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CHAPTER 22

The Electric Field I: Discrete Charge Distributions 1* · If the sign convention for charge were changed so that the charge on the electron were positive and the

charge on the proton were negative, would Coulomb's law still be written the same? Yes 2 ·· Discuss the similarities and differences in the properties of electric charge and gravitational mass. Similarities: The force between charges and masses vary as 1/r2. Differences: There are positive and negative charges but only positive masses. Like charges repel; like masses attract. The gravitational constant G is many orders of magnitude smaller than the Coulomb constant k . 3 · A plastic rod is rubbed against a wool shirt, thereby acquiring a charge of –0.8 µC. How many electrons are

transferred from the wool shirt to the plastic rod? ne = q/(–e) ne = (–0.8×10−6)/(–1.6×10–19) = 5×1012. 4 · A charge equal to the charge of Avogadro's number of protons (NA = 6.02×1023) is called a faraday.

Calculate the number of coulombs in a faraday. 1 faraday = NAe 1 faraday = 6.02×1023×1.6×10–19 C = 9.63×104 C 5* · How many coulombs of positive charge are there in 1 kg of carbon? Twelve grams of carbon contain

Avogadro's number of atoms, with each atom having six protons and six electrons. Q = 6×nC×e; nC = NA(mC/12) Q = (6×6.02×1023×103×1.6×10–19/12) C = 4.82×107 C 6 · Can insulators be charged by induction? No 7 ·· A metal rectangle B is connected to ground through a switch S that is initially closed (Figure 22-28). While

the charge +Q is near B, switch S is opened. The charge +Q is then removed. Afterward, what is the charge state of the metal rectangle B? (a) It is positively charged. (b) It is uncharged. (c) It is negatively charged. (d) It may be any of the above depending on the charge on B before the charge +Q was placed nearby.

(c) 8 ·· Explain, giving each step, how a positively charged insulating rod can be used to give a metal sphere (a) a

negative charge, and (b) a positive charge. (c) Can the same rod be used to simultaneously give one sphere a positive charge and another sphere a negative charge without the rod having to be recharged?

(a) Connect the metal sphere to ground; bring the insulating rod near the metal sphere and disconnect the sphere from ground; then remove the insulating rod. The sphere will be negatively charged. (b) Bring the insulating rod in contact with the metal sphere; some of the positive charge on the rod will be

Chapter 22 The Electric Field I: Discrete Charge Distributions

transferred to the metal sphere. (c) Yes. First charge one metal sphere negatively by induction as in (a). Then use that negatively charged

sphere to charge the second metal sphere positively by induction. 9* ·· Two uncharged conducting spheres with their conducting surfaces in contact are supported on a large

wooden table by insulated stands. A positively charged rod is brought up close to the surface of one of the spheres on the side opposite its point of contact with the other sphere. (a) Describe the induced charges on the two conducting spheres, and sketch the charge distributions on them. (b) The two spheres are separated far apart and the charged rod is removed. Sketch the charge distributions on the separated spheres. (a) On the sphere near the positively charged rod, the induced charge is negative and near the rod. On the other sphere, the net charge is positive and on the side far from the rod. This is shown in the diagram. (b) When the spheres are separated and far apart and the rod has been removed, the induced charges are distributed uniformly over each sphere. The charge distributions are shown in the diagram.

10 · Three charges, +q, +Q, and −Q, are placed at the corners of an equilateral triangle as shown in Figure 22-

29. The net force on charge +q due to the other two charges is (a) vertically up. (b) vertically down. (c) zero. (d) horizontal to the left. (e) horizontal to the right.

(e) 11 · A charge q1 = 4.0 µC is at the origin, and a charge q2 = 6.0 µC is on the x axis at x = 3.0 m. (a) Find the

force on charge q2. (b) Find the force on q1. (c) How would your answers for parts (a) and (b) differ if q2 were –6.0 µC?

(a) Use Equ. 22-2 (b) Use Newton’s third law (c) In this case the forces are in opposite directions

F12 = (8.99×109×4×6×10–12/9) N i = 2.4×10–2 N i F21 = –F12 = –2.4×10–2 N i F12 = 2.4×10–2 N i; F21 = 2.4×10–2 N i

12 · Three point charges are on the x axis: q1 = –6.0 µC is at x = –3.0 m, q2 = 4.0 µC is at the origin, and q3 =

–6.0 µC is at x = 3.0 m. Find the force on q1. Use Equ. 22-2 to find F21 and F31 F1 = F21 + F31

F21 = 2.4×10–2 N i; F31 = –0.9×10–2 N i F1 = 1.5×10–2 N I

13* ·· Two equal charges of 3.0 µC are on the y axis, one at the origin and the other at y = 6 m. A third charge q3

= 2 µC is on the x axis at x = 8 m. Find the force on q3. Use Equ. 22-2 to find F13 and F23 F3 = F13 + F23

F13 = 8.43×10–4 N i; F23 = (5.39×10–4 N)(0.8 i – 0.6 j) F3 = 1.27×10–3 N i – 3.24×10–4 N j

14 ·· Three charges, each of magnitude 3 nC, are at separate corners of a square of side 5 cm. The two charges

at opposite corners are positive, and the other charge is negative. Find the force exerted by these charges on a


fourth charge q = +3 nC at the remaining corner. The configuration of the charges and the forces on the fourth charge are shown in the figure. From the figure it is evident that the net force on q4 is along the diagonal of the square and directed away from q3. 1. Use Equ. 22-2 to find F14 and F24 2. Find i and j components of F34 3. F4 = F14 + F24 + F34

F14 = 3.24×10–5 N j F24 = 3.24×10–5 N i F34 = -1.14×10–5 N (i + j) F4 = 2.1×10–5 N (i + j)

15 ·· A charge of 5 µC is on the y axis at y = 3 cm, and a second charge of –5 µC is on the y axis at y = –3 cm.

Find the force on a charge of 2 µC on the x axis at x = 8 cm. From the geometry it is evident that the net force on the 2 µC charge is along the y axis. 1. Use Equ. 22-2 to find the y component of the force on the 2 µC exerted by the 5 µC charge. 2. F2 = 2F52y j

F52y = – (10–11×8.99×109/73)(3/ 73 ) N = –4.32×10–4 N F2 = –8.64×10–4 N j

16 ·· A point charge of –2.5 µC is located at the origin. A second point charge of 6 µC is at x = 1 m, y = 0.5 m.

Find the x and y coordinates of the position at which an electron would be in equilibrium. The positions of the charges are shown in the diagram. It is apparent that the electron must be located along the line joining the two charges. Moreover, since it is negatively charged, it must be closer to the –2.5 µC than to the 6.0 µC charge, as is indicated in the figure.

Use Equ. 22-2 to find the magnitude of the force on e acting along the line.

Set F = 0 and solve for r r < 0 is unphysical. Find x and y components.

