Section 1 Electric Charge Chapter 16 Properties of ... · PDF fileSection 1 Electric Charge Chapter 16 Properties of ... ... 23

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Chapter 16Section 1 Electric Charge

Properties of Electric Charge

• electric charge is + or -.

– like charges repel

– unlike charges attract

• Electric charge is conserved.

• Atomic Charges

– Protons (+)charged particles.

– neutronsUncharged particles.

– Electrons (-) charged particles.



Chapter 16

Electric Charge

Section 1 Electric Charge




Properties of Electric Charge, continued

• Charge is measured in coulombs (C).

• fundamental unit of charge= e, charge

of a single electron or proton.

e = -1.602 176 x 10–19 C



Chapter 16

The Milikan Experiment




Chapter 16

Milikan’s Oil Drop Experiment





Transfer of Electric Charge

• An electrical conductor is a material in which

charges can move freely.

• An electrical insulator is a material in which charges

cannot move freely. Charge will remain where it was

placed.

• Ground- often the ground reservoir of electrons, can

donate or absorb.




Transfer of Electric Charge, continued

• Insulators and conductors can be charged by contact.

• Conductors can be charged by induction.

• Induction charging a conductor by bringing it near

another charged object and grounding the conductor.



Chapter 16

Charging by Induction





Transfer of Electric Charge, continued

• A surface charge can be

induced on insulators by

polarization.

• With polarization, the

charges within individual

molecules are realigned

such that the molecule

has a slight charge

separation.



Section 2 Electric Force

Chapter 16

Objectives

• Calculate electric force using Coulomb’s law.

• Compare electric force with gravitational force.

• Apply the superposition principle to find the resultant

force on a charge and to find the position at which the

net force on a charge is zero.



Coulomb’s Law, continued

• The electrical force is a field force.

• A field force = force exerted by one object on another

with no physical contact between them

Chapter 16Section 2 Electric Force



Chapter 16

Coulomb’s Law

• Two charges near one another exert a force on one

another called the electric force.

• Kc= 8.99 E 9 n m2/C2

• Coulomb’s law


1 2

2

2

charge 1 charge 2electric force = Coulomb constant

distance

electric C

q qF k

r



Chapter 16

Coulomb’s Law, continued

• The resultant force on a charge is the vector sum of

the individual forces on that charge.

• equilibrium, the net external force acting on that body

is zero.




Chapter 16

Superposition Principle




Chapter 16

Sample Problem

The Superposition Principle

Consider three point

charges at the corners of a

triangle, as shown at right,

where q1 = 6.00 10–9 C,

q2 = –2.00 10–9 C, and

q3 = 5.00 10–9 C. Find

the magnitude and

direction of the resultant

force on q3.





Sample Problem, continued


1. Define the problem, and identify the known variables.

Given:

q1 = +6.00 10–9 C r2,1 = 3.00 m

q2 = –2.00 10–9 C r3,2 = 4.00 m

q3 = +5.00 10–9 C r3,1 = 5.00 m

q = 37.0º

Unknown: F3,tot = ? Diagram:






3. Calculate the magnitudes of the forces with Coulomb’s law.

–9 –9293 1

3,1 22 2

–8

3,1

–9 –9293 2

3,2 22 2

–9

3,1

5.00 10 C 6.00 10 CN m 8.99 10

( 3,1) C 5.00 m

1.08 10 N

5.00 10 C 2.00 10 CN m8.99 10

( 3,2) C 4.00m

5.62 10 N

C

C

q qF k

r

F

q qF k

r

F






4. Find the x and y components of each force.

At this point, the direction each component must be

taken into account.

F3,1: Fx = (F3,1)(cos 37.0º) = (1.08 10–8 N)(cos 37.0º)

Fx = 8.63 10–9 N

Fy = (F3,1)(sin 37.0º) = (1.08 10–8 N)(sin 37.0º)

Fy = 6.50 10–9 N

F3,2: Fx = –F3,2 = –5.62 10–9 N

Fy = 0 N






5. Calculate the magnitude of the total force acting

in both directions.

Fx,tot = 8.63 10–9 N – 5.62 10–9 N = 3.01 10–9 N

Fy,tot = 6.50 10–9 N + 0 N = 6.50 10–9 N






6. Use the Pythagorean theorem to find the magni-

tude of the resultant force.

