answers should be quoted to three significant figures: (f) 354 days,t = and 0.401 ly.xΔ = DOUBLE-CHECK: The motion of an object with constant proper acceleration in special relativity should be described by a hyperbola, as found in parts (d) and (e). The values found in part (f) are reasonable considering the relatively slow acceleration of 29.81 m/s .

Chapter 36: Quantum Physics

1263

Chapter 36: Quantum Physics In-Class Exercises

36.1. a 36.2. c 36.3. d 36.4. d 36.5. b 36.6. d Multiple Choice

36.1. b 36.2. e 36.3. a 36.4. c 36.5. c 36.6. c 36.7. c 36.8. b Questions

36.9. The spectral emittance as a function of wavelength is given by equation (36.13): ( ) ( )λ

πε λλ

=−

2

/( )5

2 .1Bhc K T

hce

For a given temperature the emittance depends on the wavelength of the light. Red light corresponds to a certain wavelength, but white light is made up of a distribution of visible light with different wavelengths. This is shown in Figure 36.4 of the textbook. The temperature of an object is inversely proportional to its minimum wavelength, as ε is explained by Wien’s displacement law. In the spectrum shown in Figure 36.4, you can see that the minimum wavelength of the red light corresponds to a longer wavelength than that of the white light, therefore the white-hot object is hotter than the red-hot object.

36.10. It is generally accepted that electrons exhibit both wave and particle like behavior. The wave like nature of electrons was postulated by de Broglie and experimental evidence was provided later. Details of the experimental evidence of the wave like properties of elections are discussed in the double-slit experiment for particles which starts on page 1186-1188 of the textbook. Einstein’s analysis of the photoelectric effect provides evidence of the particle like behavior of electrons.

36.11. The formula for Compton scattering is given by equation (36.20) in the textbook: ( )λ λ θ′ = + −e

1 cos .hm c

Blue light corresponds to photons with wavelength ranging from 9450 10 m−⋅ to 9495 10 m.−⋅ If you choose a value of 45θ = ° and 9450 10 m,λ −= ⋅ then the final wavelength of the photon after scattering is:

( )( )( ) ( )λ

−− −

−

⋅′ = ⋅ + − ° = ⋅

⋅ ⋅

349 9

31 8

6.626 10 Js450 10 m 1 cos45 450.00071 10 m,

9.109 10 kg 3.00 10 m/s

which is still well within the range of wavelengths corresponding to the blue light. The increase in wavelength after Compton Scattering is on the order of 1210 m,− which is too small to affect wavelengths in the visible spectrum.

36.12. The Heisenberg uncertainty relation for energy and time is given by equation (36.27) in the textbook: Δ ⋅ Δ ≥ / 2.E t Rearranging this equation to solve for tΔ gives: ( )Δ = Δ / 2 .t E For a proton-antiproton

pair ( ) ( )( )−−Δ = = ⋅ ⋅ ≈ ⋅

⋅22 27 8 6

p 19

1 V2 2 1.672 10 kg 3.00 10 m/s 1879 10 eV.1.602 10 J

eE m c Substituting this into

the equation for tΔ gives: ( )−

−⋅Δ = ≈ ⋅⋅

1625

6

6.5821 10 eV s 1.75 10 s.2 1879 10 eV

t This is the minimum lifetime of the

proto-antiproton pair.

36.13. If Planck’s constant was 5 J s the wavelength of the tennis ball, as well as other objects, would be very large relative to our universe. Consider a tennis ball with a mass, = 0.050 kgm that is travelling at a speed


1264

20 m/s.v = In the universe where 5 J sh = the tennis ball would have a wavelength of

( )( )λ′ = =5 J s 5 m.0.050 kg 20 m/s

By comparison in our universe 346.626 10 J s,h −= ⋅ so the wavelength of

the tennis ball is ( )( )λ−

−⋅= = ⋅34

346.626 10 J s 6.626 10 m.0.050 kg 20 m/s

The tennis ball with the 5 m wavelength would

appear very fuzzy and could possibly diffract through the spacing in the tennis rackets or the net.

36.14. In classical mechanics force is defined as ,F ma=

where m is the mass of the object and a

is its acceleration. If 0F =

for a massive particle it means that 0a =

which implies the particle is travelling at a

constant velocity. If you know the particles position and velocity at one point you can predict where the particle will be at some later time. In quantum mechanics the Heisenberg uncertainty relation states that you cannot instantaneously know, with absolute certainty an object’s position and momentum. Since momentum depends on the object’s mass and velocity ( =p mv ), you cannot predict with certainty the object’s trajectory from this information.

36.15. The classical physicist would expect that increasing the intensity of the UV light shining on the metal surface should increase the maximum kinetic energy of the electrons ejected from that surface. This was not observed in experiments. The experimental observations showed that increasing the intensity of the light increased the number of elections ejected from the metal surface but it did not increase their kinetic energy.

36.16. The power of the visible light source is v 60 W 60 J/s.P = = The power of the X-ray source is

x 0.002 W 0.002 J/s.P = = The wavelength of visible light, vλ is on the order of 7~1 10 m−⋅ to 61 10 m.−⋅

The wavelength of X-rays are on the order of 13~1 10 m−⋅ (hard X-rays) to 810 m− (soft X-rays). The electromagnetic spectrum is shown in Figure 31.10 of the textbook. If you consider a one second time interval, the energy of the visible light is v 60 JE = and the energy of the X-rays is x 0.002 J,E = so in terms of the total energy from the visible light source is v x30000E E= in one second. In terms of energy per photon, the X-ray photons will have higher energy and be much more damaging to the skin than a photon of visible light. For example taking values of 6

v 1 10 mλ −= ⋅ and 10x 1 10 m.λ −= ⋅ Using the

equation λ= /E hc you can obtain the ratio 4v x

x v

1 10EE

λλ

−′= = ⋅

′ or 4

v x1 10 ,E E−′ ′= ⋅ where the prime denote

that we are discussing energy per photon. It should be noted that hard X-rays ( 12~10 mλ − ) are very energetic and can cause immediate damage to human cells.

36.17. Neutrons in the neutron beam that have spins that are aligned with the spin of the neutrons in the nucleus of the polarized 3 He will not be absorbed because of the Pauli Exclusion Principle (The Pauli exclusion principle is discussed on page 1192-1195 of the textbook). Fewer neutrons in the unpolarized beam will be absorbed to form 4 He nuclei. The aligned neutrons in the beam that are not absorbed by the 3 He can be reused on a neutron beam to polarize the neutron beam.

36.18. The photocathode is made of Cesium which has a work function 2.1 eV,φ = this corresponds to a

maximum wavelength of 9max 590 nm 590 10 mλ −= = ⋅ (see Table 36.1 in textbook). Using green laser light

of wavelength 514.5 nmλ = corresponds to an energy of / 2.4 eVE hc λ= ≈ which is enough energy to cause elections to be emitted from the Cesium cathode. Doubling the power of the green laser results in an increase in intensity of the light hitting the cathode, this will increase the number of ejected electrons, but the energy per electron will remain the same.


1265

Problems

36.19. Assuming the surface temperature of the Sun is Sun 5800. KT = and the surface temperature of the Earth is

earth 300. KT = , Wien’s displacement law can be used to calculate the respective peak wavelengths.

(a) 3 3

7m,Sun

Sun

2.90 10 K m 2.90 10 K m 5.00 10 m5800. KT

λ− −

−⋅ ⋅= = = ⋅

(b) 3 3

6m,Earth

Earth

2.90 10 K m 2.90 10 K m 9.67 10 m300. KT

λ− −

−⋅ ⋅= = ≈ ⋅

36.20. The peak emission temperature can be calculated using Wien’s displacement law: λ

−⋅=3

m

2.90 10 K m .T The

total intensity of the radiation from the filament at a given temperature can be found using equation (36.1) from the textbook: 4 ,I Tσ= where σ is the Stefan-Boltzmann constant. At the short end of the visible

spectrum λ = ⋅1

-9m 380 nm=380 10 m and

−

−

⋅= =⋅

3

1 9

2.90 10 K m 7631.6 K.380 10 m

T The intensity at this

temperature is ( ) ( )−= ⋅ ⋅ = ⋅48 2 4 8 21 5.6704 10 W/ m K 7631.6 K 1.9 10 W/m .I At the long end of the visible

spectrum λ = ⋅2

-9m 780 nm=780 10 m and

−

−

⋅= =⋅

3

2 9

2.90 10 K m 3717.9 K.780 10 m

T

The intensity at this temperature is:

( ) ( )−= ⋅ ⋅ = ⋅48 2 4 7 22 5.6704 10 W/ m K 3717.9 K 1.1 10 W/m .I

36.21. A gamma ray with energy 123.5 10 eVE = ⋅ would have a very short wavelength. The wavelength of a

gamma ray with this energy is ( )( )

γλ−

− −⋅ ⋅

= = = ⋅ ≈ ⋅⋅

15 819 19

12

4.136 10 eV s 3.00 10 m/s3.545 10 m 3.5 10 m.

3.5 10 eVhcE

The rest mass energy of a proton is = 2,p p ,oE m c where = ⋅ 6 2

p 938.3 10 eV/c ,m so = ⋅ 6,p 938.3 10 eV.oE

The energy of the gamma ray is 12

6

3.5 10 eV 3700 times938.3 10 eV

⋅ ≈⋅

greater than the rest mass of a proton.

36.22. The temperature of the object is 20. °C 293 K.T = = Consider the radiation the object emits at the peak of the spectral energy density. (a) The peak wavelength can be calculated using Wien’s displacement law.

36 6

m2.90 10 m K 9.8976 10 m 9.90 10 m

293 Kλ

−− −⋅= = ⋅ ≈ ⋅

(b) The frequency at this wavelength is 8

13 136

m

3.00 10 m/s 3.0310 10 Hz 3.03 10 Hz.9.8976 10 m

cfλ −

⋅= = = ⋅ ≈ ⋅⋅

(c) The energy of one photon of light at this frequency is given by:

( )( )34 13 1 20 206.626 10 J s 3.0310 10 s 2.008366 10 J 2.01 10 J.E hf − − − −= = ⋅ ⋅ = ⋅ ≈ ⋅

Expressed in eV, 2019

1 eV2.008366 10 J 0.125366 eV 0.125 eV.1.602 10 J

E −−= ⋅ ⋅ = ≈

⋅

36.23. THINK: The temperature of your skin is approximately 35.0 °C 308.15 K.T = = Assume that it is a blackbody. Consider a total surface area of 22.00 m .A = (a) The Wien displacement law can be used to determine the peak wavelength λm of the radiation emitted by the skin, (b) the Stefan-Boltzmann radiation law can be used to determine the total power P emitted by your skin, and (c) the wavelength of the radiation needs to be considered.


1266

SKETCH:

RESEARCH: (a) The Wien displacement law is 3

m 2.90 10 K m.Tλ −= ⋅ (b) The total power is ,P IA= where I , the intensity, is given by the Stefan-Boltzmann radiation law:

4 ,I Tσ= using ( )8 2 45.6704 10 W/ m K .σ −= ⋅

(c) Power is energy per unit time. The relationship between energy and wavelength for photons is: / .E hc λ=

SIMPLIFY:

(a) λ−⋅=

3

m2.90 10 K m

T

(b) ,P IA= substituting 4P Tσ= gives: σ= 4 .P T A CALCULATE:

(a) 3

6m

2.90 10 K m 9.4111 10 m308.15 K

λ−

−⋅= = ⋅

(b) ( )( )( ) ( )48 2 4 25.6704 10 W/ m K 308.15 K 2.00 m 1022.57 WP −= ⋅ =

(c) Considering that a typical light bulb has a power of 100 W, why is it that a person does not glow with a power output of about 1000 W? The reason is because in order for you to “glow” your wavelength must be in the visible spectrum. However, the peak wavelength calculated in part (a) is λ =m 9.416 μm. This wavelength is in the infrared part of the spectrum, not the visible part. Your wavelength is not in the visible spectrum. ROUND: (a) To three significant figures, the peak wavelength is λ =m 9.42 μm. (b) To three significant figures, the total power emitted by your skin is 1.02 kW.P = DOUBLE-CHECK: The calculated values seem reasonable considering the given values. Comparing the calculated peak wavelength to the values in Figure 31.10 of the textbook shows that the wavelength of the emitted radiation is out of the visible spectrum.

36.24. THINK: The known room-temperature band-gap energies for germanium, silicon and gallium-arsenide are Ge 0.66 eV,E = Si 1.12 eV,E = and Ga-As 1.42 eV,E = respectively. The wavelength of photons can be calculated from the photon energy to (a) find the room-temperature transparency range of these three semiconductors, and (b) to explain the yellow color observed for ZnSe crystals, which have a band-gap of 2.67 eV. Semiconductors are only transparent if the energy (wavelength) of the photon is lower (higher) than the band-gap energy. (c) For a material to be used as a light detector, it must be able to absorb the incident light. This means the detecting material must have a band-gap corresponding to a longer wavelength than the incident light.


1267

SKETCH:

RESEARCH: The wavelengths associated with the given band-gap energies can be determined using the equation: / .hc Eλ = SIMPLIFY: Not required. CALCULATE:

(a) ( )( )

( )λ−

−⋅ ⋅

= = ⋅15 8

6Ge

4.136 10 eV s 3.00 10 m/s1.880 10 m

0.66 eV

( )( )( )λ

−−

⋅ ⋅= = ⋅

15 86

Si

4.136 10 eV s 3.00 10 m/s1.108 10 m

1.12 eV

( )( )( )λ

−−

⋅ ⋅= = ⋅

15 87

Ga-As

4.136 10 eV s 3.00 10 m/s8.738 10 m

1.42 eV

(b) ( )( )

( )λ−

−⋅ ⋅

= = ⋅15 8

7ZnSe

4.136 10 eV s 3.00 10 m/s4.647 10 m

2.67 eV

ROUND: The transparency range for these semiconductors is: (a) Ge 1900 nm,λ > Si 1110 nm,λ > and Ga-As 874 nm.λ > (b) The photon wavelength corresponding to the band-gap energy for ZnSe is ZnSe 465 nm.λ = Therefore, only the blue end of the visible spectrum will be absorbed by the ZnSe. This results in the yellow color that is observed for ZnSe crystals. (c) The only material that had a wavelength greater than 1550 nm was germanium ( )λ =Ge 1880 nm . This means that germanium is not transparent to the 1550 nm light and would be useful as a detector for this optical communications wavelength. DOUBLE-CHECK: The calculated wavelengths are reasonable and all had the correct units. It is expected that a material with a low band-gap energy will be able to absorb radiation with a large wavelength.

36.25. THINK: The mass of a dime is -32.268 10 kg,m = ⋅ its diameter is -317.91 10 m,d = ⋅ and its thickness is -31.35 10 m.t = ⋅ (a) The Stefan-Boltzmann radiation law can be used to determine the total radiant

energy coming from the dime. (b) Wien’s displacement law can be used to determine the wavelength of peak emission of each photon. Since each photon carries the same amount of energy, the number of photons can be determined. (c) With the temperature known, the thermal energy of air can be calculated. The Ideal Gas Law can be used to determine the volume of air required for it to have the same energy as the energy radiated from the dime in 1 second. Take room temperature to be T = 20 °C = 293 K.


1268

SKETCH:

RESEARCH: (a) The radiant energy per second can be found using the equation ,P IA= where I is the intensity and

tA is the total surface area of the dime, given by: ( )π= +t / 2 .A d d t By assuming the dime is an ideal

radiator, it is valid to use the Stefan-Boltzmann radiation law: 4 .I Tσ= (b) The energy of one photon is given by / .E hc λ= The wavelength that corresponds to peak emission

can be found using Wien’s displacement law: 3m 2.9 10 K m.Tλ −= ⋅

(c) If it is assumed that the air is made up of diatomic molecules, the energy per molecule is:

( )=air B3 / 2 .E k T Note that at at room temperature and standard pressure one mole ( 236.022 10⋅ molecules)

of air occupies a volume of 3 31 22.4 10 m .V −= ⋅

SIMPLIFY:

(a) ,P IA= substituting 4I Tσ= gives: P = σT 4πd d / 2 + t( ). The radiant energy per second is

⋅1 second.P (b) ,E nhf= where /f c λ= therefore the energy is: / .E nhc λ= The wavelength can be found from

Wien’s displacement law 3m 2.9 10 K m,Tλ −= ⋅ substitution into the energy equation gives:

( )( ) ( )

−

−

= ⋅

= ⋅

3

3

/ 2.90 10 K m

2.90 10 K m /

E nhcT

n E hcT

(c) =air B3 .2

E k T The number of molecules of air corresponding to the radiant energy emitted from the

dime in one second is: air/ .N E E= The volume of air with energy equal to one second of radiation from the dime is given by:

VT = νN = (22.4 L)N6.022 ⋅1023

CALCULATE:

(a) ( )( )( ) ( )

( )( )

48 2 4 -3 2 -3 -317.915.6704 10 W/ m K 293 K 17.91 10 m 10 m+1.35 10 m2

0.2428 W0.2428 J/s 1 sec 0.2428 J

P

E

π− = ⋅ ⋅ ⋅ ⋅

== =


1269

(b) ( )( )

( )( )( )

3

34 8

19

2.90 10 K m 0.2428 J

6.626 10 Js 3.00 10 m/s 293 K

1.208 10 photons/second

n−

−

⋅=

⋅ ⋅

= ⋅

(c)

N = 0.2428 J32

1.381⋅10−23( ) 293 K( )= 3.999 ⋅1019 molecules of air

VT =

22.4 ⋅10−3 m3( ) 3.999 ⋅1019 molecules( )6.022 ⋅1023 molecules( ) = 1.488 ⋅10−6 m3

ROUND: The calculated values should be reported to three significant figures, therefore: (a) 0.243 JE =

(b) n = 1.21⋅1019 photons per second

(c) VT = 1.49 ⋅10−6 m3 .

DOUBLE-CHECK: The calculated values all had the correct units. It is reasonable that only a very small amount of energy is radiated from the dime at room temperature. It also seems reasonable that the volume of air that has energy equal to one second of radiation from the dime is small.

36.26. The given work function is 5.8 eV.φ = The minimum light frequency necessary for the photoelectric effect to occur is given by equation (36.15) in the textbook.

15 1 15 1min 15

5.8 eV 1.402321 10 s 1.4 10 s4.136 10 eV s

fhφ − −

−= = = ⋅ ≈ ⋅⋅

36.27. The light that is incident on the sodium surface is 9470 nm 470 10 nm.λ −= = ⋅ The work function for sodium is 2.3 eVφ = (see Table 36.1 in textbook). The maximum kinetic energy of the electrons ejected from the sodium surface is max 0 .K eV hf φ= = − For photons / ,f c λ=

814 1

9

3.00 10 m/s 6.38 10 s .470 10 m

f −−

⋅= = ⋅⋅

Inserting this value into the equation for maxK gives:

( )( )15 14 1max 4.136 10 eV s 6.38 10 s 2.3 eV 0.34 eV.K − −= ⋅ ⋅ − =

36.28. The threshold wavelength is given as 9400. nm 400. 10 m.λ −= = ⋅ Frequency and wavelength for photons are related by the equation / .f c λ= The work function, φ of the alloy can be determined using equation (36.15) from the textbook:

( )( )8 -15

min 9min

3.00 10 m/s 4.136 10 eV s3.10 eV

4000 10 mhcf hφ

λ −

⋅ ⋅= = = =

⋅

36.29. The work function of Cesium is φ = 2.100 eV. The stopping potential for this material is

0 0.310 V.V = When the laser is shined on cathode made of an unknown material the stopping potential is found to be 0 0.110 V.V ′ =

(a) The wavelength of the laser light is found using equation (36.16): 0 .hceV hf φ φλ

= − = −

Where ( )( )( )

( )λφ

−⋅ ⋅

= = = ⋅+ +

-15 87

0

4.136 10 eV s 3.00 10 m/s5.15 10 m,

0.310 eV 2.100 eVhc

eV this wavelength can be used to

find the work function of the unknown material, u .φ


1270

( )( )φ

λ −

⋅ ⋅= − = − =

⋅

-15 8

u 0 7

4.136 10 eV s 3.00 10 m/s0.110 eV 2.30 eV

5.1485 10 mhc eV

(b) Work function for a number of common elements are listed in Table 36.1 in the textbook. Possible candidate materials for the unknown cathode would be potassium or sodium. They both have work functions of 2.3 eV.

