Chapter 18

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Chapter 18 I

Practice 18A, p. 669

Givens

Solutions

1. PEelectric = 6.32 × 10−19 J

q1 = q2 = 2qp + 2qn = (2)1.60 × 10−19 C + (2)(0) = 3.20 × 10−19 C kc q1q2 (8.99 × 109 N• m2/C2)(3.20 × 10−19 C)2  =  r=  PEe lectric 6.32 × 10−19 J r = 1.46 × 10−9 m

2. q1 = 6.4 µC = 6.4 × 10−6 C q2 = −3.2 µC = −3.2 × 10−6 C

(8.99 × 109 N• m2/C2)(6.4 × 10−6 C)(−3.2 × 10−6 C) kc q1q2  =  r=  −4.1 × 10−2 J PEe lectric

PEelectric = −4.1 × 10−2 J

r = 4.5 m q1 = Nqe = (1013)(−1.60 × 10−19 C) = −1.60 × 10−6 C

3. N = 103 qe = −1.60 × 10−9 −2

PEelectric = −7.2 × 10

J

q2 = −Nqe = (−1013)(−1.60 × 10−19 C) = 1.60 × 10−6 C (8.99 × 109 N• m2/C2)(−1.60 × 10−6 C)(1.60 × 10−6 C) kc q1q2  =  r=  −7.2 × 10−2 J PEe lectric r = 0.32 m

4. d = 2.0 cm = 2.0 × 10−2 m E = 215 N/C

(−6.9 × 10−19 J) ∆PE q = −  = −  = 1.60 × 10−19 C (215 N/C)(2.0 × 10−2 m) Ed

Copyright © by Holt, Rinehart and Winston. All rights reserved.

∆PEelectric = −6.9 × 10−19 J

Section Review, p. 669 5. E = 250 N/C, in the positive x direction q1 = 12 mC q1 moves from the origin to (20.0 cm, 50.0 cm). 6. q = 35 C d = 2.0 km

The displacement in the direction of the field (d) is 20.0 cm. ∆PE = −qEd = −(12 × 10−6 C)(250 N/C)(20.0 × 10−2 m) ∆PE = −6.0 × 10−4 J

∆PE = −qEd = −(35 C)(1.0 × 106 N/C)(2.0 × 103 m) = −7.0 × 1010 J

E = 1.0 × 106 N/C

Section One—Pupil’s Edition Solutions

I Ch. 18–1

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Practice 18B, p. 673

Givens

Solutions

1. r = 1.0 cm q = 1.60 × 10−19 C

I

2. q1 = 5.0 nC q2 = −3.0 nC r = 35.0 cm

k q (8.99 × 109 N • m2/C2)(1.60 × 10−19 C) = 1.4 × 10−7 V ∆V = C =  r 1.0 × 10−2 m k q (8.99 × 109 N • m2/C2)(5.0 × 10−9 C) ∆V1 = C1 =  = 260 V r1 (0.350 m/2) k q (8.99 × 109 N • m2/C2)(−3.0 × 10−9 C) ∆V2 = C2 =  = −150 V r2 (0.350 m/2) ∆Vtot = ∆V1 + ∆V2 = 260 V − 150 V = 110 V

3. q1 = 5.0 mC q2 = 3.0 mC q3 = 3.0 mC q4 = −5.0 mC Each charge is at the corner of a 2.0 m × 2.0 m square.

2 diagonal (2 .0 m )2 +( 2.0 m )2 4.0 m +4 .0 m2 r =  =  =  2 2 2

2 8. 0m r =  = 1.4 m 2

k q (8.99 × 109 N • m2/C2)(5.0 × 10−6 C) ∆V1 = C1 =  = 3.2 × 104 V r 1.4 m k q (8.99 × 109 N • m2/C2)(3.0 × 10−6 C) ∆V2 = C2 =  = 1.9 × 104 V r 1.4 m k q (8.99 × 109 N • m2/C2)(3.0 × 10−6 C) ∆V3 = C3 =  = 1.9 × 104 V r 1.4 m k q (8.99 × 109 N • m2/C2)(−5.0 × 10−6 C) ∆V4 = C4 =  = −3.2 × 104 V r 1.4 m ∆Vtot = ∆V1 + ∆V2 + ∆V3 + ∆V4 ∆Vtot = (3.2 × 104 V) + (1.9 × 104 V) + (1.9 × 104 V) + (−3.2 × 104 V) ∆Vtot = 3.8 × 104 V

