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 , w hen released from rest, w ill m ove in the negative xd irection. The w ork d one on the electron by the field is W

Fx

x

qEx

x

1.60 10

19

C 375 N C

3.20 10

2

m

1.92 10

18

J

(b) The change in the electric potential energy is the negative of the w ork d one on the particle by the field . Thus,

PE

W

1.92 10

18

J

(c) Since the Coulom b force is a conservative force, conservation of energy gives KE PE 0 , or conservation of linear momentum. , and

vf

16.2

2

2

PE me

1.92 10

9.11 10

31

18

J

kg

2.05 106 m s in the

x-direction

(a) The change in the electric potential energy is the negative of the w ork d one on the particle by the field . Thus, PE

W

qEx

q 0

x

x

qE y

5.40 10

6

y C

327 N C

32.0 10

2

m

5.65 10

4

J

(b) The change in the electrical potential is the change in electric potential energy per unit charge, or V

PE q

5.65 10 4 J +5.40 10-6 C

105 V

51

Electrical Energy and Capacitance

16.3

The w ork d one by the agent m oving the charge out of the cell is

Winput

Wfield

PEe

1.60 10

mp

v p , so q

19

C

PEe V f Vi

q

V

90 10

3

J C

1.4 10

1.92 10 17 J + 60.0 J C

16.4

v

16.5

E

16.6

Since potential d ifference is w ork per unit charge

m

V

25 000 J C 1.5 10 2 m

d

mp

v

16.7

m

(a) V

ke Q r

(b) F

qE

(c) W

F s cos

1.60 10

1.80 10

16.8

14

J

3.20 10

19

C

1.7 106 N C

1.67 10 6.64 10

vp

20

19

N

27 27

W , the w ork d one is q

V

kg 1.05 107 m s kg

C 1.13 105 N C

5.33 2.90

10

(a) Using conservation of energy,

KE

3

2.64 106 m s

1.80 10

m cos 0

14

N

4.38 10

PE 0 , w ith KE f

“ stopped ” , w e have

s 0 The required stopping p otential is then V

PE q

3.70 10 16 J 1.60 10 19 C

2.31 103 V

2.31 kV

17

J

0 since the particle is

52

Electrical Energy and Capacitance

53

(b) Being m ore m assive than electrons, protons traveling at the sam e initial speed w ill have m ore 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 sam e initial speed is

cos 16.9

0

(a) Use conservation of energy

KE PEs or

KE

rf

PEe

PEs

2 keQq m vi2

PEe

N m2 C2

6.64 10 diagonal 2

Vtotal

4Vsingle charge

Thus, 0

xmax

1 2 kxmax 2

2 QE k

a2

27

158 1.60 10 7

kg 2.00 10 m s

a2

a 2 2

2

4

ke Q r

4

2 35.0 10

a 2

ke Q a

19

C

2

Q V

m , w here

0 , giving 6

C 4.86 104 V m

V

4.36 10

2

m

4.36 cm

C Emax d

max

1.11 10

C

14

2.74 10

Q a

78.0 N m C

2

is the m axim um stretch of the spring.

4 2ke

2

QE xmax

(b) At equilibrium , Qmax

Therefore,

i

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

m vi2

rf

PEe

0

2 ke 79e 2e

2 8.99 109

r

KE PEs

f

8

F 3.0 106 N C 800 m

27.0 C 9.00 V

27 C

3.00 F

The am plitud e is the d istance from the equilibrium position to each of the turning points at x 0 and x 4.36 cm , so A 2.18 cm xmax 2

Electrical Energy and Capacitance

(c) From conservation of energy, Ceq

2.00 103 C 1.25 103 C 750 C , this gives

Q Q40 Q25 2 kxmax 2Q

V Q C

16.10

Using

y

v0 y t

0 v0 y t

25.0 F+40.0 F 65.0 F . Since

C1 C2

2

k 2A

or

2Q

V

4.00 10

6

F 1.50 V

6.00 10

6

C

6.00 C

1 2 a y t for the full flight gives 2 1 2 a y t , or C p 2 2

C2 C2

2 10.0 F

1

1 C p1

Then, using Ceq

1 Cp2

1 8.66 F

1 20.0 F

20.0 F 1

6.04 F for the upw ard part

of the flight gives

y

0 v02y max

2ay

v02y 2 2 v0 y t

Q3

16.11

C3

Thus,

V

(a) VA

ke q rA

(b)

max

V

p1

2 Q1

4

4 mg qE m

m

2.00 F 41.8 V

ymax E

g

20.6 m

qE . Equating m

1 , so the electric field strength is 83.6 C

20.6 m 1.95 103 N C

8.99 109 N m2 C2 0.250 10

1.60 10 2

previous answers will be decreased.

