Download Section 1 Measuring Electric Fields: Practice Problems

Chapter 21 Practice Problems, Review, and Assessment
Section 1 Measuring Electric Fields: Practice Problems
−6 −4 1. A positive test charge of 5.0×10 C is in an electric field that exerts a force of 2.0×10
magnitude of the electric field at the location of the test charge?
N on it. What is the
SOLUTION: −8 2. A negative charge of 2.0×10 C experiences a force of 0.060 N to the right in an electric field. What are the field’s magnitude and direction at that location?
SOLUTION: −3
4 3. Suppose that you place a 2.1×10 -N pith ball in a 6.5×10 N/C downward electric field. What net charge
(magnitude and sign) must you place on the pith ball so that the electrostatic force acting on that pith ball will
suspend it against the gravitational force?
SOLUTION: The electric force and the gravitational force algebraically sum to zero because the ball is suspended,
i.e. not in motion:
The electric force is upward (opposite the field), so the charge is negative.
4. Complete Table 2 using your understanding of electric fields.
SOLUTION: Table 2 Sample Data
Test Charge
Strength (C)
Force Exerted on
Test Charge (N)
Electric Field
Intensity (N/C)
0.30
[3.0×10 ]
eSolutions Manual - Powered by Cognero
–6
1.0×10
5
Page 1
Chapter
21 Practice Problems, Review, and Assessment
The electric force is upward (opposite the field), so the charge is negative.
4. Complete Table 2 using your understanding of electric fields.
SOLUTION: Table 2 Sample Data
Test Charge
Strength (C)
Force Exerted on
Test Charge (N)
Electric Field
Intensity (N/C)
–6
0.30
[3.0×10 ]
–6
[0.65]
3.3×10
–6
0.45
1.5×10
1.0×10
2.0×10
[3.0×10 ]
5
5
5
5. A positive charge of 3.0×10
on the charge?
−7 C is located in a field of 27 N/C directed toward the south. What is the force acting SOLUTION: 6. A negative test charge is placed in an electric field as shown in Figure 3. It experiences the force shown. What is
the magnitude of the electric field at the location of the charge?
SOLUTION: = 1.6×10–4 N/C, toward q
Manual - Powered by Cognero
eSolutions
Page 2
7. Challenge You are probing the electric field of a charge of unknown magnitude and sign. You first map the field
−6
−6
Chapter 21 Practice Problems, Review, and Assessment
6. A negative test charge is placed in an electric field as shown in Figure 3. It experiences the force shown. What is
the magnitude of the electric field at the location of the charge?
SOLUTION: = 1.6×10–4 N/C, toward q
7. Challenge You are probing the electric field of a charge of unknown magnitude and sign. You first map the field
−6
−6
with a 1.0×10 -C test charge, then repeat your work with a 2.0×10 -C test charge.
a. Would you measure the same forces at the same place with the two test charges? Explain.
b. Would you find the same field strengths? Explain.
SOLUTION: a. No. The force on the 2.0-μC charge would be twice that on the 1.0-μC charge.
b. Yes. You would divide the force by the strength of the test charge, so the results would be the same.
−6
8. What is the magnitude of the electric field at a position that is 1.2 m from a 4.2×10 -C point charge?
SOLUTION: 9. What is the magnitude of the electric field at a distance twice as far from the point charge in the previous problem?
SOLUTION: Because the field strength varies as the square of the distance from the point charge, the new field
3
strength will be one-fourth of the old field strength, or 6.5×10 N/C.
−6 10. What is the electric field at a position that is 1.6 m east of a point charge of +7.2×10
C?
SOLUTION: eSolutions Manual - Powered by Cognero
The direction of the field is east (away from the positive point charge).
Page 3
9. What is the magnitude of the electric field at a distance twice as far from the point charge in the previous problem?
SOLUTION: Because
the field
strength Review,
varies asand
the Assessment
square of the distance from the point charge, the new field
Chapter
21 Practice
Problems,
3
strength will be one-fourth of the old field strength, or 6.5×10 N/C.
−6 10. What is the electric field at a position that is 1.6 m east of a point charge of +7.2×10
C?
SOLUTION: The direction of the field is east (away from the positive point charge).
11. The electric field that is 0.25 m from a small sphere is 450 N/C toward the sphere. What is the net charge on the sphere?
SOLUTION: The charge is negative, because the field is directed toward it.
12. How far from a point charge of +2.4×10
360 N/C?
−6 C must you place a test charge in order to measure a field magnitude of
SOLUTION: 13. Explain why the strength of the electric field exerted on charge q′ by the charged body q is independent of the
charge on q′. Hint: Use mathematics to prove your point.
SOLUTION: Because the strength of the test charge (q′) and the force (F) are directly proportional, F = (Kq/r2)q′. Therefore, the electric field, which is the ratio of the force to the test charge, is independent of q': E =
F/q'.
14. What is the magnitude of the electric field exerted on the test charge shown in Figure 4?
eSolutions Manual - Powered by Cognero
Page 4
SOLUTION: Because the strength of the test charge (q′) and the force (F) are directly proportional, F = (Kq/r2)q′. Therefore,
the electric
field,
which and
is theAssessment
ratio of the force to the test charge, is independent of q': E =
Chapter
21 Practice
Problems,
Review,
F/q'.
14. What is the magnitude of the electric field exerted on the test charge shown in Figure 4?
SOLUTION: 2
or 7.5×10 N/C toward q −6 15. Challenge You place a small sphere with a net charge of 5.0×10 C at one corner of a square that measures
5.0 m on each side. What is the magnitude of the electric field at the opposite corner of the square?
SOLUTION: 3
= 6.4×10 N/C
Section 1 Measuring Electric Fields: Review
16. MAIN IDEA Suppose you are asked to measure the electric field at a point in space. How do you detect the field
at a point? How do you determine the magnitude of the field? How do you choose the magnitude of the test
charge?
SOLUTION: To detect a field at a point, place a test charge at that point and determine whether there is a force on it.
To determine the magnitude of the field, divide the magnitude of the force on the test charge by the
magnitude of the test charge.