1. ∆d = 0.060 cm E = 3.0 × 106 V/m 5. E = 8.0 × 104 V/m ∆d = 0.50 m q = 1.60 × 10−19 C

6. E = 1.0 × 106 V/m ∆d = 1.60 km

I Ch. 18–2

∆V = −E∆d = −(3.0 × 106 V/m)(0.060 × 10−2 m) = −1.8 × 103 V ∆V = 1.8 × 103 V a. ∆V = −E ∆d = −(8.0 × 104 V/m)(0.50 m) = −4.0 × 104 V b. ∆PE = −qEd = −(1.60 × 10−19 C)(8.0 × 104 V/m)(0.50 m) ∆PE = −6.4 × 10−15 J ∆V = −E∆d = −(1.0 × 106 V/m)(1.60 × 103 m) = −1.6 × 109 V ∆V = 1.6 × 109 V

Holt Physics Solution Manual


Section Review, p. 675

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Print Practice 18C, p. 681

Givens 1. C = 4.00 mF

Solutions a. Q = C∆V1 = (4.00 × 10−6 F)(12.0 V) = 4.80 × 10−5 C

∆V1 = 12.0 V ∆V2 = 1.50 V

b. PE = 2C(∆V2)2 = (0.5)(4.00 × 10−6 F)(1.50 V)2 = 4.50 × 10−6 J 1

∆V1 = 1.25 V

Q 6.0 × 10−6 C a. C =  =  = 4.8 × 10−6 F ∆V1 1.25 V

∆V2 = 1.50 V

b. PE = 2C(∆V2 )2 = (0.5)(4.8 × 10−6 F)(1.50 V)2 = 5.4 × 10−6 J

2. Q = 6.0 mC

1

Q = 18.0 pC

Q 18.0 × 10−12 C = 9.00 V a. ∆V1 =  =  C 2.00 × 10−12 F

∆V2 = 2.5 V

b. Q = C∆V2 = (2.00 × 10−12 F)(2.5 V) = 5.0 × 10−12 C

3. C = 2.00 pF

4. C = 1.00 F d = 1.00 mm

I

Cd (1.00 F)(1.00 × 10−3 m) A =  =  = 1.13 × 108 m2 8.85 × 10−12 C2/N • m2 e0

Section Review, p. 681

d = 2.0 mm

e A (8.85 × 10−12 C2/N • m2)(2.0 × 10−4 m2) a. C = 0 =  = 8.8 × 10−13 F d 2.0 × 10−3 m

∆V = 6.0 V

b. Q = C∆V = (8.8 × 10−13 F)(6.0 V) = 5.3 × 10−12 C

2. A = 2.0 cm2

3. C = 1.35 pF Copyright © by Holt, Rinehart and Winston. All rights reserved.

∆V = 12.0 V

4. d = 800.0 m A = 1.00 × 106 m2 E = 2.0 × 106 N/C

PE = 2C(∆V)2 = (0.5)(1.35 × 10−12 F)(12.0 V)2 1

PE = 9.72 × 10−11 J e A (8.85 × 10−12 C2/N • m2)(1.00 × 106 m2) a. C = 0 =  = 1.11 × 10−8 F d 800.0 m b. ∆V = −E∆d Q = C∆V = C(−E∆d) = (1.11 × 10−8 F)(−2.0 × 106 N/C)(800.0 m) = −18 C Q = ±18 C

Chapter Review and Assess, pp. 683–687 4. q1 = 9.00 × 10−9 C at the origin q2 = 3.00 × 10−9 C ∆PE = 8.09 × 10−2 J

∆PE = PEf − PEi PEi = 0 J because ri = ∞ kCq1q2 (8.99 × 109 N • m2/C2)(9.00 × 10−9 C)(3.00 × 10−9 C)  =  rf =  = 0.300 m = 30.0 cm PEf 8.09 × 10−7 J


I Ch. 18–3

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Givens

Solutions

5. ri = 55 cm ∆PE = 2.1 × 10−28 J q1 = 1.60 × 10−19 C

I

q2 = −1.60 × 10−19 C

kC q1q2 (8.99 × 109 N • m2/C2)(1.60 × 10−19 C)(−1.60 × 10−19 C)  =  PEi =  ri (55 × 10−2 m) PEi = −4.2 × 10−28 J PE f = ∆PE + PEi = (2.1 × 10−28 J) + (−4.2 × 10−28 J) = −2.1 × 10−28 J kCq1q2 (8.99 × 109 N • m2/C2)(1.60 × 10−19 C)(−1.60 × 10−19 C)  =  rf =  PEf −2.1 × 10−28 J rf = 1.1 m

