18

ELECTRIC POTENTIAL AND CAPACITANCE

Answers to Multiple-Choice Problems 1. A, C

2. C

3. A 4. A, D 5. C

6. B, C

7. B, D 8. C

9. B

10. C

11. D 12. A 13. A 14. A, C

15. C

Solutions to Problems 18.1. Set Up: Since the charge is positive the force on it is in the same direction as the electric field. Since the field is uniform the force is constant and W 5 Fs cos f. S S Solve: (a) F is upward and s is to the right, so f 5 90°and W 5 0. S S (b) F is upward and s is upward, so f 5 0°. S

W 5 Fs 5 qEs 5 1 28.0 3 1029 C 2 1 4.00 3 104 N / C 2 1 0.670 m 2 5 7.50 3 1024 J. S

(c) F is upward and s is at 45.0° below the horizontal, so f 5 135.0°. W 5 Fs cos f 5 qEs cos f 5 1 28.0 3 1029 C 2 1 4.00 3 104 N / C 2 1 2.60 m 2 cos 135.0° 5 22.06 3 1023 J. Reflect: The work is positive when the displacement has a component in the direction of the force and it is negative when the displacement has a component opposite to the direction of the force. When the displacement is perpendicular to the force the work done is zero.

18.2. Set Up: (a) The proton has charge 1e and mass 1.67 3 10227 kg. Let point a be at the negative plate and point b be at the positive plate. The electric field is directed from the positive plate toward the negative plate. The force on the positively charged proton is in this direction. The proton moves in a direction opposite to the electric force, so the work done by the electric force is negative. Solve: Wa S b 5 2qEd 5 2eEd 5 2 1 1.60 3 10219 C 2 1 2.80 3 106 N / C 2 1 0.100 m 2 5 24.48 3 10214 J q1q2 r 1 8.99 3 109 N # m2 / C2 2 1 27.20 3 1026 C 2 112.30 3 1026 C 2 kq1q2 Solve: r 5 5 5 0.372 m U 20.400 J

18.3. Set Up: U 5 k

q1q2 . ra 5 0.150 m, rb 5 " 1 0.250 m 2 2 1 1 0.250 m 2 2 5 0.354 m. r 1 8.99 3 109 N # m2 / C2 2 1 2.40 3 1026 C 2 1 24.30 3 1026 C 2 q1q2 Solve: Ua 5 k 5 5 20.619 J. ra 0.150 m

18.4. Set Up: Wa S b 5 Ua 2 Ub . U 5 k

Ub 5 k

1 8.99 3 109 N # m2 / C2 2 1 2.40 3 1026 C 2 1 24.30 3 1026 C 2 q1q2 5 5 20.262 J. rb 0.354 m Wa S b 5 Ua 2 Ub 5 20.619 J 2 1 20.262 J 2 5 20.357 J.

18-1

18-2

Chapter 18

qqr . Let r q1 5 13.00 nC and q2 5 12.00 nC. Let q3 be the electron. The potential energy of the electron is the sum of U13 and U23 . Solve: (a) r1 5 r2 5 0.250 m.

18.5. Set Up: An electron has charge q 5 2e 5 21.60 3 10219 C. For a pair of point charges, U 5 k

U13 5 k U23 5 k

1 3 21

q2q3 5 1 8.99 3 109 N # m2 / C2 r2

2.00 3 1029 C 4 3 21.60 3 10219 C 0.250 m

U 5 U13 1 U23 5 22.88 3 10217 J (b) r1 5 0.100 m and r2 5 0.400 m. U13 5 k U23 5 k

2 4 2

3 3.00 3 1029 C 4 3 21.60 3 10219 C 4 q1q3 5 1 8.99 3 109 N # m2 / C2 2 5 21.73 3 10217 J. r1 0.250 m

1 3 21

2 4 2

5 21.15 3 10217 J.

3 3.00 3 1029 C 4 3 21.60 3 10219 C 4 q1q3 5 1 8.99 3 109 N # m2 / C2 2 5 24.32 3 10217 J. r1 0.100 m q2q3 5 1 8.99 3 109 N # m2 / C2 r2

2.00 3 1029 C 4 3 21.60 3 10219 C 0.400 m

U 5 U13 1 U23 5 25.04 3 10217 J.

5 27.19 3 10218 J.

Reflect: The potential energy is negative since the charge of the electron has opposite sign from the charge of each of the other two particles. The magnitude of the potential energy increases when the electron moves toward the larger charge. qqr . In the O-H-N combination r the O2 is 0.170 nm from the H1 and 0.280 nm from the N2. In the N-H-N combination the N2 is 0.190 nm from the H1 and 0.300 nm from the other N2. U is positive for like charges and negative for unlike charges. Solve: (a) O-H-N 1 1.60 3 10219 C 2 2 O2-H 1 : U 5 2 1 8.99 3 109 N # m2 / C2 2 5 21.35 3 10218 J 0.170 3 1029 m

18.6. Set Up: For a pair of point charges the electrical potential energy is U 5 k

O2-N2: U 5 1 8.99 3 109 N # m2 / C2 2

1 1.60 3 10219 C 2 2 5 18.22 3 10219 J 0.280 3 1029 m

N-H-N N2-H 1 : U 5 2 1 8.99 3 109 N # m2 / C2 2 N2-N2: U 5 1 8.99 3 109 N # m2 / C2 2

1 1.60 3 10219 C 2 2 5 21.21 3 10218 J 0.190 3 1029 m

1 1.60 3 10219 C 2 2 5 17.67 3 10219 J 0.300 3 1029 m

The total potential energy is Utot 5 21.35 3 10218 J 1 8.22 3 10219 J 2 1.21 3 10218 J 1 7.67 3 10219 J 5 2 9.71 3 10219 J . (b) In the hydrogen atom the electron is 0.0529 nm from the proton. 1 1.60 3 10219 C 2 2 5 24.35 3 10218 J. U 5 2 1 8.99 3 109 N # m2 / C2 2 0.0529 3 1029 m The magnitude of the potential energy in the hydrogen atom is about a factor of 4 larger than what it is for the adenine-thymine bond.

Electric Potential and Capacitance

18-3

qqr . In the O-H-O combination r the O2 is 0.180 nm from the H1 and 0.290 nm from the other O2. In the N-H-N combination the N2 is 0.190 nm from the H1 and 0.300 nm from the other N2. U is positive for like charges and negative for unlike charges. In the O-H-N combination the O2 is 0.180 nm from the H1 and 0.290 nm from the N2. U is positive for like charges and negative for unlike charges. Solve: O-H-O O2-H1, U 5 21.28 3 10218 J; O2-O2, U 5 17.93 3 10219 J N-H-N N2-H1, U 5 21.21 3 10218 J; N2-N2, U 5 17.67 3 10219 J O-H-N O2-H1, U 5 21.28 3 10218 J; O2-N2, U 5 17.93 3 10219 J The total potential energy is 23.77 3 10218 J 1 2.35 3 10218 J 5 21.42 3 10218 J. Reflect: For pairs of opposite sign the potential energy is negative and for pairs of the same sign the potential energy is positive. The net electrical potential energy is the algebraic sum of the potential energy of each pair.

18.7. Set Up: For a pair of point charges the electrical potential energy is U 5 k

18.8. Set Up: The initial potential energy is converted to kinetic energy after the charges have moved far apart. qqr Apply U 5 k to each pair of charges and add to get the total electrical potential energy. In part (b) the distance r between the 4.00 mC and 3.00 mC charges is 18.0 cm. 8.99 3 109 N # m2 / C2 q1q3 q2q3 q1q2 1 3 24.00 3 1029 C 4 3 22.00 3 1029 C 4 1 1 1 5 Solve: (a) U 5 k r12 r13 r23 0.100 m

1

2 1

2

3 22.00 3 1029 C 4 3 23.00 3 1029 C 4 1 3 23.00 3 1029 C 4 3 24.00 3 1029 C 4 2 5 2.34 3 1026 J

(b) U 5 k

1

2

1

3 4.00 3 1026 C 4 3 2.00 3 1026 C 4 q1q3 q2q3 q1q2 1 1 1 5 1 8.99 3 109 N # m2 / C2 2 r12 r13 r23 0.100 m

3 2.00 3 1026 C 4 3 3.00 3 1026 C 4 3 4.00 3 1026 C 4 3 3.00 3 1026 C 4 2 5 1.68 J 1 0.150 m 0.180 m

The maximum total kinetic energy will be 1.68 J. The kinetic energy increases as the electrical potential energy decreases, as the charges move apart. The maximum kinetic energy is achieved only after a long time. q1q 1r . r2 5 2r1 5 2d. q2 5 2q1 and q 2r 5 2q 1r . d 1 2q1 2 1 2q 1r 2 q2q 2r q1q 1r Solve: U2 5 k 5k 5 2k 5 2U r2 2d d Reflect: For point charges, the electrical potential energy is proportional to 1 / r and the electrical force is proportional to 1 / r 2. The force between the charges would not change.

