Electrical Energy and Capacitance - Mosinee, WI Energy and Capacitance 53 (b) Being more massive than electrons, protons traveling at the same initial speed will have more initial

51

Chapter 16

Electrical Energy and Capacitance

Problem Solutions

16.1 (a) Because the electron has a negative charge, it experiences a force in the d irection

opposite to the field and, when released from rest, will move in the negative x-

d irection. The work done on the electron by the field is

19 2 181.60 10 C 375 N C 3.20 10 m 1.92 10 Jx xW F x qE x

(b) The change in the electric potential energy is the negative of the work done on the

particle by the field . Thus,

181.92 10 JPE W

(c) Since the Coulomb force is a conservative force, conservation of energy gives

0KE PE , or conservation of linear momentum. , and

18

6

31

2 1.92 10 J22.05 10 m s in the -direction

9.11 10 kgf

e

PEv x

m

16.2 (a) The change in the electric potential energy is the negative of the work done on the

particle by the field . Thus,

6 2 40 5.40 10 C 327 N C 32.0 10 m 5.65 10 J

x yPE W qE x qE y

q x

(b) The change in the electrical potential is the change in electric potential energy per

unit charge, or

4

-6

5.65 10 J105 V

+5.40 10 C

PEV

q

Electrical Energy and Capacitance 52

16.3 The work done by the agent moving the charge out of the cell is

input field

19 3 20J1.60 10 C 90 10 1.4 10 J

C

eW W PE q V

16.4 p

p

mv v

m, so

17191.92 10 J

3.20 10 C+60.0 J C

e

f i

PEq

V V

16.5 6

2

25 000 J C1.7 10 N C

1.5 10 m

VE

d

16.6 Since potential d ifference is work per unit charge W

Vq

, the work done is

27

7 6

27

1.67 10 kg1.05 10 m s 2.64 10 m s

6.64 10 kg

p

p

mv v

m

16.7 (a) ek QV

r

(b) 19 5 141.60 10 C 1.13 10 N C 1.80 10 NF q E

(c)

14 3 17

cos

1.80 10 N 5.33 2.90 10 m cos0 4.38 10 J

W F s

16.8 (a) Using conservation of energy, 0KE PE , with 0fKE since the particle is

“ stopped” , we have

0s

The required stopping potential is then

16

3

19

3.70 10 J2.31 10 V 2.31 kV

1.60 10 C

PEV

q


(b) Being more massive than electrons, protons traveling at the same initial speed will

have more initial kinetic energy and require a greater magnitude stopping potential.

(c) Since cos 0 , the ratio of the stopping potential for a proton to that for an electron

having the same initial speed is

cos 0

16.9 (a) Use conservation of energy

s e s ef i

KE PE PE KE PE PE

or 0s eKE PE PE

2 2

2 79 22 eef

i i

k e ek Qqr

m v m v since the block is at rest at both beginning and end.

22

9 19

2

14

227 7

N m2 8.99 10 158 1.60 10 C

C2.74 10 m

6.64 10 kg 2.00 10 m sfr , where

2 2 2

2 2 2 2

diagonal a a a ar is the maximum stretch of the spring.

total singlecharge

4 4 4 4 22

e ee

k Q k Q QV V k

r aa

Thus, 2

max max

10 0

2kx QE x , giving

6 4

2

max

2 35.0 10 C 4.86 10 V m24.36 10 m 4.36 cm

78.0 N m

QEx

k

(b) At equilibrium, max maxmax

8 61.11 10 F 3.0 10 N C 800 m 27 C

Q C V C E d

Therefore, 27.0 C

3.00 F9.00 V

QC

V

The amplitude is the d istance from the equilibrium position to each of the turning

points at 0 and 4.36 cmx x , so max2.18 cm 2A x


(c) From conservation of energy, eq 1 2 25.0 F+40.0 F 65.0 FC C C . Since

3 3

40 25 2.00 10 C 1.25 10 C 750 CQ Q Q , this gives

22

max2

2 2

k AkxV

Q Q or

6 64.00 10 F 1.50 V 6.00 10 C 6.00 CQ C V

16.10 Using 2

0

1

2y yy v t a t for the full flight gives

2

0

10

2y yv t a t , or 2 2 2 2 10.0 F 20.0 FpC C C

Then, using

1 1

eq

1 2

1 1 1 16.04 F

8.66 F 20.0 Fp p

CC C

for the upward part

of the flight gives

2 2

0 0 0

max

0

0 20.1 m s 4.10 s20.6 m

2 4 42 2

y y y

y y

v v v ty

a v t

From Newton’ s second law, y

y

F mg qE qEa g

m m m. Equating

this to the earlier result gives 1 , so the electric field strength is

3 3 12.00 F 41.8 V 83.6 C

pQ C V

Thus, 3 4

maxmax20.6 m 1.95 10 N C 4.02 10 V 40.2 kVV y E

16.11 (a)

9 2 2 19

7

2

8.99 10 N m C 1.60 10 C5.75 10 V

0.250 10 m

eA

A

k qV

r

(b) previous answers will be decreased.

2 12Q Q


(c) The original electron will be repelled by the negatively charged particle which

suddenly appears at point A . Unless the electron is fixed in place, it will move in the

opposite d irection, away from points A and B, thereby lowering the potential

d ifference between these points.