12. E = 1.7 × 106 N/C ∆d = 1.5 cm 13. F = 4.30 × 10−2 N q = 56.0 mC

∆V = −E∆d = −(1.7 × 106 N/C)(1.5 × 10−2 m) = −2.6 × 104 V ∆V = 2.6 × 104 V −F∆d −(4.30 × 10−2 N)(0.200 m) ∆V = −E∆d =  =  = −154 V q 56.0 × 10−6 C

∆d = 20.0 cm 14. q1 = +8.0 mC q2 = −8.0 mC q3 = −12 mC r1,P = 0.35 m r2,P = 0.20 m

k q (8.99 × 109 N • m2/C2)(8.0 × 10−6 C) V1 = C1 =  = 2.1 × 105 V r1,P 0.35 m k q (8.99 × 109 N • m2/C2)(−8.0 × 10−6 C) V2 = C2 =  = −3.6 × 105 V r2,P 0.20 m (r1,P )2 + (r2,P )2 = (r3,P )2

2 (r )2 = r3,P = (r 1, 2,P P )+

(0 m )2 .3 5m )2+(0. 20

(8.99 × 109 N • m2/C2)(−12 × 10−6 C) V3 =  0. m2+ m2 12 0.0 40 (8.99 × 109 N • m2/C2)(−12 × 10−6 C) = −2.7 × 105 V V3 =  2 0. 16 m Vtot = V1 + V2 + V3 = (2.1 × 105 V) + (−3.6 × 105 V) + (−2.7 × 105 V) Vtot = −4.2 × 105 V

26. ∆V = 12.0 V C = 6.0 pF 27. C = 0.20 mF

Q = C∆V = (6.0 × 10−12 F)(12.0 V) = 7.2 × 10−11 C Q = ±7.2 × 10−11 C a. Q = C∆V = (0.20 × 10−6 F)(6500 V) = 1.3 × 10−3 C

∆V = 6500 V b. PE = 2C(∆V )2 = (0.5)(0.20 × 10−6 F)(6500 V)2 = 4.2 J 1

I Ch. 18–4



k q (8.99 × 109 N • m2/C2)(−12 × 10−6 C) V3 = C3 =  r3,P (0 )2 +(0. m )2 .3 5m 20

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Solutions

28. C1 = 25 mF

PE1 = 2C 1(∆V )2 = (0.5)(25 × 10−6 F)(120 V)2 = 0.18 J

C2 = 5.0 mF ∆V = 120 V

29. ∆V = 600.0 V E = 200.0 N/C

1

PE2 = 2C 2(∆V )2 = (0.5)(5.0 × 10−6 F)(120 V)2 = 3.6 × 10−2 J 1

PEtot = PE1 + PE2 = 0.18 J + (3.6 × 10−2 J) = 0.22 J

I

kC q  ∆V r  =  = r kC q E  r2 600.0 V ∆V r =  =  = 3.000 m 200.0 N/C E ∆Vr (600.0 V)(3.000 m) q =  =  = 2.00 × 10−7 C kC 8.99 × 109 N • m2/C2

30. d = 3.0 mm E = 3.0 × 106 N/C Q = −1.0 mC

Q e A Q C = 0 =  =  d ∆V −E ∆d −e0 AE∆d = Qd −Q A =  e0 E A = pr 2 r=

−6

−(−1.0 × 10 C)  Ap = e−EQp = (8.85 × 10 C /N m )(3.0 × 10 N/C)(p) 0

−12

2

•

2

6

r = 0.11 m 31. q1 = 8.0 mC q2 = 2.0 mC


q3 = 4.0 mC r1,2 = 3.0 cm

)2 +(6. )2 r1,3 = (3 .0 cm 0cm

kC q1q2 (8.99 × 109 N • m2/C2)(8.0 × 10−6 C)(2.0 × 10−6 C)  =  PE1,2 =  = 4.8 J r1,2 3.0 × 10−2 m kC q1q3 (8.99 × 109 N • m2/C2)(8.0 × 10−6 C)(4.0 × 10−6 C)  =  PE1,3 =  r1,3 (0 m)2 +(0. .0 30 06 0m )2 (8.99 × 109 N • m2/C2)(8.0 × 10−6 C)(4.0 × 10−6 C) PE1,3 =  (9 .0 ×1 0−4 m2) +( 3.6 ×10−3 m2) (8.99 × 109 N • m2/C2)(8.0 × 10−6 C)(4.0 × 10−6 C) = 4.3 J PE1,3 =  4.5 ×10−3 m2 PE1,tot = PE1,2 + PE1,3 = 4.8 J + 4.3 J = 9.1 J