2 2 9 2 9 2

3, , ,

–9

3,

( ) ( ) (3.01 10 N) (6.50 10 N)

7.16 10 N

tot x tot y tot

tot

F F F

F






7. Use a suitable trigonometric function to find the

direction of the resultant force.

In this case, you can use the inverse tangent function:

–9,

–9

,

6.50 10 Ntan

3.01 10 N

65.2º

y tot

x tot

F

F



Section 3 The Electric Field

Chapter 16

Objectives

• Calculate electric field strength.

• Draw and interpret electric field lines.

• Identify the four properties associated with a

conductor in electrostatic equilibrium.



Chapter 16Section 3 The Electric Field

Electric Field Strength

• An electric field = region where an electric force on

a test charge would be detected.

• Test charge= + charge with no magnitude

• units of electric field, E, (N/C).

• direction of electric field vector= E, the direction the

electric force exerted on a small positive test charge.



Chapter 16

Electric Fields and Test Charges





Electric Field Strength, continued

• Electric field strength

• Electric Field Strength Due to a Point Charge

• Or E=F/q1(test charge)

2

2

charge producing the fieldelectric field strength = Coulomb constant

distance

C

qE k

r




Electric Field Diagrams, continued

• Use arrows to represent lines.

• Originate on +

• Terminate on –

• Never cross

• Number represents relative charge

ratios



Chapter 16

Rules for Drawing Electric Field Lines




Chapter 16

Rules for Sketching Fields Created by Several

Charges




Chapter 16

Calculating Net Electric Field





Sample Problem


A charge q1 = +7.00 µC is

at the origin, and a charge

q2 = –5.00 µC is on the x-

axis 0.300 m from the

origin, as shown at right.

Find the electric field

strength at point P,which is

on the y-axis 0.400 m from

the origin.



Chapter 16



1. Define the problem, and identify the knownvariables.

Given:

q1 = +7.00 µC = 7.00 10–6 C r1 = 0.400 m

q2 = –5.00 µC = –5.00 10–6 C r2 = 0.500 m

q = 53.1º

Unknown:

E at P (y = 0.400 m)

Tip: Apply the principle of superposition. You must first calculate the electric field produced by each charge individually at point P and then add these fields together as vectors.




Chapter 16



2. Calculate the electric field strength produced by each charge. Because we are finding the magnitude of the electric field, we can neglect the sign of each charge.

–69 2 2 51

1 2 2

1

–69 2 2 52

2 2 2

2

7.00 10 C8.99 10 N m /C 3.93 10 N/C

(0.400 m)

5.00 10 C8.99 10 N m /C 1.80 10 N/C

(0.500 m)

C

C

qE k

r

qE k

r




Chapter 16



3. Analyze the signs of the charges.

The field vector E1 at P due to q1 is directed vertically upward, as shown in the figure, because q1 is positive. Likewise, the field vector E2 at P due to q2 is directed toward q2 because q2 is negative.




Chapter 16



4. Find the x and y components of each electric field vector.

For E1: Ex,1 = 0 N/C

Ey,1 = 3.93 105 N/C

For E2: Ex,2= (1.80 105 N/C)(cos 53.1º) = 1.08 105 N/C

Ey,1= (1.80 105 N/C)(sin 53.1º)= –1.44 105 N/C




Chapter 16



5. Calculate the total electric field strength in both directions.

Ex,tot = Ex,1 + Ex,2 = 0 N/C + 1.08 105 N/C

= 1.08 105 N/C

Ey,tot = Ey,1 + Ey,2 = 3.93 105 N/C – 1.44 105 N/C

= 2.49 105 N/C




Chapter 16



6. Use the Pythagorean theorem to find the magnitude of the resultant electric field strength vector.

22

, ,

2 25 5

5

1.08 10 N/C 2.49 10 N/C

2.71 10 N/C

tot x tot y tot

tot

tot

E E E

E

E




Chapter 16



7. Use a suitable trigonometric function to find the direction of the resultant electric field strength vector.

In this case, you can use the inverse tangent function:

5,

5

,

2.49 10 N/Ctan

1.08 10 N/C

66.0

y tot

x tot

E

E





Conductors in Electrostatic Equilibrium

• Any charge on an isolated conductor is on the conductor’s outer surface only. (Faradays cage)

• electric field outside a charged conductor isperpendicular to the conductor’s surface.