18.9. Set Up: U1 5 U 5 k

18.10. Set Up: U 5 k

qqr . U2 5 3U1 . r1 5 R r

Solve: Ur 5 kqqr, which is constant. So, U1r1 5 U2r2 . r2 5

1 2 1 2

U1 U1 r1 5 R 5 R/3 U2 3U1

18.11. Set Up: For a pair of oppositely charged parallel metal plates, Vab 5 Ed. F 5 0 q 0 E.

Vab 360 V 5 5 8000 V / m 5 8000 N / C d 45.0 3 1023 m (b) F 5 0 q 0 E 5 1 2.40 3 1029 C 2 1 8000 N / C 2 5 1.92 3 1025 N Solve: (a) E 5

18.12. Set Up: For a pair of oppositely charged metal parallel metal plates, Vab 5 Ed. Solve: d 5

Vab 4.75 3 103 V 5 1.58 3 1023 m 5 1.58 mm 5 E 3.00 3 106 V / m

18-4

Chapter 18

18.13. Set Up: For two oppositely charged parallel plates, Vab 5 Ed, where Vab is the potential difference between the two plates, E is the uniform electric field between the plates and d is the separation of the plates. An electric field S S S E exerts a force F 5 0 q 0 E on a charge placed in the field. a 5 F / m. An electron has charge 2e and mass 9.11 3 10231 kg. Vab 25 V Solve: (a) E 5 5 5 1.25 3 104 V / m d 2.0 3 1023 m 1 1.60 3 10219 C 2 1 1.25 3 104 V / m 2 eE F (b) F 5 0 q 0 E 5 eE. a 5 5 5 5 2.20 3 1015 m / s2 m m 9.11 3 10231 kg Reflect: The electric field is the same at all points between the plates (away from the edges) so the acceleration would be the same at all points between the plates as it is for a point midway between the plates. 18.14. Set Up: For two oppositely charged sheets of charge, Vab 5 Ed. The positively charged sheet is the one at higher potential. Vab 70 3 1023 V 5 5 9.3 3 106 V / m. The electric field is directed inward, toward the interior of d 7.5 3 1029 m S the axon, since the outer surface of the membrane has positive charge and E points away from positive charge and toward negative charge. Section 18.9 explores the effects of a material other than air between the plates. (b) The outer surface has positive charge so it is at higher potential than the inner surface.

Solve: (a) E 5

18.15. Set Up: Vab 5 Ed for parallel plates. Solve: d 5

Vab 1.5 V 5 1.5 3 106 m 5 1.5 3 103 km 5 E 1.0 3 1026 V / m

Wa S b 5 Va 2 Vb . Point a is the starting and q point b is the ending point. Since the field is uniform, Wa S b 5 Fs cos f 5 E 0 q 0 s cos f. The field is to the left so the force on the positive charge is to the left. The particle moves to the left so f 5 0° and the work Wa S b is positive. Solve: (a) Wa S b 5 Kb 2 Ka 5 1.50 3 1026 J 2 0 5 1.50 3 1026 J Wa S b 1.50 3 1026 J 5 5 357 V. Point a is at higher potential than point b. (b) Va 2 Vb 5 q 4.20 3 1029 C Wa S b Va 2 Vb 357 V 5 5 5.95 3 103 V / m. 5 (c) E 0 q 0 s 5 Wa S b , so E 5 s 0q0s 6.00 3 1022 m

18.16. Set Up: The work-energy theorem says Wa S b 5 Kb 2 Ka .

18.17. Set Up: From Example 18.4, with Vb 5 0 and x rather than y as the distance from the negative plate, Vx 5 Ex. Vx 5.0 V 5 5 25 V / m Solve: E 5 x 0.200 m Reflect: Vx 5 Ex says the potential increases linearly with x and the graph of Vx versus x should be a straight line, in agreement with the figure in the problem. kq . r 1 48.0 V 2 1 0.250 m 2

18.18. Set Up: For a point charge V 5

Vr 5 5 1.33 3 1029 C k 8.99 3 109 N # m2 / C2 r1 25.0 cm 5 1 48.0 V 2 5 16.0 V. (b) Vr 5 kq 5 constant so V1r1 5 V2r2 . V2 5 V1 r2 75.0 cm Solve: (a) q 5

12

1

2

kq . r 1 8.99 3 109 N # m2 / C2 2 1 2.50 3 10211 C 2 kq 5 5 2.50 3 1023 m 5 2.50 mm Solve: (a) r 5 V 90.0 V V1 90.0 V 5 1 2.50 mm 2 5 7.50 mm. (b) Vr 5 kq 5 constant so V1r1 5 V2r2 . r2 5 r1 V2 30.0 V

18.19. Set Up: For a point charge V 5

1 2

1

2

Electric Potential and Capacitance

18-5

q r

18.20. Set Up: For a point charge, V 5 k . r2 5 3r1 . q q q 5 V/3 Solve: V1 5 k . V2 5 k 5 k r1 r2 3r1

18.21. Set Up: From Chapter 17 we know that for a spherical shell the electric field outside the shell is the same as for a point charge located at the center of the shell. The electric field determines the force on a test charge placed outside the shell, and the work done on the test charge as it moves between two points determines the potential difference between those points. Therefore, outside the shell the potential is the same as for a point charge. 1 8.99 3 109 N # m2 / C2 2 1 3.50 3 1029 C 2 q Solve: (a) V 5 k 5 5 71.5 V r 0.200 m 1 0.240 m 0.440 m 1 71.5 V 2 5 131 V. (b) Now r 5 0.240 m, so V 5 0.240 m Reflect: The potential is positive since the charge is positive and the potential increases at points closer to the surface of the sphere.

1

2

18.22. Set Up: For a single point charge V 5

kq . The total potential is the sum of the potentials due to the two r

point charges. WB S A 5 q 1 VB 2 VA 2 . 8.99 3 109 N # m2 / C2 kq1 kq2 1 2.40 3 1029 C 1 3 26.50 3 1029 C 4 2 5 2737 V 1 5 Solve: (a) VA 5 r1A r2A 0.050 m kq1 kq2 2.40 3 1029 C 26.50 3 1029 C 1 5 1 8.99 3 109 N # m2 / C2 2 1 5 2704 V (b) VB 5 r1B r2B 0.080 m 0.060 m (c) WB S A 5 q 1 VB 2 VA 2 5 1 2.50 3 1029 C 2 1 2704 V 2 1 2737 V 2 2 5 8.2 3 1028 J

1

2

qQ . Conservation of energy says Ua 1 Ka 5 Ub 1 Kb . K 5 12 mv2. r 1 1.20 3 1026 C 2 1 4.60 3 1026 C 2 5 0.198 J Solve: (a) U 5 1 8.99 3 109 N # m2 / C2 2 0.250 m (b) (i) Ua 5 10.198 J, Ka 5 0. rb 5 0.500 m so Ub 5 12 Ua 5 0.099 J.

18.23. Set Up: U 5 k

2 1 0.099 J 2 2Kb 5 5 26.6 m / s. Å m Å 2.80 3 1024 kg (ii) rb 5 5.00 m so Ub 5 0.0099 J. Kb 5 0.198 J 2 0.0099 J 5 0.188 J. vb 5 36.7 m / s. (iii) rb 5 50.0 m so Ub 5 0.00099 J. Kb 5 0.198 J 2 0.00099 J 5 0.197 J. vb 5 37.5 m / s. Reflect: As the charge q moves away from Q the repulsive force does positive work on q and its kinetic energy increases. Kb 5 Ua 1 Ka 2 Ub 5 0.198 J 2 0.099 J 5 0.099 J. vb 5

18.24. Set Up: Let a be when they are 0.750 nm apart and b when they are very far apart. A proton has charge 1e and mass 1.67 3 10227 kg. As they move apart the protons have equal kinetic energies and speeds. Solve: (a) They have maximum speed when they are far apart and all their initial electrical potential energy has been converted to kinetic energy. Ka 1 Ua 5 Kb 1 Ub . Ka 5 0 and Ub 5 0, so Kb 5 Ua 5 k

1 1.60 3 10219 C 2 2 e2 5 1 8.99 3 109 N # m2 / C2 2 5 3.07 3 10219 J. ra 0.750 3 1029 m

Kb 5 12 mvb2 1 12 mvb2 , so Kb 5 mvb2 and vb 5

Kb 3.07 3 10219 J 5 5 1.36 3 104 m / s Åm Å 1.67 3 10227 kg

(b) Their acceleration is largest when the force between them is largest and this occurs at r 5 0.750 nm, when they are closest. e2 1.60 3 10219 C 2 F 5 k 2 5 1 8.99 3 109 N # m2 / C2 2 5 4.09 3 10210 N r 0.750 3 1029 m 4.09 3 10210 N F 5 2.45 3 1017 m / s2 a5 5 m 1.67 3 10227 kg

1

2

18-6

Chapter 18

18.25. Set Up: Apply Ka 1 Ua 5 Kb 1 Ub , with Ka 5 0 and vb 5 v. Solve: (a) Kb 5 Ua 2 Ub 5 q 1 Va 2 Vb 2 5 2e 1 Va 2 Vb 2 . An electron gains energy when it moves to higher potential, so Vb 2 Va 5 V and Kb 5 eV. Kb 5 12 mv2 so v5 2 1 1.60 3 10219 C 2 1 95 V 2