The magnitude of the test charge must be chosen so that it is very small compared to the magnitudes of
the charges producing the field.
eSolutions Manual - Powered by Cognero
17. Field Strength and Direction A positive test charge of magnitude 2.40×10
Page 5
−8 C experiences a force of
21 Practice Problems, Review, and Assessment
Chapter
3
= 6.4×10 N/C
Section 1 Measuring Electric Fields: Review
16. MAIN IDEA Suppose you are asked to measure the electric field at a point in space. How do you detect the field
at a point? How do you determine the magnitude of the field? How do you choose the magnitude of the test
charge?
SOLUTION: To detect a field at a point, place a test charge at that point and determine whether there is a force on it.
To determine the magnitude of the field, divide the magnitude of the force on the test charge by the
magnitude of the test charge.
The magnitude of the test charge must be chosen so that it is very small compared to the magnitudes of
the charges producing the field.
17. Field Strength and Direction A positive test charge of magnitude 2.40×10
1.50×10
−8 C experiences a force of
−3 N toward the east. What is the electric field at the position of the test charge?
SOLUTION: 18. Field Lines How can you tell which charges are positive and which are negative by examining the electric field
lines?
SOLUTION: The arrows going away from a positive charge point away from the charge; the arrows going away from a
negative charge point toward the charge.
19. Field Versus Force How does the electric field at a test charge differ from the force on that charge?
SOLUTION: The field is a property of that region of space, and does not depend on the test charge used to measure
it. The force depends on the magnitude and sign of the test charge.
20. Critical Thinking Suppose the top charge in the field on the far right in Figure 5 is a test charge measuring the
field resulting from the two negative charges. Is it small enough to produce an accurate measurement? Explain.
eSolutions Manual - Powered by Cognero
Page 6
SOLUTION: The field is a property of that region of space, and does not depend on the test charge used to measure
it. The force depends on the magnitude and sign of the test charge.
Chapter 21 Practice Problems, Review, and Assessment
20. Critical Thinking Suppose the top charge in the field on the far right in Figure 5 is a test charge measuring the
field resulting from the two negative charges. Is it small enough to produce an accurate measurement? Explain.
eSolutions Manual - Powered by Cognero
Figure 5
SOLUTION: Page 7
Chapter 21 Practice Problems, Review, and Assessment
Figure 5
SOLUTION: No; this charge is large enough to distort the field produced by the other charges with its own field.
Section 2 Applications of Electric Fields: Practice Problems
21. The electric field intensity between two large, charged parallel metal plates is 6000 N/C. The plates are 0.05 m apart. What is the electric potential difference between them?
SOLUTION: 22. A voltmeter reads 400 V across two charged, parallel plates that are 0.020 m apart. What is the magnitude of the electric field between them?
SOLUTION: 23. What electric potential difference is between two metal plates that are 0.200 m apart if the electric field between
3 those plates is 2.50×10 N/C?
SOLUTION: ΔV = Ed 3
= (2.50×10 N/C)(0.200 m)
= 5.00×102 V
24. When you apply a potential difference of 125 V between two parallel plates, the field between them is 3 4.25×10 N/C. How far apart are the plates?
eSolutions Manual - Powered by Cognero
SOLUTION: Page 8
SOLUTION: ΔV = Ed 3
= (2.50×10
N/C)(0.200Review,
m)
Chapter
21 Practice Problems,
and Assessment
2
= 5.00×10 V
24. When you apply a potential difference of 125 V between two parallel plates, the field between them is 3 4.25×10 N/C. How far apart are the plates?
SOLUTION: 25. Challenge You apply a potential difference of 275 V between two parallel plates that are 0.35 cm apart. How large is the electric field between the plates?
SOLUTION: 26. What work is done on a 3.0-C charge when you move that charge through a 1.5-V electric potential difference?
SOLUTION: W = qΔV = (3.0 C)(1.5 V) = 4.5 J
27. What is the magnitude of the electric field between the two plates shown in Figure 12?
SOLUTION: 28. An electron in an old television picture tube passes through a potential difference of 18,000 V. How much work is done on the electron as it passes through that potential difference?
SOLUTION: W = qΔV = (1.602×10–19 C)(1.8×104 V)
–15
= 2.9×10
J
5 29. The electric field in a particle accelerator has a magnitude of 4.5×10 N/C. How much work is done to move a
proton 25 cm through that field?
eSolutions Manual - Powered by Cognero
Page 9
SOLUTION: W = qΔV = qEd
–19
5
done on the electron as it passes through that potential difference?
SOLUTION: –19
4
W =21
qΔV = (1.602×10
C)(1.8×10
V) Assessment
Chapter
Practice Problems,
Review, and
–15
= 2.9×10 J
5 29. The electric field in a particle accelerator has a magnitude of 4.5×10 N/C. How much work is done to move a
proton 25 cm through that field?
SOLUTION: W = qΔV = qEd
5
–19
= (1.602×10 C)(4.5×10 N/C)(0.25 m)
= 1.8×10–14 J
6 30. Challenge A 12-V car battery has 1.44×10 C of useable charge on one plate when it is fully energized. How
much work can this battery do before it needs to be energized again?
SOLUTION: W = qΔV = (1.44×106 C)(12 V)
7
= 1.8×10 J
31. A drop is falling in a Millikan oil-drop apparatus with no electric field. What forces are acting on the oil drop,
regardless of its acceleration? If the drop is falling at a constant velocity, describe the forces acting on it.
SOLUTION: Gravitational force (weight) downward, friction force of air upward. The two forces are equal in
magnitude if the drop falls at constant velocity.
−15 3 32. An oil drop weighs 1.9×10 N. You suspend it in an electric field of 6.0×10 N/C. What is the net charge on the
drop? How many excess electrons does it carry?
SOLUTION: 33. An oil drop carries one excess electron and weighs 6.4×10
suspend the drop so it is motionless?