• On an irregularly shaped conductor, charge accumulate at sharp points. St Elmo’s fire

• http://www.youtube.com/watch?feature=player_detailpage&v=ufKMRSkFrds



Multiple Choice

1. In which way is the electric force similar to the

gravitational force?

A. Electric force is proportional to the mass of the

object.

B. Electric force is similar in strength to gravitational

force.

C. Electric force is both attractive and repulsive.

D. Electric force decreases in strength as the

distance between the charges increases.

Standardized Test PrepChapter 16



Multiple Choice, continued

1. In which way is the electric force similar to the

gravitational force?

A. Electric force is proportional to the mass of the

object.

B. Electric force is similar in strength to gravitational

force.

C. Electric force is both attractive and repulsive.

D. Electric force decreases in strength as the

distance between the charges increases.





2. What must the charges be for

A and B in the figure so that

they produce the electric

field lines shown?

F. A and B must both be

positive.

G. A and B must both be

negative.

H. A must be negative, and

B must be positive.

J. A must be positive, and B

must be negative.





2. What must the charges be for

A and B in the figure so that

they produce the electric

field lines shown?

F. A and B must both be

positive.

G. A and B must both be

negative.

H. A must be negative, and

B must be positive.

J. A must be positive, and B

must be negative.





3. Which activity does not produce the same results as

the other three?

A. sliding over a plastic-covered automobile seat

B. walking across a woolen carpet

C. scraping food from a metal bowl with a metal

spoon

D. brushing dry hair with a plastic comb





3. Which activity does not produce the same results as

the other three?

A. sliding over a plastic-covered automobile seat

B. walking across a woolen carpet

C. scraping food from a metal bowl with a metal

spoon

D. brushing dry hair with a plastic comb





4. By how much does the electric force between two

charges change when the distance between them is

doubled?


4

2

1 2

1 4

F.

G.

H.

J.




4. By how much does the electric force between two

charges change when the distance between them is

doubled?


4

2

1 4

1

2

F.

G.

H.

J.




Use the passage below to answer questions 5–6.

A negatively charged object is brought close to the

surface of a conductor, whose opposite side is then

grounded.

5. What is this process of charging called?

A. charging by contact

B. charging by induction

C. charging by conduction

D. charging by polarization








grounded.

5. What is this process of charging called?

A. charging by contact

B. charging by induction

C. charging by conduction

D. charging by polarization








grounded.

6. What kind of charge is left on the conductor’s

surface?

F. neutral

G. negative

H. positive

J. both positive and negative








grounded.

6. What kind of charge is left on the conductor’s

surface?

F. neutral

G. negative

H. positive

J. both positive and negative





7. What is the electric field

strength 2.0 m from the

center of the conducting

sphere?

A. 0 N/C

B. 5.0 102 N/C

C. 5.0 103 N/C

D. 7.2 103 N/C


Use the graph below to answer

questions 7–10. The graph shows

the electric field strength at different

distances from the center of the

charged conducting sphere of a Van

de Graaff generator.




7. What is the electric field

strength 2.0 m from the

center of the conducting

sphere?

A. 0 N/C

B. 5.0 102 N/C

C. 5.0 103 N/C

D. 7.2 103 N/C











8. What is the strength of

the electric field at the

surface of the

conducting sphere?

F. 0 N/C

G. 1.5 102 N/C

H. 2.0 102 N/C

J. 7.2 103 N/C












the electric field at the

surface of the

conducting sphere?

F. 0 N/C

G. 1.5 102 N/C

H. 2.0 102 N/C

J. 7.2 103 N/C












the electric field inside

the conducting sphere?

A. 0 N/C

B. 1.5 102 N/C

C. 2.0 102 N/C

D. 7.2 103 N/C












the electric field inside


A. 0 N/C

B. 1.5 102 N/C

C. 2.0 102 N/C

D. 7.2 103 N/C











10. What is the radius of


F. 0.5 m

G. 1.0 m

H. 1.5 m

J. 2.0 m











10. What is the radius of


F. 0.5 m

G. 1.0 m

H. 1.5 m

J. 2.0 m










Short Response

11. Three identical

charges (q = +5.0 mC)

are along a circle with a

radius of 2.0 m at angles

of 30°, 150°, and 270°,

as shown in the

figure.What is the

resultant electric field at

the center?