2Kb 2eV 5 . Å m Å m

5 5.8 3 106 m / s Å 9.11 3 10231 kg Reflect: A positive charge gains kinetic energy when it moves from high potential to low potential but a negative charge gains kinetic energy when it moves from low potential to high potential. (b) v 5

18.26. Set Up: From Problem 18.25, v 5 "2eV / m for a particle with charge of magnitude e. The speed of light is

c 5 3.00 3 108 m / s. An electron has mass 9.11 3 10231 kg and a proton has mass 1.67 3 10227 kg. 1 9.11 3 10231 kg 2 1 3.00 3 106 m / s 2 2 mv2 5 5 26 V Solve: (a) v 5 1 0.010 2 c 5 3.00 3 106 m / s. V 5 2e 2 1 1.60 3 10219 C 2 (b) From Problem 18.25, the kinetic energy the particle gains is K 5 eV so V 5 26 V gives electrons and protons the same kinetic energy. But the protons must be accelerated by a potential decrease whereas the electrons are accelerated by a potential increase. 2 1 1.60 3 10219 C 2 1 26 V 2 2eV 5 5 7.0 3 104 m / s 5 1 0.024% 2 c (c) v 5 Å m Å 1.67 3 10227 kg

18.27. Set Up: Ka 1 Ua 5 Kb 1 Ub and U 5 qV. K 5 12 mv2.

Solve: (a) Ka 5 0 so Kb 5 Ua 2 Ub 5 q 1 Va 2 Vb 2 . v5

2q 1 Va 2 Vb 2 2Kb 5 m Å m Å

2q v1 v2 , which is constant. Let Va 2 Vb 5 V, the accelerating voltage. Then 5 . V2 5 2V1 "Va 2 Vb Å m "V1 "V2 V2 and v1 5 v. Thus v 5 v 5 "2v. Å V1 1 v

5

1 2

v22 v12 v22 5 . v2 5 3v so V2 5 V1 2 5 9V1 5 900 V. V1 V2 v1 Reflect: If the charge is positive it is accelerated by a decrease in potential. If it is negative it is accelerated by an increase in potential. (b)

18.28. Set Up: Example 18.4 shows that V 5 Ey, where y is the distance from the negative plate. V 5 0 at the DV . negative plate. For y 5 6.0 cm, V 5 12.0 V. The potential gradient is Dy 12.0 V 6.0 V V 5 3.0 cm; V 5 6.0 V when y 5 3.0 cm. The 16.0 V 5 2.0 V / cm. y 5 5 Solve: (a) E 5 6.0 cm E 2.0 V / cm equipotential surface is a flat sheet parallel to the plates and midway between them. 2.0 V V 5 1.0 cm. The 12.0 V equipotential surface is a flat sheet parallel to the plates and 1.0 cm (b) y 5 5 E 2.0 V / cm from the negative plate and 5.0 cm from the positive plate. (c)

12.0 V DV 5 5 200 V / m Dy 6.0 3 1022 m

Electric Potential and Capacitance

18-7

18.29. Set Up: Example 18.4 shows that V 5 Ey, where y is the distance from the negative plate. V 5 0 at the negative plate. For y 5 25 cm, V 5 50.0 V. 50.0 V 5 2.0 V / cm. So V 5 1 2.0 V / cm 2 y. Solve: (a) E 5 25 cm V 5 110.0 V for y 5 5 cm, V 5 120.0 V for y 5 10 cm, V 5 130.0 V for y 5 15 cm, V 5 140.0 V for y 5 20 cm and V 5 150.0 V for y 5 25 cm. The equipotential surfaces are drawn in Figure 18.29. +

positive plate

+

+

+

+

V0 1 50.0 V V0 1 40.0 V V0 1 30.0 V V0 1 20.0 V V0 1 10.0 V

–

negative plate

–

–

–

–

V0

Figure 18.29 (b) Yes, they are separated by 5 cm. (c) The equipotential surfaces are flat sheets parallel to the plates. Reflect: The electric field lines are straight lines perpendicular to the plates, so are perpendicular to the equipotential surfaces, as they must be. The electric field is uniform so the equipotential lines of constant potential difference are equally spaced. kq . r 9 2 2 1 8.99 3 10 N # m / C 2 1 5.00 3 10212 C 2 kq 4.50 V # cm Solve: (a) r 5 5 5 V V V V 5 1.00 V at r 5 4.50 cm; V 5 2.00 V at r 5 2.25 cm; V 5 3.00 V at r 5 1.50 cm; V 5 4.00 V at r 5 1.12 cm and V 5 5.00 V at r 5 0.90 cm. The equipotential surfaces are sketched in Figure 18.30.

18.30. Set Up: For a point charge, V 5

1.00 V

3.00 V

2.00 V

5.00 V 4.00 V

Figure 18.30 (b) These equipotential surfaces are not equally spaced in distance. The V 5 1.00 V and V 5 2.00 V surfaces are farther apart than the V 5 4.00 V and V 5 5.00 V surfaces. The equipotential surfaces are closer together where E is larger. (c) The equipotential surfaces are concentric spheres with the point charge at their common center.

18-8

Chapter 18

kq . r Solve: (a) and (c) The V 5 0 equipotential is the flat surface consisting of points equidistant from the two charges. The projection of this surface with a drawing in which the point charges are in the plane of the paper is a line along the perpendicular bisector of the line that connects the two charges, as shown in Figure 18.31.

18.31. Set Up: For a point charge V 5

2.00 cm 210.0 mC

2.00 cm 110.0 mC

Figure 18.31 (b) Since the point charges have equal magnitudes and opposite signs, V 5 0 at the point midway between them, since this point is equidistant from the two charges.

18.32. Set Up: The electric field lines are perpendicular to the equipotential surfaces. The electric field lines are closer together where E is larger, and where E is larger the equipotential surfaces for equal potential differences are closer together. The projection of the equipotential surfaces into the plane of the figure are lines that satisfy these two requirements. Solve: For the electric field given in the problem it is not possible to draw equipotential lines that satisfy both requirements. The electric field shown in the problem cannot be produced by an electrostatic distribution of charges.

18.33. Set Up and Solve: The electric field lines and intersections of the equipotential surfaces with the plane of the drawing are shown in Figure 18.11b in the textbook. Reflect: The figure shows that the electric field lines are always perpendicular to the equipotential surfaces. Also, the electric field lines are closer together where E is larger, and where E is larger the equipotential surfaces for equal potential differences are closer together.

18.34. Set Up: The mass of a drop is m 5 rV, where V 5 43 pr 3 (the volume of a sphere). r 5 0.820 g / cm3

5 820 kg / m3. The drops have r 5 5.00 3 1027 m. The electric field and electric force magnitudes are related by F 5 0 q 0 E, where q is the net charge of the drop. mg . 0 q 0 5 5e 5 8.00 3 10219 C. Solve: (a) F 5 w, so 0 q 0 E 5 mg and E 5 0q0 m 5 rV 5 r 43 pr 3 5 1 820 kg / m3 2 1 43 2 p 1 5.00 3 1027 m 2 3 5 4.29 3 10216 kg E5

1 4.29 3 10216 kg 2 1 9.80 m / s2 2 5 5.26 3 103 V / m 8.00 3 10219 C

(b) V 5 Ed 5 1 5.26 3 103 V / m 2 1 2.25 3 1022 m 2 5 118 V 1 4.29 3 10216 kg 2 1 9.80 m / s2 2 mg V 73.8 V 5 (c) E 5 5 5 3.28 3 103 V / m. 0 q 0 5 5 1.28 3 10218 C. 22 d E 2.25 3 10 m 3.28 3 103 V / m 0q0 1.28 3 10218 C 5 5 8. The number of electrons is N 5 e 1.60 3 10219 C

18.35. Set Up: 1 eV 5 1.60 3 10219 J. K 5 12 mv2. c 5 3.00 3 108 m / s. An electron has mass 9.11 3 10231 kg

and a proton has mass 1.67 3 10227 kg.