−15
N. What electric field strength do you need to
SOLUTION: −14 34. Challenge You suspend a positively charged oil drop that weighs 1.2×10 N between two parallel plates that are
0.64 cm apart. The potential difference between the plates is 240 V. What is the net charge on the drop? How many electrons is the drop missing?
eSolutions
Manual - Powered by Cognero
SOLUTION: Page 10
Chapter 21 Practice Problems, Review, and Assessment
−14 34. Challenge You suspend a positively charged oil drop that weighs 1.2×10 N between two parallel plates that are
0.64 cm apart. The potential difference between the plates is 240 V. What is the net charge on the drop? How many electrons is the drop missing?
SOLUTION: 35. A 27-μF capacitor has an electric potential difference of 45 V across it. What is the amount the net charge on the positively charged plate of the capacitor?
SOLUTION: 36. Suppose you connect both a 3.3-μF and a 6.8-μF capacitor across a 24-V electric potential difference. Which
capacitor has the greater net charge on its positively charged plate and what is its magnitude?
SOLUTION: q = CΔV, so the larger capacitor has a greater charge.
−6
q = (6.8×10 F)(24 V) = 1.6×10
−4
C
37. You later find that the magnitude of net charge on each of the plates for each of the capacitors in the previous
−4 problem is 3.5×10 C. Which capacitor has the larger potential difference across it? What is that potential
difference?
SOLUTION: ΔV = q/C, so the smaller capacitor has the larger potential difference.
38. Suppose that you apply an electric potential difference of 6.0 V across a 2.2-μF capacitor. What does the
magnitude of the net charge on one plate need to be to increase the electric potential difference to 15.0 V?
SOLUTION: eSolutions Manual - Powered by Cognero
Page 11
ΔV = q/C, so the smaller capacitor has the larger potential difference.
Chapter 21 Practice Problems, Review, and Assessment
38. Suppose that you apply an electric potential difference of 6.0 V across a 2.2-μF capacitor. What does the
magnitude of the net charge on one plate need to be to increase the electric potential difference to 15.0 V?
SOLUTION: 39. A sphere is charged by a 12-V battery and suspended above Earth as shown in Figure 17. What is the net charge
on the sphere?
SOLUTION: 40. Challenge You increase the potential difference across a capacitor from 12.0 V to 14.5 V. As a result, the −5 magnitude of the net charge on each plate increases by 2.5×10
C. What is the capacitance of the capacitor?
SOLUTION: Section 2 Applications of Electric Fields: Review
41. MAIN IDEA Suppose a friend asks you to explain how electric potential relates to potential energy. Write a brief
explanation that you could use to explain this concept to a friend who does not understand the relationship between
the two concepts.
SOLUTION: Answers will vary; Sample answer: Electric potential is potential energy per unit charge, and it equals
the work required to move a test charge to a location in an electric field.
42. Potential Difference What is the difference between electric potential energy and electric potential difference?
SOLUTION: Electric potential energy changes when work is done to move a charge in an electric field. It depends on
the amount of charge involved. Electric potential difference is the work done per unit charge to move a
charge in an electric field. It is independent of the amount of charge that is moved.
eSolutions
Manual Field
- Powered
by Potential
Cognero
Electric
and
43. SOLUTION: Difference Show that a volt per meter is the same as a newton per coulomb.
Page 12
42. What is the difference between electric potential energy and electric potential difference?
SOLUTION: Electric potential energy changes when work is done to move a charge in an electric field. It depends on
Chapter
Practice
Problems,
Review,
and Assessment
the 21
amount
of charge
involved.
Electric
potential difference is the work done per unit charge to move a
charge in an electric field. It is independent of the amount of charge that is moved.
43. Electric Field and Potential Difference Show that a volt per meter is the same as a newton per coulomb.
SOLUTION: V/m = J/ (C⋅ m) = (N⋅m) / (C⋅ m) = N/C
44. Millikan Experiment When the net charge on an oil drop suspended in a Millikan apparatus is changed, the drop
begins to fall. How should you adjust the potential difference between the conducting plates to bring the drop back
into balance?
SOLUTION: The potential difference should be increased.
45. Charge and Potential Difference In the previous problem, if changing the potential difference between the
conducting plates has no effect on the falling drop, what does this tell you about the new net charge on the drop?
SOLUTION: The drop is electrically neutral (no electron excess or deficiency).
46. Capacitance What is the magnitude of net charge on each conductor plate of a 0.47-μF capacitor when a potential
difference of 12 V is applied across that capacitor?
SOLUTION: q = CΔV = (4.7×10−7 F)(12 V)
−6
= 5.6×10
C
47. Charge Sharing If you touch a large, positively charged, conducting sphere with a small, negatively charged,
conducting sphere, what can be said about the
a. potentials of the two spheres;
b. charges on the two spheres.
SOLUTION: a. After touching, the charges will move until the potential differences between the spheres is zero.
b. If there were an equal number of positive and negative charges on the spheres before touching, then
both spheres would be neutral after touching. If there were an excess of positive or negative charges,
then that excess would be distributed between the two spheres based on the relative areas until the two
spheres had the same potential. The number of charges per unit area on each sphere would be the
same.
48. Critical Thinking Explain how the charge in Figure 18 continues to build up on the metal dome of a Van de
Graaff generator. In particular, why isn’t charge repelled back onto the belt at point B?
SOLUTION: The charges on the metal dome produce no field inside the dome. The charges from the belt are
transferred immediately to the outside of the dome, where they have no effect on new charges arriving
at point B.
Chapter Assessment
Section 1 Measuring Electric Fields: Mastering Concepts
49. What are the two properties that a test charge must have?
SOLUTION: Page 13
The test charge must be small in magnitude relative to the magnitudes of the charges producing the
field and be positive.
eSolutions Manual - Powered by Cognero
SOLUTION: The charges on the metal dome produce no field inside the dome. The charges from the belt are
Chapter
21 Practice
Problems,toReview,
and of
Assessment
transferred
immediately
the outside
the dome, where they have no effect on new charges arriving
at point B.
Chapter Assessment
Section 1 Measuring Electric Fields: Mastering Concepts
49. What are the two properties that a test charge must have?
SOLUTION: The test charge must be small in magnitude relative to the magnitudes of the charges producing the
field and be positive.
50. How is the direction of an electric field defined?
SOLUTION: The direction of an electric field is the direction of the force on a positive charge placed in the field.