F = ke[2.5/r2 – 6.0/(r + 1.251/2)2] µN 3.5r2 – 5.59r – 3.125 = 0; r = 2.09 m, –0.438 m x = (–2.09/1.251/2) m = –1.87 m; y = 1/2x = –0.935 m

17* ·· A charge of –1.0 µC is located at the origin, a second charge of 2.0 µC is located at x = 0, y = 0.1 m, and a

third charge of 4.0 µC is located at x = 0.2 m, y = 0. Find the forces that act on each of the three charges. Let q1 = –1.0 µC at (0, 0), q2 = 2 µC at (0, 0.1), and q3 = 4 µC at (0.2, 0). 1. Use Equ. 22-2 to find F21, F31, and F32

F21 = 1.8 N j, F31 = 0.899 N i, F32 = 0.643 N j – 1.29 N i


2. F1 = F21 + F31; F2 = F12 + F32 3. F3 + F1 + F2 = 0; F3 = -(F1 + F2)

F1 = 0.899 N i + 1.8 N j; F2 = –1.29 N i – 1.16 N j F3 = 0.391 N i – 0.643 N j

18 ·· A charge of 5.0 µC is located at x = 0, y = 0 and a charge Q2 is located at x = 4.0 cm, y = 0. The force on a

2-µC charge at x = 8.0 cm, y = 0 is 19.7 N, pointing in the negative x direction. When this 2-µC charge is positioned at x = 17.75 cm, y = 0, the force on it is zero. Determine the charge Q2. Write F on 2 µC and set it equal to 0 when x(2 µC) = 17.75 cm. Solve for Q2

10×10–12/(0.1775)2 + 2×10–6Q2/(0.1375)2 = 0 Q2 = –3.0 µC

19 ·· Five equal charges Q are equally spaced on a semicircle of radius R as shown in Figure 22-30. Find the

force on a charge q located at the center of the semicircle. By symmetry, the y component of the force on q is zero. The x components of the forces on q are kqQ/R2 for the charge Q on the x axis and 2/ 2RkqQ for each of the two charges at 45° from the x axis. The total force

on the charge q is therefore F = )21)(/( 2 +RkqQ i. 20 ··· The configuration of the NH3 molecule is approximately that of a regular tetrahedron, with three H+ ions

forming the base and an N3– ion at the apex of the tetrahedron. The length of each side is 1.64×10–10 m. Calculate the force that acts on each ion.

Let the H+ ions be in the x-y plane with H1 at (0, 0, 0), H2 at (a, 0, 0), and H3 at (a/2, 2/3a , 0). The N–3 ion

which we shall label 4, is then at )3/2,32/,2/( aaa . Here a = 1.64×10–10 m. To simplify the calculation

we shall set ke2/a2 = C = 8.56×10–9 N. 1. Use Equ. 22-2 to find F21, F31, F41 2. F1 = F21 + F31 + F41 3. By symmetry, F2 = F3 = F1 4. F1 + F2 + F3 + F4 = 0; solve for F4

F21 = -C i; F31 = -C[(1/2) i + )2/3( j];

F41 = 3C[(1/2) i + ( 32/1 ) j + ( 3/2 ) k]

F1 = C 6 k

F2 = F3 = C 6 k

F4 = –3C 6 k 21* · A positive charge that is free to move but is at rest in an electric field E will (a) accelerate in the direction perpendicular to E. (b) remain at rest. (c) accelerate in the direction opposite to E. (d) accelerate in the same direction as E. (e) do none of the above. (d) 22 · If four charges are placed at the corners of a square as shown in Figure 22-31, the field E is zero at (a) all points along the sides of the square midway between two charges. (b) the midpoint of the square. (c) midway between the top two charges and midway between the bottom two charges. (d) none of the above. (b)


23 ·· At a particular point in space, a charge Q experiences no net force. It follows that (a) there are no charges nearby. (b) if charges are nearby, they have the opposite sign of Q. (c) if charges are nearby, the total positive charge must equal the total negative charge. (d) none of the above need be true. (d) Note: In the first printing of the textbook the problem statement reads, “At … experiences no force.…” In that case, the correct answer is (a). 24 · A charge of 4.0 µC is at the origin. What is the magnitude and direction of the electric field on the x axis at

(a) x = 6 m and (b) x = -10 m? (c) Sketch the function Ex versus x for both positive and negative values of x. (Remember that Ex is negative when E points in the negative x direction.)

(a) Use Equ. 22-7 (b) At x = –10 m, E points in the –i direction

E = (8.99×109 × 4.0×10–6/36) N/C i = 999 N/C i E = –360 N/C i

(c) A plot of Ex is shown.

25* · Two charges, each +4 µC, are on the x axis, one at the origin and the other at x = 8 m. Find the electric field

on the x axis at (a) x = –2 m, (b) x = 2 m, (c) x = 6 m, and (d) x = 10 m. (e) At what point on the x axis is the electric field zero? (f) Sketch Ex versus x. (a) Use Equ. 22-7 (b) Here the fie lds due to the two charges are

oppositely directed

E = –8.99×109 × 4×10–6(1/22 + 1/102) N/C i = –9.35×103 N/C i E = 3.596×104(1/22 – 1/62) N/C i = 7.99×103 N/C i

(c) By symmetry, E(6) = E(2) (d) By symmetry, E(10) = E(–2) (e) Use symmetry argument (f) Ex versus x is shown

E = –7.99×103 N/C i E = 9.35×103 N/C I E = 0 at x = 4 m


26 · When a test charge q0 = 2 nC is placed at the origin, it experiences a force of 8.0×10–4 N in the positive y direction. (a) What is the electric field at the origin? (b) What would be the force on a charge of –4 nC placed at the origin? (c) If this force is due to a charge on the y axis at y = 3 cm, what is the value of that charge?

(a) Use Equ. 22-5 (b) F = qE (c) Use Equ. 22-2

E = (8×10–4/2×10–9) N/C j = 4×105 N/C j F = -16×10–4 N j q = (16×10–4 × 0.032/8.99×109 × 4×10–9) C = 40 nC

27 · An oil drop has a mass of 4×10–14 kg and a net charge of 4.8×10–19 C. An upward electric force just

balances the downward force of gravity so that the oil drop is stationary. What is the direction and magnitude of the electric field?

Eq = mg; Eq must point up E = (4×10–14 × 9.81/4.8×10–19 ) N/C = 8.18×105 N/C up 28 · The electric field near the surface of the earth points downward and has a magnitude of 150 N/C. (a)

Compare the upward electric force on an electron with the downward gravitational force. (b) What charge should be placed on a penny of mass 3 g so that the electric force balances the weight of the penny near the earth's surface?

(a) Ee/mg = 150×1.6×10–19/9.1×10–31 × 9.81 = 2.69×1012. The electric force is very much larger. (b) q = mg/E = 1.96×10–4 C. 29* ·· Two equal positive charges of magnitude q1 = q2 = 6.0 nC are on the y axis at y1 = +3 cm and y2 = –3 cm.

(a) What is the magnitude and direction of the electric field on the x axis at x = 4 cm? (b) What is the force exerted on a third charge q0 = 2 nC when it is placed on the x axis at x = 4 cm?