36.30. The incident light has a wavelength of 9550 nm 550 10 m.λ −= = ⋅ The work function of zinc is 4.3 eV.φ = (See table 36.1 in text) In order for the photoelectric effect to occur the energy of the incident light must be equal to or greater than the work function of zinc. The energy of a photon of light with 9550 10 mλ −= ⋅ is

given by: ( )( )

λ −

⋅ ⋅= = =

⋅

-15 8

9

4.136 10 eV s 3.00 10 m/s2.3 eV.

550 10 mhcE The energy of the incident light is not

sufficient to eject any electrons from the zinc surface so there will not be any photoelectric current and therefore no stopping voltage is required.

36.31. White light is made up of photons with wavelengths ranging from λ = ⋅ 24.00 10 nm to ⋅ 27.50 10 nm ( −⋅ 74.00 10 m to −⋅ 77.50 10 m ). The work function of barium is given as 2.48 eV.φ = (a) The maximum kinetic energy of an electron ejected from the barium surface will correspond to a photon with the minimum wavelength.

( )( )φ

λ −

⋅ ⋅= − = − =

⋅

-15 8

max 7min

4.136 10 eV s 3.00 10 m/s2.48 eV 0.622 eV

4.00 10 mhcK

(b) The longest wavelength of light that could eject electrons is given by

( )( )λ

φ−

⋅ ⋅= = = ⋅ = ⋅

-15 87 2

4.136 10 eV s 3.00 10 m/s5.00 10 m 5.00 10 nm.

2.48 eVhc This means that the

⋅ 27.50 10 nm wavelength light would not eject electrons from the barium surface.

(c) The wavelength of light that would eject electrons with zero kinetic energy is given by: /hcλ φ=

which was solved in part (b). The wavelength was λ = ⋅ 25.00 10 nm.

36.32. THINK: The maximum kinetic energy measured is =max 1.5 eVK when the wavelength is .λ When the wavelength is decreased to / 2,λ the maximum kinetic energy measured is max 3.8 eV.K ′ = By considering the photoelectric effect, (a) the work function of the material and (b) the original wavelength can be determined. SKETCH:


1271

RESEARCH: Combining equation (36.14) and (36.16) from the textbook gives: max ,K hf φ= − where the frequency is / .f c λ= SIMPLIFY:

(a) max maxhc hcK Kφ φλ λ

= − = + , ( )maxmax max/ 2 2 2 2

Khc hcK Kφ φφ φλ λ

′′ = − = − = + − ,

maxmax2 2

KK φ′

= + , so max max 2K Kφ ′= −

(b) ( )maxmax max max max2

hc hc hcKK K K K

φ λλ φ

= − = =′+ + −

, and therefore max max

hcK K

λ =′ −

.

CALCULATE: (a) ( ) ( )3.80 eV 2 1.50 eV 0.800 eVφ = − =

(b) ( )( )

( ) ( )-15 8

74.136 10 eV s 3.00 10 m/s

5.39478 10 m3.80 eV 1.50 eV

λ −⋅ ⋅

= = ⋅−

ROUND: (a) To three significant figures, the work function is 0.800 eV.φ = (b) To three significant figures, the original wavelength is 539 nm.λ = DOUBLE-CHECK: The calculated values have the correct units.

36.33. The X-rays have wavelength, 90.120 nm 0.120 10 m.λ −= = ⋅ They are scattered by the carbon. The angle between the incoming and outgoing photon is 90.0 .θ = ° The formula for Compton scattering is given by

Equation (36.20) in the textbook: ( ) ( )λ λ θ λ λ λ θ′ ′= + − − = Δ = −e e

1 cos . 1 cos .h hm c m c

Inserting the

proper values gives:

( )( )( )( )

3412

31 8

6.626 10 J s 1 cos90.02.42 10 m

9.109 10 kg 3.00 10 m/sλ

−−

−

⋅ − °Δ = = ⋅

⋅ ⋅

This is the Compton wavelength shift.

36.34. The wavelength of the incoming photon is: λ − −= = = ⋅ = ⋅⋅

4 136

1240 eV nm 6.20 10 nm 6.20 10 m.2.0 10 eV

hcE

The

outgoing photon’s wavelength can be found using the Compton scattering formula

( ) ( ) ( )( )λ λ θ − − − −′ = + − = ⋅ + ⋅ − ° = ⋅ = ⋅13 12 12 31 cos 6.20 10 m 2.426 10 m 1 cos53 1.6 10 m 1.6 10 nm.e

hm c

36.35. The wavelength of the incoming photon is λ = 0.30 nm; its original energy was:

( )( )λ

−

−

⋅ ⋅= = = =

⋅

15 8

10

4.13567 10 eV s 2.998 10 m/s4133 eV.

3.0 10 mhcE hf It rebounds at angle of 160 .θ = ° Its new

wavelength can be found using the Compton scattering formula.

( ) ( ) ( )( )( )( )λ λ θ

−− −

−

⋅ − °′ = + − = ⋅ + = ⋅

⋅ ⋅

3410 10

31 8e

6.626 10 J s 1 cos1601 cos 3.0 10 m 3.047 10 m

9.109 10 kg 2.998 10 m/sh

m c

Its new energy is: ( )( )15 8

10

4.13567 10 eV s 2.998 10 m/s4069 eV.

3.047 10 mhcEλ

−

−

⋅ ⋅′ = = =

′ ⋅ The amount of energy lost is

4133 eV 4069 eV 64 eV.E E E′Δ = − = − =


1272

36.36. THINK: The X-rays have an initial energy 54.000 10 eV.E = ⋅ They undergo Compton scattering from a target, and the scattered rays are detected at 25.0θ = ° relative to the incident rays. (a) The formula for Compton scattering can be used to find the energy of the scattered X-ray, ,E′ and (b) conservation of energy can be used to find the energy of the recoiling electron, e .E SKETCH:

RESEARCH: (a) The energy of a photon is / .E hc λ= The wavelength of the scattered X-ray is given by the Compton scattering formula:

( )e

1 cos .hm c

λ λ θ′ = + −

(b) Due to energy conservation in Compton scattering, the energy lost by the scattered photon is imparted onto the electron, that is, .eK E E′= − SIMPLIFY: (a) The energy of the scattered X-ray is:

( )( ) 1

2e

e

1 cos1 .1 cos

hc hcEhc h E m cE m c

θλ θ

− −

′ = = = + ′ + −

(b) No simplification is required. CALCULATE:

(a) ( )( )( )

( )

1

55 6

1 cos 25.01 3.7267 10 eV4.000 10 eV 0.51100 10 eV

E

− − ° ′ = + = ⋅ ⋅ ⋅

(b) ( ) ( )5 5 44.000 10 eV 3.7267 10 eV 2.7332 10 eVeK = ⋅ − ⋅ = ⋅

ROUND: (a) The energy of the scattered X-ray is 373 keV.E′ = (b) The kinetic energy of the recoiling electron is 27.3 keV.eK = DOUBLE-CHECK: The X-ray should lose energy after scattering off of the electron. As expected, this energy loss is equal to the kinetic energy of the electron: 372.7 keV 27.33 keV 400.0 keV .eE K E′ + = + = =

36.37. THINK: (a) X-rays of energy 14

0 140 keV 2.243 10 JE −= = ⋅ bounce off of a proton at θ = °90.0 . The Compton

scattering formula can be used to find their fractional change in energy, ( )= −0 0/ .f E E E (b) The equation derived in part (a) can be used to find the energy of a photon that would be necessary to cause a 1.00% change in energy at θ = °90.0 scattering.


1273

SKETCH:

RESEARCH: The energy of a photon is / .E hc λ= The wavelength of the scattered photon is found from the Compton scattering formula, but with the mass of a proton substituted for the mass of an electron:

( )λ λ θ′ = + −p

1 coshm c

The mass of a proton is −= ⋅ 27p 1.673 10 kg.m

SIMPLIFY:

(a)

( )

( )( )

0 002

0 0 0 p 00

0 p

1 cos1 1

1 cos1 cos

hcE EE E hc hcfE E E m c Ehc hE

E m c

θλλ θ

θ

− −− ′= = = − = − =′ + −

+ −

(b) Using the equation from part (a), ( )

( ) ( )( ) ( ) ( )( )θ

θ θθθ

−= + − = − =

− −+ −

2p0 2

p 0 0 02p 0

1 cos 1 cos 1 cos

1 1 cos1 cosfm cE

f f m c E E Efm c E

CALCULATE:

(a) ( )( )

( )( )

144

27 8 2 14

2.243 10 J 1 cos901.491 10

(1.673 10 kg)(2.998 10 m/s) 2.243 10 J 1 cos90f

−−

− −

⋅ − °= = ⋅

⋅ ⋅ + ⋅ − °

(b) For a 1.0% change in energy, = 0.010 :f

( )( )( )( )( )

−−

⋅ ⋅= = ⋅ =

− − °

227 812

0

0.010 1.673 10 kg 2.998 10 m/s1.519 10 J 9.481 MeV

1 (0.010) 1 cos90E

ROUND: (a) 41.49 10f −= ⋅ (b) 0 9.48 MeVE = DOUBLE-CHECK: To get a 1.00% fractional change in energy, gamma-rays would be required. These are extremely high energy photons. This is one reason why electrons are used for scattering experiments (the photons do not have to be as energetic).

36.38. THINK: The X-ray photon has an energy of 45.00 10 eV.E = ⋅ It strikes an electron which is initially at rest inside a metal and is scattered at an angle of 45 .θ = ° The Compton scattering formula can be used to find the kinetic energy eK and momentum ep (magnitude and direction) of the electron after the collision. Conservation of energy and momentum can also be used to solve the problem.


1274

SKETCH:

RESEARCH: The energy of a photon is / .E hc λ= The momentum of a photon is / .p h λ= The wavelength of the scattered photon, according to the Compton scattering formula, is

( )λ λ θ′ = + −e

1 cos .hm c

In Compton scattering, energy is conserved. The energy that is lost by the photon is imparted to the electron, that is, ′= −e .K E E Momentum is also conserved in this collision, that is, .ep p p′= +

For

scattering in two dimensions, this becomes x x exp p p′= + and .y y eyp p p′= + The magnitude of the

electron’s momentum is 2 2 ,e ex eyp p p= + and the direction is ( )θ −= 1tan / .e ey exp p

SIMPLIFY: The kinetic energy of the electron is given by:

( ) ( )λ λ θ θ

λ

′ = + − = + −

′= − = −′

e e

e

1 cos 1 cosh hc hm c E m c

hcK E E E

The momentum of the electron is given by:

2 2

1

coscos cos

0 sin sin

tan

ex x x

ey y y

e ex ey

eye

ex

h h E hp p p p pc

hp p p p

p p p

pp

θθ θλ λ λ

θ θλ

θ −

′ ′= − = − = − = −′ ′

′ ′= − = − = −′

= +

=

CALCULATE:

( )( )( )

( ) ( )( )( )( )

-15 8 3411

4 31 8

4.13567 10 eV s 2.998 10 m/s 6.626 10 J s 1 cos 452.5508 10 m

5.00 10 eV 9.109 10 kg 2.998 10 m/sλ

−−

−

⋅ ⋅ ⋅ − °′ = + = ⋅

⋅ ⋅ ⋅

( ) ( )( )( )

( )( )( )

( ) ( )( )

-15 84

e 11

4 19 3424

e 8 11

4.13567 10 eV s 2.998 10 m/s5.00 10 eV 50.0 keV 48.607 keV 1.393 keV

2.5508 10 m

5.00 10 eV 1.602 10 J/eV 6.626 10 J s cos 458.350 10 kg m/s

2.998 10 m/s 2.5508 10 mx

K

p

−

− −−

−

⋅ ⋅= ⋅ − = − =

⋅

⋅ ⋅ ⋅ °= − = ⋅

⋅ ⋅

( ) ( )( )

34 o23

e 11

6.626 10 J s sin 451.837 10 kg m/s

2.5508 10 myp−

−−

⋅= − = − ⋅

⋅


1275

( ) ( )2 224 23 23e 8.399 10 kg m/s 1.837 10 kg m/s 2.018 10 kg m/sp − − −= ⋅ + − ⋅ = ⋅

( )( )

231

e 24

1.837 10 kg m/stan 65.6

8.399 10 kg m/s

(The negative means the angle is made below the positive -axis.)x

θ−

−−

− ⋅ = = − ° ⋅

ROUND: To two significant figures: 23e e e1.4 keV, 2.0 10 kg m/s, and 66 .K p θ−= = ⋅ = − °

DOUBLE-CHECK: Using the non relativistic equation, the momentum of the electron is e e e ,p m v= when

ee

e

2.

Kv

m= Then,

( )( )( )− − −= = ⋅ ⋅ ⋅ ≈ ⋅3 19 31 23e e e2 2 1.393 10 eV 1.602 10 J/eV 9.109 10 kg 2.0 10 kg m/s.p K m

This is in agreement with the value in the solution.

36.39. (a) The wavelength of a photon is / .hc Eλ = For a photon of energy 2.00 eV,E = the wavelength is:

( )( ) ( )-15 8 74.13567 10 eV s 2.998 10 m/s / 2 eV 6.1992 10 m 620. nm.λ −= ⋅ ⋅ = ⋅ ≈

(b) The wavelength of an electron is ( )λ = = e/ / ,h p h m v and its kinetic energy is 2e / 2.K m v= In terms

of ,K the velocity v is e

2 .Kvm

= Then the wavelength of the electron is 2

e

e e

.2 2mh h

m K Kmλ = = For an

electron of kinetic energy ( ) ( ) ( )− −= ⋅ ⋅ = ⋅19 192 eV 1.602 10 J / 1 eV 3.204 10 J,K the wavelength is:

( )( )( )

234-10

19 -31

6.626 10 J s8.672 10 m 0.867 nm.

2 3.204 10 J 9.109 10 kgλ

−

−

⋅= = ⋅ ≈

⋅ ⋅

36.40. The car has a mass of = ⋅ 32.000 10 kgm and a speed ( )( )( )= ≈100.0 km/h 1000 m/km 1 hr/3600 s 27.78 m/s.v

The de Broglie Wavelength is ( )λ = =/ / .h p h mv The wavelength of this car is therefore

( ) ( )( )λ − − = ⋅ ⋅ = ⋅ 34 3 386.626 10 J s / 2.000 10 kg 27.78 m/s 1.193 10 m

36.41. The nitrogen molecule has a mass of 264.648 10 kgm −= ⋅ and a speed 300.0 m/s.v = (a) The de Broglie wavelength is ( )λ = =/ / .h p h mv The wavelength of this nitrogen molecule is

therefore ( ) ( )( )λ − − − = ⋅ ⋅ = ⋅ ≈ 34 26 116.626 10 J s / 4.648 10 kg 300.0 m/s 4.752 10 m 47.52 pm.

(b) For a double slit experiment, the fringes are 0.30 cmxΔ = apart and the screen is 70.0 cmL = in front of the slits. In a double slit experiment with particles the distance between the fringes is / .x L dλΔ = In that case, the distance d between the slits is ( )( ) ( )λ −= Δ = ⋅ =11/ 4.752 10 m 70.0 cm / 0.30 cm 11 nm.d L x

36.42. The alpha particles are accelerated through a potential difference of magnitude 20000 V.V = Alpha particles are composed of 2 protons and 2 neutrons, and therefore have a charge of 2 ,q e= where

191.602 10 C.e −= ⋅ Assuming the alpha particles are accelerated from rest, the final kinetic energy of each alpha particle is ( ) ( )= Δ = = ⋅ =2 e 20000 V 40000 eV,K U q V or 156.409 10 J.−⋅ The de Broglie

wavelength is ( )λ = =/ / .h p h mv Kinetic energy is 2 / 2.K mv= In terms of kinetic energy, the speed is


1276

e

2 .Kvm

= Substituting, the de Broglie wavelength becomes e

e e

.2 2mh h

m K Kmλ = = Note the mass of an

alpha particle is 276.645 10 kg.m −= ⋅ The de Broglie wavelength of the alpha particle is

( )( )34

-14

15 -27

6.626 10 J s 7.18 10 m.2 6.409 10 J 6.645 10 kg

λ−

−

⋅= = ⋅⋅ ⋅

36.43. The electron has a de Broglie wavelength of 550 nm.λ = (a) The de Broglie wavelength is ( )λ = =/ / .h p h mv The speed of the electron is

( )( )λ

−

−

⋅= = = ≈⋅ ⋅

34

-31 7e

6.626 10 J s 1323 m/s 1300 m/s.9.109 10 kg 5.5 10 m

hvm

(b) This speed is much less than the speed of light, so the non-relativistic approximation is sufficient. (c) In non-relativistic terms, the electron’s kinetic energy is

( )( )−−

⋅= = = ⋅

231225

9.109 10 kg 1323 m/s7.967 10 J.

2 2mvK

In eV, this becomes ( )( ) ( )−= ⋅ ⋅ = ⋅ ≈25 -19 -67.967 10 J 1 eV / 1.602 10 J 4.972 10 eV 5.0 μeV.K

36.44. THINK: The roommate wants to know if he could be diffracted when passing through a doorway. His mass is 60.0 kg.m = The width of the doorway is 0.900 m.d = The de Broglie wavelength can be used to find (a) the maximum speed maxv at which the roommate can pass through the doorway in order to be significantly diffracted and (b) the time tΔ it would take the roommate to make a step of length

0.75 mxΔ = in order to be significantly diffracted. Assume that significant diffraction occurs when the width of the diffraction aperture is less than 10.0 times the wavelength of the wave being diffracted, that is,

rm10.0 ,d λ< where rmλ is the de Broglie wavelength of the roommate. SKETCH: Not applicable. RESEARCH:

(a) The de Broglie wavelength is .h hp mv

λ = =

(b) Speed is .xvt

Δ=Δ

SIMPLIFY:

(a) For significant diffraction take rm .10.0

dλ > The speed is given by: .hvmλ

= Since v and λ are

inversely proportional, the minimum λ will yield a maximum v . Then:

maxrm

10.0 h hvm mdλ

= =

(b) max

xtvΔΔ =

CALCULATE:

(a) ( )

( )( )−

−⋅

= = ⋅34

34max

10.0 6.626 10 J s1.2270 10 m/s

60.0 kg 0.900 mv

(b) ( )

( )33

34

0.75 m6.1123 10 s

1.2270 10 m/st

−Δ = = ⋅

⋅


1277

ROUND: (a) 34

max 1.23 10 m/sv −= ⋅

(b) 336.1 10 stΔ = ⋅ (c) To achieve a de Broglie wavelength capable of diffracting through the doorway the roommate must move at a speed of 341.23 10 m/s.v −= ⋅ This would take him 336.11 10 s,⋅ or 261.94 10⋅ years! This is more than 1610 times the age of the universe. The roommate does not need to worry about diffracting through the doorway. Particles like electrons and protons can diffract because they are many orders of magnitude smaller in mass than a person. DOUBLE-CHECK: It is reasonable that the roommate would have to move extremely slow in order for him to be diffracted since his mass is so large and the doorway is so large.