This would be away from a positive object and toward a negative object.
51. BIG IDEA What are electric field lines?
SOLUTION: Electric field lines are used to represent the actual field in space around a charge. The direction of the
electric field at any point is the tangent drawn to a field line at that point.
52. How is the strength of an electric field indicated with electric field lines?
SOLUTION: The closer together the electric field lines are, the stronger the electric field.
53. Draw some of the electric field lines between each of the following:
a. two like charges of equal magnitude
b. two unlike charges of equal magnitude
c. a positive charge and a negative charge having twice the magnitude of the positive charge
d. two oppositely charged parallel plates
SOLUTION: a.
b.
c.
eSolutions Manual - Powered by Cognero
Page 14
electric field at any point is the tangent drawn to a field line at that point.
52. How is the strength of an electric field indicated with electric field lines?
SOLUTION: Chapter
21 Practice Problems, Review, and Assessment
The closer together the electric field lines are, the stronger the electric field.
53. Draw some of the electric field lines between each of the following:
a. two like charges of equal magnitude
b. two unlike charges of equal magnitude
c. a positive charge and a negative charge having twice the magnitude of the positive charge
d. two oppositely charged parallel plates
SOLUTION: a.
b.
c.
d.
54. In Figure 19, where do the electric field lines that leave the positive charge end?
eSolutions Manual - Powered by Cognero
Page 15
Chapter 21 Practice Problems, Review, and Assessment
54. In Figure 19, where do the electric field lines that leave the positive charge end?
SOLUTION: They end on distant negative charges somewhere beyond the edges of the diagram.
55. What happens to the strength of an electric field when the magnitude of the test charge is halved?
SOLUTION: Nothing. Because the force on the test charge also would be halved, the ratio F′/q′ and the electric field would remain the same.
56. You are moving a constant positive charge through an increasing electric field. Does the amount of energy required
to move it increase or decrease?
SOLUTION: Energy is proportional to the force, and the force is proportional to the electric field. Therefore, it as the
field increases, the energy required increases.
Chapter Assessment
Section 1 Measuring Electric Fields: Mastering Problems
–8 57. What charge exists on a test charge that experiences a force of 1.4×10 N at a point where the electric field
intensity is 5.0×10
−4 N/C? (Level 1)
SOLUTION: 5 58. A test charge experiences a force of 0.30 N on it when it is placed in an electric field intensity of 4.5×10 N/C.
What is the magnitude of the charge? (Level 1)
SOLUTION: eSolutions Manual - Powered by Cognero
Page 16
Chapter
21 Practice Problems, Review, and Assessment
5 58. A test charge experiences a force of 0.30 N on it when it is placed in an electric field intensity of 4.5×10 N/C.
What is the magnitude of the charge? (Level 1)
SOLUTION: −7 59. What is the electric field strength 20.0 cm from a point charge of 8.0×10
C? (Level 2)
SOLUTION: so,
−5 60. A positive charge of 1.0×10 C, shown in Figure 20, experiences a force of 0.30 N when it is located at a certain
point. What is the electric field intensity at that point? (Level 1)
SOLUTION: in the same direction as the force
61. The electric field in the atmosphere is about 150 N/C downward. (Level 1)
a. What is the direction of the force on a negatively charged particle?
−19 b. Find the electric force on an electron with charge −1.602×10 C.
eSolutions Manual - Powered by Cognero
c. Compare the force in part b with the force of gravity on the same electron (mass = 9.1×10−31 kg).
SOLUTION: Page 17
Chapter
21 Practice Problems, Review, and Assessment
in the same direction as the force
61. The electric field in the atmosphere is about 150 N/C downward. (Level 1)
a. What is the direction of the force on a negatively charged particle?
−19 b. Find the electric force on an electron with charge −1.602×10 C.
c. Compare the force in part b with the force of gravity on the same electron (mass = 9.1×10−31 kg).
SOLUTION: a. upward
b.
directed upward
−31
c. F = mg = (9.1×10 kg)(9.8 N/kg) = 8.9×10−30 N −30
F = 8.9×10
N (downward), more than one trillion times smaller
62. Carefully sketch each of the following: (Level 1)
a. the electric field produced by a +1.0-μC charge
b. the electric field resulting from a +2.0-μC charge (Make the number of field lines proportional to the change in
charge.)
SOLUTION: a. b.
−6 eSolutions
Manual - Powered by Cognero
63. A positive test charge of 6.0×10
C
is placed in an electric field of 50.0-N/C intensity, as in Figure 21. WhatPage
is 18
the strength of the force exerted on the test charge? (Level 1)
Chapter 21 Practice Problems, Review, and Assessment
−6 63. A positive test charge of 6.0×10 C is placed in an electric field of 50.0-N/C intensity, as in Figure 21. What is
the strength of the force exerted on the test charge? (Level 1)
SOLUTION: F = qE = (6.0×10
= 3.0×10−4 N
−6
C)(50.0 N/C)
64. A force of 14.005 N exists on a positive test charge (q′) that has a charge of 4.005×10
of the electric field?
−19 C. What is the magnitude
SOLUTION: = 3.497×1019 N/C
65. Charges X, Y, and Z all are equidistant from each other. X has a +1.0-μC charge. Y has a +2.0-μC charge. Z has
a small negative charge. (Level 2)
a. Draw an arrow representing the force on charge Z.
b. Charge Z now has a small positive charge on it. Draw an arrow representing the force on it.
SOLUTION: a. b.
eSolutions Manual - Powered by Cognero
Page 19
Chapter 21 Practice Problems, Review, and Assessment
= 3.497×1019 N/C
65. Charges X, Y, and Z all are equidistant from each other. X has a +1.0-μC charge. Y has a +2.0-μC charge. Z has
a small negative charge. (Level 2)
a. Draw an arrow representing the force on charge Z.
b. Charge Z now has a small positive charge on it. Draw an arrow representing the force on it.
SOLUTION: a. b.
5 66. In a television picture tube, electrons are accelerated by an electric field having a value of 1.00×10 N/C. (Level 2)
a. Find the force on an electron.
b. If the field is constant, find the acceleration of the electron (mass 9.11×10−31 kg).