(a) By symmetry, Ey=0. Find E due to q1 at x = 4cm Total Ex = 2E×4/5 (b) F = qE

E = kq1/25×10–4 = 2.158×104 N/C Ex = 3.45×104 N/C; E = 3.45×104 N/C i F = 6.9×10–5 N I

30 ·· A point charge of +5.0 µC is located at x = –3.0 cm, and a second point charge of –8.0 µC is located at x =

+4.0 cm. Where should a third charge of +6.0 µC be placed so that the electric field at x = 0 is zero? 1. Set E = 0 at x = 0; note that E due to +5 µC and –8 µC point in the i direction 1. Solve for x

(5/9) + (8/16) – (6/x2) = 0 x = 2.38 cm

31 ·· A point charge of –5 µC is located at x = 4 m, y = –2 m. A second point charge of 12 µC is located at x = 1 m, y = 2 m. (a) Find the magnitude and direction of the electric field at x = –1 m, y = 0. (b) Calculate the magnitude and direction of the force on an electron at x = –1 m, y = 0.


The diagram shows the electric field vectors at the point of interest due to the two charges. Note that the E field due to the +12 µC charge makes an angle of 225° with the x axis; the E field due to the –5 µC charge makes and angle –tan–1(0.4) = –21.8° with the x axis.

(a) 1. Find the magnitude of the two fields

2. Find the x and y components of the fields 3. Find E (b) F = –1.6×10–19E

E12 = 12×10–6k /8 E-5 = 5×10–6k /29 Ex12 = –1.06×10–6k Ex-5 = 1.6×10–7k Ey12 = –1.06×10–6k ; Ey–5 = –0.64×10–7k E = –8.09×103 N/C i –10.1×103 N/C j E = 12.9×103 N/C at ? = 231.3° F = 2.06×10–15 N at ? = 51.3°

32 ·· Two equal positive charges q are on the y axis, one at y = a and the other at y = –a. (a) Show that the

electric field on the x axis is along the x axis with Ex = 2kqx(x2 + a2)–3/2. (b) Show that near the origin, when x is much smaller than a, Ex is approximately 2kqx/a3. (c) Show that for values of x much larger than a, Ex is approximately 2kq/x2. Explain why you would expect this result even before calculating it.

(a) The distance between each charge and a point at (x, 0) is (a2 + x2)1/2. Thus, E due to each charge at that point is given by E = kq/(a2 + x2). By symmetry, the y components of the E fields cancel. The x component of E due to one charge is given by Ex = Ex/(a2 + x2)1/2. The total field at (x, 0) is therefore E = 2kqx/(a2 + x2)3/2 i.

(b) For x << a, (a2 + x2)3/2 ≅ a3, and Ex ≅ 2kqx/a3 (c) Similarly, for x >> a, Ex ≅ 2kq/x2. This is to be expected; for x >> a, the system looks like a single charge of

q. 33* ·· A 5-µC point charge is located at x = 1 m, y = 3 m, and a –4-µC point charge is located at x = 2 m, y = –2

m. (a) Find the magnitude and direction of the electric field at x = –3 m, y = 1 m. (b) Find the magnitude and direction of the force on a proton at x = –3 m, y = 1 m. The diagram shows the electric field vectors at the point of interest due to the two charges. Note that the E field due to the +5 µC charge makes an angle tan–

1(–0.5) = 206.6° with the x axis; the E field due to the –4 µC charge makes an angle tan–1 (–0.6) = –31°with the x axis


(a) 1. Find the magnitude of the two fields. 2. Find the x and y components of the fields 3. Find E (b) F = 1.6×10–19E

E5 = 5×10–6k /20 E–4 = 4×10–6k /34 Ex,5 = –0.224×10–6k , Ex,–4 = 0.101×10–6k Ey,5 = –0.112×10–6k ; Ey,–5 = –0.061×10–6k E = –1.11×103 N/C i – 1.55×103 N/C j E = 1.91×103 N/C at ? = 234.4o F = 3.06×10–16 N at ? = 234.4o

34 ·· (a) Show that the electric field for the charge distribution in Problem 32 has its greatest magnitude at the

points x = a/ 2 and x = a/ 2 by computing dEx/dx and setting the derivative equal to zero. (b) Sketch the function Ex versus x using your results for part (a) of this problem and parts (b) and (c) of Problem 32. (a) Take the derivative of Ex of Problem 32.

dEx/dx = 2kq/(a2 + x2)3/2 – 6kqx2/(a2 + x2)5/2 = 2kq(a2 – 2x2)/(a2 + x2)5/2. We see that dEx/dx = 0 when x = 2/a± . (b) A plot of Ex is shown

35 ··· For the charge distribution in Problem 32, the electric field at the origin is zero. A test charge q0 placed at

the origin will therefore be in equilibrium. (a) Discuss the stability of the equilibrium for a positive test charge by considering small displacements from equilibrium along the x axis and small displacements along the y axis. (b) Repeat part (a) for a negative test charge. (c) Find the magnitude and sign of a charge q0 that when placed at the origin results in a net force of zero on each of the three charges. (d) What will happen if any of the charges is displaced slightly from equilibrium?

(a) Since Ex is in the x direction, a positive test charge that is displaced from (0, 0) in the x direction will experience a force in the x diretion and accelerate in the x direction. Consequently, the equilibrium at (0, 0) is unstable for a small displacement along the x axis. If the positive test charge is displaced in the y direction, the charge at +a will exert a greater force than the charge at –a, and the net force is then in the –y direction; i.e., it is a restoring force. Consequently, the equilibrium at (0, 0) is stable for small displacements along the y direction.

(b) Following the same arguments as in part (a), one finds that, for a negative test charge, the equilibrium is stable at (0, 0) for displacements along the x direction and unstable for displacements along the y direction.

(c) Since the two +q charges repel, the charge Q at (0, 0) must be a negative charge. Since the force between charges varies as 1/r2, and the negative charge is midway between the two positive charges, Q = –q/4. (d) If the charge Q is displaced, the equilibrium is as discussed in part (b). If either of the +q charges are

displaced, the system is unstable. 36 ··· Two positive point charges +q are on the y axis at y = +a and y = –a as in Problem 32. A bead of mass m

carrying a negative charge –q slides without friction along a thread that runs along the x axis. (a) Show that for small displacements of x << a, the bead experiences a restoring force that is proportional to x and therefore


undergoes simple harmonic motion. (b) Find the period of the motion. (a) For x << a, (a2 + x2)3/2 ≅ a3, and Ex ≅ 2kqx/a3 (see Problem 32(b)). Since m carries a

negative charge, the force on m is directed opposite to the displacement in the x direction; i.e., it is a restoring force, proportional to x. For small displacements from equilibrium, the mass m will exhibit simple harmonic motion.

(b) Writing F = −k ′x, we see that k ′ = 2kq2/a3. The period of the oscillation is T = 2p/? = 2p 2

3

2kqma

37* · Which of the following statements about electric field lines is (are) not true? (a) The number of lines leaving a positive charge or entering a negative charge is proportional to the charge. (b) The lines begin on positive charges and end on negative charges. (c) The density of lines (the number per unit area perpendicular to the lines) is proportional to the magnitude of the field. (d) Electric field lines cross midway between charges that have equal magnitude and sign. (d) 38 · Figure 22-32 shows the electric field lines for a system of two point charges. (a) What are the relative

magnitudes of the charges? (b) What are the signs of the charges? (c) In what regions of space is the electric field strong? In what regions is it weak?