36.45. THINK: The de Broglie waves have a wavelength /h pλ = and a frequency / .f E h= A Newtonian

particle of mass ,m has momentum ,p mv= and energy ( )= 2 / 2 .E p m (a) To calculate the dispersion relation for the de Broglie waves of a Newtonian particle, the angular frequency ω needs to be found as a function of wave number, .k (b) The phase velocity pv and group velocity gv can be determined by using the dispersion relation. SKETCH: Not applicable. RESEARCH: (a) The dispersion relation is an expression for the angular frequency, 2 ,ω πν= as a function of wave number 2 / ;k π λ= that is, ( )ω ω= .k (b) The phase velocity of a wave is p / ,v kω= while the group velocity of a wave is g / .v kω= ∂ ∂

SIMPLIFY: (a) For a Newtonian particle, the dispersion relation is

( )( )

2 2 2 2 2

2 2

2 2 22 2 .2 42 2 2pE h h k hkf k

h h m h h mm mπ π πω π ω π ω

πλ π

= = = = = =

(b) For the same particle, the phase velocity is,

p ,4 2 2

phk hvk m m m

= = = =ωπ λ

while the group velocity is

g .2

pd hk hvdk m m m

= = = =ωπ λ

Since the momentum of a Newtonian particle is ,p mv= it is the group velocity that corresponds to the classical velocity of the particle. CALCULATE: Not applicable. ROUND: Not applicable. DOUBLE-CHECK: When several waves are superimposed to result in a single wave shape (the envelope of the wave) the speed of the overall wave shape is described by the group velocity. The phase velocity describes the velocity at which the peaks, or phases, of the waves propagate. The classical velocity should correspond to the group velocity of the particle.

36.46. THINK: The de Broglie waves have a wavelength /h pλ = and a frequency / .f E h= A relativistic

particle of mass m has momentum p mvγ= and energy 2 ,E mc γ= where ( )γ−

= −1/22 21 / .v c (a) To

calculate the dispersion relation for the de Broglie waves of a relativistic particle, the angular frequency ω


1278

needs to be found as a function of wave number, .k (b) The phase velocity pv and group velocity gv can be determined by using the dispersion relation. SKETCH: Not applicable. RESEARCH: (a) The dispersion relation is an expression for the angular frequency 2 ,fω π= as a function of wave number 2 / ;k π λ= that is, ( )ω ω= .k (b) The phase velocity of a wave is p / ,v kω= while the group velocity of a wave is g / .v k= ∂ ∂ω

SIMPLIFY: (a) For a relativistic particle, the dispersion relation is

( )

( )

1/2 1/22 21/22 2 4 2 4 2 4

1/22 2 42 2

2

2 2 22 22

4

E hc hkcf m c m c m cpch h h h

m ck k ch

π π πω π πλ π

πω

= = = = =+ + +

= +

(b) For the same particle, the phase velocity is, 1/2 1/22 2 4 2 2 2

2p 2 2 2 2

4 4 ,1m c m cv cck h k h kω π π = = =+ +

while the group velocity is, 2

g 1/2 1/22 2 4 2 2 22 2

2 2 2

.4 41

d kc cvdk m c m ck c

h h k

= = =

+ +

ω

π π

Using the relation, 2 2 2 24 ,h k pπ= the group velocity can be written as:

( )( )

2 2 2 42 2 22g 2 2 2 2 2 2 2 2 2 2 2 2 2 2 4

2 2 2 2 2

22 2

g 22 2 2 4

4 41 1 14

.

p c p cc c cvm c m c m c p m c p c m c

h k p pmv cpc pcv v

E mcp c m c

π ππ

γγ

= = = = =+ ++ + +

= = = =+

Therefore, the group velocity is the classical velocity of the particle. CALCULATE: Not applicable. ROUND: Not applicable. DOUBLE-CHECK: Note that the phase velocity can exceed the speed of light (this is not unusual, or worrisome, as the phase velocity does not transmit any energy or information), while the group velocity cannot. This further enforces that it must be the group velocity, and not the phase velocity, which corresponds to the classical velocity of the particle.

36.47. The mass of the particle is 50.0 kg.m = It has a de Broglie wavelength of 20.0 cm.λ = (a) The de Broglie wavelength is ( )λ = =/ / .h p h mv The speed is therefore

3435 356.626 10 J s 6.626 10 m/s 6.63 10 m/s.

(50.0 kg)(0.200 m)hv

mλ

−− −⋅= = = ⋅ ≈ ⋅

(b) From the uncertainty relation ( ) ( )1/ 2 1/ 2 ,x xx p x m vΔ ⋅ Δ ≥ Δ ⋅ Δ ≥ the uncertainty in the speed

must be ( ) ( )1/ 2 / .xv x mΔ ≥ Δ ⋅ The minimum uncertainty is

( )( )34

36 361.0546 10 J s 5.273 10 m/s 5.27 10 m/s.2 2 0.200 m 50.0 kgxv

x m

−− −⋅Δ = = = ⋅ ≈ ⋅

⋅ Δ ⋅


1279

36.48. The distance through a hydrogen atom of radius 100.53 10 mr −= ⋅ is 102 1.06 10 m.d r −= = ⋅ The time required for the light to travel through it is

1019

8

1.06 10 m 3.54 10 s.2.998 10 m/s

d dtv c

−−⋅= = = = ⋅

⋅

The largest time uncertainty cannot be greater than the actual travel time, that is 19max 3.54 10 s.t −Δ = ⋅ The

uncertainty relation between time and energy is (1/ 2) .E tΔ ⋅ Δ ≥ The uncertainty in the energy is therefore

( ) ( )34

1619

1.0546 10 J s1/ 2 1.4895 10 J.2 2 3.54 10 s

E t Et

−−

−

⋅Δ ⋅ Δ ≥ Δ ≥ = = ⋅⋅ Δ ⋅

In terms of eV, this is

( )( ) ( )16 191.4895 10 1 eV / 1.602 10 J 929.8 eV 0.930. keV.E J− −Δ ≥ ⋅ ⋅ = ≈

The smallest EΔ can be is 0.930 keV. As tΔ decreases from its maximum value, EΔ must increase according to the uncertainty relation.

36.49. The uncertainty relation between time and energy is ( )Δ ⋅ Δ ≥ 1/ 2 .E t In terms of mass, 2 .E mcΔ = Δ The

neutron’s mass is 271.67 10 kg.m −= ⋅ It has an average lifetime of 900. s.t = The largest time uncertainty cannot be greater than the actual lifetime of the particle, which is max 900. s.tΔ = The uncertainty in the mass of the neutron is therefore ( )Δ ≥ Δ1/ 2 /E t .

( ) ( )( )( )

3455 55

2 28

1/ 2 1/ 2 1.0546 10 J s6.519 10 kg 6.52 10 kg

900. s 2.998 10 m/sm

t c

−− −⋅

Δ ≥ = = ⋅ ≈ ⋅Δ ⋅ ⋅

As the uncertainty in the time tΔ decreases from its maximum value, the uncertainty in the mass increases, according to the uncertainty relation.

36.50. Fuzzy lives in a universe where 1.00 J s.= Fuzzy’s mass is 0.500 kgm = and he lives somewhere within a 0.750 m wide pond. The uncertainty relation between position and momentum (in one dimension) is ( )Δ ⋅ Δ ≥ 1/ 2 .x p In terms of velocity, ,p m vΔ = Δ and so the uncertainty relation becomes

( )Δ ⋅ Δ ≥ 1/ 2 / .x v m Since the largest uncertainty in Fuzzy’s position is the width of the pond,

max 0.75 m,xΔ = the minimum uncertainty in his speed is

( ) ( ) ( )( ) ( )( ) Δ = Δ = = ≈ min max1/ 2 / 1/ 2 1.00 J s / 0.500 kg 0.750 m 1.3333 m/s 1.33 m/s.v m x As the uncertainty in his position xΔ decreases from its maximum value, the uncertainty in his velocity increases, according to the uncertainty relation. If the uncertainty prevails for 5.00 s,t = Fuzzy could move ( )( )Δ = Δ = = ≈1.3333 m/s 5.00 s 6.6666 m 6.67 mx vt away from his pond.

36.51. The uncertainty relation between position and momentum (in one dimension) is ( )Δ ⋅ Δ ≥ 1/ 2 .x p In

terms of velocity, ,p m vΔ = Δ and so the uncertainty relation becomes ( )Δ ⋅ Δ ≥ 1/ 2 / .x v m The electron is confined to a box of dimensions 20.0 μm.L = The maximum uncertainty in the (one-dimensional) position of the electron is the dimension of the box, that is max 20.0 μm.x LΔ = = The minimum uncertainty in the speed of the electron is

( ) ( ) ( )( ) ( )( )34 31min max1/ 2 / 1/ 2 1.05457 10 J s / 9.109 10 kg 20.0 μm 2.894 m/s.v m x − − Δ = Δ = ⋅ ⋅ =

This uncertainty could be due to the change in direction of the electron after colliding with a wall inside the box, that is ( ) ( )Δ = − = final initial 2x x x xv v v v (where the average magnitudes of the initial and final velocities are the same). Then the minimum speed at which the electron could be moving inside the box is


1280

( )min min min min2 / 2 2.894 m/s /2 1.45 m/s.v v v v= Δ = Δ = =

36.52. THINK: The dust particle has mass 161.00 10 kgm −= ⋅ and diameter 5.00 μm.d = It is confined to a box of length 15.0 μm.L = The Heisenberg uncertainty relation can be used to determine (a) if the particle can be at rest, (b) the range of its velocity, and (c) how long it will take for it to move a distance of

31.00 10 mx −= ⋅ at the lower range of the velocity. SKETCH:

RESEARCH: The uncertainty relation between position and momentum (in one dimension) is

1 .2xx pΔ Δ ≥

In terms of velocity, ,p m vΔ = Δ and so the uncertainty relation becomes

( ) 1 .2xx m vΔ Δ ≥

The equation for velocity is / .xv x t= SIMPLIFY: (a) If the particle is at rest, then there is no uncertainty in the momentum, 0.pΔ = Then Heisenberg’s uncertainty relation,

1 1 ,2 2x

x

x p xp

Δ Δ ≥ Δ ≥Δ

would required that .xΔ = ∞ However, the particle is known to be contained in the box, so x LΔ = (the length of the box). Therefore, due to Heisenberg’s uncertainty relation, we cannot know if the particle is at rest. (b) With the particle confined to the box, the uncertainty in position is Δ = − .x L d The uncertainty in the speed is:

Δ ≥Δ

2xvm x

Therefore, the particle’s velocity must be somewhere in the range

2 2x x

xv v

vΔ Δ

− ≤ ≤

.4 4xv

m x m x− ≤ ≤

Δ Δ

(c) / xt x v= CALCULATE:

(b)( )

( )( )−

−− −

⋅Δ= = ⋅

⋅

3414

16 6

1.0546 10 J s2.6365 10 m/s

2 4 10 kg 10 10 mxv


1281

(c) ( )

( )−

−

⋅= = ⋅

⋅

310

14

1 10 m3.79 10 s

2.6365 10 m/st

ROUND: To three significant figures: (b) 14 142.64 10 m/s 2.64 10 m/sxv− −− ⋅ ≤ ≤ ⋅ (c)

10 33.79 10 s 1.20 10 yearst = ⋅ ≈ ⋅ DOUBLE-CHECK: For all intent, the dust particle is at rest since it would take it 2400 years to move just 1 mm. However, by the Heisenberg uncertainty principle, one cannot be sure that at any given time the particle is truly at rest.

36.53. THINK: A quantum state of energy E can be occupied by any number n of bosonic particles (including 0).n = At absolute temperature ,T the probability of finding n particles in this state is

B

exp ,nnEP Nk T

= −

where Bk is Boltzmann’s constant and N is the normalization factor. Calculate the

mean or expected value of n, ,n i.e. the occupancy of this state, given this probability distribution. SKETCH: Not applicable.

RESEARCH: The expectation value of n is 0

.nn

n nP∞

=

= The value of the constant N is determined by

the requirement that all the probabilities sum to one, that is 0

1.nn

P∞

=

= To simplify the notation, let

Bexp .Ez

k T

= −

With this, .nnP Nz=

SIMPLIFY: In order to evaluate the normalization factor N: 0 0

1 1 .1n

n

n n

NP N z N zz

∞ ∞

= =

= = = = −− Then

simplify the expected value to

( )0 1 1 1

0 exp 1 .nn n n n

n n

B

nEnP N n N nz z nzn k T

∞ ∞ ∞ ∞

= = = =

−= = + = = −

There are several ways to evaluate the sum in this expression. If 1,z < then the original series 1n

nnz∞

= is

absolutely convergent, and it is okay to interchange the order of the sums. One way to evaluate it is as a

sequence of sequences: 1 1 2 3 0

... .n n n n

n n n n

k

n

n

k knz z z z z z

∞ ∞ ∞ ∞ ∞ ∞

= = = = = =

= + + + =

Next,

0

1 .1

j

n k n k j

n k n k k kz z z z z zz

∞ ∞ ∞

= = =

−= = = ⋅− Substituting this into the sequence of sequences formula:

( )1 02

0 0

1 1 .1 1 1 1 1

k

n k n k k k

n k n k z z znz z z z zz z z z z

∞ ∞ ∞ ∞ ∞

= = = = =

= = = = ⋅ = − − − − −

Altogether, this makes the occupancy:

( )( )

∞

=

− = = − = =

− −− −

1

2

exp1 .

11 1 exp

Bn

B

n

Ek Tz zN nz zn z Ez

k T

CALCULATE: Not applicable. ROUND: Not applicable.


1282

DOUBLE-CHECK: The expectation value of n calculated above is an expected result as it is the Bose-Einstein distribution, which describes the distribution of identical (and therefore indistinguishable) bosons in an energy state E at thermal equilibrium.

36.54. THINK: The quantum state of energy E and temperature T has a probability distribution

B

exp ,nnEP Nk T

= −

(as the preceding problem), but with fermionic particles. Here Bk is Boltzmann’s constant and N is the normalization factor. Due to the Pauli exclusion principle, the only possible occupation numbers are 0n = and 1.n = Calculate the mean occupancy n of the state in this case. SKETCH: Not applicable. RESEARCH: The expectation value of n is

1

0.n

nn nP

=

=

The normalization factor N is determined by the requirement that all the probabilities sum to unity: 1

01 .n

nP

=

=

SIMPLIFY: 1

n 0 1 10

(0) (1) .n

n nP P P P=

= = + = The normalization factor N is determined from

1

0 10

1 .nn

P P P=

= = +

From the probability distribution:

( )0 exp 0P N N= = and 1B

exp .EP Nk T

= −

Therefore,

B B

B

11 exp 1 exp .1 exp

E EN N N NEk T k T

k T

= + − = + − =

−+

The occupancy of the state is,

B1

B B

exp1 .

1 exp exp 1

Ek T

Pn E Ek T k T

− = = = −+ +

CALCULATE: Not applicable. ROUND: Not applicable. DOUBLE-CHECK: The expectation value of n calculated above for fermions is an expected result as it is the Fermi-Dirac distribution, which describes the distribution of identical (and therefore indistinguishable) fermions in an energy state E at thermal equilibrium.

36.55. THINK: The system is made up of N particles. The average energy per particle is given by B/

,iE k

iTE e

EZ

−

=

where Z is the partition function, i B/ ,E k T

ii

Z g e−=


1283

and ig is the degeneracy of the state with energy .iE This system is a 2-state system with 1 0E = and

2E E= and 1 2 1.g g= = Calculate the heat capacity of the system, ( )/ ,C N d E dT= and approximate its

behavior at very high and very low temperatures (i.e. B 1k T and B 1k T ). SKETCH: Not applicable. RESEARCH: Not applicable as the necessary equations were all given in the problem. SIMPLIFY: The average energy per particle is for 1 0E = and 2E E= is:

( ) ( ) ( ) ( )

( ) ( ) ( ) ( )BB B

BB B

00 exp exp exp

0 1 exp1 exp 1 exp

E EE Ek Tk T k TE EE

k Tk T k T

+ −− − = = −++− −

Therefore,

B B

.1 exp 1 exp

E NEE N EE E

k T k T

= =

+ +

The heat capacity of the system is,

( ) BB

B

2

B

2

exp.

exp 1

Ed N k TEd EEC N Nk

k TdT dT Ek T

= = =

+

For Bk T >> 1, B

exp 1:E

k T

≈

B

B

2

4ENkC

k T

≈

For 0 < Bk T << 1, ( )Bexp 1:/E k T >>

BB B

B B

2

B

2

B

2

expexp

exp

Ek TE E E

C Nk C Nkk T k T k TE

k T

−≈ ≈

For each temperature extreme, the heat capacity approaches zero. CALCULATE: Not applicable. ROUND: Not applicable. DOUBLE-CHECK: In general, in the extremely low temperature limit, the heat capacity must approach zero to be consistent with the third law of thermodynamics.

Additional Problems

36.56. The work function of tungsten is 4.55 eV.φ = For a photon of wavelength 360 nm,λ = its energy is

( )( )λ

−

−

⋅ ⋅= = = = ≈

⋅

15 8

ph 7

4.13567 10 eV s 2.998 10 m/s3.44409 eV 3.4 eV

3.6 10 mhcE hf

These photons are not energetic enough to overcome the work function of tungsten, and so no electrons are ejected from the tungsten cathodes. No stopping potential is required ( 0 0v = ).


1284

36.57. The de Broglie wavelength is / .h pλ = The proton and the electron have the same kinetic energy. In

terms of kinetic energy, momentum p can be written as: 2

2 .2pK p mKm

= = Then, the de Broglie

wavelength becomes: .2

h hp mk

λ = = The ratio of the de Broglie wavelengths of a proton and an electron

of the same kinetic energy K is:

λλ

−

−

⋅= = = =⋅

31pp e

27e p

e

2 9.109 10 kg 0.02333.1.673 10 kg

2

hm k mh mm k

36.58. In one Einstein of light there are 236.02 10N = ⋅ photons. If these photons have a wavelength of 400 nm,λ = the energy contained in one Einstein of photons is

( )( )( )23 34 8

tot ph5

7

6.02 10 6.626 10 J s 2.998 10 m/s2.99 10 J.