Electric Potential and Capacitance

Solve: (a) K 5 1.00 eV 5 1.60 3 10219 J. v 5 2 1 1.60 3 10219 J 2

18-9

2K Åm

5 5.93 3 105 m / s Å 9.11 3 10231 kg mp 2 1 1.60 3 10219 J 2 ve proton: vp 5 5 1.38 3 104 m / s. 5 5 "1836 227 vp Å me Å 1.67 3 10 kg (b) K 5 1.00 keV 5 1.60 3 10216 J electron: ve 5 1.87 3 107 m / s; proton: vp 5 4.38 3 105 m / s (c) ve 5 vp 5 1 0.0100 2 c 5 3.00 3 106 m / s electron: ve 5

Ke 5 12 m eve2 5 12 1 9.11 3 10231 kg 2 1 3.00 3 106 m / s 2 2 5 4.10 3 10218 J 5 0.0256 keV Kp 5 12 mpvp2 5 12 1 1.67 3 10227 kg 2 1 3.00 3 106 m / s 2 2 5 7.52 3 10215 J 5 47.0 keV

The proton energy is larger by a factor of mp / m e . Reflect: When we use K 5 12 mv2 we should express all quantities in SI units. Q Vab 1.25 3 1026 C 5 1.11 3 1027 F 5 0.111 mF Solve: (a) C 5 11.3 V (b) Q 5 CVab 5 1 7.28 3 1026 F 2 1 25.0 V 2 5 1.82 3 1024 C 5 182 mC

18.36. Set Up: C 5

1

21

2

18.37. Set Up: 1 F 5 1 C2 / N # m and 1 N 5 1 kg # m / s2. Solve: 1 F 5

1 C2 1N C2 # s2 5 N # m 1 kg # m / s2 kg # m2

Q Q . and C 5 P0 A Vab 6 23 4 Solve: (a) Vab 5 Ed 5 1 4.00 3 10 V / m 2 1 2.50 3 10 m 2 5 1.00 3 10 V Q 80.0 3 1029 C 5 5 2.26 3 1023 m2 5 22.6 cm2 (b) A 5 6 EP0 1 4.00 3 10 V / m 2 1 8.854 3 10212 C2 / 1 N # m2 2 Q 80.0 3 1029 C 5 5 8.00 3 10212 F 5 8.00 pF (c) C 5 Vab 1.00 3 104 V

18.38. Set Up: For a parallel-plate capacitor, Vab 5 Ed, E 5

18.39. Set Up: For a parallel-plate capacitor C 5

P0 A Q Q .C5 . Vab 5 Ed. The surface charge density is s 5 . Vab d A

Q 0.200 3 1026 C 5 5 400 V C 500.0 3 10212 F 1 500.0 3 10212 F 2 1 0.600 3 1023 m 2

Solve: (a) Vab 5

Cd 5 5 0.0339 m2 P0 8.854 3 10212 C2 / 1 N # m2 2 Vab 400 V 5 5 6.67 3 105 V / m (c) E 5 d 0.600 3 1023 m Q 0.200 3 1026 C (d) s 5 5 5 5.90 3 1026 C / m2 A 0.0339 m2 Reflect: If the plates are square, each side is about 18 cm in length. We could also calculate s from E 5 s / P0 . (b) A 5

A d 1 8.854 3 10212 F / m 2 1 3.0 3 1022 m 2 2

18.40. Set Up: C 5 P0 when there is air between the plates. Solve: C 5

5.0 3 1023 m

5 1.59 3 10212 F 5 1.59 pF

P0 A Q .C5 . Vab d Solve: (a) Q 5 CVab 5 1 10.0 3 1026 F 2 1 12.0 V 2 5 1.20 3 1024 C 5 120 mC (b) When d is doubled C is halved, so Q is halved. Q 5 60 mC.

18.41. Set Up: C 5

18-10

Chapter 18

(c) If r is doubled, A increases by a factor of 4. C increases by a factor of 4 and Q increases by a factor of 4. Q 5 480 mC. Reflect: When the plates are moved apart, less charge on the plates is required to produce the same potential difference. With the separation of the plates constant, the electric field must remain constant to produce the same potential difference. The electric field depends on the surface charge density, s. To produce the same s, more charge is required when the area increases.

18.42. Set Up: C 5

P0 A Q .C5 . Vab d

Solve: (a) 12.0 V Q and Q is constant, so V doubles. V 5 24.0 V. C (ii) When r is doubled, A increases by a factor of 4. V decreases by a factor of 4 and V 5 3.0 V. (b) (i) When d is doubled, C is halved. Vab 5

A d 1 8.854 3 10212 F / m 2 p 1 12.2 3 1023 m 2 2

18.43. Set Up: C 5 P0 . We measure that the diameter of a quarter is 24.5 mm, so r 5 12.2 3 1023 m. C 5

Q V

5 2.07 3 10212 F 5 2.07 pF 2.00 3 1023 m (b) Q 5 CV 5 1 2.07 3 10212 F 2 1 12 V 2 5 2.48 3 10211 C 5 24.8 pC Solve: (a) C 5

18.44. Set Up: C 5 Solve: (a) d 5

P0 A Q . V 5 Ed. C 5 . Vab ab d

Vab 1.00 3 102 V 5 1.00 3 1022 m 5 1.00 cm. 5 E 1.00 3 104 N / C A5

1 5.00 3 10212 F 2 1 1.00 3 1022 m 2 Cd 5 5 5.65 3 1023 m2. P0 8.854 3 10212 C2 / 1 N # m2 2

A 5 4.24 3 1022 m 5 4.24 cm. Åp (b) Q 5 CVab 5 1 5.00 3 10212 F 2 1 1.00 3 102 V 2 5 5.00 3 10210 C 5 500 pC A 5 pr 2 so r 5

18.45. Set Up: C 5 P0

P0pr 2 A 5 d d

P0p 1 3r0 2 2 P0pr 2 P0pr02 P0pr02 . r 5 3r0; d 5 3d0 . C 5 53 5 3C0 5 d0 d 3d0 d0 Reflect: C for a capacitor depends only on the dimensions of the capacitor. Solve: C0 5

A d

18.46. Set Up: Since the capacitor is disconnected from the battery, Q does not change. C 5 P0 . C 5 A A A C 5 P0 5 . Solve: (a) C 5 P0 . Cnew 5 P0 d dnew 2d 2 (b) Q 5 Q0 Q0 Q0 Q0 Q0 .V 5 5 5 2V0 (c) V0 5 52 C new Cnew C C/2

Q . V

18.47. Set Up: The capacitors between b and c are in parallel. This combination is in series with the 15 pF capacitor. C1 5 15 pF, C2 5 9.0 pF and C3 5 11 pF. Solve: (a) For capacitors in parallel, Ceq 5 C1 1 C2 1 c so C23 5 C2 1 C3 5 20 pF 1 1 1 5 1 1 c so (b) C1 5 15 pF is in series with C23 5 20 pF. For capacitors in series, Ceq C1 C2 1 15 pF 2 1 20 pF 2 C1C23 1 1 1 5 1 5 5 8.6 pF. and C123 5 C123 C1 C23 C1 1 C23 15 pF 1 20 pF

Reflect: For capacitors in parallel the equivalent capacitance is larger than any of the individual capacitors. For capacitors in series the equivalent capacitance is smaller than any of the individual capacitors.

Electric Potential and Capacitance

18-11

18.48. Set Up: For capacitors in series the voltage across the combination equals the sum of the voltages in the individual capacitors. For capacitors in parallel the voltage across the combination is the same as the voltage across each individual capacitor. Solve: (a) Connect the capacitors in series so their voltages will add. (b) V 5 V1 1 V2 1 V3 1 c5 NV1 , where N is the number of capacitors in the series combination, since the capacitors are identical. V 500 V N5 5 5 5000. V1 0.10 V 1 1 1 5 1 1 c For capacitors in parallel, Ceq 5 C1 1 C2 1 c Ceq C1 C2 Solve: The possible combinations and their equivalent capacitance are shown in Figure 18.49.

18.49. Set Up: For capacitors in series,

5 mF

10 mF

3.3 mF 20 mF

15 mF

6.7 mF

30 mF

Figure 18.49 1 1 1 5 1 1 c For N equivalent capacitors in series, Ceq 5 C / N. Ceq C1 C2 For capacitors in parallel, Ceq 5 C1 1 C2 1 c For N equivalent capacitors in parallel, Ceq 5 NC. Solve: There are many ways to achieve the required equivalent capacitance. In each case one simple solution is shown in Figure 18.50.

18.50. Set Up: For capacitors in series,

450 nF

25 nF

50 nF

(c)

(a)

75 nF

(d)

Figure 18.50

(b)

18-12

Chapter 18

18.51. Set Up: In parallel, the potential difference is the same for each capacitor and is equal to the applied Q . Vab Solve: (a) and (b) The potential difference across each capacitor is 48.0 V. Q 5 CVab . For the 4.00 mF capacitor, Q 5 192 mC. For the 6.00 mF capacitor, Q 5 288 mC. potential difference. C 5

18.52. Set Up: In series, the charges on the capacitors are the same and the sum of the potential differences across the capacitors is the applied potential difference. C 5 Solve: (a) V1 1 V2 5 48.0 V. Q5

1

2

Q . Let C1 5 4.00 mF and C2 5 6.00 mF. Q1 5 Q2 5 Q. Vab

Q Q 1 5 48.0 V. C1 C2

1

2

3 4.00 3 1026 mF 4 3 6.00 3 1026 mF 4 C1C2 1 48.0 V 2 5 1 48.0 V 2 5 1.15 3 1024 C. C1 1 C 2 4.00 3 1026 mF 1 6.00 3 1026 mF

The charge on each capacitor is 115 mC. Q Q 1.15 3 1024 C 1.15 3 1024 C 5 28.8 V. V2 5 5 19.2 V. 5 5 (b) V1 5 26 C1 C2 4.00 3 10 F 6.00 3 1026 F The potential difference across the 4.00 mF capacitor is 28.8 V and the potential difference across the 6.00 mF capacitor is 19.2 V.