(a) There are 32 lines emanating from the positive charge and 8 lines terminating on the negative charge. The relative magnitudes of the charges are 4:1. (b) The charge on the left is positive; that on the right is negative. (c) The field is strong near the positive charge. It is weak to the right of the negative charge. 39 · Two charges +4q and –3q are separated by a small distance. Draw the electric field lines for this system.

A sketch of the electric field lines is shown. We assign 2 lines per charge q.

40 · Two charges +q and –3q are separated by a small distance. Draw the electric field lines for this system.


A sketch of the field lines is shown. We assign 2 lines per charge q.

41* · Three equal positive point charges are situated at the corners of an equilateral triangle. Sketch the electric

field lines in the plane of the triangle.

A sketch of the field lines is shown. Here we have assigned 7 field lines to each charge q.

42 · The acceleration of a particle in an electric field depends on the ratio of the charge to the mass of the

particle. (a) Compute e/m for an electron. (b) What is the magnitude and direction of the acceleration of an electron in a uniform electric field with a magnitude of 100 N/C? (c) When the speed of an electron approaches the speed of light c, relativistic mechanics must be used to calculate its motion, but at speeds significantly less than c, Newtonian mechanics applies. Using Newtonian mechanics, compute the time it takes for an electron placed at rest in an electric field with a magnitude of 100 N/C to reach a speed of 0.01c. (d) How far does the electron travel in that time?

(a) See textbook endpaper (b) a = F/m = –eE/m (c) t = v/a (d) x = vavt

e/m = (1.6×10–19/9.11×10–31) C/kg = 1.76×1011 C/kg a = 1.76×1013 m/s2, directed opposite to E. t = (3×106/1.76×1013) s = 1.71×10–7 s = 0.171 µs x = 1.5×106×1.71×10–7 m = 0.257 m = 25.7 cm

43 · (a) Compute e/m for a proton, and find its acceleration in a uniform electric field with a magnitude of 100


N/C. (b) Find the time it takes for a proton initially at rest in such a field to reach a speed of 0.01c (where c is the speed of light).

(a), (b) Proceed as in Problem 42. One obtains e/mp = 9.58×107 C/kg; a = 9.58×109 m/s2, in the direction of E; t = 3.13 × 10–4 s = 313 µs. 44 · An electron has an initial velocity of 2×106 m/s in the x direction. It enters a uniform electric field E =

(400 N/C)j, which is in the y direction. (a) Find the acceleration of the electron. (b) How long does it take for the electron to travel 10 cm in the x direction in the field? (c) By how much and in what direction is the electron deflected after traveling 10 cm in the x direction in the field?

(a) a = –eE/m (b) t = x/vx (c) y = 1/2ayt2

a = (–1.76×1011 × 400) m/s2 j = –7.04×1013 m/s2 j t = 0.1/2×106 s = 0.05 µs y = (–7.04×1013 × 25×10–16/2) m = –8.8 cm

45* ·· An electron, starting from rest, is accelerated by a uniform electric field of 8×104 N/C that extends over a

distance of 5.0 cm. Find the speed of the electron after it leaves the region of uniform electric field. v2 = 2ax; a = eE/m; meExv /2= 05.01076.11082 114 ×××××=v m/s = 3.75×107 m/s 46 ·· An electron moves in a circular orbit about a stationary proton. The centripetal force is provided by the

electrostatic force of attraction between the proton and the electron. The electron has a kinetic energy of 2.18×10–18 J. (a) What is the speed of the electron? (b) What is the radius of the orbit of the electron? (a) mKv /2= (b) mv2/r = ke2/r2; r = ke2/2K

3118 1011.9/1018.22 −− ×××=v m/s = 2.19×106 m/s r = 8.99×109 × (1.6×10–19)2/(2×2.18×10–18) m = 5.28×10–11 m = 52.8 pm

47 ·· A mass of 2 g located in a region of uniform electric field E = (300 N/C)i carries a charge Q. The mass,

released from rest at x = 0, has a kinetic energy of 0.12 J at x = 0.50 m. Determine the charge Q . K = QEx Q = 0.12/150 C = 8×10–4 C = 800 µC 48 ·· A particle leaves the origin with a speed of 3×106 m/s at 35° to the x axis. It moves in a constant electric

field E = Ey j. Find Ey such that the particle will cross the x axis at x = 1.5 cm if the particle is (a) an electron, and (b) a proton.

(a) 1. Write expressions for x and y 2. Set y = 0 and solve for t ≠ 0 3. Solve for Ey and evaluate (b) For the proton, change sign of e and mass m

x = (v cos ?)t; y = (v sin ?)t – 1/2(eEy/m)t2 t = (2mv sin ?)/eEy = x/(v cos ?) Ey = (2mv2 sin ? cos ?)/ex = 3.21×103 N/C Ey = –5.88×106 N/C

49* ·· An electron starts at the position shown in Figure 22-33 with an initial speed v0 = 5×106 m/s at 45° to the x

axis. The electric field is in the positive y direction and has a magnitude of 3.5×103 N/C. On which plate and at what location will the electron strike?

1. Note that a = eE/m downward. Use Equ. 3-22 2. Find ymax; ymax = m(v0 sin ?)2/2eEy

x = (mv02/eE)sin 2? = 4.07 cm on the lower plate

ymax = 1.02 cm; electron does not hit the upper plate


50 ·· An electron with kinetic energy of 2×10–16 J is moving to the right along the axis of a cathode-ray tube as shown in Figure 22-34. There is an electric field E = (2×104 N/C)j in the region between the deflection plates. Everywhere else, E = 0. (a) How far is the electron from the axis of the tube when it reaches the end of the plates? (b) At what angle is the electron moving with respect to the axis? (c) At what distance from the axis will the electron strike the fluorescent screen?

(a) 1. Find the time between plates 2. Find y(t) = 1/2ayt2 = 1/2(–eEy/m)t2 (b) ? = tan–1(vy/v0) = tan–1(–eEyx/2K) (c) For stright line motion, y/x = vy/vx

t = x/v0 = x/(m/2K)1/2 y = –eEyx2/4K = –1.6×10–19 × 2×104 ×16×10–4/4×2×10–16 = –6.4×10–3 m = –6.4 mm ? = tan–1(–0.32) = –17.7o y = –12×0.32 cm = –3.84; ytotal = –(3.86 + 0.64) cm = –4.48 cm

51 · Two point charges, q1 = 2.0 pC and q2 = –2.0 pC, are separated by 4 µm. (a) What is the dipole moment of

this pair of charges? (b) Sketch the pair, and show the direction of the dipole moment.

(a) p = qa = 2×10–12×4×10–6 C.m = 8×10–18 C.m (b) The dipole moment is shown in the figure.