16.12 (a) At the origin, the total potential is

1 2 1 2 10.0 CQ Q Q Q

(b) At point B located at 1 12 10.0 CQ Q , the needed d istances are

1

10 C

3Q

and

2 2 2 2

2 2 2 1.50 cm 1.80 cm 2.34 cmB Br x x y y

giving

66

9 2 61 2

2 2

1 2

2.24 10 C4.50 10 C8.99 10 N m C 1.21 10 V

1.95 10 m 2.34 10 m

e eB

k q k qV

r r

16.13 (a) Calling the 4.00 F charge 6.00 F ,

31 2

2 21 2 1 2

2 6 6 69

2 2 2

6

N m 8.00 10 C 4.00 10 C 2.00 10 C8.99 10

0.060 0 m 0.030 0 mC 0.060 0 0.030 0 m

2.67 10 V

e ie

i i

k q qq qV k

r r r r r

V

(b) Replacing eq 1.79 F 6.00 V 10.7 CQ C V in part (a) yields

eq

1 1 1 2 1

18.0 F 36.0 F 36.0 FC


16.14 f iW q V q V V , and

0fV since the 8.00 C is infinite d istance from other charges.

2 6 6

91 2

2 2 21 2

6

N m 2.00 10 C 4.00 10 C8.99 10

0.030 0 mC 0.030 0 0.060 0 m

1.135 10 V

i e

q qV k

r r

Thus, 6 68.00 10 C 0 1.135 10 V 9.08 JW

16.15 (a)

2 9 99

2

N m 5.00 10 C 3.00 10 C8.99 10 103 V

0.175 m 0.175 mC

e i

i i

k qV

r

(b) 2

12

9 929 7

2

5.00 10 C 3.00 10 CN m8.99 10 3.85 10 J

0.350 mC

e ik q qPE

r

The negative sign means that positive work must be done to separate the charges

(that is, bring them up to a state of zero potential energy).

16.16 The potential at d istance 0.300 mr from a charge 99.00 10 CQ is

9 2 2 98.99 10 N m C 9.00 10 C270 V

0.300 m

ek QV

r

Thus, the work required to carry a charge 93.00 10 Cq from infinity to this location

is

9 73.00 10 C 270 V 8.09 10 JW qV


16.17 The Pythagorean theorem gives the d istance from the midpoint of the base to the charge

at the apex of the triangle as

2 2 2

3 4.00 cm 1.00 cm 15 cm 15 10 mr

Then, the potential at the midpoint of the base is e i i

i

V k q r , or

9 9 929

2 2

4

7.00 10 C 7.00 10 C 7.00 10 CN m8.99 10

0.010 0 m 0.010 0 mC 15 10 m

1.10 10 V 11.0 kV

V

16.18 Outside the spherical charge d istribution, the potential is the same as for a point charge

at the center of the sphere,

eV k Q r , where 91.00 10 CQ

Thus, 1 1

e e

f i

PE q V ek Qr r

and from conservation of energy eKE PE ,

or 21 1 10

2e e

f i

m v ek Qr r

This gives 2 1 1e

e f i

k Qev

m r r, or

29 9 19

2

31

N m2 8.99 10 1.00 10 C 1.60 10 C

C 1 1

0.020 0 m 0.030 0 m9.11 10 kgv

67.25 10 m sv


16.19 (a) When the charge configuration consists of only the

two protons 1 2 and in the sketchq q , the potential

energy of the configuration is

2

9 2 2 19

1 2

15

12

8.99 10 N m C 1.60 10 C

6.00 10 m

ea

k q qPE

r

or 143.84 10 JaPE

(b) When the alpha particle 3 in the sketchq is added to the configuration, there are

three d istinct pairs of particles, each of which possesses potential energy. The total

potential energy of the configuration is now

2

1 2 1 3 2 3

12 13 23 13

22

ee e eb a

k ek q q k q q k q qPE PE

r r r r

where use has been made of the facts that 2

1 3 2 3 2 2q q q q e e e and

2 2 15

13 23 3.00 fm 3.00 fm 4.24 fm 4.24 10 mr r . Also, note that the first

term in this computation is just the potential energy comp uted in part (a). Thus,

29 2 2 192

14 13

15

13

4 8.99 10 N m C 1.60 10 C43.84 10 J 2.55 10 J

4.24 10 m

eb a

k ePE PE

r

(c) If we start with the three-particle system of part (b) and allow the alpha particle to

escape to infinity [thereby returning us to the two-particle system of part (a)], the

change in electric potential energy will be

14 13 133.84 10 J 2.55 10 J 2.17 10 Ja bPE PE PE

(d) Conservation of energy, 0KE PE , gives the speed of the alpha particle at

infinity in the situation of part (c) as 2 2 0m v PE , or

13

6

27

2 2.17 10 J28.08 10 m s

6.64 10 kg

PEv

m


(e) When, starting with the three-particle system, the two protons are both allowed to

escape to infinity, there will be no remaining pairs of particles and hence no

remaining potential energy. Thus, 0 b bPE PE PE , and conservation of energy

gives the change in kinetic energy as bKE PE PE . Since the protons are

identical particles, this increase in kinetic energy is split equally between them

giving

21 1

2 2proton p p bKE m v PE

or 13

7

-27

2.55 10 J1.24 10 m s

1.67 10 kg

bp

p

PEv

m

16.20 (a) If a proton and an alpha particle, initially at rest 4.00 fm apart, are released and

allowed to recede to infinity, the final speeds of the two particles will d iffer because

of the d ifference in the masses of the particles. Thus, attempting to solve for the

final speeds by use of conservation of energy alone leads to a situation of having

one equation with two unknowns , and does not permit a solution.