32. ∆V = 12 V ∆d = 0.30 cm

∆V 12 V = 4.0 × 103 V/m E =  =  ∆d 0.30 × 10−2 m


I Ch. 18–5

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Givens

Solutions

33. A = 5.00 cm2

Q a. ∆V =  C e0 A C =  d Qd Q (400.0 × 10−12 C)(1.00 × 10−3 m) ∆V =  =  =  = 90.4 V ε0 A e0 A (8.85 × 10−12 C2/N • m2)(5.00 × 10−4 m2)  d

d = 1.00 mm Q = 400.0 pC

I

90.4 V ∆V b. E =  =  = 9.04 × 104 V/m ∆d 1.00 × 10−3 m e A (8.85 × 10−12 C2/N • m2)(175 × 10−4 m2) = 3.87 × 10−9 F a. C = 0 =  d (0.0400 × 10−3 m)

34. A = 175 cm2 d = 0.0400 mm

Q 500.0 × 10−12 C = 0.129 V b. ∆V =  =  C 3.87 × 10−9 F

Q = 500.0 pC

35. KE = 1.00 × 10−19 J me = 9.109 × 10−31 kg mp = 1.673 × 10−27 kg

36. ∆V = 25 700 V

(2)(1.00 × 10−19 J)  = 4.69 × 105 m/s 9.109 × 10−31 kg

a. ve =

b. vp =

2KE  = me

2KE  = mp

(2)(1.00 × 10−19 J)  = 1.09 × 104 m/s 1.673 × 10−27 kg

a. KEf = ∆PE = q∆V = (1.60 × 10−19 C)(25 700 V) = 4.11 × 10−15 J

vi = 0 m/s q = 1.60 × 10−19 C mp = 1.673 × 10−27 kg

b. vf =

(2)(4.11 × 10−15 J)  = 2.22 × 106 m/s 1.673 × 10−27 kg

2KEf  = mp

1 m v 2 2 p f

vi = 0 m/s −27

mp = 1.673 × 10

−19

q = 1.60 × 10

kg

C

38. ∆d = 5.33 mm

vf =

= q∆V −19

(2)(1.60 × 10 C)(120 V)  = 1.5 × 10 m/s 2qm∆V = 1.673 × 10 kg p

−27

5

∆V = 600.0 V

600.0 V ∆V a. E =  =  = 1.13 × 105 V/m ∆d 5.33 × 10−3 m

q = −1.60 × 10−19 C

b. F = qE = (−1.60 × 10−19 C)(1.13 × 105 V/m) = −1.81 × 10−14 N F = 1.81 × 10−14 N

∆d = (5.33 mm − 2.90 mm) = 2.43 mm

I Ch. 18–6

c. ∆PE = −qEd = −(−1.60 × 10−19 C)(1.13 × 105 V/m)(2.43 × 10−3 m) ∆PE = 4.39 × 10−17 J



KEf = ∆PE = q∆V

37. ∆V = 120 V

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Solutions

39. q1 = 5.0 × 10−9 C

r12 + (0.010 m)2 = (0.040 m)2

q2 = −5.0 × 10−9 C q3 = −5.0 × 10−9 C r1,2 = r1,3 = 4.0 cm r2,3 = 2.0 cm

r1 = (0 m)2 −(0. .0 40 01 0m )2 (8.99 × 109 N • m2/C2)(5.0 × 10−9 C) k q V1 = C1 =  r1 (0.0 m )2 −(0. )2 40 01 0m