52 · A dipole of moment 0.5 e⋅nm is placed in a uniform electric field with a magnitude of 4.0×104 N/C. What is

the magnitude of the torque on the dipole when (a) the dipole is parallel to the electric field, (b) the dipole is perpendicular to the electric field, and (c) the dipole makes an angle of 30° with the electric field? (d) Find the potential energy of the dipole in the electric field for each case.

(a), (b), (c) t = p × E; t = pE sin ? (d) U = –p .E = –pE cos ?

? = 0°, t = 0; ? = 90°, t = 3.2×10–24 N.m; ? = 30°, t = 1.6×10–24 N.m ? = 0°, U = –3.2×10–24 J; ? = 90o, U = 0; ? = 30°, U = –2.77×10–24 J

53* ·· For a dipole oriented along the x axis, the electric field falls off as 1/x3 in the x direction and 1/y3 in the y

direction. Use dimensional analysis to prove that, in any direction, the field far from the dipole falls off as 1/r3. Dimensionally, we can write [E] = [kQ]/[L]2 and [p] = [Q][L], where p represents the dipole. Thus the dimension of charge [Q] is [p]/[L], and the electric field has the dimension [kp]/[L]3. This shows that the field E due to a dipole p falls off as 1/r3.

54 ·· A water molecule has its oxygen atom at the origin, one hydrogen nucleus at x = 0.077 nm, y = 0.058 nm and the other hydrogen nucleus at x = –0.077 nm, y = 0.058 nm. If the hydrogen electrons are transferred completely to the oxygen atom so that it has a charge of –2e, what is the dipole moment of the water molecule?


(Note that this characterization of the chemical bonds of water as totally ionic is simply an approximation that overestimates the dipole moment of a water molecule.) From the symmetry of the system, it is evident that the x component of the dipole moment is zero. The y component is 2e×0.058×10–9 C.m = 1.86×10–29 C.m.

55 ·· An electric dipole consists of two charges +q and –q separated by a very small distance 2a. Its center is on the x axis at x = x1, and it points along the x axis in the positive x direction. The dipole is in a nonuniform electric field, which is also in the x direction, given by E = Cxi, where C is a constant. (a) Find the force on the positive charge and that on the negative charge, and show that the net force on the dipole is Cpi. (b) Show that, in general, if a dipole of moment p lies along the x axis in an electric field in the x direction, the net force on the dipole is given approximately by (dEx/dx)pi.

(a) The force on the negative charge is F–q = –Cq(x1 – a) i, that on the positive charge is Fq = Cq(x1 + a) i. The net force is F = Fq + F–q = 2Cqa i = Cp, where p = 2qa i.

(b) Fx = –dU/dx; U = –pxEx. So Fx = px(dEx/dx). 56 ··· A positive point charge +Q is at the origin, and a dipole of moment p is a distance r away and in the radial

direction as in Figure 22-26. (a) Show that the force exerted by the electric field of the point charge on the dipole is attractive and has a magnitude of approximately 2kQp/r3 (see Problem 55). (b) Now assume that the dipole is centered at the origin and that a point charge Q is a distance r away along the line of the dipole. From your result for part (a) and Newton's third law, show that the magnitude of the electric field of the dipole along the line of the dipole a distance r away is approximately 2kp/r3.

(a) Let p = p r, where r is the unit vector in the radial direction. Then U = –pEr and F = –dU/dr = p(dEr/dr). The field Er = kQ/r2 and dEr/dr = –2kQ/r3. So the force on the dipole is F = –2kQp/r3 r.

(b) By Newton’s third law, the force on the charge Q is equal and opposite to the force on the dipole. But the force on a charge Q is given by EQ. From the result of part (a) it follows that the magnitude of the field is E = 2kp/r3.

57* ··· A quadrupole consists of two dipoles that are close together, as shown in Figure 22-35. The effective charge at the origin is –2q and the other charges on the y axis at y = a and y = –a are each +q. (a) Find the electric field at a point on the x axis far away so that x >> a. (b) Find the electric field on the y axis far away so that y >> a.

(a) We have, in effect, three charges: +q at (0, a), +q at (0, –a), and –2q at (0, 0). From the symmetry of the system it is evident that the field E along the x axis has no y component. The x component of E due to one of the charges +q is

2/3222222 )( axkqx

ax

xax

kqE qx +

=++

=+ . For the –2q charge, 222x

kqE qx

−=− . The total

field along the x axis is Ex = 2E+qx + E–2qx. For x >> a, (x2 + a2)–3/2 ≅ (1 – 3a2/2x2)/x3, and Ex = –3kqa2/x4. (b) Along the y axis, Ex = 0 by symmetry. Ey = kq/(y – a)2 + kq/(y + a)2 – 2kq/y2. Again using the binomial expansion one finds that for y >> a, Ey = 6kqa2/y4. 58 · A charged insulator and an uncharged metal (a) always repel one another. (b) exert no electrostatic force on one another. (c) always attract one another. (d) may attract or repel, depending on the sign of the charge on the insulator.


(c) 59 · Which of the following statements are true? (a) A positive charge experiences an attractive electrostatic force toward a nearby neutral conductor. (b) A positive charge experiences no electrostatic force near a neutral conductor. (c) A positive charge experiences a repulsive force, away from a nearby conductor. (d) Whatever the force on a positive charge near a neutral conductor, the force on a negative charge is then

oppositely directed. (e) None of the above is correct. (a) 60 · The electric field lines around an electrical dipole are best represented by which, if any, of the diagrams in

Figure 22-36? (d) 61* ·· A molecule with electric dipole moment p is oriented so that p makes an angle ? with a uniform electric field

E. The dipole is free to move in response to the force from the field. Describe the motion of the dipole. Suppose the electric field is nonuniform and is larger in the x direction. How will the motion be changed?

The dipole experiences a torque t = pE sin ?. In a uniform electric field, it will oscillate about its equilibrium orientation, ? = 0. If the field is nonuniform and dE/dx > 0, the dipole will accelerate in the x direction as it oscillates about ? = 0. 62 ·· True or false: (a) The electric field of a point charge always points away from the charge. (b) All macroscopic charges Q can be written as Q = ±Ne, where N is an integer and e is the charge of the electron. (c) Electric field lines never diverge from a point in space. (d) Electric field lines never cross at a point in space. (e) All molecules have electric dipole moments in the presence of an external electric field. (a) False (b) True (c) False, if that point does not carry a charge; True, if it carries a positive charge (d) True (e) True 63 ·· A small, nonconducting ball with no net charge is suspended from a thread. When a positive charge is

brought near the ball, the ball is attracted toward the charge. How does this come about? How would the situation be different if the charge brought near the ball were negative instead of positive?

The charge induces a dipole in the nonconducting ball. Since the field of the charge is nonuniform, the dipole is attracted to the charge. The same effect is observed for either sign of the charge. 64 ·· Two metal balls have charges +q and –q. How will the force on one of them change if (a) the balls are

placed in water, the distance between them being unchanged, and (b) a third uncharged metal ball is placed between the first two? Explain.

(a) The force between the balls is diminished because the field produced by the two charges creates a dipolar field that opposes that of the two charges when they are out of the water (see Section 25-5).