(b) In the situation described in part (a) above, one can obtain a second equation with

the two unknown final speeds by using conservation of linear momentum. Then, one

would have two equations which could be solved simultaneously both unknowns.

(c) From conservation of energy: 2 21 1

0 0 02 2

e p

p p

i

k q qm v m v

r

or

9 2 2 19 19

2 2

15

2 8.99 10 N m C 3.20 10 C 1.60 10 C2

4.00 10 m

e p

p p

i

k q qm v m v

r

yield ing 2 2 132.30 10 Jp pm v m v [1]

From conservation of linear momentum,

0p pm v m v or p

p

mv v

m [2]

Substituting Equation [2] into Equation [1] gives

2

2 2 132.30 10 Jp

p p p

mm v m v

m or

2 131 2.30 10 Jp

p p

mm v

m


and

13 137

27 27 27

2.30 10 J 2.30 10 J1.05 10 m s

1 1.67 10 6.64 10 +1 1.67 10 kgp

p p

vm m m

Then, Equation [2] gives the final speed of the alpha particle as

27

7 6

27

1.67 10 kg1.05 10 m s 2.64 10 m s

6.64 10 kg

p

p

mv v

m

16.21 ek QV

r so

9 2 2 98.99 10 N m C 8.00 10 C 71.9 V mek Qr

V V V

For 100 V, 50.0 V, and 25.0 V, V 0.719 m, 1.44 m, and 2.88 mr

The radii are inversely proportional to the potential.

16.22 By definition, the work required to move a charge from one point to any other point on

an equipotential surface is zero. From the definition of work, cosW F s , the work

is zero only if 0s or cos 0F . The d isplacement s cannot be assumed to be zero in

all cases. Thus, one must require that cos 0F . The force F is given by F qE and

neither the charge q nor the field strength E can be assumed to be zero in all cases.

Therefore, the only way the work can be zero in all cases is if cos 0 . But if cos 0 ,

then 90 or the force (and hence the electric field) must be perpendicular to the

d isplacement s (which is tangent to the surface). That is, the field must be

perpendicular to the equipotential surface at all points on that surface.

16.23 From conservation of energy, e ef iKE PE KE PE , which gives

21

0 02

ei

f

k Qqm v

r or

2 2

2 79 22 eef

i i

k e ek Qqr

m v m v

22

9 19

2

14

227 7

N m2 8.99 10 158 1.60 10 C

C2.74 10 m

6.64 10 kg 2.00 10 m sfr


16.24 (a) The d istance from any one of the corners of the square to the point at the center is

one half the length of the d iagonal of the square, or

2 2 2

2 2 2 2

diagonal a a a ar

Since the charges have equal magnitudes and are all the same distance from the

center of the square, they make equal contributions to the total potential. Thus,

total singlecharge

4 4 4 4 22

e ee

k Q k Q QV V k

r aa

(b) The work required to carry charge q from infinity to the point at the center of the

square is equal to the increase in the electric potential energy of the charge, or

center total 0 4 2 4 2e e

Q qQW PE PE qV q k k

a a

16.25 (a)

6 2212 8

0 2

1.0 10 mC8.85 10 1.1 10 F

800 mN m

AC

d

(b) max maxmax

8 61.11 10 F 3.0 10 N C 800 m 27 C

Q C V C E d

16.26 (a) 27.0 C

3.00 F9.00 V

QC

V

(b) Q C V

16.27 (a) The capacitance of this air filled dielectric constant, 1.00 parallel plate capacitor

is

3

40 40.0 F 50.0 V 2.00 10 C 2.00 mCQ

(b) eq 1 2 25.0 F+40.0 F 65.0 FC C C

(c) 3 3

40 25 2.00 10 C 1.25 10 C 750 CQ Q Q


16.28 (a) eq

750 C11.5 V

65.0 F

QV

C

(b) 6 64.00 10 F 1.50 V 6.00 10 C 6.00 CQ C V

16.29 (a) 40 25 750 C 288 C 462 CQ Q Q d irected toward the negative plate

(b)

12 2 2 4 2

0

-3

8.85 10 C N m 7.60 10 m

1.80 10 m

AC

d

1 3 2 3.33 F 2.00 F 8.66 Fp s sC C C C

(c) 2 2 2 2 10.0 F 20.0 FpC C C on one plate and

1 1

eq

1 2

1 1 1 16.04 F

8.66 F 20.0 Fp p

CC C

on the other plate.