Q2

20.1 m s 4.10 s

Fy

From N ew ton’ s second law , a y this to the earlier result gives

v0 y t

m

19

C

4.02 104 V

5.75 10

7

40.2 kV

V

54

Electrical Energy and Capacitance

55

(c) The original electron w ill be repelled by the negatively charged particle w hich sud d enly appears at point A . Unless the electron is fixed in place, it w ill move in the opposite d irection, aw ay from points A and B, thereby low ering the potential d ifference betw een these points.

16.12

(a) At the origin, the total potential is

Q1 Q2

Q1 Q2 10.0 C

(b) At point B located at Q1 2 Q1 10.0 C , the need ed d istances are Q1

10 3

C

xB

x2

and r2

2

yB

y2

2

1.50 cm

2

1.80 cm

2

2.34 cm

giving

VB

16.13

ke q1 r1

ke q2 r2

9

8.99 10 N m C

2

4.50 10 6 C 1.95 10 2 m

2.24 10 2.34 10

2

6

C

1.21 106 V

m

(a) Calling the 4.00 F charge 6.00 F ,

V i

ke qi ri

8.99 109

V

q1 r1

ke

q3

q2 r2

N m2 C2

2 1

r

r22

8.00 10 6 C 0.060 0 m

4.00 10 6 C 0.030 0 m

2.00 10 0.060 0

2

6

C

0.030 0

2.67 106 V

(b) Replacing Q Ceq 1 Ceq

1 18.0 F

V

1.79 F 6.00 V

1 36.0 F

2 1 36.0 F

10.7 C in part (a) yield s

2

m

Electrical Energy and Capacitance

16.14

W

q

V

Vi , and

q Vf

Vf

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

Vi

ke

q1 r1

q2 r2

8.99 109

N m2 C2

2.00 10 6 C 0.030 0 m

4.00 10 0.030 0

2

6

C

0.060 0

2

m

1.135 106 V

Thus, W

16.15

8.00 10

(a) V i

C 0 1.135 106 V

9.08 J

ke qi ri N m2 C2

8.99 109

(b) PE

6

5.00 10 9 C 0.175 m

3.00 10 9 C 0.175 m

103 V

ke qi q2 r12 8.99 109

N m2 C2

5.00 10

9

C

3.00 10

9

C

3.85 10

0.350 m

7

J

The negative sign m eans that positive work must be done to separate the charges (that is, bring them up to a state of zero potential energy).

16.16

9.00 10 9 C is

The potential at d istance r 0.300 m from a charge Q

V

ke Q r

8.99 109 N m 2 C 2 9.00 10

9

C

270 V

0.300 m

Thus, the w ork required to carry a charge q 3.00 10 is W

qV

3.00 10

9

C

270 V

8.09 10

7

J

9

C from infinity to this location

56

Electrical Energy and Capacitance

16.17

57

The Pythagorean theorem gives the d istance from the m id point of the base to the charge at the apex of the triangle as r3

4.00 cm

2

1.00 cm

2

15 cm

15 10

Then, the potential at the m id point of the base is V

2

m

ke qi ri , or i

V

1.10 104 V

16.18

7.00 10

N m2 8.99 10 C2 9

9

C

7.00 10

0.010 0 m

9

C

0.010 0 m

7.00 10 15 10

2

9

C

m

11.0 kV

Outsid e the spherical charge d istribution, the potential is the sam e as for a point charge at the center of the sphere,

V

Thus,

keQ r , w here Q 1.00 10 9 C

PEe

q

V

ekeQ

1 rf

1 ri

and from conservation of energy

or

1 me v 2 0 2

ekeQ

2 8.99 109 v

v

1 rf

N m2 C2

1 ri

This gives v

1.00 10

9.11 10

7.25 106 m s

PEe ,

KE

31

9

kg

2 keQe 1 me rf

C 1.60 10

19

C

1 , or ri

1 0.020 0 m

1 0.030 0 m

Electrical Energy and Capacitance

16.19

58

(a) When the charge configuration consists of only the tw o protons q1 and q2 in the sketch , the potential energy of the configuration is