SOLUTION: a. F = Eq
−19
5
= (−1.602×10 C)(1.00×10 N/C)
= −1.60×10−14 N
b. F = ma
16
2
= –1.76×10 m/s
eSolutions
Manual - Powered by Cognero
67. The nucleus of a lead atom has a charge of 82 protons. (Level 2)
a. What are the direction and the magnitude of the electric field at 1.0×10−10 m from the nucleus?
Page 20
b. F = ma
21 Practice Problems, Review, and Assessment
Chapter
16
2
= –1.76×10 m/s
67. The nucleus of a lead atom has a charge of 82 protons. (Level 2)
a. What are the direction and the magnitude of the electric field at 1.0×10−10 m from the nucleus?
b. What are the direction and the magnitude of the force exerted on an electron located at this distance?
SOLUTION: a. Q = (82 protons)
−19
(1.602×10 C/proton) = 1.31×10−17 C
b. F = Eq
13
−19
= (1.2×10 N/C)(−1.602×10
C)
−6
= −1.9×10 N, toward the nucleus
Chapter Assessment
Section 2 Applications of Electric Fields: Mastering Concepts
68. What SI unit is used to measure electric potential energy? What SI unit is used to measure electric potential
difference?
SOLUTION: electric potential energy: joule; electric potential: volt
69. Define volt in terms of the change in potential energy of a charge moving in an electric field.
SOLUTION: A volt is the change in electric potential energy, ΔPE, resulting from moving a unit test charge, q, a
distance, d, of 1 m in an electric field, E, of 1 N/C. ΔV = ΔPE/q = Ed
70. Why does a charged object lose its charge when it is touched to the ground?
SOLUTION: The charge is shared with the surface of Earth, which is an extremely large object.
71. A charged rubber rod that is placed on a table maintains its charge for some time. Why is the charged rod not
discharged immediately?
SOLUTION: The table is an insulator, or at least a very poor conductor.
72. Computers Delicate parts in electronic equipment, such as those pictured in Figure 22, are contained within a
metal box inside a plastic case. Why?
SOLUTION: The metal box shields the parts from external electric fields, which do not exist inside a hollow
conductor.
eSolutions Manual - Powered by Cognero
Chapter Assessment
Section 2 Applications of Electric Fields: Mastering Problems
Page 21
71. A charged rubber rod that is placed on a table maintains its charge for some time. Why is the charged rod not
discharged immediately?
SOLUTION: Chapter
21 Practice Problems, Review, and Assessment
The table is an insulator, or at least a very poor conductor.
72. Computers Delicate parts in electronic equipment, such as those pictured in Figure 22, are contained within a
metal box inside a plastic case. Why?
SOLUTION: The metal box shields the parts from external electric fields, which do not exist inside a hollow
conductor.
Chapter Assessment
Section 2 Applications of Electric Fields: Mastering Problems
73. If 120 J of work is performed to move 2.4 C of charge from the positive plate to the negative plate shown in Figure 23, what potential difference exists between the plates? (Level 1)
SOLUTION: 74. How much work is done to transfer 0.15 C of charge through an electric potential difference of 9.0 V? (Level 1)
SOLUTION: W = qΔV = (0.15)(9.0 V) = 1.4 J
75. An electron is moved through an electric potential difference of 450 V. How much work is done on the electron? (Level 1)
SOLUTION: W = qΔV
–19
= (–1.602×10 C)(450 V)
= –7.2×10–17 J
76. A 12-V battery does 1200 J of work transferring charge. How much charge is transferred? (Level 1)
SOLUTION: eSolutions Manual - Powered by Cognero
Page 22
W = qΔV
–19
= (–1.602×10
C)(450
V)
Chapter
21 Practice Problems,
Review,
and Assessment
–17
= –7.2×10 J
76. A 12-V battery does 1200 J of work transferring charge. How much charge is transferred? (Level 1)
SOLUTION: 3 77. The electric field intensity between two charged plates is 1.5×10 N/C. The plates are 0.060 m apart. What is the electric potential difference, in volts, between the plates? (Level 1)
SOLUTION: ΔV = Ed
= (1.5×103 N/C)(0.060 m)
1
= 9.0×10 V
78. A voltmeter indicates that the electric potential difference between two plates is 70.0 V. The plates are 0.020 m apart. What electric field intensity exists between them? (Level 1)
SOLUTION: ΔV = Ed
= 3500 V/m
= 3500 N/C
79. A capacitor that is connected to a 45.0-V source has a charge of 90.0 μC. What is the capacitor’s capacitance?
(Level 1)
SOLUTION: = 2.00 μF
−15 80. The oil drop shown in Figure 24 is negatively charged and weighs 4.5×10
N. The drop is suspended in an
3 electric field intensity of 5.6×10 N/C. (Level 1)
a. What is the charge on the drop?
b. How many excess electrons does it carry?
eSolutions Manual - Powered by Cognero
SOLUTION: a.
Page 23
SOLUTION: 21 Practice Problems, Review, and Assessment
Chapter
= 2.00 μF
−15 80. The oil drop shown in Figure 24 is negatively charged and weighs 4.5×10
N. The drop is suspended in an
3 electric field intensity of 5.6×10 N/C. (Level 1)
a. What is the charge on the drop?
b. How many excess electrons does it carry?
SOLUTION: a.
b.
81. What electric potential difference exists across a 5.4-μF capacitor that has a charge of 8.1×10
−4 C? (Level 1)
SOLUTION: 82. What is the charge of a 15.0-pF capacitor when it is connected across a 45.0-V source? (Level 1)
SOLUTION: q = CΔV
–12
= (15.0×10 F)(45.0 V)
= 6.75×10–10 C
83. A force of 0.065 N is required to move a charge of 37 μC a distance of 25 cm in a uniform electric field, as in Figure 25. What is the size of the electric potential difference between the two points? (Level 2)
eSolutions Manual - Powered by Cognero
Page 24
q = CΔV
–12
F)(45.0 V) Review, and Assessment
= (15.0×10
Chapter
21 Practice Problems,
–10
= 6.75×10 C
83. A force of 0.065 N is required to move a charge of 37 μC a distance of 25 cm in a uniform electric field, as in Figure 25. What is the size of the electric potential difference between the two points? (Level 2)
SOLUTION: 84. Photoflash The potential energy of a capacitor with capacitance (C) and an electric potential difference (ΔV) is
2
represented by PE = ½ CΔV . One application of this is in the electronic photoflash of a strobe light, such the one in
Figure 26. In such a unit, a capacitor of 10.0 μF has a charge of 3.0×102 V. Find the electrical energy stored.