I

(8.99 × 109 N • m2/C2)(5.0 × 10−9 C) V1 =  (1 .6 ×10−3m 2)−(1. 0×10−4m 2) (8.99 × 109 N • m2/C2)(5.0 × 10−9 C) V1 =  = 1200 V 1. 5 ×10−3 m2 0.020 m r2 = r3 =  = 0.010 m 2 kC q2 (8.99 × 109 N • m2/C2)(−5.0 × 10−9 C) V2 =  =  = −4500 V r2 0.010 m k q (8.99 × 109 N • m2/C2)(−5.0 × 10−9 C) V3 = C3 =  = −4500 V r3 0.010 m Vtot = V1 + V2 + V3 = (1200 V) + (−4500 V) + (−4500 V) = −7800 V

40. q1 = −3.00 × 10−9 C at the origin −9

q2 = 8.00 × 10 C at x = 2.00 m, y = 0.00 m

For the location between the two charges, Vtot = V1 + V2 = 0 V k q V1 = C1 P

V1 = −V2 k q2 V2 = C (2.00 m − P) kC q1 −k q2  = C P (2.00 m − P) −Pq 2 = (2.00 m − P)(q1) −Pq 2 = (2.00 m)(q1) − Pq1


P(q1 − q2 ) = (2.00 m)(q1) (2.00 m)(q1) (2.00 m)(−3.00 × 10−9 C) P =  =  = 0.545 m q1 − q 2 (−3.00 × 10−9 C) − (8.00 × 10−9 C) P is 0.545 m to the right of the origin, at x = 0.545 m . For the location to the left of the y-axis, k q V1 = C1 P k q2 V2 = C (2.00 m + P) −k q2 kC q1  = C P (2.00 m + P) −Pq 2 = (2.00 m + P)(q1) −Pq 2 = (2.00 m)(q1) + Pq1 P(q1 + q2) = −(2.00 m)(q1) −(2.00 m)(q1) −(2.00 m)(−3.00 × 10−9 C) P =  =  = 1.20 m q1 + q 2 (−3.00 × 10−9 C) + (8.00 × 10−9 C) P is 1.20 m to the left of the origin, at x = −1.20 m .


I Ch. 18–7

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Givens

Solutions

41. d = 5.0 cm

F qE q∆V a. a =  =  =  m m md

∆V = 550 V

1

∆x = 2a∆t 2

vi,e = 0 m/s

I

vi,p = 0 m/s −27

mp = 1.673 × 10

−31

me = 9.109 × 10

md 2∆ax = 2∆ qx∆ V 2∆x m d ∆t =  q∆ V e

∆te = kg kg

q = 1.60 × 10−19 C

e e

p p

p

∆te = ∆tp md  =  2∆ qx∆ V q∆V

2∆xpmpd

e e

∆xeme = ∆xpmp ∆xe + ∆xp = d ∆xe = d − ∆xp (d − ∆xp)me = ∆xpmp dme − ∆xpme = ∆xpmp dme = ∆xpmp + ∆xpme dme ∆xp =  mp + me ∆te = ∆tp =

2∆xpmpd  = q∆V

dme mpd 2  mp + me  = q∆V

2d 2memp  (mp + me)q∆V

mp + me = (1.673 × 10−27 kg) + (9.109 × 10−31 kg) = 1.674 × 10−27 kg ∆te = ∆tp =

−2

−31

−27

(2)(5.0 × 10 m) (9.109 × 10 kg)(1.673 × 10 kg)  (1.674 × 10 kg)(1.60 × 10 C)(550 V) 2

−27

−19

q∆V (1.60 × 10−19 C)(550 V)(7.2 × 10−9 s) = 1.4 × 107 m/s b. ve = ae ∆te =  ∆te =  me d (9.109 × 10−31 kg)(5.0 × 10−2 m)

q∆V (1.60 × 10 C)(550 V)(7.2 × 10 s) v = a ∆t =  ∆t =  = 7.6 × 10 m/s m d (1.673 × 10 kg)(5.0 × 10 m) −19