16.30 total eq 6.04 F 60.0 V 362 Cab

Q C V , so

12 2 2 12 2

90

-15

8.85 10 C N m 21.0 10 m3.10 10 m

60.0 10 F

Ad

C

1

11

362 C41.8 V

8.66 F

p

pp

QV

C

16.31 (a) Assuming the capacitor is air-filled 1 , the capacitance is

3 3 12.00 F 41.8 V 83.6 C

pQ C V

(b) Q C V

(c) 1 1.00 F 10.0 V 10.0 CQ

(d) 2 2.00 F 0 0Q

(e) Increasing the d istance separating the plates decreases the capacitance, the charge

stored , and the electric field strength between the plates. This means that all of the

previous answers will be decreased.


16.32 2 12Q Q

0 sin15.0 tan15.0xF qE T mg

or 1 12 10.0 CQ Q

tan15.0mgd

V Edq

6 2

3

9

350 10 kg 9.80 m s 0.040 0 m tan15.01.23 10 V 1.23 kV

30.0 10 CV

16.33 (a) Capacitors in a series combination store the same charge, 7.00 F and the 5.00 F,

where eqC is the equivalent capacitance and 4.00 F is the potential d ifference

maintained across the series combination. The equivalent capacitance for the given

series combination is eq 1 2

1 1 1

C C C, or 1 2

eq

1 2

C CC

C C, giving

2

2 26 41 1Energy stored 4.50 10 F 12.0 V 3.24 10 J

2 2 2

QC V

C

so the charge stored on each capacitor in the series combination is

eq 1.79 F 6.00 V 10.7 CQ C V

(b) When connected in parallel, each capacitor has the same potential difference,

eq

1 1 1 2 1

18.0 F 36.0 F 36.0 FC, maintained across it. The charge stored on each

capacitor is then

For 1 2.50 FC : 1 1 2.50 F 6.00 V 15.0 CQ C V

For 2 6.25 FC : 2 2 6.25 F 6.00 V 37.5 CQ C V

16.34 (a) When connected in series, the equivalent capacitance is eq 1 2

1 1 1

C C C, or

1 2eq

1 2

4.20 F 8.50 F2.81 F

4.20 F 8.50 F

C CC

C C


(b) When connected in parallel, the equivalent capacitance is

if the computed equivalent capacitance is truly equivalent to the original combination.

16.35 (a) First, we replace the parallel combination

between points b and c by its equivalent

capacitance, 2.00 F 6.00 F 8.00 FbcC .

Then, we have three capacitors in series

between points a and d . The equivalent

capacitance for this circuit is therefore

eq ab bc cd

1 1 1 1 3

8.00 FC C C C

giving eq

8.00 F2.67 F

3C

(b) The charge store on each capacitor in the series combination is

ab bc cd eq ad 2.67 F 9.00 V 24.0 CQ Q Q C V

Then, note that bcbc

bc

24.0 C3.00 V

8.00 F

QV

C. The charge on each capacitor in the

original circuit is:

On the 8.00 F between a and b: 8 ab 24.0 CQ Q

On the 8.00 F between c and d : 8 cd 24.0 CQ Q

On the 2.00 F between b and c: 2 2 bc 2.00 F 3.00 V 6.00 CQ C V

On the 6.00 F between b and c: 6 6 bc 6.00 F 3.00 V 18.0 CQ C V

(c) Note that 2 2

3

1 1Energy stored 1.50 F 6.00 V 27.0 J

2 2f i

C V , and that

8

750.0 C 1.00 10 V1 1

2.50 10 J100 2 200

W Q V . We earlier found that

bc 3.00 VV , so we conclude that the potential d ifference across each capacitor in

the circuit is

8 2 6 8 3.00 VV V V V


16.36 0Q [1]

V Q C

Thus, using Equation [1], 2 2

series

2 2

9.00 pF 2.00 pF

9.00 pF

C CC

C C which reduces to

22

2 29.00 pF 18.0 pF 0C C , or nylon

Therefore, either 2 6.00 pFC and , from Equation [1], 3.40

or 85.0 Vi

V and 1 6.00 pFC .

We conclude that the two capacitances are 3.00 pF and 6.00 pF .

16.37 (a) The equivalent capacitance of the series combination

in the upper branch is

370 pCfQ

or 80 See Table 16.1

Likewise, the equivalent capacitance of the series combination in the lower branch

is

0 80 1.48 pF 118 pFfC C or lower 1.33 FC

These two equivalent capacitances are connected in parallel with each other, so the

equivalent capacitance for the entire circuit is

eq upper lowerC 2.00 F 1.33 F 3.33 FC C


(b) Note that the same potential d ifference, equal to the potential d ifference of the

battery, exists across both the upper and lower branches. The charge stored on each

capacitor in the series combination in the upper branch is

3 6 upper upper 2.00 F 90.0 V 180 CQ Q Q C V

and , the charge stored on each capacitor in the series combination in the lower

branch is

12 2 2 4 2

0

-3

2.1 8.85 10 C N m 175 10 m

0.040 0 10 m

AC

d

(c) The potential d ifference across each of the capacitors in the circuit is:

22

2

120 C60.0 V

2.00 F

QV

C 4

4

4

120 C30.0 V

4.00 F

QV

C

33

3

180 C60.0 V

3.00 F

QV

C

00w LA

Cd d

16.38 (a) The equivalent capacitance of the series

combination in the rightmost branch of the

circuit is

8 3

12 2 2 20

9.50 10 F 0.025 0 10 m1.04 m

3.70 8.85 10 C N m 7.00 10 m

C dL

w

or right 6.00 FC

Figure P16.38

(b) The equivalent capacitance of the three

capacitors now connected in parallel with

each other and with the battery is

eq 4.00 F 2.00 F 6.00 F 12.0 FC

Diagram 1

(c) The total charge stored in this circuit is

12.0 F 36.0 Vtotal eqQ C V

or total 432 CQ

Diagram 2


(d) The charges on the three capacitors shown in Diagram 1 are:

13 -3 142.01 10 F 100 10 V 2.01 10 CQ C V

14

5

-19

2.01 10 C1.26 10

1.60 10 C

Qn

e

right right 6.00 F 36.0 V 216 CQ C V

4 2Yes. as it should.right totalQ Q Q Q

(e) The charge on each capacitor in the series combination in the rightmost branch of

the original circuit (Figure P16.38) is

0eq 1 2 3 where

AC d d d d

d

(f) eqC 3.33 F

(g) 88

8

216 C27.0 V

8.00 F

QV

C Note that

8 24 36.0 VV V V as it should .

16.39

The circuit may be reduced in steps as shown above.

Using the Figure 3, 4.00 F 24.0 V 96.0 CacQ

Then, in Figure 2, 96.0 C

16.0 V6.00 F

ac

abab

QV

C

and 24.0 V 16.0 V 8.00 Vbc ac ab

V V V


Finally, using Figure 1, 1 1 1.00 F 16.0 V 16.0 Cab

Q C V

262

34

4 64

120 10 CEnergy stored 1.80 10 J 1.80 mJ

2 2 4.00 10 F

Q

C, 6.00 F

and 4 4.00 F 32.0 Cbc

Q V

16.40 From Q C V , the initial charge of each capacitor is

10 10.0 F 12.0 V 120 CQ and

2C

After the capacitors are connected in parallel, the potential d ifference across each is

3.00 VV , and the total charge of 10 120 CxQ Q Q is d ivided between the two

capacitors as

10 10.0 F 3.00 V 30.0 CQ and

10 120 C 30.0 C 90.0 CxQ Q Q

Thus, 90.0 C

30.0 F3.00 V

xx

QC

V

16.41 (a) From Q C V , 3

25 25.0 F 50.0 V 1.25 10 C 1.25 mCQ

and 3

40 40.0 F 50.0 V 2.00 10 C 2.00 mCQ

(b) When the two capacitors are connected in parallel, the equivalent capacitance is

eq 1 2 25.0 F+40.0 F 65.0 FC C C .

Since the negative plate of one was connected to the positive plate of the other, the

total charge stored in the parallel combination is

3 3

40 25 2.00 10 C 1.25 10 C 750 CQ Q Q

The potential d ifference across each capacitor of the parallel combination is

eq

750 C11.5 V

65.0 F

QV

C


and the final charge stored in each capacitor is

25 1 25.0 F 11.5 V 288 CQ C V

and 40 25 750 C 288 C 462 CQ Q Q

16.42 (a) The original circuit reduces to a single equivalent capacitor in the steps shown

below.

1 1

1 2

1 1 1 13.33 F

5.00 F 10.0 FsC

C C

1 3 2 3.33 F 2.00 F 8.66 Fp s sC C C C

2 2 2 2 10.0 F 20.0 FpC C C

1 1

eq

1 2

1 1 1 16.04 F

8.66 F 20.0 Fp p

CC C


(b) The total charge stored between points a and b is

total eq 6.04 F 60.0 V 362 Cab

Q C V

Then, looking at the third figure, observe that the charges of the series capacitors of

that figure are 1 2 total 362 Cp pQ Q Q . Thus, the potential d ifference across the

upper parallel combination shown in the second figure is

1

11

362 C41.8 V

8.66 F

p

pp

QV

C

Finally, the charge on 3C is

3 3 12.00 F 41.8 V 83.6 C

pQ C V

16.43 From Q C V , the initial charge of each capacitor is

1 1.00 F 10.0 V 10.0 CQ and

2 2.00 F 0 0Q

After the capacitors are connected in parallel, the potential d ifference across one is the

same as that across the other. This gives

1 2

1.00 F 2.00 F

Q QV or

2 12Q Q [1]