(b) The force is again reduced because a dipole is induced on the third metal ball. 65* ·· A metal ball is positively charged. Is it possible for it to attract another positively charged ball? Explain. Yes. A positively charged ball will induce a dipole on the metal ball, and if the two are in close proximity, the net force can be attractive.


66 · In interstellar space, two charged point-like objects, each of mass m and charge q, are separated by a distance d and released. They remain motionless at that separation. Find an expression for q in terms of m, G, and k.

F = kq2/d2 - Gm2/d2 = 0. Solve for q

kG

mq =

67 ·· Point charges of –5.0 µC, +3.0 µC, and +5.0 µC are located along the x axis at x = –1.0 cm, x = 0, and x =

+1.0 cm, respectively. Calculate the electric field at x = 3.0 cm and at x = 15.0 cm. Is there some point on the x axis where the magnitude of the electric field is zero? Locate that point. The location of the three charges and points of interest are shown in the diagram. From the diagram it is evident that E along the x axis has no y component.

1. Find Ex at x = 3 cm; use Equ. 22-8 2. Find Ex at x = 15 cm; use Equ. 22-8

3. There are two points where E = 0. One is between x = 0 and x = 1 cm, one is at x < –1 cm.

4. For x < –1 cm, let y = –x. Set E = 0

Ex = (–5/16 + 3/9 + 5/4)(8.99×107) N/C = 1.14×108 N/C Ex = (–5/256 + 3/225 + 5/196)(8.99×107) N/C = 1.74×106 N/C For 0 < x < 1, E = 0 if 3/x2 – 5/(1+x)2 – 5/(1–x)2 = 0. 7x4 + 16x2 – 3 = 0; x = 0.417 cm. 5/(y–1)2 – 3/y2 – 5/(y+1)2 = 0; 20y3 = 3(y2 – 1)2. Solve for y numerically: y = 6.95 cm; x = –6.95 cm

68 ·· For the charge distribution of Problem 67, find the electric field at x = 15.0 cm as the vector sum of the

electric field due to a dipole formed by the two 5.0-µC charges and a point charge of 3.0 µC, both located at the origin. Compare your result with the result obtained in Problem 67 and explain any difference between these two.

For the two 5 µC charges, p = 10–8 C.m. 1. Find Ex at x = 15 cm, using Epx = 2kp/x3

Ex = k(2×10–8/0.15 + 3×10–6)/0.0225 = 1.25×106 N/C

This result is only a rough approximation because the separation between the two charges of the dipole is more than 10% of the distance to the point of interest, i.e., x is not much greater than a. The correct result is that of Problem 67, namely Ex = 1.74×106 N/C.

69* ·· In copper, about one electron per atom is free to move about. A copper penny has a mass of 3 g. (a) What percentage of the free charge would have to be removed to give the penny a charge of 15 µC? (b) What would be the force of repulsion between two pennies carrying this charge if they were 25 cm apart? Assume that the pennies are point charges.

(a) Find the number of free electrons, N = Na Find ne for a charge q = –15 µC Fraction to be removed = ne/N (b) Use Equ. 22-2

From Example 22-1, N = 2.84×1022 ne = 15×10–6/e ne/N = 15×10–6/(2.84×1022 × 1.6×10–19) = 3.3×10–7 % F = 225×10–12 × 8.99×109/0.0625 N = 32.4 N


70 ·· Two charges q1 and q2 have a total charge of 6 µC. When they are separated by 3 m, the force exerted by

one charge on the other has a magnitude of 8 mN. Find q1 and q2 if (a) both are positive so that they repel each other, and (b) one is positive and the other is negative so that they attract each other.

(a) Given: q1 + q2 = 6 µC; kq1q2/9 = 8×10–3 N Solve quadratic equation for q1 (b) Now q1 – q2 = 6 µC; proceed as in (a) Solve for q1 and q2

q2 = 6×10–6 – q1; 8.99×109q1(6×10–6 – q1)/9 = 8×10–3

q12 – 6×10–6q1 + 9 × 8×10–3/8.99×109 = 0

q1 = 4 µC, q2 = 2 µC; or q1 = 2 µC, q2 = 4 µC q1

2 – 6×10–6q1 – 9 × 8×10–3/8.99×109 = 0 q1 = 7.12 µC, q2 = –1.12 µC

71 ·· Three charges, +q, +2q, and +4q, are connected by strings as shown in Figure 22-37. Find the tensions T1

and T2. 1. Find T2 = sum of Coulomb forces on +4q 2. Find T1 = sum of Coulomb forces on +q

T2 = k(8q2/d2 + 4q2/4d2) = 9kq2/d2 T1 = k(2q2/d2 + 4q2/4d2) = 3kq2/d2

72 ·· A positive charge Q is to be divided into two positive charges q1 and q2. Show that, for a given separation D,

the force exerted by one charge on the other is greatest if q1 = q2 = Q21

.

Write F = kq1q2/D2 = kq1(Q – q1)/D2. Take the derivative of q1(Q – q1) with respect to q1 and set it equal to zero to determine the value of q1 for which F is a maximum: Q – 2q1 = 0, or q1 = 1/2Q = q2.

73* ·· A charge Q is located at x = 0 and a charge 4Q is at x = 12.0 cm. The force on a charge of –2 µC is zero if that charge is placed at x = 4.0 cm and is 126.4 N in the positive x direction if placed at x = 8.0 cm. Determine the charge Q.

1. Write F on the –2 µC charge when at x = 4 cm 2. Solve for Q

126.4 N = Q×2×10–6 × 8.99×109(4/16×10–4 – 1/64×10–4) Q = 3 µC

74 ·· Two small spheres (point charges) separated by 0.60 m carry a total charge of 200 µC. (a) If the two

spheres repel each other with a force of 80 N, what are the charges on each of the two spheres? (b) If the two spheres attract each other with a force of 80 N what are the charges on the two spheres? Except for a change in the data, this problem is identical to Problem 70. Using the same procedure as in Problem 70 one finds the following results; (a) q1 = 17.5 µC, q2 = 182.5 µC. (b) q1 = –15 µC, q2 = 215 µC.

75 ·· A ball of known charge q and unknown mass m, initially at rest, falls freely from a height h in a uniform electric field E that is directed vertically downward. The ball hits the ground at a speed ghv 2= . Find m in

terms of E, q, and g. 1. Use conservation of energy 2. Solve for m

1/2mv2 = 2mgh = mgh + qEh m = qE/g

76 ·· Charges of 3.0 µC are located at x = 0, y = 2.0 m and at x = 0, y = –2.0 m. Charges Q are located at x =

4.0 m, y = 2.0 m and at x = 4.0 m, y = –2.0 m (Figure 22-38). The electric field at x = 0, y = 0 is (4.0×103 N/C)i. Determine Q.

Note that the electric field due to the two 3 µC charges at (0, 0) is zero.