4.00 10 mhcE NE Nλ

−

−

⋅ ⋅ ⋅= = = = ⋅

⋅

36.59. The de Broglie wavelength is / .h pλ = The momentum of the baseball is:

( )( )( )( )0.100 kg 100. mi/h 1609 m/mi 1 h/3600 s 4.469 kg m/s.p mv= = = The de Broglie wavelength of the baseball is:

( ) ( )34 34/ 6.626 10 J s / 4.469 kg m/s 1.48 10 m.h pλ − −= = ⋅ = ⋅

The momentum of the spacecraft is: ( )( )( )( ) 6250. kg 125000 km/h 1000 m/km 1 h/3600 s 8.681 10 kg m/s.p mv= = = ⋅

The de Broglie wavelength of the spacecraft is:

( ) ( )34 6 41/ 6.626 10 J s / 8.681 10 kg m/s 7.63 10 m.h pλ − −= = ⋅ ⋅ = ⋅

36.60. The Heisenberg uncertainty relation can be used to find the uncertainty in the velocity. In one dimension it is stated as: / 2.x pΔ Δ ≥ Writing pΔ as ,p m vΔ = Δ the uncertainty relationship becomes

/ 2.xm vΔ Δ ≥ The uncertainty in the velocity is therefore ( )Δ ≥ ⋅ Δ ⋅ / 2 .v x m The minimum uncertainty in the velocity corresponds to the maximum uncertainty in the position. In this case, if the particle of mass

121.0 ng 1.0 pkg 1.0 10 kgm −= = = ⋅ is restricted to be somewhere on the pinhead, the maximum uncertainty in its position is the width of the pinhead, max 1.0 mm 0.0010 m.xΔ = = The minimum uncertainty in the velocity of the particle is

( )( )−

− −⋅Δ = = = ⋅ ≈ ⋅Δ ⋅ 34

20 20min -12

max

1.055 10 J s 5.273 10 m/s 5.3 10 m/s.2 2 0.0010 m 1.0 10 kg

vx m

36.61. The wavelength of light is 7700. nm 7.00 10 m.λ −= = ⋅ The energy of each photon is therefore

( )( )34 8-19

ph 7

6.626 10 J s 2.998 10 m/s2.8378 10 J.

7.00 10 mhcEλ

−

−

⋅ ⋅= = = ⋅

⋅

The light intensity on the surface of area 210.0 cmA = is 20.300 W/cm .I = The total power incident on the surface is therefore ( )( )2 20.300 W/cm 10.0 cm 3.00 W 3.00 J/s.P IA= = = = The photon flux ,Φ or

number of photons per unit time, through the surface A is 19 1

19ph

3.00 J/s 1.06 10 s .2.8378 10 J

PE

−−Φ = = = ⋅

⋅


1285

36.62. The intensity of the Sun is measured to be about 21400 W/m .I = The peak of the wavelength spectrum emitted by the Sun is at 500 nm.λ = (a) The corresponding photon frequency is

814 14

7

2.998 10 m/s 5.996 10 Hz 6.00 10 Hz.5.00 10 m

cfλ −

⋅= = = ⋅ ≈ ⋅⋅

(b) The corresponding energy per photon is

( )( )34 14 -19 -19ph 6.626 10 J s 5.996 10 Hz 3.973 10 J 3.97 10 J.E hf −= = ⋅ ⋅ = ⋅ ≈ ⋅

(c) The number flux of photons Φ arriving at the Earth (assuming all light emitted by the Sun has the same peak wavelength) is

221 2 1

19ph

1400. W/m 3.52 10 m s .3.973 10 J

IE

− −−Φ = = = ⋅

⋅

That is, about 213.52 10⋅ photons hit one meter-squared of surface area of the Earth per second.

36.63. The plates have a potential difference of 20. kVV = between them. The magnitude of the stopping potential is therefore 0 20. kV.V = The work function of silver is 4.7 eV.φ = The largest wavelength (lowest frequency and energy) of light maxλ that can be shined on the cathode to produce a current through the anode is found from the equation, 0 .eV hf φ= − The wavelength of light is

( ) ( )( )

( )( )( ) ( )

( )( )( ) ( )

0 0min max

max 0

15 8 15 8

max 4

4.136 10 eV s 2.998 10 m/s 4.136 10 eV s 2.998 10 m/s61.853 pm 62 pm.

20. kV 4.7 eV 2.0 10 eV 4.7 eV

eV eVc hcfh h eV

e

φ φλ

λ φ

λ− −

+ += = =

+

⋅ ⋅ ⋅ ⋅= = = ≈

+ ⋅ +

36.64. The surface has an area 210.0 m .A = A force of 0.100 NF = is exerted on the surface by photons of wavelength 600. nm.λ = In general, force is the rate of change of momentum (by Newton’s Second Law). By the conservation of momentum, the momentum supplied to the plate must be the momentum of the incoming photons. The momentum of a single photon is =ph / .p E c Since the photon energy is given by

,E hf= the momentum of each photon is:

( ) ( )34 7 27ph / / 6.626 10 J s / 6.00 10 m 1.1043 10 kg m/s.p hf c h λ − − −= = = ⋅ ⋅ = ⋅

The total momentum transferred by the photons is total ph ,p np= where n is the total number of photons. The total momentum per second transferred to the surface must be equal to the force F exerted on the surface, that is total / .p s F= The number of photons required per second is

ph 25 1 25 127

ph

0.100 N 9.05551 10 s 9.06 10 s .1.1043 10 kg m/s

np n FFs s p

− −−= = = = ⋅ ≈ ⋅

⋅

About 259.06 10⋅ photons per second must strike the surface to exert a force of 0.100 N.F =

36.65. The wave function describing an electron predicts a statistical spread of 410 m/sv −Δ = in the electron’s velocity. The corresponding statistical spread in its position xΔ is found from the Heisenberg uncertainty principle, / 2.x pΔ Δ ≥ In terms of velocity and electron mass, this is Δ ⋅ Δ ≥ e / 2.x m v Solving for xΔ gives:

( )( ) ( )( )

e

34 31 4

/ 2

1.0546 10 J s / 2 9.109 10 kg 10 m/s

0.579 m

x m v

x

x

− − −

Δ ≥ Δ

Δ ≥ ⋅ ⋅ Δ ≥

The uncertainty in the electron’s position is at least 0.579 m.xΔ =


1286

36.66. Wien’s displacement law states 32.90 10 K m.Tλ −= ⋅ For a blackbody whose peak emitted wavelength is in the X-ray portion of the spectrum, that is, 13 810 m 10 m,λ− −< < the temperature of the blackbody ranges from:

( ) ( )( ) ( ) ( ) ( )

λ λ− −

− − − −

⋅ < < ⋅

⋅ < < ⋅

⋅ < < ⋅

3 3max min

3 8 3 13

5 10

2.90 10 K m / 2.90 10 K m /

2.90 10 K m / 10 m 2.90 10 K m / 10 m

2.90 10 K 2.90 10 K

T

T

T

or, approximately, 5 1010 K 10 K,T< < depending on the exact wavelength of the emitted light.

36.67. A nocturnal bird’s eye can detect monochromatic light of frequency 145.8 10 Hzf = ⋅ with a power as small as 172.333 10 W.P −= ⋅ The energy of each detected photon is

( )( )−= = ⋅ ⋅ = ⋅34 14 -19ph 6.626 10 J s 5.8 10 Hz 3.843 10 J.E hf

The number of photons, ,n detected by the bird per second is:

( ) ( )ph

17 19

/ /

/ 2.333 10 W / 3.843 10 J 61 photons/s.

n s P E

n s − −

=

= ⋅ ⋅ ≈

That is, the minimum number of photons that this bird can detect in one second is about 61 photons.

36.68. The UV light wavelength is 355 nm.λ = The work function of calcium is 2.9 eV.φ = The stopping potential is found from 0eV .hf φ= − The stopping potential in this case is

( ) ( )( )( ) ( )

( )

0

15 8 70

0

/ / /

4.13567 10 eV s 2.998 10 m/s / 3.55 10 m 2.9 eV /

3.493 eV 2.9 eV / 0.593 V 0.59 V.

V hf e hc e

V e

V e

φ λ φ− −

= − = −

= ⋅ ⋅ ⋅ − = − = ≈

36.69. The electron is accelerated from rest through a potential difference of 51.00 10 V.V −= ⋅ From energy conservation, ( )Δ = −Δ Δ = 2

e f 1/ 2 .U K e V m v The electron’s final velocity is then

( ) ( )( ) ( ) 1/21/2 9 5 31f e2 / 2 1.60 10 C 1.00 10 V / 9.109 10 1874.30 m/s.v e V m − − − = Δ = ⋅ ⋅ ⋅ =

From the de Broglie wavelength formula, the wavelength of the electron is ( )

( ) ( )( )34 31

7

/ /

6.626 10 J s / 9.109 10 kg 1874.30 m/s

3.88098 10 m 388 nm.

h p h mvλ

λ

λ

− −

−

= =

= ⋅ ⋅ = ⋅ ≈

36.70. THINK: Compton used photons of wavelength 0.0711 nm.λ = The formula for Compton scattering can

be used to find (a) the wavelength eλ ′ of the photons scattered at o180θ = from an electron, (b) the

energy of these photons, and (c) the wavelength pλ ′ of the photons scattered at o180θ = from a proton.


1287

SKETCH:

RESEARCH: For an electron, the formula for Compton scattering is

( )ee

1 cos .hm c

λ λ θ′ = + −

The energy of a photon is / .E hc λ= If the target were a proton and not an electron, the electron mass em in the Compton scattering formula would need to be replaced with the mass of a proton, p .m

SIMPLIFY:

(b) hcEλ

′ =′

(c) ( )p

1 coshm c

λ λ θ′ = + −

CALCULATE:

(a) ( )( ) ( )( )

( )( )34 o

e 31 8

6.626 10 J s 1 cos 1800.0711 nm

9.109 10 kg 2.998 10 m/s0.0711 nm 0.0049 nm0.07600 nm

λ−

−

⋅ −′ = +

⋅ ⋅

= +=

(b) ( )( )

( )34 8

159

6.626 10 J s 2.998 10 m/s2.6138 10 J

0.07600 10 mE

−−

−

⋅ ⋅′ = = ⋅

⋅

(c) For a proton target,

( ) ( ) ( )( )( )( )

34

p 27 8

6

6.626 10 J s 1 cos 1800.0711 nm

1.673 10 kg 2.998 10 m/s

0.0711 nm 2.6 10 nm0.0711026 nm.

λ−

−

−

⋅ − °′ = +

⋅ ⋅

= + ⋅=

ROUND:

(a) To four decimal places, e 0.0760 nm.λ ′ =

(b) To three significant figures, 152.61 10 J.E −′ = ⋅

(c) To four decimal places, p 0.0711 nm.λ ′ = Therefore, the wavelength of the photon will be smaller if the target electron is replaced by a proton. DOUBLE-CHECK: Since some of the initial photon’s energy is imparted on the electron upon scattering, it is expected that the wavelength of the photon will increase. Since a proton is about 1000 times more


1288

massive than an electron, it is expected that the wavelength of the photon will change very little (in this case, by a negligible amount).

36.71. THINK: To estimate the number of photons that impact the Earth, it is useful to know that the intensity of the Sun’s radiation on the Earth is 21370 W/m .I = Use the peak wavelength of the light emitted by the Sun, 500 nm,λ = as stated in section 36.2. Note that the Earth’s upper atmosphere, the ionosphere, is roughly 300 kmd = above the Earth’s surface. The radius of the Earth is 6378 km.R = Finally, keep in mind that only half of the Earth’s surface can face the Sun at any given time. Note that one year has approximately ( )( )( )= =1 year 365.24 days/yr 24 hr/day 3600 s/hr 31556736 s.t SKETCH:

RESEARCH: The energy of a photon is ph / .E hf hc λ= = The photon flux rate Φ (the number of

photons per unit area per unit time) is found from ph/ .I EΦ = The number of photons N that strike the

Earth’s upper atmosphere per year is atmyear .

2A

N t= Φ ⋅ ⋅ The surface area of a sphere is 24 .A rπ=

SIMPLIFY: ph

I IE hc

λΦ = = , ( )( ) ( )22 year

atm year year

21 1 42 2

I R d tIN A t R d thc hc

π λλ π+ = Φ ⋅ = + =

CALCULATE: ( )( )( ) ( )( )

( )( )

22 7 643

34 8

2 1370 W/m 5 10 m 6.678 10 m 1 yr 31556736 s/yr3.049 10

6.626 10 J s 2.998 10 m/sN

π −

−

⋅ ⋅= = ⋅

⋅ ⋅

ROUND: Accuracy is limited by the peak wavelength value, 500 nm. To one significant figure, the number of photons received by Earth’s upper atmosphere in one year is 433 10 .N = ⋅ DOUBLE-CHECK: This is a huge number, but is expected for the number of photons from the Sun to hit the Earth in one full year. Dimensional analysis confirms that the calculation yields a dimensionless result.

( ) ( )2

2W m m s J/s m s J/s m sm 1

J s m/s J s m/s J mN = = = =

36.72. THINK: An X-ray has an initial wavelength of 8.5 nm.λ = Its wavelength is increased by 1.5 pmλΔ = in a collision with an electron. Some of the energy of the photon will be imparted to the electron, giving it a velocity. To solve this problem, the conservation of energy is used. It is assumed that the electron is initially at rest.


1289

SKETCH:

RESEARCH: Since energy is conserved in this scattering event, the kinetic energy that the electron receives is simply equal to the energy loss endured by the photon:

′= −e .K E E Before the collision, the energy of the photon is / .E hc λ= After the collision, the energy of the X-ray is

.hc hcEλ λ λ

′ = =′ + Δ

SIMPLIFY: The kinetic energy of the electron after the collision is: ( )

( )

λ λ λ λλ λ λ λ λ λ λ λ λ

λλ λ λ

+ Δ − Δ= − = = = + Δ + Δ + Δ Δ=+ Δ

2e 2 2

2e

12

2

ehc hc hcK hc m v

hcvm


( ) ( ) ( )( )

34 8 12

231 9 9 12

2 6.626 10 J s 2.998 10 m/s 1.5 10 m95150 m/s

9.109 10 kg 8.5 10 m 8.5 10 m 1.5 10 mv

− −

− − − −

⋅ ⋅ ⋅= =

⋅ ⋅ + ⋅ ⋅

ROUND: Rounding the result to two significant figures gives 95 km/s.v = DOUBLE-CHECK: Momentum must also be conserved: e .p p p′= + The initial momentum of the photon is / .p h λ= Since the x-direction is chosen to be the initial direction of the photon,

xp p= and 0.yp = The final direction of the photon is given by the Compton scattering formula,

λλ λ θ θ − Δ ′ = + − = − =

1 oe

e1 cos cos 1 67.56 .

m chm c h

The components of the final momentum of the photon are

cos cosxh hp θ θλ λ λ

′ = =′ + Δ

and sin sin .yh hp θ θλ λ λ

′ = =′ + Δ

The difference between the final and initial momentum of the photon must be equal to the final momentum of the electron.

( )( )

( ) ( )

θλ λ λ

θλ λ

− −−

− −

−−

−

⋅ ⋅′= − = − = − = ⋅+ Δ ⋅ ⋅

⋅′= − = − = − = − ⋅+ Δ ⋅

= +

34 34o 26

e, 9 9

34o 26

e, 9

2e e, e,

6.626 10 J s 6.626 10 J scos cos 67.56 4.82 10 kg m/s8.5 10 m 8.5015 10 m

6.626 10 J s0 sin sin 67.56 7.20 10 kg m/s8.5015 10 m

x x x

y y y

x y

h hp p p

hp p p

p p p ( ) ( )− − −= ⋅ + − ⋅ = ⋅2 22 26 26 264.82 10 J s/m 7.20 10 J s/m 8.7 10 kg m/s


1290

The momentum of the electron from the original calculation is

( )( )− −= = ⋅ = ⋅31 26e e 9.109 10 kg 95150 m/s 8.7 10 kg m/s.p m v

Since the calculated momentum using two methods is the same, the speed of the electron found is correct.

36.73. THINK: The definition of the de Broglie wavelength is needed to compute the momentum of the proton. The wavelength 3.5 fmλ = is given in the question. SKETCH: Not required. RESEARCH: The de Broglie wavelength of the proton is given by / .h pλ = The energy of a relativistic

particle is ( ) ( )22 2 .E pc mc+= Heisenberg’s energy-time uncertainty principle states that / 2.E tΔ Δ ≥

The mass of a proton is 27p 1.673 10 kg.m −= ⋅

SIMPLIFY: The momentum of the proton is / .p h λ= The energy of the proton is

( )2

22p .hcE m c

λ =

+

The interval of time for which a low-energy proton could have the same energy as the accelerated proton is

( )2p

.2 2

t tE E m c

Δ ≥ Δ ≥Δ −

CALCULATE: Substituting the numerical values yields: ( )

( )34

1915

6.626 10 J s1.893 10 kg m/s,

3.5 10 mp

−−

−

⋅= = ⋅

⋅

( )( )( ) ( )( )( )

34 827 8 10

15

2226.626 10 J s 2.998 10 m/s

1.673 10 kg 2.998 10 m/s 1.6072 10 J,3.5 10 m

E−

− −−

⋅ ⋅ = ⋅ ⋅ = ⋅ ⋅

+ and

( )( ) ( )( )( )

3424

10 27 28

1.055 10 J s5.0963 10 s.

2 1.6072 10 J 1.673 10 kg 2.998 10 m/st

−−

− −

⋅Δ ≥ = ⋅

⋅ − ⋅ ⋅

ROUND: Rounding the results to two significant figures gives 191.9 10 kg m / s,p −= ⋅ 101.6 10 JE −= ⋅ and 245.1 10 s.t −≥ ⋅

DOUBLE-CHECK: The values for the momentum and energy are small, but reasonable for a proton. It is expected that the time a low-energy proton could have the same energy as the accelerated proton be very small.

36.74. THINK: The energy of the backscatter peak corresponds to the energy of the gamma-ray after Compton scattering at an angle of o180 .θ = The Compton edge energy is the energy cut off or the maximum energy that can be transferred to an electron. The Compton scattering formula can be used to determine the energies of the Compton edge and the back scatter peak for a gamma-ray photon of energy 511 keV.E = The mass of an electron is = 2

e 511 keV/ .m c


1291

SKETCH:

RESEARCH: The Compton scattering formula is given by

( )1 cos.

hmc

θλ λ

−′ = +

Using ,hcE

λ = it becomes ( )1 coshhc hc

E E mcθ−

= +′

or ( )

2

1 cos1 1 .E E mc

θ−= +

′

SIMPLIFY: For the backscatter peak energy, substituting o180θ = yields

= + =+

2e

bs2 2bs e e

1 1 2 .2

Em cE

E E m c E m c

The maximum energy transferred to an electron occurs when the scattered photon energy is a minimum, which occurs when o180 ,θ = or when bs .E E′ =

′= − = − = − =+ +

2 2e

edge bs edge2 2e e

2 .2 2

Em c EE E E E E E EE m c E m c

CALCULATE: ( )( )( ) ( )

2 2

bs 2 2

511 keV 511 keV/170.333 keV

2 511 keV 511 keV/

c cE

c c= =

+

( )( ) ( )

2

edge 2 2

2 511 keV340.667 keV

2 511 keV 511 keV/E

c c= =

+

ROUND: Rounding to three significant figures, =bs 170. keVE and edge 341 keV.E =

DOUBLE-CHECK: Since the energy of the incident gamma-ray is the same as the rest energy of an electron, it is reasonable that the energy of the Compton edge is exactly twice the energy of the backscatter peak.