PEa

8.99 109 N m2 C2 1.60 10

ke q1q2 r12

or

6.00 10

PEa

3.84 10

14

15

19

C

2

m

J

(b) When the alpha particle q3 in the sketch is ad d ed to the configuration, there are three d istinct pairs of particles, each of w hich possesses potential energy. The total potential energy of the configuration is now

PEb

ke q1q2 r12

ke q1q3 r13

ke q2 q3 r23

PEa

2

w here use has been m ad e of the facts that q1q3 r13

r23

3.00 fm

2

3.00 fm

2

ke 2e2 r13

q2 q3

4.24 fm 4.24 10

e 2e 15

2e2 and

m . Also, note that the first

term in this com putation is just the potential energy com p uted in part (a). Thus,

PEb

4ke e2 r13

PEa

3.84 10

14

4 8.99 109 N m2 C2 1.60 10

J

4.24 10

15

19

C

2

2.55 10

m

(c) If w e start w ith the three-particle system of part (b) and allow the alpha particle to escape to infinity [thereby returning us to the tw o -particle system of part (a)], the change in electric potential energy w ill be

PE

PEa

PEb

3.84 10

14

J 2.55 10

13

J

2.17 10

13

J

(d ) Conservation of energy, KE PE 0 , gives the speed of the alpha particle at infinity in the situation of part (c) as m v2 2 0 PE , or

v

2

PE m

2

2.17 10

6.64 10

27

13

kg

J

8.08 106 m s

13

J

Electrical Energy and Capacitance

59

(e) When, starting w ith the three-particle system , the tw o protons are both allow ed to escape to infinity, there w ill be no rem aining pairs of particles and hence no rem aining potential energy. Thus, PE 0 PEb PEb , and conservation of energy gives the change in kinetic energy as KE PE PEb . Since the protons are id entical particles, this increase in kinetic energy is split equally betw een them giving KE proton

or

16.20

1 m p v 2p 2

2.55 10 13 J 1.67 10-27 kg

PEb mp

vp

1 PEb 2

1.24 107 m s

(a) If a proton and an alpha particle, initially at rest 4.00 fm apart, are released and allow ed to reced e to infinity, the final speed s of the tw o particles w ill d iffer because of the d ifference in the m asses of the particles. Thus, attem pting to solve for the final speed s by use of conservation of energy alone lead s to a situation of having one equation with two unknowns , and d oes not perm it a solution. (b) In the situation d escribed in part (a) above, one can obtain a second equation w ith the tw o unknow n final speed s by using conservation of linear momentum. Then, one w ould have tw o equations w hich could be solved sim ultaneously both unknow ns. 1 m v2 2

(c) From conservation of energy:

or m v2

mp v 2p

0

0

2 8.99 109 N m2 C2 3.20 10

2ke q q p ri m v2

yield ing

1 m p v 2p 2

4.00 10 mp v2p

2.30 10

13

15

ke q q p ri 19

0

C 1.60 10

19

m [1]

J

From conservation of linear m om entum ,

mv

mp v p

0

v

or

mp m

vp

[2]

Substituting Equation [2] into Equation [1] gives

m

mp m

2

v2p

mp v 2p

2.30 10

13

J

or

mp m

1 m p v 2p

2.30 10

13

J

C

Electrical Energy and Capacitance

60

and

2.30 10 13 J mp m 1 mp

vp

13

2.30 10 1.67 10

27

6.64 10

27

J

+1 1.67 10

27

kg

1.05 107 m s

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

16.21

V

ke Q r

r

ke Q V

mp

1.67 10 6.64 10

vp

m

27 27

kg 1.05 107 m s kg

2.64 106 m s

so 8.99 109 N m 2 C 2 8.00 10

9

C

71.9 V m V

V

For V 100 V, 50.0 V, and 25.0 V,

r

0.719 m, 1.44 m, and 2.88 m

The rad ii are inversely proportional to the potential.

16.22

By d efinition, the w ork required to m ove a charge from one point to any other point on an equipotential surface is zero. From the d efinition of w ork, W F cos s , the w ork is zero only if s 0 or F cos 0 . The d isplacem ent s cannot be assum ed to be zero in all cases. Thus, one m ust require that F cos 0 . The force F is given by F qE and neither the charge q nor the field strength E can be assum ed to be zero in all cases. Therefore, the only w ay the w ork can be zero in all cases is if cos 0 . But if cos 0, then 90 or the force (and hence the electric field) m ust be perpend icular to the d isplacem ent s (w hich is tangent to the surface). That is, the field m ust be perpend icular to the equ ipotential surface at all points on that surface.