18.53. Set Up: C 5

Q . For two capacitors in parallel, Ceq 5 C1 1 C2 . For two capacitors in series, V C1C2 1 1 1 and Ceq 5 5 1 . Ceq C1 C2 C1 1 C2

For capacitors in parallel, the voltages are the same and the charges add. For capacitors in series, the charges are the same and the voltages add. Let C1 5 3.00 mF, C2 5 5.00 mF and C3 5 6.00 mF. C12 a

a C

b

b C3

(a)

(b)

Figure 18.53 Solve: (a) The equivalent capacitance for C1 and C2 in parallel is C12 5 C1 1 C2 5 8.00 mF. This gives the circuit shown in Figure 18.53a. In that circuit the equivalent capacitance is

1 8.00 mF 2 1 6.00 mF 2 C12C3 5 3.43 mF. 5 C12 1 C3 8.00 mF 1 6.00 mF This gives the circuit shown in Figure 18.53b. In Figure 18.53b, Q 5 CV 5 1 3.43 3 1026 F 2 1 24.0 V 2 5 8.23 3 1025 C. In Figure 18.53a each capacitor therefore has charge 8.23 3 1025 C. The potential differences are C5

V3 5

Q3 Q12 8.23 3 1025 C 8.23 3 1025 C 5 13.7 V and V12 5 5 10.3 V. 5 5 26 C3 C12 6.00 3 10 F 8.00 3 1026 F

Note that V3 1 V12 5 24.0 V. Then in the original circuit, V1 5 V2 5 V12 5 10.3 V.

Q1 5 V1C1 5 1 10.3 V 2 1 3.00 3 1026 F 2 5 3.09 3 1025 C.

Q2 5 V2C2 5 1 10.3 V 2 1 5.00 3 1026 F 2 5 5.15 3 1025 C. Q1 5 30.9 mC, Q2 5 51.5 mC and Q3 5 82.3 mC. Note that Q1 1 Q2 5 Q3 . (b) V1 5 10.3 V, V2 5 10.3 V and V3 5 13.7 V Reflect: Note that Q1 1 Q2 5 Q3 , V1 5 V2 and V1 1 V3 5 24.0 V

Electric Potential and Capacitance

18.54. Set Up: C 5

18-13

Q . For two capacitors in parallel, Ceq 5 C1 1 C2 . For two capacitors in series, V C1C2 1 1 1 5 1 . and Ceq 5 Ceq C1 C2 C1 1 C2

For capacitors in parallel, the voltages are the same and the charges add. For capacitors in series, the charges are the same and the voltages add. Label the capacitors as shown in Figure 18.54a. C1

C12

C2 a

a

C3

C3 b

b C4

(a)

C4

(b) C123

a

a

b

b

Ceq C4

(c)

(d)

Figure 18.54 Solve: The equivalent capacitance for C1 and C2 in series is C1C2 C C12 5 5 . C1 1 C2 2 This gives the circuit shown in Figure 18.54b. The equivalent capacitance for C12 and C3 in parallel is 3C C123 5 C12 1 C3 5 . 2 This gives the circuit shown in Figure 18.54c. The equivalent capacitance for C123 and C4 in series is Ceq 5

1 3C / 2 2 1 C 2 C123C4 5 5 0.600C 5 2.40 3 1026 F. C123 1 C4 3C / 2 1 C

This gives the circuit shown in Figure 18.54d. In Figure 18.54d, Q 5 CeqVab 5 1 2.40 3 1026 F 2 1 28.0 V 2 5 6.72 3 1025 C. This is the charge on each capacitor in Figure 18.54c. Q4 5 67.2 mC and V4 5

67.2 mC 67.2 mC Q4 5 16.8 V. V123 5 5 5 11.2 V. C4 4.00 mF 1 3 / 2 2 1 4.00 mF 2

Note that V4 1 V123 5 Vab . Then in Figure 18.54b, V3 5 V12 5 11.2 V. Q3 5 C3V3 5 1 4.00 mF 2 1 11.2 V 2 5 44.8 mC.

Q12 5 C12V12 5 12 1 4.00 mF 2 1 11.2 V 2 5 22.4 mC. Q1 5 Q2 5 22.4 mC. V1 5

22.4 mC 22.4 mC Q1 Q2 5 5 5.6 V. V2 5 5 5 5.6 V. C1 4.00 mF C2 4.00 mF

In summary, Q1 5 22.4 mC, Q2 5 22.4 mC, Q3 5 44.8 mC, Q4 5 67.2 mC, V1 5 5.6 V, V2 5 5.6 V, V3 5 11.2 V and V4 5 16.8 V.

18.55. Set Up: For capacitors in series the voltages add and the charges are the same; 1 5 1 1 1 1 c For Ceq

C1 Q capacitors in parallel the voltages are the same and the charges add; Ceq 5 C1 1 C2 1 c C 5 . V

C2

18-14

Chapter 18

Solve: (a) The equivalent capacitance of the 5.0 mF and 8.0 mF capacitors in parallel is 13.0 mF. When these two capacitors are replaced by their equivalent we get the network sketched in Figure 18.55. The equivalent capacitance of these three capacitors in series is 3.47 mF. (b) Qtot 5 CtotV 5 1 3.47 mF 2 1 50.0 V 2 5 174 mC (c) Qtot is the same as Q for each of the capacitors in the series combination shown in Figure 18.55, so Q for each of the capacitors is 174 mC. 10.0 mF

9.0 mF

a

b

13.0 mF

Figure 18.55 Qtot Qtot 5 17.4 V, V13 5 5 13.4 V and Reflect: The voltages across each capacitor in Figure 18.55 are V10 5 C10 C13 Qtot V9 5 5 19.3 V. V10 1 V13 1 V9 5 17.4 V 1 13.4 V 1 19.3 V 5 50.1 V. The sum of the voltages equals the C9 applied voltage, apart from a small difference due to rounding. 1 1 1 5 1 1 c For Ceq C1 C2 Q capacitors in parallel the voltages are the same and the charges add; Ceq 5 C1 1 C2 1 c C 5 . V Solve: (a) The equivalent capacitance of the 18.0 nF, 30.0 nF and 10.0 nF capacitors in series is 5.29 nF. When these capacitors are replaced by their equivalent we get the network sketched in Figure 18.56. The equivalent capacitance of these three capacitors in parallel is 19.3 nF, and this is the equivalent capacitance of the original network.

18.56. Set Up: For capacitors in series the voltages add and the charges are the same;

7.5 nF

5.29 nF a

b

6.5 nF

Figure 18.56

(b) Qtot 5 CeqV 5 1 19.3 nF 2 1 25 V 2 5 482 nC (c) The potential across each capacitor in parallel network of Figure 18.56 is 25 V. Q6.5 5 C6.5V6.5 5 1 6.5 nF 2 1 25 V 2 5 162 nC.

(d) 25 V

18.57. Set Up: The energy stored in a charged capacitor is 12 CV 2. Solve:

1 2 2 CV

5 12 1 450 3 1026 F 2 1 295 V 2 2 5 19.6 J

Vab 400 V 5 5 8.00 3 104 V / m. d 5.00 3 1023 m u 5 12 P0 E 2 5 12 1 8.854 3 10212 C2 / 1 N # m2 2 2 1 8.00 3 104 V / m 2 2 5 0.0283 J / m3.

18.58. Set Up: The energy density is u 5 12 P0 E 2. Vab 5 Ed. E 5

Q . U 5 12 CV 2. V Solve: (a) Q 5 CV 5 1 5.0 mF 2 1 1.5 V 2 5 7.5 mC. U 5 12 CV 2 5 12 1 5.0 mF 2 1 1.5 V 2 2 5 5.62 mJ (b) U 5 12 CV 2 5 12 C 1 Q / C 2 2 5 Q 2 / 2C. Q 5 "2CU 5 "2 1 5.0 3 1026 F 2 1 1.0 J 2 5 3.2 3 1023 C.

18.59. Set Up: C 5

V5

Q 3.2 3 1023 C 5 640 V. 5 C 5.0 3 1026 F

Electric Potential and Capacitance

18.60. Set Up: C 5

Q V

18-15

1 2

Q Q2 Q 5 so U 5 12 QV 5 12 Q C C 2C (b) Q 5 CV so U 5 12 1 CV 2 V 5 12 CV 2 Solve: (a) V 5

Q Q2 . The stored energy is 12 QV 5 5 1 CV 2. V 2C 2 1 3.33 3 1029 C 2 2 Q2 Solve: (a) Energy 5 5 5 8.53 3 10213 J 5 0.853 pJ. 2C 2 1 6.50 3 1026 F 2

18.61. Set Up: C 5

V5

Q 3.33 3 1029 C 5 5.12 3 1024 V 5 0.512 mV. 5 C 6.50 3 1026 F

(b) Energy 5 12 CV 2 5 12 1 12.7 3 1029 F 2 1 18.5 V 2 2 5 2.17 3 1026 J 5 2.17 mJ.

Q 5 CV 5 1 12.7 3 1029 F 2 1 18.5 V 2 5 2.35 3 1027 C 5 0.235 mC.

Reflect: There are three equivalent expressions for the stored energy and we can choose the one most appropriate to the information we are given. P0 A Q2 . . The charge of an electron is 1.60 3 10219 C. C 5 d 2C Solve: (a) Q 5 "2C 1 25.0 J 2 5 "2 1 5.00 3 1029 F 2 1 25.0 J 2 5 5.00 3 1024 C. The number of electrons with 5.00 3 1024 C 5 3.12 3 1015 electrons. this magnitude of charge is 1.60 3 10219 C / electron (b) To double the stored energy, halve the capacitance. To do this, either double the plate separation or halve the plate area.