1292

Chapter 37: Quantum Mechanics

In-Class Exercises

37.1. e 37.2. b 37.3. d 37.4. b Multiple Choice

37.1. c 37.2. b 37.3. d 37.4. b 37.5. b 37.6. e 37.7. d 37.8. a, e 37.9. a 37.10. a

Questions

37.11. The answer can be true or false depending on the system. Let us consider the case of a particle in an

infinite, potential well. The wave function for this potential is given by ( ) 2 sinnn xx

a aπψ =

where a is

the width of the infinite potential well. The kinetic energy is given by 2 2 2

2 .2n

nEmaπ= It can be seen that if

the amplitude of ,nψ 2 / ,a is larger, a must be smaller. As a consequence, the kinetic energy is larger as long as n is the same. Therefore, the statement is true. However, if n is not the same, the kinetic energy cannot be compared from the amplitude of wave functions.

37.12. Since the wave functions of a particle in an infinite potential well have symmetric property for odd n and an antisymmetric property for an even n, where n is the quantum number. Therefore, the probability of the particle is symmetric about / 2.c L= This means that the probability of finding the particle in the interval between 0 and / 2L is the same as for the interval between / 2L and .L This does not depend on the energy of the particle. Therefore, the probability of finding the particle between 0 and / 2L stays the same regardless the value of the energy of the particle.

37.13. The wave function for a particle in an infinite square wall is given by n2 sin ,n xL L

πψ =

where L is the

width of the potential well. The probability of finding the particle is ( ) π ∏ =

22 sin .n xxL L

As n

increases, the probability density fluctuates around an average probability density given by

( ) π ∏ =

22 sin .n xxL L

Since 2 1sin ,2

n xLπ =

it becomes ( )∏ = 1 .x

L This is exactly the classical

probability distribution. Therefore, it does obey the correspondence principle.

37.14. It is known that the wave functions for a particle in a one dimensional harmonic oscillator are symmetric for even-n states. It can be shown that the first derivative of the wave functions with respect to the spatial variable is antisymmetric for even-n states. Since the expectation value of the momentum is defined as ( ) ( ) ( ) ( )* *

n nn ndP x P x dx i x x dx

dxψ ψ ψ ψ

∞ ∞

−∞ −∞= = − and ( )*

nxψ is symmetric and nd

dxψ

is

antisymmetric. Therefore, the above integral is zero, and thus 0.P =

37.15. The expectation value of the position is defined as 2 .x xdxψ= Since x is an odd function

(antisymmetric), the integral becomes zero where 2ψ is symmetric (or even function). The probability


1293

( ) ψ∏ = 2x is symmetric when the wave function is a symmetric function or an antisymmetric function.

For an antisymmetric wave function, the probability at 0x = is zero. Therefore, 0x = and ( )∏ =0 0 for an antisymmetric or odd wave function. As an example, the first excited state of a particle in a harmonic oscillator.

37.16. The two lowest energies for an electron in an infinite potential wall are

( )( )( )

234 22 222

1 2 231 9

1.0546 10 J s1.506 10 J 0.00094 eV

2 2 9.109 10 kg 20 10 mE

ma

ππ−

−

− −

⋅= = = ⋅ ≈

⋅ ⋅

and 22 12 0.0038 eV.E E= =

Since 0V is much larger than 1E or 2 ,E the two lowest energies for the particle in a finite well is approximately the same as 1E and 2E for the infinite well. However, since the electron in the finite well can penetrate into the classically forbidden region, the effective wavelength for the finite well is larger than the wavelength for the infinite well. Since energy is proportional to κ 2 or 21/ ,λ the energy of the particle in the finite well is lower than the energy of the particle in the infinite well.

37.17.

The Coulomb potential energy of the central nucleus due to only two adjacent nuclei is

( ) 2 2

0

1 1 1 .4

U x z ea x a xπε = + − +

For small oscillation about an equilibrium point ( )0 ,x = the potential

energy can be approximated by a simply harmonic oscillator potential. Expanding the potential energy in


1294

Taylor series about 0x = and keeping only up to the term 2x yield

( ) ( ) ( ) ( ) ( ) 20 ' 0 1/ 2 " 0 ,U x U U x U x= + + where

( )2 2

0

10 ,2

z eUaπε

=

( )' 0 0U = and

( )( ) ( )

2 22 2

3 3 30 00

1 1 1" 0 .4

x

z eU z eaa x a xπε πε

=

= + = − +

Thus ( ) ( ) ( ) ( ) ( )2 20 1/ 2 " 0 1/ 2U x U U x kx− = = with 2 2

30

.z ekaπε

= The energy ( )0U is just a shift in energy

and it can be neglected. The energy states of this harmonic oscillator is

( ) ( )( ) ( )( )

22 1934

327 12 2 15

14

6 1.6 10 C1 1.054 10 J s2 12 1.66 10 kg 8.85 10 C / (N m) 20 10 m

14.8 10 J .2

nE n

n

π

−−

− − −

−

⋅ = + ⋅ ⋅ ⋅ ⋅

= ⋅ +

The maximum energy allowed is given by ( )23 4 19max B

3 3 1.38 10 J/K 10 K 2 10 J.2 2

E k T − −= = ⋅ ⋅ = ⋅ Therefore,

the central nucleus is in its ground state 0.n =

37.18. The wave functions for a finite square well is in the form of ( ) ( )

( ) ( )0

0

exp exp if sin cos if

A x B x E UC kx D kx E U

α αψ

− + + <= + >

where ( )0

2

2m E Uk

−=

and

( )02

2.

m U Eα

−=

If 0 ,U U= both solutions are equal, a constant.ψ =

This corresponds to a wavelength .λ = ∞ This is impossible.

37.19. Since the potential ( )U x = ∞ for 0,x < the condition for the solution is ( )0 0.xψ = = For 0,x < the

solution is known to be ( ) 0.xψ = For 0,x > the wave function must satisfy a harmonic oscillation potential. Therefore, the solution of the potential should be the wave functions of the harmonic oscillator with the requirement ( )0 0.xψ = = This requirement is satisfied by all wave functions with odd .n Thus

the energies of the states are ( )( ) 01/ 2nE n w= + where n is an odd number.

37.20. The probability density is given by ( ) ( )* .P x xψ ψ= The new probability is

( ) ( ) ( ) ( ) ( ) ( )0 0new new new* * * .i iP x x x e e x x x Pψ ψ ψ ψ ψ ψ− += = = = The probability is the same. An

additional phase does not change the probability.

37.21. The ground state is approximated by approximating the potential of a harmonic oscillator potential about the equilibrium position. The equilibrium position of the potential in Taylor series up to 2x yields

( ) ( ) ( ) ( ) ( ) 20 ' 0 1/ 2 " 0 ,U x U U x U x≈ + + where ( )0 0,U =

( )' 0 0U = and ( ) 0 02 2" 0 cos .

U UxU haa a

= =

Since ( ) 20" 0 / ,k U U a= = ω =0 /k m and the ground state energy for the harmonic oscillator is

( ) ω= osc 01/ 2 ,E the ground state energy of the particle in ( )U x is therefore,


1295

( )120

osc 0 210 .2

UE U E U

ma = + = +

37.22. The operator for energy is ( )/i t∂ ∂ and the operator for momentum is .i− ∇ Replacing the energy and

the momentum in the relativistic energy-momentum relation, 2 2 2 2E p c mc− = yields

( ) ( ) ( ) ( )

( ) ( )

( )

22 2 2 2 2 4

2

22 2 2 2 4

2 2

2 2 22

2 2 2

1

1 0.

i i c r m c rt

c r m c rc t

m c rc t

ψ ψ

ψ ψ

ψ

∂ − − ∇ = ∂

− ∂ + ∇ = ∂

∂∇ − + = ∂

This is known as the Klein-Gordon equation.

Problems

37.23. The kinetic energy of a neutron is 1210.0 MeV 1.60 10 J.−= ⋅ The size of an object that is necessary to observe diffraction effects is on the order of the de Broglie wavelength of the neutron. The (relativistic) de Broglie wavelength is given by

2 2.

2

h hcp K Kmc

λ = =+

( )( )( ) ( )( )( )

34 815

2 212 12 27 8

6.63 10 J s 3.00 10 m/s9.0454 10 m = 9.05 fm.

1.60 10 J 2 1.60 10 J 1.67 10 kg 3.00 10 m/sλ

−−

− − −

⋅ ⋅= = ⋅

⋅ + ⋅ ⋅ ⋅

Since protons and neutrons have a diameter of about 1.00 fm, they would be useful targets to demonstrate the wave nature of 10.0-MeV neutrons.

37.24. Given ( ) ( ) ( )8 3 7 2 (8 7 ) (3 2 ) ,f x i i x x x i= + + − = + + −

( ) ( ) ( )

( )

2 *

2 2 2

22 2

2 2

2 2

2

(8 7 ) (3 2 ) (8 7 ) (3 2 )

(8 7 ) (8 7 )(3 2 ) (8 7 )(3 2 ) (3 2 )

(8 7 ) (3 2 ) 1

(8 7 ) (3 2 )

64 112 49 9 12 4

53 100 73.

f x f x f x x x i x x i

x x x i x x i x i

x x

x x

x x x x

x x

= = + − − + + − = + + + − − + − − −

= + − − −

= + + −

= + + + − +

= + +

37.25. The energies of an electron in a box are given by 2 2

22 .

2nE nma

π= The two lowest energies are:

( )( )( )

( )π−

−

− −

⋅= = ⋅ =

⋅ ⋅

234 22 20

1 231 9

1.055 10 J s1 1.5 10 J 0.094 eV

2 9.11 10 kg 2.0 10 mE and

( )2 202 12 6.0 10 J 0.38 eV.E E −= = ⋅ =


1296

37.26. The energies of a proton in a box are given by 2 2

22 .

2nE nma

π= The three lowest energies are

( )( )( )

π−−

− −

⋅= = ⋅ =

⋅ ⋅

234 221

1 227 10

1.055 10 J s3.3 10 J 0.020 eV,

2 1.67 10 kg 1.0 10 mE ( )2 20

2 12 1.3 10 J 0.082 eVE E −= = ⋅ = and

( )2 203 13 3.0 10 J 0.18 eV.E E −= = ⋅ =

37.27. The energies for a particle in an infinite square well are 2 2

222nE n

mLπ= for a square well of length L and

( )2 2

222 2

nE nm L

π= for a square well of length 2 .L Therefore, ( )( )

( ) ( )( ) ( )

2 2 2 2 22 1

2 2 2 2 22 1

2 1 / 24.

2 2 1 / 8

mLE E LE E L mL

π

π

−−= =

− −

37.28. THINK: The second excited state is the state with 3.n = SKETCH:

RESEARCH: The energy state of an electron in a one-dimensional infinite well is given by 2 2

22 .

2nE nma

π=

The wavelength of light emitted by the transition from the second excited state to the ground state is found

by .hcEλ

Δ =

SIMPLIFY: (a) The energy difference between the second excited state and the ground state is:

( )2 2 2 2

2 23 1 2 23 1 8 .

2E E E

ma maπ πΔ = − = − =

(b) The wavelength is .hcE

λ =Δ

CALCULATE:

(a) ( )

( )( )

234 219

231 9

1.055 10 J s8 4.823 10 J 3.0 eV

2 9.11 10 kg 1.0 10 mE

π−−

− −

⋅Δ = = ⋅ =

⋅ ⋅


1297

(b) ( )( )

( )34 8

719

6.63 10 J s 3.00 10 m/s4.12 10 m

4.823 10 Jλ

−−

−

⋅ ⋅= = ⋅

⋅

ROUND: Round to two significant figures. (a) The energy difference between the second excited state and the ground state is 3.0 eV.EΔ = (b) The wavelength of light emitted is 410 nm.λ = DOUBLE-CHECK: The wavelength of falls in the visible part of the electromagnetic spectrum. The energy and wavelength are reasonable results.

37.29. THINK: In order to get the solution of the Schrödinger equation for a given potential, the continuity conditions need to be satisfied. SKETCH:

RESEARCH: In regions I and III the potential energy is infinite, so the wave function is ( ) 0xψ = for these regions. In region II, the potential energy is zero. Therefore, the potential energy is given by:

for / 2( ) 0 for / 2 / 2

for / 2

x aU x a x a

x a

∞ < −= − ≤ ≤∞ >

The wave function must satisfy the Schrödinger equation,

( ) ( )22

2 .2

d xE x

m dxψ

ψ− =

The solution of this equation has the form of ( ) ( ) ( )sin cos ,x A x B xψ κ κ= + where 22 /mEκ = for an infinite square well. Since the wave function must be continuous at the boundaries, the wave function must satisfy ( ) ( )/ 2 / 2 0.a aψ ψ= − = SIMPLIFY: Continuity at the boundaries gives:

( ) ( ) ( )/ 2 sin / 2 cos / 2 0a A a B aψ κ κ= + = (1)

( ) ( )( ) ( )( ) ( ) ( )/ 2 sin / 2 cos / 2 0 sin / 2 cos / 2 0a A a B a A a B aψ κ κ κ κ− = − + − = − + = (2)

Subtracting (1) with (2) yields ( )2 sin / 2 0.A aκ = This implies that / 2 / 2 / ,a n n aκ π κ π= = with n

even. Adding equations (1) and (2) yields ( )2 cos / 2 0.B aκ = This implies that / 2 / 2 / ,a n n aκ π κ π= = with n odd. Therefore, there are two sets of solutions:

( )sin , with even

cos , with odd

n xA nax

n xB na

π

ψπ

=


1298

The normalization condition can be used to determine the constants A and B. The result is the same as that shown in the text: 2 / .A B a= = Therefore, the solution to the Schrödinger equation for this potential is:

( )

0 for / 2 and / 2

2 sin for / 2 / 2 with even

2 cos for / 2 / 2 with odd

x a x a

n xx a x a na a

n x a x a na a

< − > = − ≤ ≤

− ≤ ≤

πψ

π

CALCULATE: Not required. ROUND: Not required. DOUBLE-CHECK: The above solutions can be found from the solutions for the infinite square well with interval ( )0,a by replacing the variable x with / 2.x a+ Doing so yields:

( ) 2/ 2 sin .2

n ax a xa a

πψ + = +

Using the trigonometric identity ( )sin sin cos cos sin ,α β α β α β+ = + the above equation becomes:

( ) 2/ 2 sin cos cos sin .2 2

n x n n x nx aa a a

π π π πψ + = +

Here if n is odd then ( ) ( )cos / 2 0, sin / 2 1n nπ π= = ± and if n is even then ( ) ( )sin / 2 0, cos / 2 1.n nπ π= = ± Therefore, the wave function is given by:

( ) 0 for / 2 or / 2

2 sin for / 2 / 2 with even

2 cos for / 2 / 2 with odd

x x a x a

n x a x a na a

n x a x a na a

ψ

π

π

= < − >

± − ≤ ≤ ± − ≤ ≤

This matches the solution above to within a minus sign, which is physically insignificant.

37.30. THINK: The three dimensional Schrödinger equation can be used and separation of variables can be assumed in order to solve the problem. SKETCH:

RESEARCH: Separation of variables allows us to write the potential as 1 2 3( , , ) ( ) ( ) ( )U x y z U x U y U z= ⋅ ⋅ with:

1 2 3

for 0 for 0 for 0( ) 0 for 0 , ( ) 0 for 0 , ( ) 0 for 0

for for for x y z

x zy

x y zU x x L U y y L U z z L

x L z Ly L

∞ < ∞ < ∞ < = ≤ ≤ = ≤ ≤ = ≤ ≤

∞ > ∞ >∞ >


1299

The three dimensional Schrödinger equation is given by

( ) ( ) ( ) ( )2 2 2 2

2 2 2 , , , , , , , , .2

x y z U x y z x y z E x y zm x y z

ψ ψ ψ − ∂ ∂ ∂+ + + = ∂ ∂ ∂

SIMPLIFY: The wave function is also a product of three separable functions, 1 2 3( , , ) ( ) ( ) ( )x y z x y zψ ψ ψ ψ= ⋅ ⋅ with:

( )1

0 for 0

2 sin with 1,2,3... for 0

0 for

xx x

x x

x

xn x

x n x LL L

x L

πψ

<

= = ≤ ≤

>

( )2

0 for 0


0 for

yy y

y y

y

yn y

y n y LL L

y L

πψ

<

= = ≤ ≤ >

( )3

0 for 0


0 for

zz z

z z

z

zn z

z n z LL L

z L

πψ

<

= = ≤ ≤

>

(a) Therefore, the solution of the wave function of an electron in a potential rectangle is:

( ) 1 2 38, , ( ) ( ) ( ) sin sin sin .yx z

x y z x y z

n yn x n zx y z x y z

L L L L L Lππ π

ψ ψ ψ ψ

= ⋅ ⋅ =

In the same fashion that the allowed energies were derived in the text for the one-dimensional infinite potential well, the allowed energies are:

π π π= + + 2 2 2 2 2 22 2 2

, , 2 2 2 .2 2 2x y zn n n x y z

x y zE n n n

mL mL mL

(b) The lowest energy for a potential cube with side L occurs when 1,x y zn n n= = = and is given by: 2 2

1,1,1 23 .2

EmL

π=

CALCULATE: (b) For a potential cube with side 101.00 10 m,−⋅ the lowest allowed energy for the electron is:

( )( )( )

234 217

1,1,1 231 10

3 1.055 10 J s1.81 10 J 112.9 eV.

2 9.11 10 kg 1.00 10 mE

π−−

− −

⋅= = ⋅ =

⋅ ⋅

ROUND: (b) To 3 significant figures, the lowest energy is 1,1,1 113 eV.E = DOUBLE-CHECK: This is a reasonable amount of energy for an electron to have in such a small volume.

37.31. The potential energy for the well is given by:

1

for 0( ) 0 for 0

for

xU x x a

U x a

∞ <= ≤ ≤ >

This is illustrated in the diagram:


1300

Since the question states that the electron is confined to the potential well, 1.E U< As shown in the text, the wave function for this finite potential well can be written as:

( ) ( )0 for 0

sin for 0

for x

xx A x x a

Be x aγ

ψ κ−

<= ≤ ≤ >

Where 22mEκ =

and ( )1

2

2.

m U Eγ

−=

The wave function ( )xψ must satisfy the boundary conditions

at :x a= ( ) ( )( ) ( )1 sin

2 cos .

a

a

A a Be

A a Be

γ

γ

κκ κ γ

−

−

=

= −

Dividing (1) and (2) yields ( )tan 1aκκ γ

= − or ( )tan .a κκγ

= − Since κ and γ are positive, ( )tan aκ must

be negative. This is satisfied when

( )2 1 , 1,2,3...2

n a n nπ κ π− < < =

For the third state ( )3 :n = 2 2 2

2 2 2 22

5 25 25 23 9 9 .2 4 4

mEaa aπ π πκ π κ π π< < < < < <

Therefore, 2 2 2 2

3 1 3 12 225 259 9 ,4 42 2

E E E Ema ma

π π< < < <

where 1E is the ground state energy for the infinite square well:

( )( )( )

π−−

− −

⋅= = ⋅ =

⋅ ⋅

234 220

1 231 9

1.055 10 J s6.03 10 J 0.376 eV.

2 9.11 10 kg 1.0 10 mE

Therefore,

( ) ( )325 0.376 eV 9 0.376 eV4

E< < 32.4 eV 3.4 eV.E< <

Since 1 32.0 eV ,U E= < the third state is not a bound state.