From conservation of charge, 1 2 1 2 10.0 CQ Q Q Q . Then, substituting from

Equation [1], this becomes

1 12 10.0 CQ Q , giving 1

10 C

3Q

Finally, from Equation [1], 2

20 C

3Q


16.44 Recognize that the 7.00 F and the 5.00 F of the

center branch are connected in series. The total

capacitance of that branch is

11 1

2.92 F5.00 7.00

sC

Then recognize that this capacitor, the 4.00 F

capacitor, and the 6.00 F capacitor are all connected

in parallel between points a and b. Thus, the equivalent

capacitance between points a and b is

eq 4.00 F 2.92 F+6.00 F 12.9 FC

16.45 2

2 26 41 1Energy stored 4.50 10 F 12.0 V 3.24 10 J

2 2 2

QC V

C

16.46 (a) The equivalent capacitance of a series combination of 1 2 and C C is

eq

1 1 1 2 1

18.0 F 36.0 F 36.0 FC or eq 12.0 FC

When this series combination is connected to a 12.0-V battery, the total stored

energy is

2 26 4

eq

1 1Total energy stored 12.0 10 F 12.0 V 8.64 10 J

2 2C V

(b) The charge stored on each of the two capacitors in the series combination is

4

1 2 total eq 12.0 F 12.0 V 144 C 1.44 10 CQ Q Q C V

and the energy stored in each of the individual capacitors is

242

411 6

1

1.44 10 CEnergy stored in 5.76 10 J

2 2 18.0 10 F

QC

C

and

242

422 6

2

1.44 10 CEnergy stored in 2.88 10 J

2 2 36.0 10 F

QC

C


4 4 4

1 2Energy stored in Energy stored in 5.76 10 J 2.88 10 J 8.64 10 JC C , which

is the same as the total stored energy found in part (a). This must be true

if the computed equivalent capacitance is truly equivalent to the original combination.

(c) If 1 2 and C C had been connected in parallel rather than in series, the equivalent

capacitance would have been eq 1 2 18.0 F 36.0 F 54.0 FC C C . If the total

energy stored 2

eq

1

2C V in this parallel combination is to be the same as was

stored in the original series combination, it is necessary that

4

6

eq

2 8.64 10 J2 Total energy stored5.66 V

54.0 10 FV

C

Since the two capacitors in parallel have the same potential d iffer ence across them,

the energy stored in the individual capacitors 21

2C V is d irectly proportional

to their capacitances. 2The larger capacitor, , stores the most energy in this case.C

16.47 (a) The energy initially stored in the capacitor is

2

2 2

1


2 2 2

ii i

i

QC V

C

(b) When the capacitor is d isconnected from the battery, the stored charge becomes

isolated with no way off the plates. Thus, the charge remains constant at the value

iQ as long as the capacitor remains d isconnected . Since the capacitance of a parallel

plate capacitor is e0C A d , when the d istance d separating the plates is doubled ,

the capacitance is decreased by a factor of 2 2 1.50 Ff iC C . The stored energy

(with Q unchanged) becomes

2 2 2

2 1Energy stored 2 2 Energy stored 108 J

2 2 2 2

i i i

f i f

Q Q Q

C C C

(c) When the capacitor is reconnected to the battery, the potential d ifference between

the plates is reestablished at the original value of 6.00 Vi

V V , while the

capacitance remains at 2 1.50 Ff iC C . The energy stored under these

conditions is

2 2

3


2 2f i

C V


16.48 The energy transferred to the water is

8

750.0 C 1.00 10 V1 1

2.50 10 J100 2 200

W Q V

Thus, if m is the mass of water boiled away,

vW m c T L becomes

7 6J2.50 10 J 4186 100 C 30.0 C 2.26 10 J kg

kg Cm

giving 72.50 10 J

9.79 kg2.55 J kg

m

16.49 (a) Note that the charge on the plates remains constant at the original value, 0Q , as the

d ielectric is inserted . Thus, the change in the potential d ifference, V Q C , is due

to a change in capacitance alone. The ratio of the final and initial capacitances is

0

0

f

i

C A d

C A d and

0

0

85.0 V3.40

25.0 V

f f i

i i f

Q V VC

C Q V V

Thus, the d ielectric constant of the inserted material is 3.40 , and the material is

probably nylon (see Table 16.1).

(b) If the d ielectric only partially filled the space between the plates, leaving the

remaining space air-filled , the equivalent d ielectric constant would be somewhere

between 1.00 (air) and 3.40 . The resulting potential d ifference would then

lie somewhere between 85.0 Vi

V and 25.0 Vf

V .


16.50 (a) The capacitance of the capacitor while air-filled is

12 2 2 4 2

1200 2

8.85 10 C N m 25.0 10 m1.48 10 F 1.48 pF

1.50 10 m

AC

d

The original charge stored on the plates is

-12 2 12

0 0 01.48 10 F 2.50 10 V 370 10 C 370 pCQ C V

Since d istilled water is an insulator, introducing it between the isolated capacitor

plates does not allow the charge to change. Thus, the final charge is 370 pCfQ .

(b) After immersion d istilled water 80 See Table 16.1 , the new capacitance is

0 80 1.48 pF 118 pFfC C

and the new potential d ifference is 370 pC

3.14 V118 pF

f

ff

QV

C

(c) The energy stored in a capacitor is: 2Energy stored 2Q C . Thus, the change in the

stored energy due to immersion in the d istilled water is

2122 2 2

0 0

12 12

8 9

370 10 C1 1 1 1E

2 2 2 2 118 10 F 1.48 10 F

4.57 10 J 45.7 10 J 45.7 nJ

f

f i f i

Q Q Q

C C C C

16.51 (a) The d ielectric constant for Teflon ® is 2.1 , so the capacitance is

12 2 2 4 2

0

-3

2.1 8.85 10 C N m 175 10 m

0.040 0 10 m

AC

d

98.13 10 F 8.13 nFC

(b) For Teflon®, the d ielectric strength is 660.0 10 V mmaxE , so the maximum voltage

is

6 -3

max max 60.0 10 V m 0.040 0 10 m V E d

3

max 2.40 10 V 2.40 kVV


16.52 Before the capacitor is rolled , the capacitance of this parallel plate capacitor is

00

w LAC

d d

where A is the surface area of one side of a foil strip. Thus, the required length is