18.62. Set Up: The energy stored in a capacitor is

18.63. Set Up: The two capacitors are in series. For capacitors in series the voltages add and the charges are the Q 1 1 1 5 1 1 cC 5 . U 5 12 CV 2 . Ceq C1 C2 V 1 150 nF 2 1 120 nF 2 C1C2 1 1 1 Solve: (a) so Ceq 5 5 1 5 5 66.7 nF. Ceq C1 C2 C1 1 C2 150 nF 1 120 nF same.

Q 5 CV 5 1 66.7 nF 2 1 36 V 2 5 2.4 3 1026 C 5 2.4 mC

(b) Q 5 2.4 mC for each capacitor. (c) U 5 12 CeqV 2 5 12 1 66.7 3 1029 F 2 1 36 V 2 2 5 43.2 mJ (d) We know C and Q for each capacitor so rewrite U in terms of these quantities. U 5 12 CV 2 5 12 C 1 Q / C 2 2 5 Q 2 / 2C

150 nF: U 5

1 2.4 3 1026 C 2 2 1 2.4 3 1026 C 2 2 5 19.2 mJ; 5 24.0 mJ 120 nF: U 5 2 1 150 3 1029 F 2 2 1 120 3 1029 F 2

Note that 19.2 mJ 1 24.0 mJ 5 43.2 mJ, the total stored energy calculated in part (c). 150 nF: V 5

Q Q 2.4 3 1026 C 2.4 3 1026 C 5 5 16 V; 120 nF: V 5 5 5 20 V 29 C C 150 3 10 F 120 3 1029 F

Note that these two voltages sum to 36 V, the voltage applied across the network. Reflect: Since Q is the same the capacitor with smaller C stores more energy 1 U 5 Q 2 / 2C 2 and has a larger voltage 1 V 5 Q/C 2 .

18.64. Set Up: For capacitors in parallel, the voltages are the same and the charges add. Ceq 5 C1 1 C2 . C 5 1 2 2 CV .

U5 Solve: (a) Ceq 5 C1 1 C2 5 35 nF 1 75 nF 5 110 nF. Qtot 5 CeqV 5 1 110 3 1029 F 2 1 220 V 2 5 24.2 mC (b) V 5 220 V for each capacitor. 35 nF: Q35 5 C35V 5 1 35 3 1029 F 2 1 220 V 2 5 7.7 mC; 75 nF: Q75 5 C75V 5 1 75 3 1029 F 2 1 220 V 2 5 16.5 mC.

Q . V

18-16

Chapter 18

Note that Q35 1 Q75 5 Qtot . (c) Utot 5 12 CeqV 2 5 12 1 110 3 1029 F 2 1 220 V 2 2 5 2.66 mJ (d) 35 nF: U35 5 12 C35V 2 5 12 1 35 3 1029 F 2 1 220 V 2 2 5 0.85 mJ; 75 nF: U75 5 12 C75V 2 5 12 1 75 3 1029 F 2 1 220 V 2 2 5 1.81 mJ. Since V is the same the capacitor with larger C stores more energy.

(e) 220 V for each capacitor. 18.65. Set Up: C 5

Q . Let C1 5 20.0 mF and C2 5 10.0 mF. The energy stored in a capacitor is Vab 2 1 1 1Q 2 . 2 QVab 5 2 CVab 5 2 Vab

Solve: (a) The initial charge on the 20.0 mF capacitor is

Q 5 C1 1 800 V 2 5 1 20.0 3 1026 F 2 1 800 V 2 5 0.0160 C.

(b) In the final circuit, charge Q is distributed between the two capacitors and Q1 1 Q2 5 Q. The final circuit contains only the two capacitors, so the voltage across each is the same, V1 5 V2 . V5

Q1 Q2 C1 Q 5 . Q 5 Q 5 2Q2 . so V1 5 V2 gives C C1 C2 1 C2 2

Using this in Q1 1 Q2 5 0.0160 C gives 3Q2 5 0.0160 C and Q2 5 5.33 3 1023 C. Q 5 2Q2 5 1.066 3 1022 C. V1 5

Q1 Q2 1.066 3 1022 C 5.33 3 1023 C 5 533 V. V2 5 5 533 V. 5 5 26 C1 C2 20.0 3 10 F 10.0 3 1026 F

The potential differences across the capacitors are the same, as they should be. (c) Energy 5 12 C1V 2 1 12 C2V 2 5 12 1 C1 1 C2 2 V 2 5 12 1 20.0 3 1026 F 1 10.0 3 1026 F 2 1 533 V 2 2 5 4.26 J. (d) The 20.0 mF capacitor initially has energy 5 12 C1V 2 5 12 1 20.0 3 1026 F 2 1 800 V 2 2 5 6.40 J. The decrease in stored energy that occurs when the capacitors are connected is 6.40 J 2 4.26 J 5 2.14 J. Reflect: The decrease in stored energy is because of conversion of electrical energy to other forms during the motion of the charge when it becomes distributed between the two capacitors. Thermal energy is generated by the current in the wires and energy is emitted in electromagnetic waves. 1 1 1 5 1 1 c For Ceq C1 C2 Q capacitors in parallel the voltages are the same and the charges add; Ceq 5 C1 1 C2 1 c C 5 . U 5 12 CV 2. V Solve: (a) Find Ceq for the network by replacing each series or parallel combination by its equivalent. The successive simplified circuits are shown in Figure 18.66a-c.

18.66. Set Up: For capacitors in series the voltages add and the charges are the same;

Utot 5 12 CeqV 2 5 12 1 2.19 3 1026 F 2 1 12.0 V 2 2 5 1.58 3 1024 J 5 158 mJ 4.06 mF

a

b

8.60 mF

4.80 mF 3.50 mF

(a)

7.56 mF

8.60 mF

2.19 mF b

a

a

4.80 mF (b)

Figure 18.66

(c)

b

Electric Potential and Capacitance

18-17

(b) From Figure 18.66c, Qtot 5 CeqV 5 1 2.19 3 1026 F 2 1 12.0 V 2 5 2.63 3 1025 C. From Figure 18.66b, Qtot 5 2.63 3 1025 C. V4.8 5

Q4.8 2.63 3 1025 C 5 5.48 V. 5 C4.8 4.80 3 1026 F

U4.8 5 12 CV 2 5 12 1 4.80 3 1026 F 2 1 5.48 V 2 2 5 7.21 3 1025 J 5 72.1 mJ This one capacitor stores nearly half the total stored energy. Q A . For a parallel-plate capacitor, C 5 P0 . d V Solve: (a) U 5 12 CV 2 5 12 C 1 Q / C 2 2 5 Q 2 / 1 2C 2 . U is proportional to Q 2 so increase the charge by a factor of "2 A to double the stored energy. We didn’t use C 5 P0 so this result applies to any capacitor. d P0 A 2 (b) Still connected to the battery means V stays constant and Q changes. U 5 12 CV 2 5 12 V . d S d / 2 means d U S 2U0 . (c) U 5 12 CV 2. V 5 25 V goes to V 5 75 V. V S 3V so U S 9U 5 90 J. Reflect: In parts (b) and (c) the change causes the charge on the capacitor to increase and this increases the electric field between the plates and therefore increases the stored energy.

18.67. Set Up: U 5 12 CV 2. C 5

1 2

18.68. Set Up: The field when there is air between the plates is E0 5 the plates is E 5 Solve: E0 5

E0 . K

Q . The field when the dielectric is between P0 A

Q 0.180 3 1026 C 5 5 2.03 3 106 V / m. 212 P0 A 1 8.854 3 10 C2 / 1 N # m2 2 2 1 100 3 1024 m2 2 K5

2.03 3 106 V / m E0 5 5 6.15. E 0.330 3 106 V / m

A d equation applies whether or not a dielectric is present.

18.69. Set Up: C 5 KC0 5 KP0 . A 5 1.0 cm2 5 1.0 3 1024 m2. V 5 Ed for a parallel plate capacitor; this

1 8.85 3 10212 F / m 2 1 1.0 3 1024 m2 2 5 1.18 mF per cm2. 7.5 3 1029 m 85 mV V 5 5 1.13 3 106 V / m. (b) E 5 1 10 2 1 7.5 3 1029 m 2 Kd Reflect: The dielectric material increases the capacitance and decreases the electric field that corresponds to a given potential difference. Solve: (a) C 5 1 10 2

18.70. Set Up: The capacitance with the dielectric present is C 5 KC0 5 is the maximum allowed field between the plates. Vab 4.00 3 103 V 5 2.00 3 1024 m. 5 Solve: d 5 E 20.0 3 106 V / m A5

KP0 A . Vab 5 Ed. The dielectric strength d

1 1.50 3 1029 F 2 1 2.00 3 1024 m 2 Cd 5 5 0.0106 m2. KP0 1 3.20 2 1 8.854 3 10212 C2 / 1 N # m2 2 2

18.71. Set Up: C 5 KC0 . U 5 12 CV 2.

Solve: (a) With the dielectric, C 5 1 3.75 2 1 12.5 mF 2 5 46.9 mF. before: U 5 12 C0V 2 5 12 1 12.5 3 1026 F 2 1 24.0 V 2 2 5 3.60 mJ after: U 5 12 CV 2 5 12 1 46.9 3 1026 F 2 1 24.0 V 2 2 5 13.5 mJ (b) DU 5 13.5 mJ 2 3.6 mJ 5 9.9 mJ. The energy increased. Reflect: The power supply must put additional charge on the plates to maintain the same potential difference when the dielectric is inserted. U 5 12 QV, so the stored energy increases.