(Level 2)
SOLUTION: 85. Suppose it took 25 s to energize the capacitor in the previous problem. (Level 2)
a. Find the average power required to energize the capacitor in this time.
−4 b. When the plates are discharged through the strobe lamp, it transfers all its energy in 1.0×10
delivered to the lamp.
c. How is such a large amount of power possible?
s. Find the power
SOLUTION: a.
b.
= 4.5×103 W
c. Power is inversely proportional to the time. The shorter the time for a given amount of energy to be
expended, the greater the power.
eSolutions Manual - Powered by Cognero
Page 25
86. The plates of a 0.047-μF capacitor are 0.25 cm apart and are charged to a potential difference of 120 V. How much charge is on one plate of the capacitor? (Level 1)
Chapter 21 Practice Problems, Review, and Assessment
85. Suppose it took 25 s to energize the capacitor in the previous problem. (Level 2)
a. Find the average power required to energize the capacitor in this time.
−4 b. When the plates are discharged through the strobe lamp, it transfers all its energy in 1.0×10
delivered to the lamp.
c. How is such a large amount of power possible?
s. Find the power
SOLUTION: a.
b.
= 4.5×103 W
c. Power is inversely proportional to the time. The shorter the time for a given amount of energy to be
expended, the greater the power.
86. The plates of a 0.047-μF capacitor are 0.25 cm apart and are charged to a potential difference of 120 V. How much charge is on one plate of the capacitor? (Level 1)
SOLUTION: 8
= (4.7×10 F)(120 V)
= 5.6×106 C = 5.6 μC
87. Lasers Lasers are used to try to produce controlled fusion reactions. These lasers require brief pulses of energy
that are stored in large rooms filled with capacitors. One such room has a capacitance of 6.1×10
energized to a potential difference of 10.0 kV. (Level 2)
−2 F and is
a. Given that PE = ½ CΔV2, find the energy stored in the capacitors.
−8 b. The capacitors’ plates are discharged in 10 ns (1×10 s). What power is produced?
c. If the capacitors are energized by a generator with a power capacity of 1.0 kW, how many seconds will be required to energize the capacitors?
SOLUTION: a.
b.
c. Manual - Powered by Cognero
eSolutions
Page 26
8
F)(120
V)
= (4.7×10
Chapter
21 Practice
Problems,
Review, and Assessment
6
= 5.6×10 C = 5.6 μC
87. Lasers Lasers are used to try to produce controlled fusion reactions. These lasers require brief pulses of energy
that are stored in large rooms filled with capacitors. One such room has a capacitance of 6.1×10
energized to a potential difference of 10.0 kV. (Level 2)
−2 F and is
a. Given that PE = ½ CΔV2, find the energy stored in the capacitors.
−8 b. The capacitors’ plates are discharged in 10 ns (1×10 s). What power is produced?
c. If the capacitors are energized by a generator with a power capacity of 1.0 kW, how many seconds will be required to energize the capacitors?
SOLUTION: a.
b.
c.
Chapter Assessment: Applying Concepts
88. What will happen to the electric potential energy of a charged particle in an electric field when the particle is
released and free to move?
SOLUTION: The electric potential energy of the particle will be converted into kinetic energy of the particle.
89. Figure 27 shows three spheres with charges of equal magnitude, with their signs as shown. Spheres y and z are held in place, but sphere x is free to move. Initially, sphere x is equidistant from spheres y and z. Which path will sphere x will begin to follow? Assume that no other forces act on the spheres.
SOLUTION: Sphere x will follow path C. It will experience forces shown by D and B. The vector sum is C.
Rank
the following
90. Ranking
eSolutions
Manual - Task
Powered
by Cognero
point charges according to the magnitude of the electric force experienced,Page 27
from greatest to least. Specifically indicate any ties.
SOLUTION: Chapter
21 Practice Problems, Review, and Assessment
Sphere x will follow path C. It will experience forces shown by D and B. The vector sum is C.
90. Ranking Task Rank the following point charges according to the magnitude of the electric force experienced,
from greatest to least. Specifically indicate any ties.
A. a charge of 3 nC at a point where the electric field is 70 N/C
B. a charge of 5 nC at a point where the electric field is 600 N/C
C. a charge of 3 nC at a point where the electric field is 20 N/C
D. a charge of 6 nC at a point where the electric field is 35 N/C
E. a charge of 8 nC at a point where the electric field is 10 N/C
SOLUTION: B>A=D>E>C
91. Two unlike-charged oil drops are held motionless in a Millikan oil-drop experiment at the same time.
a. Can you be sure that the charges are the same?
b. The ratios of which two properties of the oil drops have to be equal?
SOLUTION: a. No. Their masses could be different.
b. charge to mass ratio, q/m (or m/q)
92. José and Sue are standing on an insulating platform and holding hands when they are given a charge, as in Figure 28. José is larger than Sue. Who has the larger amount of charge from the machine, or do they both have the same amount?
SOLUTION: José has a larger surface area, so he will have a larger amount of charge.
93. Reverse Problem Write a physics problem for which the following equation would be part of the solution:
SOLUTION: Answers will vary, but a correct form of the answer is, “In a region of space with a uniform electric field,
the potential changes by 9 V over a distance of 0.85 cm. What is the magnitude of the electric field in
this region?”
94. How can you store different amounts of electrical energy in a capacitor?
eSolutions Manual - Powered by Cognero
SOLUTION: Change the voltage across the capacitor.
Page 28
SOLUTION: Answers will vary, but a correct form of the answer is, “In a region of space with a uniform electric field,
the 21
potential
changes
by 9 Review,
V over aand
distance
of 0.85 cm. What is the magnitude of the electric field in
Chapter
Practice
Problems,
Assessment
this region?”