1. Write Ex due to the charges Q 2. Solve for Q with x = 4, y2 = 4

Ex = –2kQx/(x2 + y2)3/2 = 4×103 N/C Q = –4.97 µC

77* ·· Two identical small spherical conductors (point charges), separated by 0.60 m, carry a total charge of 200

µC. They repel one another with a force of 120 N. (a) Find the charge on each sphere. (b) The two spheres are placed in electrical contact and then separated so that each carries 100 µC. Determine the force exerted by one sphere on the other when they are 0.60 m apart.

(a) Given: q1 + q2 = 200 µC; kq1q2/0.36 =120 N Solve quadratic equation for q1 (b) Now q1 = q2 = 100 µC; find F

q2 = 2×10–4 – q1; 8.99×109q1(2×10–4 – q1)/0.36 = 120 q1

2 – 2×10–4q1 + 43.2 /8.99×109 = 0 q1 = 28 µC, q2 = 172 µC; or q1 = 172 µC, q2 = 28 µC F = 8.99×109 × 10–8/0.36 N = 250 N

78 ·· Repeat Problem 77 if the two spheres initially attract one another with a force of 120 N. (a) The problem is identical to Problem 70(b) except that the data are different. Following the same procedure,

one obtains q1 = –21.7 µC, q2 = 221.7 µC. (b) Since the final system configuration is the same as in Problem 77(b) the result is the same, i.e., F = 250 N 79 ·· A charge of –3.0 µC is located at the origin; a charge of 4.0 µC is located at x = 0.2 m, y = 0; a third charge

Q is located at x = 0.32 m, y = 0. The force on the 4.0-µC charge is 240 N, directed in the positive x direction. (a) Determine the charge Q. (b) With this configuration of three charges, where, along the x direction, is the electric field zero?

The charge configuration is shown in the diagram. Here we also indicate the approximate locations, labeled x1 and x2, where the electric field is zero.

(a) 1. Write the force on charge +4 µC 2. Solve for Q (b) By inspection, the points where E = 0 must be between the –3 µC and +4 µC charges. Write the condition for E = 0 and solve numerically for x.

240 = k(Q × 4×10–6/0.122 – 12×10–12/0.22) Q = –97.2 µC

03

)2.0(4

)32.0(2.97

222=−

−−

− xxx

x1 = 0.0508 m, x2 = 0.169 m 80 ·· Two small spheres of mass m are suspended from a common point by threads of length L. When each

sphere carries a charge q, each thread makes an angle ? with the vertical as shown in Figure 22-39. (a) Show that the charge q is given by

k

mgLq

θθ tansin2=

where k is the Coulomb constant. (b) Find q if m = 10 g, L = 50 cm, and ? = 10°. (a) The forces acting on one sphere are mg acting downward, FE = kq2/(2L sin ?)2 acting horizontally, and the tension T in the string. The angle ? is therefore given by tan ? = FE/mg = kq2/(2L sin ?)2mg. Solving for q, one obtains the result given in the problem statement.


(b) Substitute m = 0.010 kg, L = 0.5 m, ? = 10°, q = 0.241 µC. 81* ·· (a) Suppose that in Problem 80, L = 1.5 m, m = 0.01 kg, and q = 0.75 µC. What is the angle that each string

makes with the vertical? (b) Find the angle that each string makes with the vertical if one mass carries a charge of 0.50 µC, the other a charge of 1.0 µC.

(a) 1. Use the expression given in Problem 80 2. Since sin2? tan? << 1, sin? ≅ tan? ≅ ? Solve for ? (b) Repeat part (a) replacing q2 by q1q2

sin2? tan? = 81.901.025.24

)1075.0(1099.84

269

2

2

××××××

=−

mgLkq

= 5.73×10–3 ? 3 = 5.73×10–3; ? = 0.179 rad = 10.25o ? = 9.86o

82 ·· Four charges of equal magnitude are arranged at the corners of a square of side L as shown in Figure 22-40.

(a) Find the magnitude and direction of the force exerted on the charge in the lower left corner by the other charges. (b) Show that the electric field at the midpoint of one of the sides of the square is directed along that side toward the negative charge and has a magnitude E given by

−=

255

18

2Lq

kE

Let the origin be at the lower left-hand corner (a) Find the forces acting on the charge F = F-q,+q + F+q,+q (b) Consider the midpoint along the y axis; write E(0, L/2); note that the x components due to the charges at (0, L) and (L, L) cancel

F–q,+q = kq2/L2 (i + j); F+q,+q = kq2/2 2 L2(–i –j) F = (kq2/L2)(1 – 1/2 2 )(i + j) E+q = E–q = 4kq/ L2 j for charges along y axis E–qy = E+qy = –kq/(5L2/4)(1/ 5 ) for charges at x = L.

Ey = (8kq/L2)(1 – 5 /25) 83 ·· Figure 22-41 shows a dumbbell consisting of two identical masses m attached to the ends of a thin

(massless) rod of length a that is pivoted at its center. The masses carry charges of +q and –q and the system is located in a uniform electric field E. Show that for small values of the angle ? between the direction of the dipole and the electric field, the system displays simple harmonic motion, and obtain an expression for the period of that motion.

Note that the torque acting on the system is a restoring torque. For ? << 1, t = –pE sin ? ≅ –pE? = –qaE?. Apply Equs. 9-20, 9-3, and 9-17. –qaE? = (ma2/2)(d 2?/dt2), d 2?/dt2 = –(2qE/ma)?. This is the differential equation for a harmonic oscillator with angular frequency ? = (2qE/ma)1/2 and period T = 2p(ma/2qE)1/2. 84 ·· For the dumbbell in Figure 22-41, let m = 0.02 kg, a = 0.3 m, and E = (600 N/C)i. Initially the dumbbell is at

rest and makes an angle of 60° with the x axis. The dumbbell is then released, and when it is momentarily aligned with the electric field, its kinetic energy is 5×10–3 J. Determine the magnitude of q.

1. Write the change in potential energy, ?U 2. Use energy conservation; K + ?U = 0

?U = pE(cos 60° – 1) = –q(0.3 × 600 × 0.5) 90q = 5×10–3 ; q = 55.6 µC

85* ·· An electron (charge –e, mass m) and a positron (charge +e, mass m) revolve around their common center of

mass under the influence of their attractive coulomb force. Find the speed of each particle v in terms of e, m, k , and their separation r.


The force on each particle is ke2/r2. The centripetal acceleration of each particle is v2/(r/2). Using F = ma one obtains v = (ke2/2mr)1/2. 86 ·· The equilibrium separation between the nuclei of the ionic molecule KBr is 0.282 nm. The masses of the two

ions, K+ and Br –, are very nearly the same, 1.4×10–25 kg, and each of the two ions carries a charge of magnitude e. Use the result of Problem 83 to determine the frequency of oscillation of a KBr molecule in a uniform electric field of 1000 N/C.

From Problem 83, f =maqE2

21π . Inserting the values q = 1.6×10–19 C, m = 1.4×10–25 kg, and a = 2.82×10–10 m,

one obtains f = 4.53×108 Hz. 87 ··· A small (point) mass m, which carries a charge q, is constrained to move vertically inside a narrow,

frictionless cylinder (Figure 22-42). At the bottom of the cylinder is a point mass of charge Q having the same sign as q. (a) Show that the mass m will be in equilibrium at a height y0 = (kqQ/mg)1/2. (b) Show that if the mass m is displaced by a small amount from its equilibrium position and released, it will exhibit simple harmonic motion with angular frequency ? = (2g/y0)1/2.