37.32. The tunneling probability or transmission coefficient is given by:

( )12 ( )2

2where .b a m U E

T e γ γ− − −= =


1301

( )( )( )( )

27 1314 1

234

2 1.67 10 kg 29.8 MeV 18.0 MeV 1.602 10 J/MeV7.53 10 m

1.055 10 J sγ

− −−

−

⋅ − ⋅= = ⋅

⋅

The tunneling probability is ( )( )− −− ⋅ ⋅= =

14 1 152 7.53 10 m 1.00 10 m0.222.T e Therefore, there is a 22.2% chance that

the proton will tunnel through the barrier.

37.33. THINK: The equation for the transmission coefficient can be used to calculate the tunneling probability. The factor that the neutron’s probability of tunneling through the barrier increases by can be found by taking a ratio of the tunneling probabilities. The potential barrier is 8.4 fmb a− = wide and

1 36.2 MeVU = high. Originally, the neutron has a kinetic energy of 1 22.4 MeVE = and this is increased to 2 11.15 .E E= SKETCH:

RESEARCH: The tunneling probability for a square barrier is given by

( )12 ( )2

2where .b a m U E

T e γ γ− − −= =

SIMPLIFY: The ratio of the two tunneling probabilities for the two energies 2E and 1E is

( )22 1

1

2 ( )2 ( )2

2 ( )1

.b a

b ab a

T e eT e

γγ γ

γ

− −− − −

− −= =

CALCULATE: Since 2 11.15 :E E=

( ) ( )( )( )( )( )

( )( )( )( )

27 13

2 1 234

27 13

234

14 1

2 1.67 10 kg 36.2 MeV 1.15 22.4 MeV 1.6 10 J/MeV

1.055 10 J s

2 1.67 10 kg 36.2 MeV 22.4 MeV 1.6 10 J/MeV

1.055 10 J s

1.060 10 m .

γ γ− −

−

− −

−

−

⋅ − ⋅− =

⋅ ⋅

⋅ − ⋅−

⋅ ⋅

= − ⋅

Therefore, ratio is ( )( )14 1 152 1.060 10 m 8.4 10 m2

15.935.

Te

T

− −− − ⋅ ⋅= =

ROUND: To two significant figures, the neutron’s probability of tunneling through the barrier increases by 5.9 times. DOUBLE-CHECK: Due to the exponential equation it is reasonable that a small increase in energy leads to a large increase in the probability of tunneling.


1302

37.34. THINK: The rate of tunneling TI is proportional to the tunneling probability and the rate of incidence 1000. electrons/s,II = and the rate of reflection RI is the rate of incidence minus the rate of tunneling.

The width and height of the potential barrier are 1.00 nmb a− = and 1 2.51 eV,U = respectively. Each electron has kinetic energy = 2.50 eV.E SKETCH:

RESEARCH: The tunneling probability is given by:

( )12 ( )2

2where .b a m U E

T e γ γ− − −= =

The reflection probability is given by 2 ( )1 1 .b aR T e γ− −= − = − The wavelength of the electron is calculated

using 2 2

.2

h hcp K Kmc

λ = =+

SIMPLIFY: The rate of tunneling is given by T II I T= and the rate of reflection is = = −I I T .RI I R I I


( )γ

− −−

−

⋅ − ⋅= = ⋅

⋅

31 198 1

234

2 9.11 10 kg 2.51 eV 2.50 eV 1.602 10 J/eV5.121 10 m ,

1.055 10 J s

( ) ( )( )8 1 92 5.126 10 m 1.00 10 mT 1000. electrons/s 358.7 electrons/s,I e

− −− ⋅ ⋅= = and

( )R 1000. 358.7 electrons/s 641.3 electrons/s.I = − = The wavelength of an electron is

( )( )( )( )( ) ( )( )( )( )

34 8

2 219 19 31 8

10

6.63 10 J s 3.00 10 m/s

2.50 eV 1.602 10 J/eV 2 2.50 eV 1.602 10 J/eV 9.11 10 kg 3.00 10 m/s

7.761 10 m.

λ−

− − −

−

⋅ ⋅=

⋅ + ⋅ ⋅ ⋅

= ⋅

The wavelength of an electron before and after passing the barrier is the same because ( ) 0U x = on either side of the barrier. ROUND: To three significant figures, the rate at which electrons pass through the barrier is

=T 359 electrons/s,I the rate at which electrons reflect back from the barrier is =R 641 electrons/s,I and the wavelength of the electrons before and after they pass through the barrier is 0.776 nm.λ = DOUBLE-CHECK: Since there is such a small difference in energy between the energy of the incident electrons and the potential energy of the barrier, it is reasonable that a large portion of the electrons tunnel through the barrier.


1303

37.35. THINK: Given that the tunneling probability is 0.100,T = the equation for the transmission coefficient can be used to calculate the energy of the electron. The potential barrier is 2.00 nmb a− = wide and

1 7.00 eVU = high. SKETCH:

RESEARCH: The tunneling probability of the electron is given by:

( )12 ( )2

2where .b a m U E

T e γ γ− − −= =

SIMPLIFY: Solving for E gives:

( ) ( )

( )( )

( )

( )

12

2

12

22

1

22

1

2ln( ) 2

2 ln( )2

ln( )2 2

ln( ) .2 2

m U ET b a

m U E Tb a

TU Em b a

TE Um b a

−= − −

−= − −

− = − −

= − − −

CALCULATE:

( )( ) ( )( )

( )( )

−− −

− −

⋅ = ⋅ − − = ⋅ = ⋅ ⋅

223419 18

31 9

1.055 10 J s ln 0.1007.00 eV 1.602 10 J/eV 1.119 10 J 6.987 eV

2 9.11 10 kg 2 2.00 10 mE

ROUND: To three significant figures, the energy of the electron is 6.99 eV.E = DOUBLE-CHECK: Since only 10.0% of the electrons are transmitted through the barrier, it is reasonable that the energy of the electron is only slightly less than the height of the potential barrier.

37.36. THINK: The three dimensional Schrödinger equation can be used and separation of variables can be assumed in order to solve the problem. The infinite potential box has dimensions 1.00 nm,xL =

2.00 nmyL = and 3.00 nm.zL =


1304

SKETCH:

RESEARCH: Separation of variables allows us to write the potential as 1 2 3( , , ) ( ) ( ) ( )U x y z U x U y U z= ⋅ ⋅ with:

1 2 3

for 0 for 0 for 0( ) 0 for 0 , ( ) 0 for 0 , ( ) 0 for 0

for for for x y z

x zy

x y zU x x L U y y L U z z L

x L z Ly L

∞ < ∞ < ∞ < = ≤ ≤ = ≤ ≤ = ≤ ≤

∞ > ∞ >∞ >

The three dimensional Schrödinger equation is given by

( ) ( ) ( ) ( )2 2 2 2

2 2 2 , , , , , , , , .2

x y z U x y z x y z E x y zm x y z

ψ ψ ψ − ∂ ∂ ∂+ + + = ∂ ∂ ∂

SIMPLIFY: The wave function is also a product of three separable functions, 1 2 3( , , ) ( ) ( ) ( )x y z x y zψ ψ ψ ψ= ⋅ ⋅ with:

( )1

0 for 0


0 for

xx x

x x

x

xn x

x n x LL L

x L

πψ

<

= = ≤ ≤

>

( )2

0 for 0


0 for

yy y

y y

y

yn y

y n y LL L

y L

πψ

<

= = ≤ ≤ >

( )3

0 for 0


0 for

zz z

z z

z

zn z

z n z LL L

z L

πψ

<

= = ≤ ≤

>

Therefore, the solution of the wave function of an electron in a potential rectangle is:

( ) 1 2 38, , ( ) ( ) ( ) sin sin sin .yx z

x y z x y z

n yn x n zx y z x y z

L L L L L Lππ π

ψ ψ ψ ψ

= =

In the same fashion that the allowed energies were derived in the text for the one-dimensional infinite potential well, the allowed energies are:

22 22 2

, , 2 2 2 .2x y z

yx zn n n

x y z

nn nE

m L L Lπ

= + +


1305

CALCULATE: By trial and error one finds from the term 2 22 2 2 2

2 2 2 2 2 2 ,1.00 nm 4.00 nm 9.00 nm

y yx z x z

x y z

n nn n n nL L L

+ + = + +

that the six lowest energy levels correspond to:

( ) ( ) ( ) ( ) ( ) ( ) ( ), , 1,1 1 , 1,1,2 , 1,2,1 , 1,1,3 , 1,2,2 , 1,2,3 .x y zn n n =

The energy is given by:

( )( ) ( )

( )

( )

π−

−

−

⋅ = + + ⋅ = ⋅ + +

= + +

234 2 22 2 29, , 2 2 231

22 220 2

2 2 2

22 22

2 2 2

1.055 10 J s10

1.00 m 4.00 m 9.00 m2 9.11 10 kg

6.02915 10 J m1.00 m 4.00 m 9.00 m

0.37635 eV m1.00 m 4.00 m 9.00 m

x y z

yx zn n n

yx z

yx z

nn nE

nn n

nn n

The six lowest energy states are given by

, , x y zn n n ( ), , eVx y zn n nE

(1,1,1) 0.51225(1,1,2) 0.63770(1,2,1) 0.79452(1,1,3) 0.84679(1,2,2) 0.91997(1,2,3) 1.12905

Since none of the quantum states have the same energy, none of the levels are degenerate. ROUND: The answers should be rounded to three significant figures:

, , x y zn n n ( ), , eVx y zn n nE

(1,1,1) 0.512(1,1,2) 0.638(1,2,1) 0.795(1,1,3) 0.847(1,2,2) 0.920(1,2,3) 1.13

DOUBLE-CHECK: These are reasonable energy values for an electron confined to a small infinite potential box. Any other combination of ( ), ,x y zn n n leads to a larger energy, so these are the six lowest

energy states.

37.37. THINK: The work function is given by 1 .W U E= − The equation for the transmission coefficient can be used to find the work function of the probe given that the width of the barrier is 0.100 nmb a− = and the tunneling probability is 0.100% or 0.00100.T = Use the conversion factor: 181.000 J = 6.242 10 eV.⋅


1306

SKETCH:

RESEARCH: The tunneling probability of the electron is given by:

( )12 ( )2

2where .b a m U E

T e γ γ− − −= =

SIMPLIFY: Solving for the work function W gives:

( )

( )

( )

2

2

2

22

2ln( ) 2

2 ln( )2

ln( ) .2 2

mWT b a

mW Tb a

TWm b a

= − −

= − −

= − −

CALCULATE: ( )

( )( )

( ) ( )2234

1831 9

181.055 10 J s ln 0.00100

7.287 10 J 6.242 10 eV/J 45.5 eV2 9.11 10 kg 2 0.100 10 m

W−

−− −

⋅ = − = ⋅ ⋅ = ⋅ ⋅

ROUND: To 3 significant figures, the work function of the probe of the scanning tunneling microscope is 45.5 eV.W =

DOUBLE-CHECK: The unit of the work function is electron volts, as expected.

37.38. THINK: The attractive square well potential is given by the function:

( ) 0

0 for for

0 for 0

xU x U x

x

αα α

< −= − < < >

The one-dimensional Schrödinger equation and the boundary conditions can be used to determine the reflection amplitude, R. SKETCH:

RESEARCH: The solution to the Schrödinger equation for each region is given by:


1307

( ) for

for

for

i x i x

i x i x

i x

e Re x

x Ae Be x

Te x

κ κ

κ κ

κ

αψ α α

α

−

′ ′−

+ < −= + − < < >

where R is the amplitude of the reflected wave, T is the amplitude of the transmitted, and

( ) ( )κ κ+

′= =

2 022 2

22 ,m E UmE

As is suggested in the question, boundary conditions at x α= − and x α= are required in order to find an expression for .R Boundary conditions require that the wave function and its derivative are continuous at

:x α= − i i i ie Re Ae Beκα κα κ α κ α′ ′− −+ = + (1)

( ) ( )'i i i ii e Re i Ae Beκα κα κ α κ ακ κ ′ ′− −− = − (2)

At :x α= ' 'i i iAe Be Teκ α κ α κα−+ = (3)

( )' i i ii Ae Be i Teκ α κ α κακ κ′ ′−− = (4)

SIMPLIFY: There are four equations and four unknown coefficients, so an expression for R can be found. Equations (3) and (4) can be used to eliminate T:

i i i i

i i i i

Ae Be Ae Be

Ae Be Ae Be

κ α κ α κ α κ α


κ κκ κ

κ κ κ κ

′ ′ ′ ′− −

′ ′ ′ ′− −

′ ′+ = −

′ ′+ = −

( )( )

2iA Be κ ακ κκ κ

′−′ +

=′ −

(5)

Substituting (5) into (1) and solving for B gives: ( )( )

3 ,i i i ie Re B e Beκα κα κ α κ ακ κκ κ

′ ′− −′ +

+ = +′ −

which implies:

( )( )

3

.i i

i i

e ReBe e

κα κα

κ α κ ακ κκ κ

−

′ ′−

+=′ +

+′ −

(6)

Substituting (5) into (2) and solving for B gives: ( )( )

3 ,i i i ie Re B e Beκα κα κ α κ ακ κκ κ κ κ

κ κ′ ′− −

′ +′ ′− = −

′ − which

implies:

( )( )

3

i i

i i

e ReBe e

κα κα

κ α κ α

κ κκ κ

κ κκ κ

−

′ ′−

−=′ +

′ ′−′ −

(7)

Setting (6) and (7) equal and solving for R gives: ( )( )

( )( )

3 3

.i i i i

i i i i

e Re e Re

e e e e

κα κα κα κα


κ κκ κ κ κ

κ κκ κ κ κ

− −

′ ′ ′ ′− −

+ −=′ ′+ +

′ ′+ −′ ′− −

( )( )

( ) ( ) ( )( )

( ) ( ) ( )( )

( ) ( )

( )( )

( ) ( )

3 3 3

3...

i i i i i i

i i

e e R e R e e e

R e R e

κ κ α κ κ α κ κ α κ κ α κ κ α κ κ α

κ κ α κ κ α

κ κ κ κ κ κκ κ κ κ κ κ

κ κ κ κ κ κκ κ

κ κκ κ

′ ′ ′ ′ ′ ′− + − − + − + −

′ ′− +

′ ′ ′+ + +′ ′ ′ ′− + − = + −

′ ′ ′− − −′ +

− −′ −

Gathering like terms and simplifying gives:


1308

( ) ( )( )

( ) ( ) ( ) ( ) ( ) ( ) ( )( )

( )

( ) ( ) ( ) ( ) ( )( ) ( ) ( )

( ) ( ) ( ) ( ) ( ) ( ) ( )

3 3

2 23 3

2 23 32 2

i i i i

i i i i

i i i i

R e e e e

R e e e e

R e e e e

κ κ α κ κ α κ κ α κ κ α



κ κ κ κκ κ κ κ κ κ κ κ

κ κ κ κ

κ κ κ κ κ κ κ κ

κ κ κ κ κ κ

′ ′ ′ ′− + − − +

′ ′ ′ ′− + − − +

′ ′ ′ ′− + − − +

′ ′ + +′ ′ ′ ′+ − − = + − − ′ ′− −

′ ′ ′ ′+ − − = + − − ′ ′ ′+ − − = − −

At this point it is convenient to multiply both sides by ( ) .ie κ κ α′− Doing so and solving for R gives:

( ) ( ) ( ) ( ) ( )

( )

2 2 2 22 2 2 2

2 2 2 2 2

i ii i

i i i

R e e e e

e e e

κ κ α κ κ ακ α κ α

κα κ α κ α

κ κ κ κ κ κ

κ κ

′ ′− − +′ ′−

′ ′− −

′ ′ ′+ − − = − − ′= − −

( )( ) ( )

2 2 2 2 2

2 22 2.

i i i

i i

e e eR

e e

κα κ α κ α

κ α κ α

κ κ

κ κ κ κ

′ ′− −

′ ′−

′ − − =′ ′+ − −

(8)

Using Euler`s formula, ϑ ϑ ϑ= +cos sin ,ie i the exponential terms become: ( ) ( )( ) ( ) ( ) ( )

2

2

cos 2 sin 2cos 2 sin 2 cos 2 sin 2

i

i

e ie i i

κ α

κ α

κ α κ ακ α κ α κ α κ α

′

′−

′ ′= +′ ′ ′ ′= − + − = −

Substitution of these expressions into (8) and further simplification gives:

( ) ( ) ( )( ) ( ) ( )( )( ) ( ) ( )( ) ( ) ( ) ( )( )

( ) ( )( ) ( ) ( )( ) ( ) ( ) ( )( )

( ) ( )

2 2 2

2 2

2 2 2

2 2 2 2

2 2 2

cos 2 sin 2 cos 2 sin 2

cos 2 sin 2 cos 2 sin 2

2 sin 2

2 cos 2 sin 2 2 cos 2 sin 2

sin 2

2 co

i

i

i

e i iR

i i

ie

i i

ie

κα

κα

κα

κ κ κ α κ α κ α κ α

κ κ κ α κ α κ κ κ α κ α

κ κ κ α

κ κκ κ κ α κ α κ κκ κ κ α κ α

κ κ κ α

κκ

−

−

−

′ ′ ′ ′ ′− + − − =′ ′ ′ ′ ′ ′+ − − − +

′ ′−=

′ ′ ′ ′ ′ ′ ′ ′ ′ ′+ + − − − + +

′ ′−=

′ ( ) ( ) ( )2 2s 2 sin 2iκ α κ κ κ α′ ′ ′− +

No reflected wave, 0,R = occurs when:

( ) ( ) ( )( )

2 2 2 22 02 2 2 2 2

02 2 2

2sin 2 0 2 2 .

2 2 2n

m E U nn n E U nm

π πκ α κ α π κ α π αα

+′ ′ ′= = = = + =

CALCULATE: Not required. ROUND: Not required. DOUBLE-CHECK: The energies,

( )2 2

20 2 ,

2 2nE U n

mπα

+ =

are the allowed energies for the infinite square well of width 2 .α Remarkably, perfect transmission occurs when the energy of the particle plus the potential of the well is equal to the allowed energies of an infinite square well.