94. How can you store different amounts of electrical energy in a capacitor?
SOLUTION: Change the voltage across the capacitor.
Chapter Assessment: Mixed Review
95. How much work does it take to move 0.25 μC between two parallel plates that are 0.40 cm apart if the field between the plates is 6400 N/C? (Level 1)
SOLUTION: W = qΔV = qEd
= (2.5×10−7 C)(6400 N/C)(4.0×10−3 m)
−6
= 6.4×10
J
96. How much charge is on a 0.22-μF parallel plate capacitor if the plates are 1.2 cm apart and the electric field between them is 2400 N/C? (Level 1)
SOLUTION: q = CΔV = CEd
= (2.2×10−7 F)(2400 N/C)(1.2×10−2 m)
= 6.3 μC
97. Two identical small spheres 25 cm apart carry equal but opposite charges of 0.060 μC, as in Figure 29. If the
potential difference between them is 300 V, what is the capacitance of the system? (Level 1)
SOLUTION: –10
= 2.0×10
F
98. Problem Posing Complete this problem so that it must be solved using the concept indicated: “ A point charge of
4.0 mC is at rest…”
a. electric field
b. electric potential difference
SOLUTION: a. Answers will vary. Possible form of the correct answer, . “…in an electric field. If it experiences a
force of 0.60 N, what is the magnitude of the electric field?”
b. Answers will vary. Possible form of the correct answer, “…and then moved to another location. If
0.35 J of work were done on the charge in moving it, what is the electric potential difference between
the two locations?”
99. The plates of a 0.047-μF capacitor are 0.25 cm apart and are charged to a potential difference of 120 V.
eSolutions Manual - Powered by Cognero
a. How much charge is on the capacitor? (Level 1)
b. What is the strength of the electric field between the plates of the capacitor? (Level 1)
Page 29
a. Answers will vary. Possible form of the correct answer, . “…in an electric field. If it experiences a
force of 0.60 N, what is the magnitude of the electric field?”
b. Answers will vary. Possible form of the correct answer, “…and then moved to another location. If
0.3521
J of
work were
done on
the charge
in moving it, what is the electric potential difference between
Chapter
Practice
Problems,
Review,
and Assessment
the two locations?”
99. The plates of a 0.047-μF capacitor are 0.25 cm apart and are charged to a potential difference of 120 V.
a. How much charge is on the capacitor? (Level 1)
b. What is the strength of the electric field between the plates of the capacitor? (Level 1)
c. An electron is placed between the plates of the capacitor, as in Figure 30. What force is exerted on that
electron? (Level 1)
SOLUTION: a.
b.
c.
100. How much work would it take to move an additional 0.010 μC between the plates at 120 V in the previous problem? (Level 1)
SOLUTION: eSolutions Manual - Powered by Cognero
Page 30
101. The graph in Figure 31 represents the amount of charge stored on one plate of a capacitor as a function of the
Chapter 21 Practice Problems, Review, and Assessment
100. How much work would it take to move an additional 0.010 μC between the plates at 120 V in the previous problem? (Level 1)
SOLUTION: 101. The graph in Figure 31 represents the amount of charge stored on one plate of a capacitor as a function of the
charging potential. (Level 2)
a. What does the slope of the line represent?
b. What is the capacitance of the capacitor?
c. What does the area under the graph line represent?
SOLUTION: a. capacitance of the capacitor
b. C = slope = 0.50 μF
c. work done to charge the capacitor
Chapter Assessment: Thinking Critically
102. Analyze and Conclude Two small spheres, A and B, lie on the x-axis, as in Figure 32. Sphere A has a charge of
−6 −6 +3.00×10 C. Sphere B is 0.800 m to the right of sphere A and has a charge of −5.00×10 C. Find the magnitude
and the direction of the electric field strength at a point above the x-axis that would form the apex of an equilateral
triangle with spheres A and B. eSolutions Manual - Powered by Cognero
Page 31
102. Analyze and Conclude Two small spheres, A and B, lie on the x-axis, as in Figure 32. Sphere A has a charge of
−6 −6 +3.00×10 C. Sphere B is 0.800 m to the right of sphere A and has a charge of −5.00×10 C. Find the magnitude
and 21
the Practice
direction Problems,
of the electric
field strength
at a point above the x-axis that would form the apex of an equilateral
Chapter
Review,
and Assessment
triangle with spheres A and B. SOLUTION: Draw the spheres and vectors representing the fields due to each charge at the given point.
Now do the math:
EAx = EA cos 60.0° = (4.22×104 N/C)(cos 60.0°) 4
= 2.11×10 N/C
EAy = EA sin 60.0° 4
= (4.22×10 N/C)(sin 60.0°) = 3.65×104 N/C
EBx = EB cos (−60.0°) eSolutions Manual - Powered by Cognero
= (7.03×104 N/C)(cos −60.0°) 4
= 3.52×10 N/C
Page 32
EAy = EA sin 60.0° 4
= (4.22×10 N/C)(sin 60.0°) Chapter 21 Practice
Problems, Review, and Assessment
= 3.65×104 N/C
EBx = EB cos (−60.0°) = (7.03×104 N/C)(cos −60.0°) 4
= 3.52×10 N/C
EBy = EB sin (−60.0°) 4
= (7.03×10 N/C)(sin −60.0°) = −6.09×104 N/C
Ex = EAx + EBx = (2.11×104 N/C) + (3.52×104 N/C) 4
= 5.63×10 N/C
Ey = EAy + EBy 4
4
= (3.65×10 N/C) + (−6.09×10 N/C) = −2.44×104 N/C
103. Analyze and Conclude In an ink-jet printer, drops of ink are given a certain amount of charge before they move
between two large, parallel plates. The plates deflect the charged ink particles as shown in Figure 33. The plates
6
have an electric field of E = 1.2×10 N/C between them and are 1.5 cm long. Drops with a mass m = 0.10 ng and a
−16 charge q = 1.0×10 C are moving horizontally at a speed, v = 15 m/s, parallel to the plates. What is the vertical displacement of the drops when they leave the plates? To answer this question, complete the following steps.