8 3

12 2 2 20

9.50 10 F 0.025 0 10 m1.04 m

3.70 8.85 10 C N m 7.00 10 m

C dL

w

16.53 (a) 12

16 3

3

1.00 10 kg9.09 10 m

1100 kg m

mV

Since 34

3

rV , the radius is

1 33

4

Vr , and the surface area is

2 316 32 3

2 10 23 9.09 10 m3

4 4 4 4.54 10 m4 4

VA r

(b) 0 AC

d

12 2 2 10 2

13

9

5.00 8.85 10 C N m 4.54 10 m2.01 10 F

100 10 m

(c) 13 -3 142.01 10 F 100 10 V 2.01 10 CQ C V

and the number of electronic charges is

14

5

-19

2.01 10 C1.26 10

1.60 10 C

Qn

e

16.54 Since the capacitors are in parallel, the equivalent capacitance is

0 1 2 30 1 0 2 0 3

eq 1 2 3

A A AA A AC C C C

d d d d

or 0eq 1 2 3 where

AC A A A A

d


16.55 Since the capacitors are in series, the equivalent capacitance is given by

3 1 2 31 2

eq 1 2 3 0 0 0 0

1 1 1 1 d d d dd d

C C C C A A A A

or 0eq 1 2 3 where

AC d d d d

d

16.56 (a) Please refer to the solution of Problem 16.37 where the

following results were obtained:

eqC 3.33 F 3 6 180 CQ Q

2 4 120 CQ Q

The total energy stored in the full circuit is then

2 26

eqtotal

2 3

1 1Energy stored C 3.33 10 F 90.0 V

2 2

1.35 10 J 13.5 10 J 13.5 mJ

V

(b) The energy stored in each individual capacitor is

For 2.00 F:

262

32

2 62


2 2 2.00 10 F

Q

C

For 3.00 F :

262

33

3 63


2 2 3.00 10 F

Q

C

For 4.00 F:

262

34

4 64


2 2 4.00 10 F

Q

C

For 6.00 F :

262

36

6 66


2 2 6.00 10 F

Q

C

(c) The total energy stored in the individual capacitors is

totalEnergy stored= 3.60 5.40 1.80 2.70 mJ 13.5 mJ Energy stored

Thus, the sums of the energies stored in the individual capacitors equals the total

energy stored by the system.



16.57 In the absence of a d ielectric, the

capacitance of the parallel plate

capacitor is 00

AC

d

With the d ielectric inserted , it fills

one-third of the gap between the

plates as shown in sketch (a) at the

right. We model this situation as

consisting of a pair of capacitors, 1C

(a)

(b)

and 2C , connected in series as shown in sketch (b) at the right.

In reality, the lower plate of 1C and the upper plate of

2C are

one and the same, consisting of the lower surface of the

dielectric shown in sketch (a). The capacitances in the model of sketch (b) are given by:

0 01

3

3

A AC

d d and 0 0

2

3

2 3 2

A AC

d d

and the equivalent capacitance of the series combination is

0 0 0 0 0 0

1 2 1 2 1 2 1 2 1 12

3 3 3 3 3 3eq

d d d d d

C A A A A A C

and eq 0

3

2 1C C

16.58 For the parallel combination: 1 2pC C C which gives 2 1pC C C [1]

For the series combination: 1

1 2 2 1 1

1 1 1 1 1 1 or s

s s s

C C

C C C C C C C C

Thus, we have 12

1

s

s

C CC

C C and equating this to Equation [1] above gives

11

1

sp

s

C CC C

C C or 2

1 1 1p p s sC C C C C C C 1sC C

We write this result as : 2

1 1 0p p sC C C C C

and use the quadratic formula to obtain 2

1

1 1

2 4p p p sC C C C C


Then, Equation [1] gives 2

2

1 1

2 4p p p sC C C C C

16.59 The charge stored on the capacitor by the battery is

1

100 VQ C V C

This is also the total charge stored in the parallel combination when this charged

capacitor is connected in parallel with an uncharged 10.0- F capacitor. Thus, if 2

V is

the resulting voltage across the parallel combination, 2pQ C V gives

100 V 10.0 F 30.0 VC C or 70.0 V 30.0 V 10.0 FC

and 30.0 V

10.0 F 4.29 F70.0 V

C

16.60 (a) The 1.0- C is located 0.50 m from point P, so its contribution to the potential at P is

6

9 2 2 411

1

1.0 10 C8.99 10 N m C 1.8 10 V

0.50 me

qV k

r

(b) The potential at P due to the 2.0- C charge located 0.50 m away is

6

9 2 2 422

2

2.0 10 C8.99 10 N m C 3.6 10 V

0.50 me

qV k

r

(c) The total potential at point P is 4 4

1 2 1.8 3.6 10 V 1.8 10 VPV V V

(d) The work required to move a charge 3.0 Cq to point P from infinity is

6 4 23.0 10 C 1.8 10 V 0 5.4 10 JPW q V q V V


16.61 The stages for the reduction of this circuit are shown below.

Thus, 6.25 FeqC

16.62 (a) Due to spherical symmetry, the charge on each of the concentric spherical shells will

be uniformly d istributed over that shell. Inside a spherical surface having a uniform

charge d istribution, the electric field due to the charge on that surface is zero. Thus ,

in this region, the potential due to the charge on that surface is constant and equal

to the potential at the surface. Outside a spherical surface having a uniform charge

d istribution, the potential due to the charge on that surface is given by ek qV

r

where r is the d istance from the center of that surface and q is the charge on that

surface.