37.39. THINK: (a) The Schrödinger equation and the relevant boundary conditions can be used to find the wave function and the energy levels. (b) The solution to the Schrödinger equation can be used to find the penetration distance η for a decrease in the wave function by a factor of 1/ .e (c) This particular quantum well has width 1 nm and depth 0.300 eV with energy 0.125 eV. The finite well potential is given by the function:


1309

( ) ≤ −= − ≤ ≤ ≥

0

0

for / 20 for / 2 / 2

for / 2

U x aU x a x a

U x a

SKETCH:

RESEARCH: (a) The solution to the Schrödinger equation for each region is given by:

( ) ( ) ( )γ γ

γ γ

ψ κ κ

−

−

+ ≤ −= + − ≤ ≤ + ≥

for / 2cos sin for / 2 / 2

for / 2

x x

x x

Ae Be x ax C x D x a x a

Ee Fe x a

where,

κ γ −= =

2 2 02 2

2 ( )2 , .m U EmE

Combining these expressions gives

γ κ γ κ−= = − + =

2 2 2 20 0 0

2 2 2

2 ( ) 2 2 ,

m U E mU mU

which represents circles in the κγ-plane of radius

02

2.

mU However, in the region < − / 2,x a as x → −∞

the second term blows up and in the region > / 2,x a as x → ∞ the first term blows up. Therefore, the physical solution is given by:

( ) ( ) ( )γ

γ

ψ κ κ−

≤ −= + − ≤ ≤ ≥

for / 2cos sin for / 2 / 2

for / 2

x

x

Ae x ax C x D x a x a

Fe x a

b) For ≥ / 2x a the solution requires that

( ) ( )ηψ ψ

= + ==

/2 /21

x a x ax x

e

SIMPLIFY: (a) At / 2,x a= − continuity of the function and its derivative requires:

γ κ κ κ κ− = − + − = −/2 cos( / 2) sin( / 2) cos( / 2) sin( / 2)aAe C a D a C a D a (1) /2 sin( / 2) cos( / 2) sin( / 2) cos( / 2)aA e C a D a C a D aγγ κ κ κ κ κ κ κ κ− =− − + − = + (2)

At / 2,x a= continuity of the function and its derivative requires: /2 cos( / 2) sin( / 2)aFe C a D aγ κ κ− = + (3) /2 sin( / 2) cos( / 2)aF e C a D aγγ κ κ κ κ−− =− + (4)

These four equations can be simplified: Adding (1) and (3):


1310

γ κ−+ =/2( ) 2 cos( / 2)aA F e C a (5) Subtracting (4) from (2):

/2( ) 2 sin( / 2)aA F e C aγγ κ κ−+ = (6) Adding (2) and (4):

/2( ) 2 cos( / 2)aA F e D aγγ κ κ−− = (7) Subtracting (1) from (3):

/2( ) 2 sin( / 2)aF A e D aγ κ−− = (8) If ≠ 0C and ≠ − ,A F dividing (6) by (5) yields:

γ κ κ κ γ κ= =tan( / 2) tan( / 2) /a a If ≠ 0D and ≠ ,A F dividing (7) by (8) yields:

γ κ κ κ κ γ− = = −cot( / 2) tan( / 2) /a a

If these two equations are simultaneously valid then they imply that 2tan ( / 2) 1aκ = − which cannot be true for real values of the energy (i.e. κ must be real). This means that solutions can be divided into two separate classes. The wave functions split into even and odd parity solutions are given by: (i) For even parity solutions where κΨ =( ) cos( )x C x in the well, = =0 and .D A F The wave function is given by:

γ

γ

ψ κ−

≤ −

= − ≤ ≤ ≥

for / 2( ) cos( ) for / 2 / 2

for / 2

x

x

Ae x ax C x a x a

Ae x a

This leads to the solution κ κ γ=tan( / 2) .a (ii) For odd parity solutions where ( ) sin( )x D xκΨ = in the well, = = −0 and .C A F The wave function is given by:

γ

γ

ψ κ−

≤ −

= − ≤ ≤− ≥

for / 2( ) sin( ) for / 2 / 2

for / 2

x

x

Ae x ax D x a x a

Ae x a This leads to the solution κ κ γ= −cot( / 2) .a The energy levels can be found by solving numerically or

graphically each of these solutions with the required relation between κ and γ : γ κ= −

2 202

2mU. Solving

κ κ γ=tan( / 2)a and κ κ γ= −cot( / 2)a graphically (intersection points) gives discrete values for and κ γ

and hence the allowed energy levels are obtained from the κ values at the intersection points and κ=

2 2

2E

m. A sketch of such a graph is shown:


1311

(b) ( ) ( )( )γ η γ γη γη− + − − −= = =/2 /2 11 1a aFe Fe e ee

The penetration distance is given by ( ) ( )η

γ= = =

− − 2

0 0

1 .2 2m U E m U E

CALCULATE:

(c) ( )

( )( )( )η

−−

− −

⋅= = ⋅

⋅ − ⋅

3410

GaAs-GaAlAs 31 19

1.055 10 J s4.668 10 m

2 9.109 10 kg 0.300 eV 0.125 eV 1.602 10 J/eV

ROUND: (c) To three significant figures, the penetration distance is η =GaAs-GaAlAs 467 pm. DOUBLE-CHECK: It is reasonable that the penetration depth is independent of the width of the well. A unit analysis of the units for the penetration depth provides the correct unit of length:

( )2 2 222

kg m / s sJ s J s m m.kg kgJ kg

= = = =

37.40. The energy states of a harmonic oscillator are given by: 01 ,2nE n ω = +

14

0 2.99 10 rad/s.ω = ⋅ The

energy of the ground state and the first two excited states are:

( )( )( )( )( )( )

ω

ω

ω

− −

− −

− −

= = ⋅ ⋅ = ⋅ =

= = ⋅ ⋅ = ⋅ =

= = ⋅ ⋅ = ⋅ =

34 14 200 0

34 14 201 0

34 14 202 0

1 1 1.055 10 J s 2.99 10 rad/s 1.58 10 J 0.0984 eV,2 23 3 1.055 10 J s 2.99 10 rad/s 4.73 10 J 0.295 eV,2 25 5 1.055 10 J s 2.99 10 rad/s 7.89 10 J 0.492 eV.2 2

E

E

E

37.41. The energy levels of a harmonic oscillator are given by: 01 .2nE n ω = +

The energy of a photon is given

by / .E hc λ= The energy of the photon with wavelength 3 1λ → is given by:

3 1 3 1 0 07 3 2 .2 2

E E E ω ω→ Δ = − = − =

The energy of a photon with wavelength 3 2λ → is given by:

3 2 3 2 0 07 5 .2 2

E E E ω ω→ Δ = − = − =

Then


1312

( )( ) ( )03 1 3 1 3 2 3 1

3 2 3 13 2 3 2 3 1 3 2 0

2/360 nm 720 nm.

/E hc EE hc E

ωλ λλ λ

λ λ ω→ → → →

→ →→ → → →

Δ Δ= = = = =

Δ Δ

37.42. The spacing of two energy levels in a harmonic oscillator is given by: 201 0 9 10 J.n nE E E ω −

+Δ = − = = ⋅

For a spring, the frequency is given by: 0 /k mω = where − −= ⋅ ⋅ = ⋅27 272 1.67 10 3.34 10 kgm is the mass of a diatomic hydrogen molecule. Therefore,

( ) ( ) ( )( )

220 272

2 234

9 10 J 3.34 10 kg2432 N/m 2000 N/m.

1.055 10 J s

E mkE km

− −

−

⋅ ⋅ΔΔ = = = = ≈

⋅

37.43. THINK: Since the electron is confined to a cube, the electron can be treated as if it was inside a three-dimensional infinite potential well. In the text, the equation for the energy states for a two dimensional infinite potential is derived. An analogous form for the three dimensional case can be used to determine the ground state energy of the electron in the cube of side length 2R, where 0.0529 nm.R = The spring constant can be found by setting the ground state energy for a potential well equal to the ground state energy for a harmonic oscillator. SKETCH:

RESEARCH: The three dimensional energy states (analogous to equation (37.16)) for the electron are:

2 2 2 2 2 22 2 2

, , 2 2 2e e e

,2 2 2x y zn n n x y zE n n n

m a m a m aπ π π= + +

where a is the side length of the cube and em is the mass of an electron. The ground state of a harmonic oscillator is given by:

0 0e

10 ,2 2

kEm

ω = + =

where k is the spring constant. SIMPLIFY: The ground state, ( ) ( ), , 1, 1, 1 ,x y zn n n = energy for a three dimensional infinite potential well

of side length 2a R= is: 2 2

1, 1, 1 2e

3 .8

Em R

π=

For the case 1, 1, 1 0 :E E= 2 2 2 2 4

1, 1, 1 2 2 4e ee e e

3 3 9 .28 4 16

k kE km mm R m R m R

π π π= = = =


1313

CALCULATE: ( )

( )( )ππ

−−

− −

⋅= = ⋅ =

⋅ ⋅

234 22 2

171, 1, 1 2 231 9e

3 1.055 10 J s3 1.6159 10 J 100.87 eV8 8 9.11 10 kg 0.0529 10 m

Em R

( )( )( )

234 44

431 9

9 1.055 10 J s8.5484 10 N/m

16 9.11 10 kg 0.0529 10 mk

π−

− −

⋅= = ⋅

⋅ ⋅

ROUND: To three significant figures, the ground state energy for an electron confined to a cube of twice the Bohr radius is 101 eVE = and the spring constant that would give the same ground state energy for a harmonic oscillator is 85.5 kN/m.k = DOUBLE-CHECK: The ionization energy of an electron in a hydrogen atom is 13.6 eV and is comparable to the energy calculated.

37.44. THINK: The normalization condition,

( ) 21,x dxψ

∞

−∞

=

can be used to normalize the given wave function, ( ) ( ) ( )0 1,0 .x A x xψ ψ ψ = +

SKETCH: Not required.

RESEARCH: The oscillator wave functions are given by: ( ) 2 2/20 1/4

1 ,xx e σψσπ

−= and

( ) 2 2/21 1/4

1 1 2 ,2

xxx e σψσσπ

− =

where0

.mw

σ = Normalization of the wave function requires that,

( ) ( ) 220 1 1.A x x dxψ ψ

∞

−∞+ =

SIMPLIFY: 2 2 2 2

2 2

2 2

2 2

2 2

22 /2 /2

1/4 1/4

222 /2

1/4

22/

1/2

2 2/

1/2 2

2 2/

1/2 2

1 1 11 22

1 11 22

21

2 2 21

1

x x

x

x

x

x

xA e e dx

xA e dx

A xe dx

A x xe dx

A xe

σ σ

σ

σ

σ

σ

σσπ σπ

σσπ

σσπ

σσπ σ

σπ σ

∞ − −

−∞

∞ −

−∞

∞ −

−∞

∞ −

−∞

−

= +

= +

= +

= + +

= +

dx∞

−∞

The 2 2/ 2 2x xe σ

σ−

term vanishes because this is an odd function, so the result will be zero when

integrating from to .−∞ ∞ Using integral tables, the Gaussian integrals are evaluated: 3

2 21/2 2

1 1 1 21 1 .2 2 3

A A Aσ πσ πσπ σ

= + = + =

CALCULATE: Not required. ROUND: Not required.


1314

DOUBLE-CHECK: As expected, the coefficient A does not depend on ,σ so it is unitless.

37.45. THINK: The normalization condition,

( ) 21,x dxψ

∞

−∞

=

can be used to normalize the given wave function, ( ) 2 2

2/2

0 .x bx A e−=ψ SKETCH:

RESEARCH: (a) The oscillator wave function is given by:

( ) 2 2

2/2

0 .x bx A e−=ψ Normalization of the wave function requires that,

( ) 2

0 1.x dxψ∞

−∞=

(b) As seen from the sketch, the probability that the quantum harmonic oscillator will be found in the classically forbidden region is given by:

( ) ( ) ( )2 2 20 0 02 .

b

b bx dx x dx x dxψ ψ ψ

∞ − ∞

−∞Π = + =

SIMPLIFY:

(a) ( ) 2 2 2 222 /2 2 / 20 2 2 2 2 4

11 x b x bx dx A e dx A e dx A b Ab

∞ ∞ ∞− −

−∞ −∞ −∞= = = = = ψ π

π

(b) Consider the equation: 2 2 2 22

/2 /0

22 .x b x bb b

A e dx e dxbπ

∞ ∞− −Π = = With the substitution

/ ,u x b dx bdu= = the expression becomes:


1315

2

1

2 .ue duπ

∞ −Π =

CALCULATE:

(b) An integration table provides 2

10.139,ue du

∞ − = so ( )12 0.139 0.157Π = =π

ROUND: No rounding is required. DOUBLE-CHECK: The ratio is less than one, as it must be.

37.46. The wave function for an infinite square well is derived in the text. For a well of width L and for the 3n = state, the wave function inside the well is given by:

( ) 2 3sin xxL L

πψ =

The probability that the particle is found in the rightmost 10.0% of the well is given by:

( ) 2 20.9 0.9

2 3sinL L

L L

xx dx dxL L

πψ Π = =

The identity 22sin 1 cos2θ θ= − can be used to simplify the integrand:

( )

0.9

0.9

1 61 cos

1 6sin6

10.100 sin6 sin5.4 0.495 4.95%.6

L

L

L

L

x dxL L

L xxL L

π

ππ

π ππ

Π = −

= −

= − − = =

37.47. (a) Normalization requires that ( ) 2 1.x dxψ∞

−∞= Given that the wave function of the electron in the

region 0 x L< < is ( ) ( )sin 2 / ,x A x Lψ π=

2 20

2sin 1L xA dx

Lπ =

The identity 22sin 1 cos2θ θ= − can be used to simplify the integrand: 2

0

2

02

41 1 cos2

4sin 2 4

2 . 2

L

L

A x dxL

A L xxL

A L AL

π

ππ

= −

= −

= =

(b) The probability of finding the electron in the region 0 / 3x L< < is:

( )2

/3 /3 /32 20 0 0

/3

0

2 2 2 2sin sin

1 4 1 1 4sin sin 0.402.4 3 4 3

L L L

L

x xx dx dx dxL L L L

L xxL L

Π = = =

= − = − =

π πψ

π ππ π

37.48. The wave function for an infinite square well is derived in the text. For a well of width 2.00 nmL = and for the 2n = state, the wave function inside the well is given by:


1316

( ) 2 2sin .2.00 nm 2.00 nm

xx πψ

=

The probability that the particle is found between 0.800 nmx = and 0.900 nmx = is given by:

( )2

0.900 nm 0.900 nm 0.900 nm2 20.800 nm 0.800 nm 0.800 nm

2 2 2 2sin sin .2.00 nm 2.00 nm 2.00 nm 2.00 nm

x xx dx dx dxπ πψ Π = = =

The identity 22sin 1 cos2θ θ= − can be used to simplify the integrand: 0.900 nm

0.800 nm

0.900 nm

0.800 nm

1 41 cos2 nm 2.00 nm

1 2.00 nm 4sin2.00 nm 4 2.00 nm

0.100 nm 0.900 nm 0.800 nm1 sin 4 sin 42.00 nm 4 2.00 nm 2.00 nm

0.02109 2

x dx

xx

π

ππ

π ππ

Π = −

= −

= − − = =

.11%.

37.49. THINK: An electron is trapped in a one dimensional infinite potential well of width 300. pm.L = The square wave function for a particle in an infinite potential well can be integrated over the range ( )0.500 L, 0.750 L to find the probability that the electron in its first excited state is within this range.

SKETCH:

RESEARCH: The wave function for an infinite square well is derived in the text. For a well of width L and for the first excited state =( 2)n , the wave function inside the well is given by:

( ) 2 2sin .xxL L

πψ =

SIMPLIFY: The probability that the particle is found in the range 0.5 0.75L x L< < is given by:

( )2

0.750 0.750 0.7502 20.500 0.500 0.500

2 2 2 2sin sin .L L L

L L L

x xx dx dx dxL L L L

π πψ Π = = =

The identity 22sin 1 cos2θ θ= − can be used to simplify the integrand: 0.750

0.500

0.750

0.500

1 41 cos

1 4sin4

L

L

L

L

x dxL L

L xxL L

π

ππ

Π = −

= −

CALCULATE: ( ) ( )( )10.250 sin 4 0.750 sin 4 0.500 0.24999 0.2504

π ππ

Π = − ⋅ − ⋅ = =


1317

ROUND: Therefore, the probability that the electron in the first excited state is found in the range 0.500 0.750L x L< < is 0.250. DOUBLE-CHECK: It is reasonable that the actual length, 300 pm,L = is irrelevant in finding the probability since the range was given as in terms of L. The diagram agrees with this probability that was found.

37.50. THINK: The relationship shown in the text for the uncertainty in position can be used for the wave function,

( ) 2, .x i tx t Ae eλ ω− −Ψ =

Leave the normalization constant A as a variable, and do not attempt to determine it numerically. SKETCH: Not required.

RESEARCH: The uncertainty is given by ( ) ( )22 .x x xΔ = − Since ( ),x tΨ is symmetric about

0,x = 0,x = and so

( ) ( ) ( )2 2 2, , .x x x t x x t dx∞ ∗

−∞Δ = = Ψ Ψ

SIMPLIFY: ( ) 2 2 2 2 22 2 2 2 2 2 2x i t x i t x x xx Ae e x Ae e dx A e x e dx A x e dxλ ω λ ω λ λ λ∞ ∞ ∞− − − − − −

−∞ −∞ −∞Δ = = =

CALCULATE: Using integral tables, the uncertainty of x for the given wave function is:

43 .

32x A π

λΔ =

ROUND: Not required. DOUBLE-CHECK: The expression of the uncertainty in x states that the larger λ is, the smaller the uncertainty is. This is logical since ( ),x tΨ decays more rapidly for larger λ making ( ),x tΨ more localized.

37.51. THINK: A one dimensional plane-wave wave function can be generalized for three dimensions to find ( ), r tΨ for a non relativistic particle of mass m and momentum .p For a free particle, ( ) 0U r =

identically. It is constantly zero. SKETCH:

RESEARCH: A plane-wave wave function in one dimension is given by:

( ), , where / and / .i x i tx t Ae e p Eκ ω κ ω−Ψ = = = The wave function can be assumed separable into spatial and time dependent parts. Here p is the

momentum of the particle and E is the energy. The probability density function is ( ) 2, .r tΨ SIMPLIFY: The spatial wave function for such a particle can be written as the product of three plane waves. Hence, the wave function takes the form

( ) ( )/ /, ,i p r iEtr t Ae e⋅ −Ψ =


1318

where .= + + r x y z κ and ω have been rewritten as / and / .κ ω= = p E Since ( )2 / 2E p m= is the energy of a non relativistic particle, the full wave function can also be written as

( ) ( ) 2/ /2, .i p r ip t mr t Ae e⋅ −Ψ =

The probability density is ( ) ( )( ) ( )( )2 22 / //2 /2 2, .i p r i p rip t m ip t mr t Ae e Ae e A− ⋅ ⋅ −Ψ = =

CALCULATE: Not applicable. ROUND: Not applicable. DOUBLE-CHECK: The spatial part of this wave function clearly represents a plane wave as ,k r c⋅ =

where c is a constant, is the general form of a plane perpendicular to k

. The wave function can also be substituted into the time dependent Schrödinger Equation satisfying the equation:

( ) ( )2 2 2( / 2 ) ( , ) / ( ) ( , ) ( , ) /m r t r U r r t i r t t− ∂ Ψ ∂ + Ψ = ∂ Ψ ∂ Since U = 0 for a free particle Schrödinger’s Equation becomes:

( ) ( )2 2 2( / 2 ) ( , ) / ( , ) /m r t r i r t t− ∂ Ψ ∂ = ∂ Ψ ∂

Substituting for ( ) ( ) ( ) ( ) 2/ /2, i p r ip t mr t r t Ae e⋅ −Ψ = Ψ Ψ = and differentiating, the left side is:

( ) 2/2 2 /2

2

( )2

i p r ip t mp Ae em

⋅ −− −

and the right side is: ( ) 2/2 /2

2

⋅ −−

i p r ip t mi Aip e em

After cancelling like terms and recalling that i2 = -1 these are equal and so the wave function does satisfy the time dependent Schrödinger Equation.

37.52. THINK: Separation of variables can be used to write the wave function as a product of two functions that depend on only one variable. The equation for the expectation value of x is given in the text. The derivative of this expression provides / .d x dt SKETCH: Not required.