Short Response, continued

11. Three identical

charges (q = +5.0 mC)

are along a circle with a

radius of 2.0 m at angles

of 30°, 150°, and 270°,

as shown in the

figure.What is the

resultant electric field at

the center?

Answer: 0.0 N/C






12. If a suspended object is attracted to another object

that is charged, can you conclude that the suspended

object is charged? Briefly explain your answer.





12. If a suspended object is attracted to another object

that is charged, can you conclude that the suspended

object is charged? Briefly explain your answer.

Answer: not necessarily; The suspended object might

have a charge induced on it, but its overall charge

could be neutral.





13. One gram of hydrogen contains 6.02 1023 atoms,

each with one electron and one proton. Suppose that

1.00 g of hydrogen is separated into protons and

electrons, that the protons are placed at Earth’s north

pole, and that the electrons are placed at Earth’s

south pole. Assuming the radius of Earth to be 6.38

106 m, what is the magnitude of the resulting

compressional force on Earth?





13. One gram of hydrogen contains 6.02 1023 atoms,

each with one electron and one proton. Suppose that

1.00 g of hydrogen is separated into protons and

electrons, that the protons are placed at Earth’s north

pole, and that the electrons are placed at Earth’s

south pole. Assuming the radius of Earth to be 6.38

106 m, what is the magnitude of the resulting

compressional force on Earth?

Answer: 5.12 105 N





14. Air becomes a conductor when the electric field

strength exceeds 3.0 106 N/C. Determine the

maximum amount of charge that can be carried by a

metal sphere 2.0 m in radius.





14. Air becomes a conductor when the electric field

strength exceeds 3.0 106 N/C. Determine the

maximum amount of charge that can be carried by a

metal sphere 2.0 m in radius.

Answer: 1.3 10–3 C




Extended Response

Use the information

below to answer

questions 15–18.

A proton, which has a

mass of 1.673 10–27 kg,

accelerates from rest in a

uniform electric field of

640 N/C. At some time

later, its speed is 1.2 106

m/s.

15. What is the magnitude

of the acceleration of

the proton?




Extended Response, continued

Use the information

below to answer

questions 15–18.


mass of 1.673 10–27 kg,





m/s.

15. What is the magnitude

of the acceleration of

the proton?

Answer: 6.1 1010 m/s2





Use the information

below to answer

questions 15–18.


mass of 1.673 10–27 kg,





m/s.

16. How long does it take

the proton to reach this

speed?





Use the information

below to answer

questions 15–18.


mass of 1.673 10–27 kg,





m/s.

16. How long does it take

the proton to reach this

speed?

Answer: 2.0 10–5 s





Use the information

below to answer

questions 15–18.


mass of 1.673 10–27 kg,





m/s.

17. How far has it moved

in this time interval?





Use the information

below to answer

questions 15–18.


mass of 1.673 10–27 kg,





m/s.

17. How far has it moved

in this time interval?

Answer: 12 m





Use the information

below to answer

questions 15–18.


mass of 1.673 10–27 kg,





m/s.

18. What is its kinetic

energy at the later

time?





Use the information

below to answer

questions 15–18.


mass of 1.673 10–27 kg,





m/s.

18. What is its kinetic

energy at the later

time?

Answer: 1.2 10–15 J





19. A student standing on a piece of insulating material

places her hand on a Van de Graaff generator. She

then turns on the generator. Shortly thereafter, her

hairs stand on end. Explain how charge is or is not

transferred in this situation, why the student is not

shocked, and what causes her hairs to stand up after

the generator is started.





19. (See previous slide for question.)

Answer: The charge on the sphere of the Van de Graaff generator is transferred to the student by means of conduction. This charge remains on the student because she is insulated from the ground. As there is no path between the student and the generator and the student and the ground by which charge can escape, the student is not shocked. The accumulation of charges of the same sign on the strands of the student’s hair causes the strands to repel each other and so stand on end.




Charging By Induction




Transfer of Electric Charge




Electric Field Lines

Section 1 Electric Charge Chapter 16 Properties of ... · PDF fileSection 1 Electric Charge Chapter 16 Properties of ... ... 23

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