In the region between a pair of concentric spherical shells, with the inner shell

having charge Q and the outer shell having radius b and charge Q , the total

electric potential is given by

due to due toinner shell outer shell

1 1eee

k Qk QV V V k Q

r b r b

The potential d ifference between the two shells is therefore,

1 1 1 1

e e er a r b

b aV V V k Q k Q k Q

a b b b ab

The capacitance of this device is given by

e

Q abC

V k b a


(b) When b a , then b a b . Thus, in the limit as b , the capacitance found

above becomes

04e e

ab aC a

k b k

16.63 The energy stored in a charged capacitor is 21

2W C V . Hence,

3

-6

2 300 J24.47 10 V 4.47 kV

30.0 10 F

WV

C

16.64 From Q C V , the capacitance of the capacitor with air between the plates is

00

150 CQC

V V

After the d ielectric is inserted , the potential d ifference is held to the original value, but

the charge changes to 0 200 C=350 CQ Q . Thus, the capacitance with the d ielectric

slab in place is

350 CQ

CV V

The d ielectric constant of the d ielectric slab is therefore

0

350 C 3502.33

150 C 150

C V

C V

16.65 The charges initially stored on the capacitors are

3

1 1 6.0 F 250 V 1.5 10 Ci

Q C V

and 2

2 2 2.0 F 250 V 5.0 10 Ci

Q C V

When the capacitors are connected in parallel, with the negative plate of one connected

to the positive plate of the other, the net stored charge is

3 2 3

1 2 1.5 10 C 5.0 10 C=1.0 10 CQ Q Q


The equivalent capacitance of the parallel combination is eq 1 2 8.0 FC C C . Thus, the

final potential d ifference across each of the capacitors is

3

eq

1.0 10 C125 V

8.0 F

QV

C

and the final charge on each capacitor is

1 1 6.0 F 125 V 750 C 0.75 mCQ C V

and 2 2 2.0 F 125 V 250 C 0.25 mCQ C V

16.66 The energy required to melt the lead sample is

Pb

6 36.00 10 kg 128 J kg C 327.3 C 20.0 C 24.5 10 J kg

0.383 J

fW m c T L

The energy stored in a capacitor is 21

2W C V , so the required potential d ifference is

-6

2 0.383 J2121 V

52.0 10 F

WV

C

16.67 When excess charge resides on a spherical surface that is far removed from any other

charge, this excess charge is uniformly d istributed over the spherical surface, an d the

electric potential at the surface is the same as if all the excess charge were concentrated

at the center of the spherical surface.

In the given situation, we have two charged spheres, initially isolated from each other,

with charges and potentials of: 1 6.00 CQ ,

1 1 1eV k Q R where 1 12.0 cmR ,

2 4.00 CQ , and 2 2 2eV k Q R with

2 18.0 cmR .


When these spheres are then connected by a long conducting thread , the charges are

redistributed 1 2yielding charges of and respectivelyQ Q until the two surfaces come to a

common potential 1 1 1 2 2 2V kQ R V kQ R . When equilibrium is established , we have:

From conservation of charge: 1 2 1 2 Q Q Q Q

1 2 2.00 CQ Q [1]

From equal potentials: 1 2 22 1

1 2 1

kQ kQ R

Q QR R R

or 2 11.50Q Q [2]

Substituting Equation [2] into [1] gives: 1

2.00 C0.800 C

2.50Q

Then, Equation [2] gives: 2 1.50 0.800 C 1.20 CQ

16.68 The electric field between the plates is d irected downward with magnitude

4

-3

100 V5.00 10 N m

2.00 10 my

VE

d

Since the gravitational force experienced by the electron is negligible in comparison to

the electrical force acting on it, the vertical acceleration is

19 4

15 2

31

1.60 10 C 5.00 10 N m8.78 10 m s

9.11 10 kg

y y

y

e e

F qEa

m m

(a) At the closest approach to the bottom plate, 0yv . Thus, the vertical displacement

from point O is found from 2 2

0 2y y yv v a y as

262

0 0

15 2

5.6 10 m s sin 450 sin0.89 mm

2 2 8.78 10 m sy

vy

a

The minimum distance above the bottom plate is then

1.00 mm 0.89 mm 0.11 mm2

Dd y


(b) The time for the electron to go from point O to the upper plate is found from

2

0

1

2y yy v t a t as

3 6 15 2

2

m 1 m1.00 10 m 5.6 10 sin 45 8.78 10

s 2 st t

Solving for t gives a positive solution of 91.11 10 st . The horizontal

d isplacement from point O at this time is

6 9

0 5.6 10 m s cos45 1.11 10 s 4.4 mmxx v t

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