18-18

Chapter 18

18.72. Set Up: The capacitance with the dielectric present is C 5 KC0 5 is the maximum allowed field between the plates. 1 0.200 3 1026 F 2 1 0.0800 3 1023 m 2 Cd 5 5 0.723 m2 Solve: (a) A 5 KP0 1 2.50 2 1 8.854 3 10212 C2 / 1 N # m2 2 2 (b) Vab 5 12 1 50.0 3 106 V / m 2 1 0.0800 3 1023 m 2 5 2.00 3 103 V

KP0 A . Vab 5 Ed. The dielectric strength d

E0 . K 3.20 3 105 V / m

18.73. Set Up: E 5 Solve: K 5

E0 5 1.28. 5 E 2.50 3 105 V / m

18.74. Set Up: The potential of a point charge is V 5

kq . r

y

2q

x P 1q

Figure 18.74 Solve: (a) The positions of the two charges are shown in Figure 18.74. (b) Any point on the x axis is the same distance from each charge, so V5

k 1 2q 2 kq 1 5 0. r r

The potential is zero at all points on the x axis.

18.75. Set Up: For a point charge, E 5 and away from a positive charge.

k0q0 r

2

and V 5

1 21 2

kq . The electric field is directed toward a negative charge r

kq / r kq r 2 4.98 V V 5 5 5 r. r 5 5 0.415 m. 2 r kq E k 0 q 0 /r 12.0 V / m 1 0.415 m 2 1 4.98 V 2 rV 5 5 2.30 3 10210 C (b) q 5 k 8.99 3 109 N # m2 / C2 (c) q . 0, so the electric field is directed away from the charge. Reflect: The ratio of V to E due to a point charge increases as the distance r from the charge increases, because E falls off as 1 / r 2 and V falls off as 1 / r. Solve: (a) V . 0 so q . 0.

18.76. Set Up: U 5 k

qqr . Ua 1 Ka 5 Ub 1 Kb . The charge of the alpha particle is 12e and the charge of the radon r

nucleus is 186e. Solve: (a) The final energy of the alpha particle, 4.79 MeV, equals the electrical potential energy of the alpha-radon combination just before the decay. U 5 4.79 MeV 5 7.66 3 10213 J. 1 8.99 3 109 N # m2 / C2 2 1 2 2 1 86 2 1 1.60 3 10219 C 2 2 kqqr 5 5 5.17 3 10214 m (b) r 5 U 7.66 3 10213 J

Electric Potential and Capacitance

18-19

Q s 5 and V0 5 E0 d. The energy density in the electric field P0 P0 A is u 5 12 P0 E 2. The volume of a shell of thickness t and average radius R is 4pR 2t. The volume of a solid sphere of E0 V0 radius R is 43 pR 3. With the dielectric present, E 5 and V 5 . K K 0.50 3 1023 C / m2 s 5 5.6 3 107 V / m Solve: (a) E0 5 5 P0 8.854 3 10212 C2 / 1 N # m2 2 (b) V0 5 E0d 5 1 5.6 3 107 V / m 2 1 5.0 3 1029 m 2 5 0.28 V. The outer wall of the cell is at higher potential, since it has positive charge. (c) For the cell, Vcell 5 43 pR 3 and

18.77. Set Up: With air between the layers, E0 5

R5

1 2/ 1 3Vcell 4p

13

5

2

3 1 10216 m3 2 1/3 5 2.9 3 1026 m. 4p

The volume of the cell wall is Vwall 5 4pR 2t 5 4p 1 2.9 3 1026 m 2 2 1 5.0 3 1029 m 2 5 5.3 3 10219 m3. u0 5 12 P0 E02 5 12 1 8.854 3 10212 C2 / 1 N # m2 2 2 1 5.6 3 107 V / m 2 2 5 1.39 3 104 V / m3 .

The total electric field in the cell wall is 1 1.39 3 104 V / m3 2 1 5.3 3 10219 m3 2 5 7 3 10215 V. 5.6 3 107 V / m V0 E0 0.28 V 5 5 1.0 3 107 V / m and V 5 5 5 0.052 V. (d) E 5 K 5.4 K 5.4

18.78. Set Up: Let a be the initial situation, where the alpha particle is very far from the gold nucleus and has kinetic energy K 5 10.0 MeV 5 1.60 3 10212 J. At a the gold nucleus has zero kinetic energy. Let b be at the distance of closest approach, when the distance between the two particles is rb . Conservation of energy says q1q2 Ka 1 Ua 5 Kb 1 Ub . U 5 k . The alpha particle has charge q1 5 12e and the gold nucleus has charge r q2 5 179e. Solve: Ka 5 1.60 3 10212 J and Kb 5 0, since at the distance of closest approach the alpha particle has momentarily come to rest. Ua 5 0, since ra is very large. Ub 5 k

1 12e 2 1 179e 2 . rb

Conservation of energy gives Ka 5 Ub . rb 5

1 2 2 1 79 2 1 1.60 3 10219 C 2 2 1 8.99 3 109 N # m2 / C2 2 1 2 2 1 79 2 e 2k 5 5 2.27 3 10214 m Ka 1.60 3 10212 J

18.79. Set Up: Use conservation of energy Ua 1 Ka 5 Ub 1 Kb to find the distance of closest approach rb . Kb 5 0. Initially the two protons are far apart, so Ua 5 0. The maximum force is at the distance of closest approach, 0 q1q2 0 F 5 k 2 . A proton has mass 1.67 3 10227 kg and charge q 5 1e 5 11.60 3 10219 C. rb q1q2 e2 Solve: Ka 5 Ub . 2 1 12 mva2 2 5 k . mva2 5 k and rb rb 9 2 1 8.99 3 10 N # m2 / C2 2 1 1.60 3 10219 C 2 2 ke rb 5 5 5 1.38 3 10213 m. 2 mva 1 1.67 3 10227 kg 2 1 1.00 3 106 m / s 2 2 F5k

1 1.60 3 10219 C 2 2 e2 9 2 2 # 1 2 5 8.99 3 10 N m C / 1 1.38 3 10213 C 2 2 5 0.012 N. rb2

Reflect: The acceleration a 5 F / m of each proton produced by this force is extremely large.

18.80. Set Up: The proton has charge qp 5 1e and mass mp 5 1.67 3 10227 kg. The alpha particle has charge qa 5 14e and mass ma 5 4mp 5 6.68 3 10227 kg. We can apply both conservation of energy and conservation of 0 q1q2 0 F linear momentum to the system. a 5 , where F 5 k 2 . m r

18-20

Chapter 18

Solve: acceleration: The maximum force and hence the maximum acceleration occurs just after they are released, when r 5 0.225 nm. F 5 1 8.99 3 109 N # m2 / C2 2 ap 5

1 2 2 1 1.60 3 10219 C 2 2 5 9.09 3 1029 N 1 0.225 3 1029 m 2 2

F 9.09 3 1029 N F 9.09 3 1029 N 5 5 5 5.44 3 1018 m / s2; aa 5 5 1.36 3 1018 m / s2 227 mp ma 1.67 3 10 kg 6.68 3 10227 kg

The acceleration of the proton is larger by a factor of ma / mp . speed: Conservation of energy says Ui 1 Ki 5 Uf 1 Kf . Ki 5 0 and Uf 5 0, so Kf 5 Ui . Ui 5 k

1 2 2 1 1.60 3 10219 C 2 2 qqr 5 1 8.99 3 109 N # m2 / C2 2 5 2.05 3 10218 J, r 0.225 3 1029 m

so the total kinetic energy of the two particles when they are far apart is Kf 5 2.05 3 10218 J. Conservation of linear momentum says how this energy is divided between the proton and alpha particle. pi 5 pf . 0 5 m pvp 2 mava and va 5

1 2 mp

ma

vp .

Kf 5 12 mpvp2 1 12 mava2 5 12 mpvp2 1 12 m a vp 5

1 2 mp

ma

2

1

vp2 5 12 mpvp2 1 1

mp ma

2

2 1 2.05 3 10218 J 2 2Kf 5 5 4.43 3 104 m / s Å mp 1 1 1 1 mp / ma 2 2 Å 1 1.67 3 10227 kg 2 1 1 1 14 2 va 5

1 2 mp

ma

vp 5 14 1 4.43 3 104 m / s 2 5 1.11 3 104 m / s

The maximum acceleration occurs just after they are released. The maximum speed occurs after a long time.