a. What is the vertical force on the drops?
b. What is their vertical acceleration?
c. How long are they between the plates?
d. How far are they displaced?
eSolutions Manual - Powered by Cognero
Page 33
Chapter
21 Practice Problems, Review, and Assessment
103. Analyze and Conclude In an ink-jet printer, drops of ink are given a certain amount of charge before they move
between two large, parallel plates. The plates deflect the charged ink particles as shown in Figure 33. The plates
6
have an electric field of E = 1.2×10 N/C between them and are 1.5 cm long. Drops with a mass m = 0.10 ng and a
−16 charge q = 1.0×10 C are moving horizontally at a speed, v = 15 m/s, parallel to the plates. What is the vertical displacement of the drops when they leave the plates? To answer this question, complete the following steps.
a. What is the vertical force on the drops?
b. What is their vertical acceleration?
c. How long are they between the plates?
d. How far are they displaced?
SOLUTION: a. F = Eq
−16
6
= (1.0×10 C)(1.2×10 N/C) = 1.2×10−10 N
b. c.
Chapter Assessment: Writing in Physics
104. Choose the name of an electric unit, such as coulomb, volt, or farad, and research the life and work of the scientist
for whom it was named. Write a brief essay on this person and include a discussion of the work that justified the
honor of having a unit named for him.
SOLUTION: Student answers will vary. Some examples of scientists they could choose are Volta, Coulomb, Ohm,
and Ampère.
Chapter Assessment: Cumulative Review
105. Michelson measured the speed of light by sending a beam of light to a mirror on a mountain 35 km away.
a. How long does it take light to travel the distance to the mountain and back?
b. Assume Michelson used a rotating octagon with a mirror on each face of the octagon. Also assume the light
eSolutions Manual - Powered by Cognero
Page 34
reflects from one mirror travels to the other mountain, reflects off of a fixed mirror on that mountain, and returns to
the rotating mirrors. If the rotating mirror has advanced so that when the light returns, it reflects off of the next
mirror in the rotation, how fast is the mirror rotating?
honor of having a unit named for him.
SOLUTION: Student answers will vary. Some examples of scientists they could choose are Volta, Coulomb, Ohm,
Chapter
21 Practice Problems, Review, and Assessment
and Ampère.
Chapter Assessment: Cumulative Review
105. Michelson measured the speed of light by sending a beam of light to a mirror on a mountain 35 km away.
a. How long does it take light to travel the distance to the mountain and back?
b. Assume Michelson used a rotating octagon with a mirror on each face of the octagon. Also assume the light
reflects from one mirror travels to the other mountain, reflects off of a fixed mirror on that mountain, and returns to
the rotating mirrors. If the rotating mirror has advanced so that when the light returns, it reflects off of the next
mirror in the rotation, how fast is the mirror rotating?
1 1 c. If each mirror has a mass of 1.0×10 g and rotates in a circle with an average radius of 1.0×10 cm, what is the
approximate centripetal force needed to hold the mirror while is it rotating?
SOLUTION: a. (35 km/trip)(2 trips)(1000 m/1 km) / 3.00×108 m/s = 2.3×10−4 s
b. Note that if students carry extra digits from part a to prevent rounding errors, they will get an answer of
2
5.4×10 rev/s.
c.
106. Mountain Scene You can see an image of a distant mountain in a smooth lake just as you can see a mountain
biker next to the lake because light from each strikes the surface of the lake at about the same angle of incidence
and is reflected to your eyes. If the lake is about 100 m in diameter, the reflection of the top of the mountain is about in the middle of the lake, the mountain is about 50 km away from the lake, and you are about 2 m tall, then approximately how high above the lake does the top of the mountain reach?
SOLUTION: Since
the- Powered
angle ofbyincidence
of the light from the top of the mountain is equal to its angle of reflection
eSolutions
Manual
Cognero
Page 35
from the lake, you and the reflection of the top of the mountain form a triangle that is similar to a
triangle formed by the mountain and the top of its reflection in the lake. Your height makes up one side,
Chapter
21 Practice Problems, Review, and Assessment
106. Mountain Scene You can see an image of a distant mountain in a smooth lake just as you can see a mountain
biker next to the lake because light from each strikes the surface of the lake at about the same angle of incidence
and is reflected to your eyes. If the lake is about 100 m in diameter, the reflection of the top of the mountain is about in the middle of the lake, the mountain is about 50 km away from the lake, and you are about 2 m tall, then approximately how high above the lake does the top of the mountain reach?
SOLUTION: Since the angle of incidence of the light from the top of the mountain is equal to its angle of reflection
from the lake, you and the reflection of the top of the mountain form a triangle that is similar to a
triangle formed by the mountain and the top of its reflection in the lake. Your height makes up one side,
hyou = 2 m and the top of the mountain is halfway across the lake, d you = 50 m. The mountain is a
distance d mountain = 50,000 m from its reflection. Find hmountain by equating the ratios of the sides of
the two similar triangles.
107. A converging lens has a focal length of 38.0 cm. If it is placed 60.0 cm from an object, at what distance from the lens will the image be?
SOLUTION: = 104 cm
The image is 104 cm from the lens.
108. A force (F) is measured between two charges (Q and q) separated by a distance (r). What would the new force
be for the following?
a. r is tripled
b. Q is tripled
c. both r and Q are tripled
d. both r and Q are doubled
e. all three, r, Q, and q, are tripled
SOLUTION: a. F/9
b. 3F
c. F/3
eSolutions
Manual - Powered by Cognero
d. F/2
e. F
Page 36
Chapter
21 Practice Problems, Review, and Assessment
= 104 cm
The image is 104 cm from the lens.
108. A force (F) is measured between two charges (Q and q) separated by a distance (r). What would the new force
be for the following?
a. r is tripled
b. Q is tripled
c. both r and Q are tripled
d. both r and Q are doubled
e. all three, r, Q, and q, are tripled
SOLUTION: a. F/9
b. 3F
c. F/3
d. F/2
e. F
eSolutions Manual - Powered by Cognero
Page 37