(a) Use SF = 0 and solve for y0 mg = kqQ/y02;

mgkqQ

y =0

(b) Let y = y0 + ?y; then 040

200

20

/22

2yymg

yykqQ

ykqQ

yyykqQ

F ∆−=∆

−=−∆+

= to lowest order in ?y.

From F = ma we now have (d2?y/dt2) = –2g?y/y0. This is the SHO differential equation, and comparing it with the expression for ?, Equ. 14-8, we find ? = 0/2 yg .

88 ··· A small bead of mass m and carrying a negative charge –q is constrained to move along a thin frictionless rod (Figure 22-43). A distance L from this rod is a positive charge Q. Show that if the bead is displaced a distance x, where x << L, and released, it will exhibit simple harmonic motion. Obtain an expression for the period of this motion in terms of the parameters L, Q, q, and m.

The x component of the force on m is 32/322 )( LkqQx

xLkqQx

Fx−

≅+

−= for x << L. F=ma =m(d2x/dt2) yields the

SHO equation. Comparison with Equs. 14-2 and 14-12 gives T = 2pLkqQmL

.

89* ··· Repeat Problem 81 with the system located in a uniform electric field of 1.0×105 N/C that points vertically

downward. (a) Note that if the two charges are equal, each mass experiences an equal downward force of qE in addition

to its weight mg. Thus, we may use the expression in Problem 80 provided we replace mg by (mg + qE). As derived

in Problem 81, sin2? tan? = )075.081.901.0(25.24

)1075.0(1099.8)(4

269

2

2

+×××××

=+

−

qEmgLkq

= 3.25×10–3 and ? = 8.48°.

(b) The downward forces on the two masses are not equal. Let the mass carrying the charge of 0.5 µC be m1,

and that carying the charge of 1.0 µC be m2. Since we already know from part (a) that the angles are small, we


shall make the small angle approximation sin? = tan? = ?.

1. Write the horizontal and vertical forces on m1 due to g, the charges q1 and q2, and tension T

2. T2x and T2y are similar except for the subscripts 3. ?1 = T1x/T1y; ?2 = T2x/T2y; find ?1/?2 4. Write the expression for ?1 + ?2 5. Solve for ?1 + ?2 6. Substitute numerical values for m1 = m2 = m to determine ?1 + ?2, ?1/?2, and ?1 and ?2

yyyx TEqgmFTL

qkqF 111112

12

211 ;

)( 2

=+==+

=θθ

yyxx TEqgmFTL

qkqF 222222

12

212 ;

)( 2

=+==+

=θθ

?1/?2 = (m2g + q2E)/(m1g + q1E)

+

+++

=+EqgmEqgmL

qkq

22112

212

2121

11)( θθ

θθ

3/1

22112

2121

11

+

++

=+EqgmEqgmL

qkqθθ

?1 + ?2 = 0.287 rad = 16.4°; ?1/?2 = 1.34 ?1 = 9.4°, ?2 = 7.0°.

90 ··· Suppose that the two masses in Problem 80 are not equal. One mass is 0.01 kg, the other is 0.02 kg. The

charges on the two masses are 2.0 µC and 1.0 µC, respectively. Determine the angle that each of the strings supporting the masses makes with the vertical.

See Problem 89 for the general solution. Substitute the numerical values for m1, m2, q1, q2, and L into the expressions for ?1 + ?2 and ?1/?2 given in Problem 89, setting E = 0. One obtains ?1 + ?2 = 0.496 rad = 28.4°, ?1/?2 = 1/2, and ?1 = 9.47°, ?2 = 18.9°. ( Note: The small angle approximation is not as good here as in the preceding problems; however, the error introduced is less than 3%.) 91 ··· A simple pendulum of length L = 1.0 m and mass M = 5.0×10–3 kg is placed in a uniform, vertically directed

electric field E. The bob carries a charge of –8.0 µC. The period of the pendulum is 1.2 s. What is the magnitude and direction of E?

1. Write the force on the mass M due to g and E 2. Use Equ. 14-27, replacing g by g′ 3. Solve for E with q = –8 µC

F = Mg + qE = M(g + qE/M) = Mg′ g′ = 4π2L/T = 27.4 m/s2; qE/M = 17.6 N/kg downward E = 1.1×104 N/C upward

92 ··· Two neutral polar molecules attract each other. Suppose that each molecule has a dipole moment p and that

these dipoles are aligned along the x axis and separated by a distance d. Derive an expression for the force of attraction in terms of p and d. The potential energy of the dipole p1 is U1 = –p1E1, where E1 is the field at p1 due to p2. E1 = 2kp2/x3, where x is the separation between the two dipoles. So U1 = –2kp1p2/x3. F = –dU/dx = 6kp1p2/x4. For p1 = p2 = p and x = d we have F = 6kp2/d4.

93* ··· A small bead of mass m, carrying a charge q, is constrained to slide along a thin rod of length L. Charges Q are fixed at each end of the rod (Figure 22-44). (a) Obtain an expression for the electric field due to the two charges Q as a function of x, where x is the distance from the midpoint of the rod. (b) Show that for x << L, the magnitude of the field is proportional to x. (c) Show that if q is of the same sign as Q, the force that acts on the object of mass m is always directed toward the center of the rod and is proportional to x. (d) Find the period of


oscillation of the mass m if it is displaced by a small distance from the center of the rod and then released. (a) Write the expression for Ex (b) For x << L, neglect x in denominator of (a) (c) Fx = qEx (d) d2x/dt2 = –(16kQq/mL3)x; use Equs. 14-8, 14-12

Ex = kQ/(1/2L + x)2 – kQ/(1/2L – x)2 Ex = –32kQx/L3 Fx = –32kQqx/L3; note that Fx is proportional to –x.

T = (p/2) kQqmL 2/3

94 ··· Two equal positive charges Q are on the x axis at x = L21

and x = L21− (a) Obtain an expression for the

electric field as a function of y on the y axis. (b) A ring of mass m, which carries a charge q, moves on a thin frictionless rod along the y axis. Find the force that acts on the charge q as a function of y; determine the sign of q such that this force always points toward y = 0. (c) Show that for small values of y the ring exhibits simple harmonic motion. (d) If Q = 5 µC, |q| = 2 µC, L = 24 cm, and m = 0.03 kg, what is the frequency of the oscillation for small amplitudes?

(a) Find Ey; see Problem 32(b) (b) Fy = qEy; to point in -y direction, the charge q must be negative. (c) Write equation of motion for y << a = L/2

(d) Use Equs. 14-8 and 14-11 with numerical values

2/322 )(2

yakQy

Ey+

= , where a = L/2.

2/322 )(2

yakqQy

Fy+

=

ymLkQq

dtyd

32

2 16−= ; this is the SHO equation of motion

f = 1.1/L3/2 = 9.36 Hz

ch22

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