RESEARCH: The expectation value of the particle’s position is given by: ( ) ( ), , .x x t x x t dx∞ ∗

−∞= Ψ Ψ

As shown in the text, the wave function can be written as: ( ) ( ) ( ) ( ) /, , where iEtx t x t t Aeψ χ χ −Ψ = =

using separation of variables. Therefore, the expectation value of x is:

( ) ( ) ( ) ( ) .x x t x x t dxψ χ ψ χ∞ ∗ ∗

−∞=

SIMPLIFY: The expectation value of x can be simplified as:

( ) ( ) ( ) ( )

( ) ( )

( ) ( )

2 / /

2

iEt iEt

x x x x t t dx

A x x x e e dx

A x x x dx

ψ ψ χ χ

ψ ψ

ψ ψ

∞ ∗ ∗

−∞∞ ∗ −

−∞∞ ∗

−∞

=

=

=

Since the time dependence vanishes,

( ) ( )2 0d x d A x x x dx

dt dtψ ψ

∞ ∗

−∞

= =

CALCULATE: Not required.


1319

ROUND: Not required. DOUBLE-CHECK: It is reasonable that for a stationary state, the expectation value of the position of the particle does not depend on time (i.e. it remains stationary).

37.53. THINK: A quantum particle of mass m is in an infinite one dimensional potential well and has a wave

function given by: ( ) ( ) ( )1 21, , , .2

x t x t x t Ψ = Ψ + Ψ The time-independent wave function for an

infinite potential well is derived in the text. Since the wave function is separable, ( ) ( ) ( ) ( ) /, , with .iEtx t x t t eψ χ χ −Ψ = =

The probability density distribution is just ( ) 2, .x tΨ

SKETCH:

RESEARCH: The probability density distribution is given by:

( ) ( ) ( ) ( ) ( )21 2 1 2

1, , , , ,2

x t x t x t x t x t∗ ∗ Ψ = Ψ + Ψ Ψ + Ψ

The wave functions 1Ψ and 2Ψ are:

( )

( )

1

2

/1

/2

2, sin

2 2, sin

iE t

iE t

xx t ea a

xx t ea a

π

π

−

−

Ψ = Ψ =

with 2 2

1 22E

maπ= and

2 2

2 22 .Ema

π=

SIMPLIFY: The probability density distribution is:

( ) ( ) ( ) ( ) ( ) ( ) ( )( )

( )

1 2

1 2

2 2 21 2 1 2 1 2

/2 2

/

2

1, , , , , , ,2

2 2 2 2 2sin sin sin sin . . .12 2 2. . . sin sin

1 sin s

i E E t

i E E t

x t x t x t x t x t x t x t

x x x x ea a a a a a a

x x ea a a

xa a

π π π π

π π

π

∗ ∗

−

− −

Ψ = Ψ + Ψ + Ψ Ψ + Ψ Ψ + + +

= +

= +

( ) ( )1 2 1 2/ /2 2 2in sin sin i E E t i E E tx x x e ea a aπ π π − − − + +

Using Euler`s formula, cos sin ,ie iθ θ θ= + the exponential terms become:

( ) ( ) ( )

( ) ( ) ( ) ( ) ( )

1 2

1 2

/ 1 2 1 2

/ 1 2 1 2 1 2 1 2

cos sin

cos sin cos sin

i E E t

i E E t

E E t E E te i

E E t E E t E E t E E te i i

−

− −

− −= +

− − − −

= − + − = −

Therefore, the imaginary terms cancel out to give:


1320

( ) ( )2 1 22 2

2 22 2

2

2 2

1 2 2, sin sin 2sin sin cos

1 2 2 3sin sin 2sin sin cos2

1 2sin sin

E E tx x x xx ta a a a a

x x x x ta a a a a ma

xa a

π π π π

π π π π π

π

− Ψ = + + = + + −

= +

2

22 32sin sin cos

2x x x t

a a a maπ π π π +

Using trig identities, this reduces to:

( )22 2 22

1 3, sin 1 4cos 4cos cos .2

x x x tx ta a a a ma

π π π π Ψ = + +

CALCULATE: Not required. ROUND: Not required. DOUBLE-CHECK: The probability density function is real and has units of inverse length, as expected.

37.54. The energy released by the annihilation of a proton and an antiproton is

( )( )22 2 27 8 10 9 9p2 2 1.6726 10 kg 3.00 10 m/s 3.01 10 J 1.88 10 eV 1.9 10 eV.E mc m c − −= = = ⋅ ⋅ = ⋅ = ⋅ ≈ ⋅

The energy released from the annihilation is about 4500 times greater than that for a nuclear-fusion reaction.

37.55. The energy time uncertainty relation is given by:

.2 2

E t tE

Δ Δ ≥ Δ ≥Δ

(a) For an electron/positron pair:

( )( )

( )( )

2

34

231 8

22

2 2

1.055 10 J s

2 2 9.11 10 kg 3.00 10 m/s

3.22 10 s.

e

tm c

−

−

−

Δ ≥

⋅≥

⋅ ⋅

≥ ⋅

(b) For a proton/anti-proton pair:

( )( )

( )( )34

2522 27 8

1.055 10 J s1.75 10 s.

2 2 2 2 1.67 10 kg 3.00 10 m/sp

tm c

−−

−

⋅Δ ≥ = = ⋅

⋅ ⋅

37.56. The positron-electron annihilation releases two 2.0 MeV gamma rays or a total of tot 4.0 MeV.E = Since energy must be conserved, the kinetic energy of the two particles and the energy created due to the annihilation must be equal to 4.0 MeV. The energy released when the positron and electron annihilate is:

( )( )− −= = = ⋅ ⋅ = ⋅ =22 2 31 8 13

e2 2 9.11 10 kg 3.00 10 m/s 1.64 10 J 1.02 MeV.E mc m c

Therefore, the total kinetic energy of the particles is + = −p e tot .K K E E Since =e p / 2,K K the kinetic energy of the electron is:

( )

( )

+ = − = − = −

= − =

pp tot p tot p tot

p

3 22 2 3

2 4.0 MeV 1.02 MeV 2.0 MeV3

KK E E K E E K E E

K


1321

Finally, = = =e p / 2 2.0 MeV / 2 1.0 MeV.K K

Additional Problems

37.57. The energy needed to promote the electron from the ground state to the first excited state is given by

2 1,E E E= − 2 2

2n 2 ,

2E n

maπ=

( ) ( )( )( )

234 22 2 2 218

2 2 231 10

1.0546 10 J s3 34 1 4.52 10 J 28.2 eV.22 2 9.109 10 kg 2.00 10 m

Ema ma

ππ π−

−

− −

⋅= − = = = ⋅ =

⋅ ⋅

37.58. The probability of tunneling is given by:

( ) ( )22

2, where b a em U E

T e γ γ− − −= =

The factor by which the tunneling current changes is: ( )

( )

( )( )( )( )

( )( )

2i

2 0.1 nmf

31 199

234

exp 2 0.1 nm

2 9.11 10 kg 4.0 eV 1.602 10 J/eVexp 2 0.1 10 m

1.055 10 J s

7.8.

b a

b a

T eT e

γ

γ

γ

− −

− − +

− −−

−

=

= ⋅ ⋅ = ⋅ ⋅

=

Therefore, the tunneling current decreases by a factor of 7.8 when the tip moves 0.10 nm farther from the surface.

37.59. The normalized solution of the wave function in the ground state ( )1n = for an electron in an infinite cubic potential well of side length L is given by:

(a) ( ) ( ) ( )3

2 sin sin sin ; 0 , ,x y z Lyx zx y z x y z

L L L L<

= = <

ππ πψ ψ ψ ψ

(b) Since the energies are given by

( )2 2

2 2 2 ,2 x y zE n n n

mL= + + π

the different energies depend on the energy state, ( )2 2 2 .x y zn n n+ + The ground state is for

( ) ( ), , 1,1,1 ,x y zn n n = the first excited state is for ( ) ( ) ( ) ( ), , 1,2,1 , 2,1,1 , 1,1,2 ,x y zn n n = and the second

excited state is for ( ), ,x y zn n n ( ) ( ) ( )1,2,2 , 2,1,2 , 2,2,1 .= Since an electron has two spin states (up or

down), there are a total of 14 possible energy states.

37.60. The energy of a harmonic oscillator is given by ( )0 1/ 2 .nE nω= + The quantum excitation number is then

( )( )ω − −= − = − = ⋅

⋅33

34 10

1 1.00 J 1 2.13 10 .2 21.055 10 J s 4.45 s

nEn


1322

37.61. The allowed energies for a proton in a one dimensional infinite potential well of width α are 2 2

2n 2 .

2E n

mπα

= For the first excited state, 2.n = Therefore, the energy of the first excited state of a proton

is:

( )( )( )

234 222

2 24

27 9

4 1.055 10 J s1.31558 10 J 8.21 10 eV.

2 1.67 10 kg 1.00 10 mE

π−− −

− −

⋅= = ⋅ = ⋅

⋅ ⋅

37.62. The probability of tunneling is given by:

( ) ( )γ γ− − −= =

2

2

2, where b a m U E

T e

The probability that the alpha particle will tunnel through the barrier is:

( ) ( )

( )( )( )( )

( )

2

27 1315

234

49

2exp 2

2 6.64 10 kg 17 MeV 5.6 MeV 1.602 10 J/MeVexp 2 38 10 m

1.055 10 J s

1.9 10

m U ET b aα

− −−

−

−

− = − − ⋅ − ⋅

= − ⋅ ⋅

= ⋅

The probability is very small.

37.63. The distance between fringes (central maximum and first order peak) for a double slit setup is given by

.Lyd

λΔ = The wavelength is given by:

( )( )( )λ

−

−

⋅= = =

⋅ ⋅

34

27 8

6.626 10 J s1.32 fm

1.67 10 kg 3.00 10 m/sh

mc

The distance between interference peaks is:

( )( )( )

15

3

1.32 10 m 1.5 m4.0 pm.

0.50 10 my

−

−

⋅Δ = =

⋅

37.64. The ground state ( )1n = energy of an electron in a one dimensional quantum box (infinite well) of length 0.100 nmL = is:

37.65. The ground state ( )1n = energy of an electron in a one dimensional infinite well of length L is: 2 2

1 22E

mLπ=

(a) For 2 GaAs layers, 0.56 nm,L = so the energy is:

( )( )( )

π−−

− −

⋅= = ⋅ =

⋅ ⋅

234 219

1 231 9

1.055 10 J s1.9 10 J 1.2 eV.

2 9.11 10 kg 0.56 10 mE

(b) For 5 GaAs layers, =1.4 nm,L so the energy is:

( )( )( )

234 22 218

1 2 231 9e

1.055 10 J s6.02915 10 J 37.6 eV.

2 2 9.11 10 kg 0.100 10 nmE

m L

ππ−

−

− −

⋅= = = ⋅ =

⋅ ⋅


1323

( )( )( )

π−−

− −

⋅= = ⋅ =

⋅ ⋅

234 220

1 231 9

1.055 10 J s3.1 10 J 0.19 eV.

2 9.11 10 kg 1.4 10 mE

37.66. (a) The ground state ( )1n = energy of a water vapor molecule in a room (an infinite potential well) is:

( ) ( ) ( )

( )( ) ( ) ( ) ( )

2 2

1 2 2 2

234 2

2 2 226

43 25

1 1 12 10.0 m 10.0 m 4.00 m

1.055 10 J s 1 1 12 2.992 10 kg 10.0 m 10.0 m 4.00 m

1.5145 10 J 9.45 10 eV.

Emπ

π−

−

− −

= + +

⋅ = + + ⋅

= ⋅ = ⋅

(b) The average kinetic energy of a molecule is given by:

avg3 ,2

K kT=

where k is the Boltzmann constant and T is the temperature. Therefore,

( )( )23 21avg

3 1.38 10 J/K 300. K 6.21 10 J 0.0388 eV.2

K − −= ⋅ = ⋅ =

(c) Since avg ,K E>> thermal energies are so great on a macroscopic scale that quantum effects cannot be observed.

37.67. The fundamental state ( )1n = energy of a neutron between rigid walls (a one dimensional infinite potential well) 8.4 fmL = apart is:

( )( )( )

ππ−

−

− −

⋅= = = ⋅ =

⋅ ⋅

234 22 2

131 2 227 15

1.055 10 J s4.7 10 J 2.9 MeV.

2 2 1.67 10 kg 8.4 10 mE

mL

37.68. THINK: Since the tunneling current proportional to the tunneling probability, the ratio of the current is found by using the given wave function dependence and the two working gap distances. SKETCH:

RESEARCH: The electron wave function falls off exponentially as:

( ) 110.0 nm .aeψ−

−=

SIMPLIFY: Equation 37.23 shows that the ratio of tunneling currents is:( ) ( )

( ) ( )ψ

ψ

−

−

−

−=

1

1

2 2 10.0 nm 0.400 nm2

2 2 10.0 nm 0.420 nm1

.e

e


1324

CALCULATE: ( ) ( )12

2 10.0 nm 0.020 nm22

1

1.49eψ

ψ

−

= =

ROUND: To three significant figures, the ratio of the current when the STM tip is 0.400 nm above a surface feature to the current when the tip is 0.420 nm above the surface is 1.49. DOUBLE-CHECK: It is expected that the tunneling current is greater when the STM is closer to the surface since tunneling probability is greater.

37.69. THINK: The equation for the allowed energy states of a particle in an infinite square well can be found in the text. The energy difference between the 4n = state and the 2n = state is the energy of the resulting radiation. The wavelength of the radiation can be found from this energy. SKETCH:

RESEARCH: The energy of a particle in a one dimensional infinite potential well of width L is given by:

2 22

2 .2nE n

mLπ=

The wavelength of a photon with energy E is given by: .hcE

λ =

SIMPLIFY: For an electron transition from the 4n = state to the 2n = state the change in energy is

( )2 2 2 2

4 2 2 2616 4 .

2 e e

Em L m L

π π→ = − =

Therefore, the corresponding wavelength of the radiation is given by: 2

4 24 2

.3

ecm LhcE

λπ→

→= =


1325

CALCULATE: The wavelength of the radiation for a transition from the 4n = state to the 2n = state is:

( )( )( )( )

28 31 96

4 2 34

3.00 10 m/s 9.11 10 kg 2.00 10 m1.099 10 m 1099 nm.

3 1.055 10 J sλ

π

− −−

→ −

⋅ ⋅ ⋅= = ⋅ =

⋅

ROUND: To 3 significant figures, the wavelength is 34 2 1.10 10 nm.λ → = ⋅

DOUBLE-CHECK: The units work out to get a length for the wavelength, as it should.

37.70. THINK: This scenario can be modeled as a tunneling problem with a potential barrier height of 1.00 eVUΔ = and a width of 2.00 nm.b a− =

SKETCH:

RESEARCH: The tunneling probability for an electron is given by:

( )2 ,b aT e γ− −= where 22

.em Uγ

Δ=

SIMPLIFY: ( )22

exp 2 .em UT b a

Δ= − −


( )( )

31 199 9

234

2 9.11 10 kg 1.00 eV 1.602 10 J/eVexp 2 2.00 10 m 1.270 10 .

1.055 10 J sT

− −− −

−

⋅ ⋅ = − ⋅ = ⋅ ⋅

ROUND: To 3 significant figures, the probability that a conduction electron in one wire will be found in the other wire after arriving at the gap is 91.27 10 .T −= ⋅ DOUBLE-CHECK: Classically, the probability that an electron in one wire can be found in the other wire is zero. However, quantum mechanically it is expected that there is a small probability that this can happen.

37.71. THINK: In the text, the equations for the energy states for a one and two dimensional infinite potential are derived. An analogous form for the three dimensional case can be used to determine the ground state energy of the electron in the potential cube of side length 0.10 nm.a =


1326

SKETCH:

RESEARCH: The allowed energies for the one dimensional infinite potential well are given by:

2 22

, 1D 2 .2nE n

maπ=

The allowed energies for the three dimensional infinite potential cube in its ground state are given by:

( )2 2

2 2 2, 3D 2 .

2n x y zE n n nma

π= + +

SIMPLIFY: The electron confined to the cube is in its ground state, so: 2 2

1, 3D 23 .2

Ema

π=

, 1DnE is closest to 1, 3DE for 2n = (the first excited state), so the smallest energy difference is given by: 2 2 2 2 2 2

min 2, 1D 1,3D 2 2 24 3 .2 2 2

E E Ema ma ma

π π π= − = − =

CALCULATE: ( )

( )( )

234 218

min 231 9

1.055 10 J s6.029 10 J 37.6 eV

2 9.11 10 kg 0.10 10 mE

π−−

− −

⋅= = ⋅ =

⋅ ⋅

ROUND: To two significant figures, the minimum energy difference is 38 eV. DOUBLE-CHECK: The energy is of the same order of magnitude with the ionization energy of an electron ( )13.6 eV in a hydrogen atom. Therefore, the answer is reasonable.

37.72. THINK: This scenario can be modeled as a tunneling problem with a potential barrier height of 1U and width 2116.8 fm 529.2 fmb a− = − for an electron with energy 129 keV.E = Given that he probability of tunneling is 10%, the equation for the tunneling probability can be used to determine the height of the potential barrier 1.U


1327

SKETCH:

RESEARCH: The probability of tunneling is given by:

( ) ( )γ γ− − −= =

2 e 1

2

2, where b a m U E

T e

SIMPLIFY: ( ) ( )

( ) ( ) ( )

( )( )

( )

( )( )

12

e 12

2e 1

2

22

1e

2exp 2

2ln 2

ln 214

ln8

em U ET b a

m U ET b a

T m U Eb a

TU E

m b a

− = − −

−= − −

−= −

= + −

CALCULATE:

( )( )

( )( )( ) ( )( )

22343 19

1 31 15

14

1.055 10 J s ln 0.100129 10 eV 1.602 10 J/eV

2 9.11 10 kg 2 2116.8 fm 529.2 fm 10 m/fm

2.388 10 J 149.1 keV

U−

−− −

−

⋅ = + ⋅ ⋅ ⋅ −

= ⋅ =

ROUND: To three significant figures, the height of the potential barrier is 1 149 keV.U = DOUBLE-CHECK: It is expected that the potential barrier is larger than the energy of the particle in order to allow for tunneling. Since the tunneling probability is 10.0% it is reasonable that the potential barrier is comparable to the kinetic energy of the particle.

37.73. THINK: The equation for the allowed energies of a two dimensional infinite potential well is given in the text.


1328

SKETCH:

RESEARCH: The allowed energies for an electron in an infinite potential rectangle of dimensions

andx yL L are given by: 222 2

, 2 2 .2x y

yxn n

x y

nnE

m L Lπ

= +

SIMPLIFY: For and 2 ,x yL w L w= =

( )222 2 2 2

2 2, 2 2 2 4 .

2 4 8x y

yxn n x y

nnE n n

m w w mwπ π

= + = +

CALCULATE: The lowest energy for which degeneracy occurs is for:

( ) ( ), 2,2x yn n = and ( ) ( ), 1,4 .x yn n =

ROUND: Not required.

DOUBLE-CHECK: ( ) ( )( )2 2 2 2

2 22, 2 2 2

54 2 28 2

Emw mw

π π= + = and ( ) ( )( )2 2 2 2

2 21,4 2 2

54 1 4 .8 2

Emw mw

π π= + = These

values are the same, as required.

Bauer Solucionario Tomo2

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