p

p

p

p

p

c. ∆tp,tot =

−9

−27

−2

 q∆V 2∆xtot mpd

∆xtot = d ∆tp,tot =

2mpd 2  = q∆V

−27

−19

∆tp,tot = 3.1 × 10−7 s

42. ∆V = 60.0 V ∆PE = 1.92 × 10−17 J

I Ch. 18–8

∆PE 1.92 × 10−17 J q =  =  = 3.20 × 10−19 C ∆V 60.0 V


−2

(2)(1.673 × 10 kg)(5.0 × 10 m)  (1.60 × 10 C)(550 V) 2

3


∆te = ∆tp = 7.2 × 10−9 s

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43. ∆V = 100.0 V

Q 400.0 × 10−6 C C =  =  = 4.000 × 10−6 F ∆V 100.0 V

Q = 400.0 mC 44. ∆V = 4.5 × 106 V

a. KEf = ∆PE = ∆Vq = (4.5 × 106 V)(1.60 × 10−19 C) = 7.2 × 10−13 J

vi = 0 m/s

I

q = 1.60 × 10−19 C mp = 1.673 × 10−27 kg

b. vf =

−13

(2)(7.2 × 10 J) m =  1.673×1 0kg = 2.9 × 10 m/s 2KEf p

KE = ∆PE

45. vf,positron = 9.0 × 107 m/s −31

mpositron = 9.109 × 10 −19

q = 1.60 × 10

7

−27

kg

C

mproton = 1.673 × 10−27 kg

1 mv 2 2

= ∆Vq

mpositron(vf,positron)2 (9.109 × 10−31 kg)(9.0 × 107 m/s)2 ∆V =  =  2q (2)(1.60 × 10−19 C) ∆V = 2.3 × 104 V vf,proton =

−19

(2)(2.3 × 10 V)(1.60 × 10 C)   m2∆Vq = 1.673 × 10 kg 4

−27

proton

vf,proton = 2.1 × 106 m/s 46. vf = (0.600)(3.00 × 108 m/s) −31

me = mp = 9.109 × 10 qe = −1.60 × 10−19 C qp = 1.60 × 10−19 C

kg

a. ∆PE = KEf 1

∆Vq = 2mvf 2 me vf 2 (9.109 × 10−31 kg)[(0.600)(3.00 × 108 m/s)]2 ∆Ve =  =  (2)(−1.60 × 10−19 C) 2qe


∆Ve = −9.22 × 104 V mp vf 2 (9.109 × 10−31 kg)[(0.600)(3.00 × 108 m/s)]2 b. ∆Vp =  =  2qp (2)(1.60 × 10−19 C) ∆Vp = 9.22 × 104 V a. KEf = ∆PE

47. ∆V = 2200 V −19

q = 1.60 × 10

1 mv 2 f 2

C

−31

me = 9.109 × 10

−27

mp = 1.673 × 10

kg kg

vf,e =

= ∆Vq −19

(2)(2200 V)(1.60 × 10 C)  = 2.8 × 10 m/s 2∆mVq = 9.109 × 10 kg e

b. vf,p =

(2)(2200 V)(1.60 × 10−19 C)

Q = 1.75 × 10−8 C d = 6.50 × 10−4 m

7

 = 6.5 × 10 m/s 2∆mVq = 1.673 × 10 kg p

48. C = 3750 pF

−31

−27

5

Q 1.75 × 10−8 C a. ∆V =  =  = 4.67 V C 3750 × 10−12 F ∆V 4.67 V = 7180 V/m b. E =  =  d 6.50 × 10−4 m


I Ch. 18–9

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Givens

Solutions

2.50 × 10−3 m 49. r =  2 d = 1.40 × 10−4 m

e A e0p r 2 (8.85 × 10−12 C2/N • m2)(p)(2.50 × 10−3 m/2)2  =  a. C = 0 =  d d 1.40 × 10−4 m C = 3.10 × 10−13 F

I ∆V1 = 0.12 V

b. Q = C∆V1 = (3.10 × 10−13 F)(0.12 V) = 3.7 × 10−14 C

c. PEelectric = 2Q∆V1 = (0.5)(3.7 × 10−14 C)(0.12 V) = 2.2 × 10−15 J 1

∆d1 = 1.40 × 10−4 m

d. ∆V1 = E∆d1 ∆V 0.12 V E = 1 =  = 860 V/m ∆d1 1.40 × 10−4 m 1.40 × 10−4 m ∆d2 = 1.10 × 10−4 m −  = 4.00 × 10−5 m 2

∆V2 = E∆d2 = (860 V/m)(4.00 × 10−5 m) = 3.4 × 10−2 V Q2 = (0.707)Q

Q e. ∆V3 = 2 C Because the capacitance has not changed, ∆V3 = (0.707)(∆V1).


∆V3 = (0.707)(∆V1) = (0.707)(0.12 V) = 8.5 × 10−2 V

I Ch. 18–10