18.81. Set Up: C 5

P0 A Q2 Q Q .C5 . Vab 5 Ed and E 5 . The stored energy is 12 QVab 5 12 CVab2 5 12 . d Vab P0 A C

1 8.854 3 10212 C2 / 1 N # m2 2 2 1 0.200 m 2 2 P0 A 5 5 4.43 3 10211 F d 0.800 3 1022 m (b) Q 5 CVab 5 1 4.43 3 10211 F 2 1 120 V 2 5 5.32 3 1029 C Vab 120 V 5 5 1.50 3 104 V / m (c) E 5 d 0.800 3 1022 m (d) Energy 5 12 QV 5 12 1 5.32 3 1029 C 2 1 120 V 2 5 3.19 3 1027 J Solve: (a) C 5

(e) Since the battery is disconnected, the charge Q on the capacitor stays constant. P0 A (a) C 5 so C 5 12 1 4.43 3 10211 F 2 5 2.22 3 10211 F. d (b) The charge can’t change, so Q 5 5.32 3 1029 C. Q . Since Q doesn’t change, E doesn’t change and E 5 1.50 3 104 V / m. (c) E 5 P0 A Q2 . 2C Q doesn’t change and C changes by a factor of 12 , so the stored energy doubles and becomes 6.38 3 1027 J. Reflect: Since the stored energy increases, work must be done by the force that pulls the plates apart. (d) Energy 5

18.82. Set Up: Since the battery remains connected, V remains constant and Q changes.

1 8.854 3 10212 C2 / 1 N # m2 2 2 1 0.200 m 2 2 P0 A 5 5 2.21 3 10211 F d 1.60 3 1022 m (b) Q 5 CVab 5 1 2.21 3 10211 F 2 1 120 V 2 5 2.65 3 1029 C Vab 120 V 5 5 7.50 3 103 V / m (c) E 5 d 1.60 3 1022 m (d) Energy 5 12 CV 2 5 12 1 2.21 3 10211 F 2 1 120 V 2 2 5 1.59 3 1027 J Solve: (a) C 5

Electric Potential and Capacitance

18-21

KP0 A Q2 Q .C5 . Energy 5 12 QVab 5 12 CVab2 5 12 . d Vab C 1 8.854 3 10212 C2 / 1 N # m2 2 2 1 16.0 3 1024 m2 2 Solve: (a) C 5 1 5.00 2 5 3.54 3 10211 F 0.200 3 1022 m (b) Q 5 CVab 5 1 3.54 3 10211 F 2 1 300 V 2 5 1.06 3 1028 C (c) Energy 5 12 CV 2 5 12 1 3.54 3 10211 F 2 1 300 V 2 2 5 1.59 3 1026 J

18.83. Set Up: C 5

P0 A Q .C5 . Vab 5 Ed. The stored energy is 12 QV. Vab d 0.0180 3 1026 C Solve: (a) C 5 5 9.00 3 10211 F 5 90.0 pF 200 V 1 9.00 3 10211 F 2 1 1.50 3 1023 m 2 P0 A Cd 5 5 0.0152 m2. (b) C 5 so A 5 P0 d 8.854 3 10212 C2 / 1 N # m2 2 (c) V 5 Ed 5 1 3.0 3 106 V / m 2 1 1.50 3 1023 m 2 5 4.5 3 103 V (d) Energy 5 12 QV 5 12 1 0.0180 3 1026 C 2 1 200 V 2 5 1.80 3 1026 5 1.80 mJ

18.84. Set Up: C 5

18.85. Set Up: For capacitors in series, the equivalent resistance Ceq is given by capacitors in parallel, the equivalent capacitance Ceq is given by Ceq 5 C1 1 C2 1 c C1

C1

C1

1 1 1 5 1 1 c. For Ceq C1 C2

C1

a

a C2

6.9 mF

C2

2.3 mF

C2

b

b C1

C1

C1

(a)

C1

(b)

C1

C1 a

a C2

6.9 mF

2.3 mF b

b

C1

C1

(c)

(d)

a

2.3 mF b (e)

Figure 18.85 Solve: (a) Using the rules for combining capacitors in series and in parallel gives the sequence of equivalent networks shown in Figure 18.85. The equivalent capacitance of the network is 2.3 mF. (b) In Figure 18.85d the three capacitors in series have the same capacitance, so the voltage across each is 1 420 V 2 / 3 5 140 V. Q1 5 C1V1 5 1 6.9 mF 2 1 140 V 2 5 966 mC. The voltage across C2 in Figure 18.85c is 140 V, so Q2 5 C2V2 5 1 4.6 mF 2 1 140 V 2 5 644 mC. Reflect: We could continue to analyze the networks in Figure 18.85 and find V and Q for each capacitor in the network.

18-22

Chapter 18

Q . When the two capacitors are connected to each other, V1 5 V2 and Q1 1 Q2 5 0 q1 2 q2 0 , V where q1 and q2 are the charges on the two capacitors when they are connected to the line in parallel. Let C1 5 4.00 mF and C2 5 6.00 mF.

18.86. Set Up: C 5

Solve: (a) In parallel, the voltage across each capacitor equals the line voltage.

q1 5 C1V 5 1 4.00 mF 2 1 660 V 2 5 2.64 3 1023 C and q2 5 C2V 5 1 6.00 mF 2 1 660 V 2 5 3.96 3 1023 C.

(b) Q1 1 Q2 5 3.96 3 1023 C 2 2.64 3 1023 C 5 1.32 3 1023 C. V1 5 V2 so

Q1 Q2 C2 5 . Q2 5 Q1 5 1.50Q1 . C1 C2 C1

Q1 1 1.50Q1 5 1.32 3 1023 C and Q1 5 5.28 3 1024 C. Q2 5 1.50Q1 5 7.92 3 1024 C. V1 5

Q1 Q2 5.28 3 1024 C 7.92 3 1024 C 5 132 V. V2 5 5 132 V. 5 5 26 C1 C2 4.00 3 10 F 6.00 3 1026 F

1 1 1 1 5 1 1 . In series, Ceq C1 C2 C3 the charge on each capacitor equals the charge on the equivalent capacitor. The energy stored in the three capacitors is equal to the energy stored in the equivalent capacitor. 8.4 mF 1 1 1 1 5 1 1 5 2.1 3 1026 F. Solve: (a) gives Ceq 5 Ceq 8.4 mF 8.4 mF 4.2 mF 4

18.87. Set Up: For capacitors in series, the equivalent capacitance Ceq is given by

Q 5 CeqV 5 1 2.1 3 1026 F 2 1 36 V 2 5 7.56 3 1025 C 5 75.6 mC.

(b) The total stored energy is 12 QV 5 12 1 7.56 3 1025 C 2 1 36 V 2 5 1.36 3 1023 J. Reflect: The three capacitors each have the same Q, so the 4.2 mF capacitor has twice the voltage across it than is across each of the 8.4 mF capacitors. We could also find the total stored energy by finding the energy stored in each of the three capacitors.

18.88. Set Up: The capacitor is equivalent to two capacitors in parallel, as shown in Figure 18.88. Each of these two capacitors have plates that are 12.0 cm by 6.0 cm. For a parallel-plate capacitor with dielectric filling the volume A between the plates, C 5 KP0 . d

C1

C2

Figure 18.88 Solve: (a) C 5 C1 1 C2 . C2 5 P0

1 8.854 3 10212 F / m 2 1 0.120 m 2 1 0.060 m 2 A 5 5 1.42 3 10211 F d 4.50 3 1023 m

C1 5 KC2 5 1 3.40 2 1 1.42 3 10211 F 2 5 4.83 3 10211 F. C 5 C1 1 C2 5 6.25 3 10211 F 5 62.5 pF.

(b) U 5 12 CV 2 5 12 1 6.25 3 10211 F 2 1 18.0 V 2 2 5 1.01 3 1028 J (c) Now C1 5 C2 and C 5 2 1 1.42 3 10211 F 2 5 2.84 3 10211 F.

U 5 12 CV 2 5 12 1 2.84 3 10211 F 2 1 18.0 V 2 2 5 4.60 3 1029 J.

The plexiglass increases the capacitance and that increases the energy stored for the same voltage across the capacitor.

Electric Potential and Capacitance

18-23

18.89. Set Up: For a single capacitor with dielectric of dielectric constant K completely filling the volume between KP0 A . The capacitor is sketched in Figure 18.89a. It is equivalent to two capacitors in series as shown d 1 1 1 1 . in Figure 18.89b. For two capacitors C1 and C2 in series the equivalent capacitance C is given by 5 C C1 C2 its plates, C 5

K1

d2

/

C1

K2

d2

/

C2

(a)

(b)

Figure 18.89 Solve: (a)

K1P0 A K2P0 A 2K1P0 A 2K2P0 A C1 1 C2 1 1 1 5 1 5 . C1 5 . and C2 5 5 5 C C1 C2 C1C2 d d d/2 d/2

1

21

2

1

2K1P0 A 2K2P0 A d d 2P0 A K1K2 5 C5 2K2P0 A 2K1P0 A d K1 1 K2 1 d d

2

KP0 A 2P0 A K 2 , which is the correct expression. 5 d 2K d P0 A 2P0 A 1 5 , which is the correct expression. (c) For K1 5 K2 5 1.0, C S d 2 d Reflect: Our result applies only when each dielectric fills half the space between the plates. When K1 2 K2 the electric field has a different value in each dielectric. (b) For K1 5 K2 , C S

12