16.23

From conservation of energy, KE PEe

0

ke Qq rf

1 m vi2 2

2 8.99 109 rf

0 or rf

N m2 C2

6.64 10

27

KE PEe i , w hich gives

f

2 keQq m vi2

2 ke 79e 2e

158 1.60 10 7

kg 2.00 10 m s

m vi2 19

2

C

2

2.74 10

14

m

Electrical Energy and Capacitance

16.24

61

(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 a2

diagonal 2

r

a2

a 2 2

2

a 2

Since the charges have equal m agnitud es and are all the sam e distance from the center of the square, they m ake equal contributions to the total potential. Thus, Vtotal

4Vsingle

4

charge

ke Q r

4

ke Q a

2

4 2ke

Q a

(b) The w ork 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 W

16.25

(a) C

(b)

0

A d

Qmax

PEcenter

8.85 10

C

V

16.26

(a) C

Q V

(b) Q C 16.27

C2 N m2

qVtotal

0 q 4 2k e

1.0 106 m2 800 m

Q a

1.1 10

8

27.0 C 9.00 V

F 3.0 106 N C 800 m

(b) Ceq (c)

8

qQ a

F

27 C

3.00 F

V

(a) The capacitance of this air filled dielectric constant, is

Q40

4 2k e

C Emax d

max

1.11 10

12

PE

40.0 F 50.0 V

C1 C2

Q Q40 Q25

2.00 103 C

2.00 mC

25.0 F+40.0 F 65.0 F

2.00 103 C 1.25 103 C 750 C

1.00 parallel plate capacitor

Electrical Energy and Capacitance

16.28

Q Ceq

V

(a)

(b) Q C

16.29

750 C 65.0 F

V

4.00 10

(a) Q40

Q Q25

(b) C

0

8.85 10

A

12

F 1.50 V

6.00 10

C3 Cs

C2 C2

16.30

Qtotal

1 C p1

C 2 N m 2 7.60 10

2 3.33 F

2 10.0 F

d

V

A

8.85 10

6.04 F 60.0 V

ab

12

362 C 8.66 F

C p1

Q3

C3

(b) Q C

V

Q1

(d ) Q2

m2

C2 N m 2

V

p1

6.04 F on the other plate.

362 C , so

21.0 10

12

m2

3.10 10 9 m

41.8 V

2.00 F 41.8 V

1.00 F 10.0 V 2.00 F 0

1

1 20.0 F

(a) Assum ing the capacitor is air-filled

(c)

4

2.00 F 8.66 F

60.0 10-15 F Q p1

p1

6.00 C

20.0 F on one plate and

1 8.66 F

C V

16.31

1 Cp2

Ceq 0

C

462 C d irected tow ard the negative plate

1

Ceq

6

1.80 10-3 m

d

Cp2

6

750 C 288 C

C p1 Cs (c)

11.5 V

1 , the capacitance is 83.6 C

10.0 C

0

(e) Increasing the d istance separating the plates d ecreases the capacitance, the charge stored , and the electric field strength betw een the plates. This m eans that all of the previous answers will be decreased.

62

Electrical Energy and Capacitance

16.32

Q2 Fx or

16.33

2 Q1 0

qE T sin15.0

mg tan15.0

Q1 2 Q1 10.0 C V

V

63

mgd tan15.0 q

Ed

350 10

6

kg 9.80 m s 2 0.040 0 m tan15.0 30.0 10

9

C

1.23 103 V

1.23 kV

(a) Capacitors in a series com bination store the sam e charge, 7.00 F and the 5.00 F , w here Ceq is the equivalent capacitance and 4.00 F is the potential d ifference m aintained across the series com bination. The equivalent capacitance for the given C1C2 1 1 1 series com bination is , or Ceq , giving Ceq C1 C2 C1 C2 Energy stored

Q2 2C

1 C 2

V

1 4.50 10 2

2

6

F 12.0 V

2

3.24 10

4

J

so the charge stored on each capacitor in the series com bination is

Q Ceq

V

1.79 F 6.00 V

10.7 C

(b) When connected in parallel, each capacitor has the sam e potential difference, 1 1 1 2 1 , m aintained across it. The charge stored on each Ceq 18.0 F 36.0 F 36.0 F capacitor is then

16.34

For C1

2.50 F :

Q1 C1

V

2.50 F 6.00 V

15.0 C

For C2

6.25 F :

Q2

V

6.25 F 6.00 V

37.5 C

C2

(a) When connected in series, the equivalent capacitance is

Ceq

C1C2 C1 C2

4.20 F 8.50 F 4.20 F 8.50 F

2.81 F

1 Ceq

1 C1

1 , or C2

Electrical Energy and Capacitance

64

(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, w e replace the parallel combination betw een points b and c by its equivalent capacitance, Cbc 2.00 F 6.00 F 8.00 F . Then, w e have three cap acitors in series betw een points a and d . The equivalent capacitance for this circuit is therefore 1 Ceq

giving

1 Cab

1 Cbc

1 Ccd

3 8.00 F

8.00 F 3

Ceq

2.67 F

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

Qab

Qbc

Qcd

Then, note that

Ceq Vbc

Vad Qbc Cbc

2.67 F 9.00 V 24.0 C 8.00 F

24.0 C

3.00 V . The charge on each capacitor in the

original circuit is: On the 8.00 F betw een a and b:

Q8

Qab

24.0 C

On the 8.00 F betw een c and d :

Q8

Qcd

24.0 C

On the 2.00 F betw een b and c:

Q2

C2

Vbc

2.00 F 3.00 V

6.00 C

On the 6.00 F betw een b and c:

Q6

C6

Vbc

6.00 F 3.00 V

18.0 C

1 1 2 2 Cf V i 1.50 F 6.00 V 27.0 J , and that 2 2 50.0 C 1.00 108 V 1 1 W Q V 2.50 107 J . We earlier found that 100 2 200 Vbc 3.00 V , so w e conclud e that the potential d ifference across each capacitor in the circuit is

(c) N ote that Energy stored

V8

V2

3

V6

V8

3.00 V

Electrical Energy and Capacitance

16.36

[1]

Q0 V

QC

Cseries

Thus, using Equation [1],

C22

9.00 pF C2 18.0 pF

Therefore, either C2 or

65

V

2

9.00 pF C2 C2 9.00 pF C2

C2

2.00 pF w hich red u ces to

0 , or nylon

6.00 pF and , from Equation [1],

3.40

85.0 V and C1 6.00 pF .

i

We conclud e that the tw o capacitances are 3.00 pF and 6.00 pF .

16.37

(a) The equivalent capacitance of the series com bination in the upper branch is

Qf or

370 pC

80

See Table 16.1

Likew ise, the equivalent capacitance of the series com bination in the low er branch is

Cf

C0

80 1.48 pF

118 pF

or

Clower 1.33 F

These tw o equivalent capacitances are connected in parallel w ith each other, so the equivalent capacitance for the entire circuit is

Ceq

Cupper

Clower

2.00 F 1.33 F

3.33 F

Electrical Energy and Capacitance

66

(b) N ote that the sam e potential d ifference, equal to the potential d ifference of the battery, exists across both the upper and low er branches. The charge stored on each capacitor in the series com bination in the upper branch is

Q3

Q6

Qupper

Cupper

V

2.00 F 90.0 V

180 C

and , the charge stored on each capacitor in the series com bination in the low er branch is

C

2.1 8.85 10

A

0

12

C2 N m2 175 10

4

m2

0.040 0 10-3 m

d

(c) The potential d ifference across each of the capacitors in the circuit is: V2

Q2 C2

120 C 2.00 F

60.0 V

V3

Q3 C3

180 C 3.00 F

60.0 V

Q4 C4

V4

0

C

120 C 4.00 F

A

0

d

30.0 V

w L d

16.38 (a) The equivalent capacitance of the series com bination in the rightm ost branch of the circuit is

L

C d 0w

or Cright

9.50 10

8

3.70 8.85 10

12

F 0.025 0 10 2

C N m

2

3

m

7.00 10

2

m

1.04 m Figure P16.38

6.00 F

(b) The equivalent capacitance of the three capacitors now connected in parallel w ith each other and w ith the battery is

Ceq

4.00 F 2.00 F 6.00 F

12.0 F

D iagram 1

(c) The total charge stored in this circuit is

Qtotal or Qtotal

Ceq

V

432 C

12.0 F 36.0 V

D iagram 2

Electrical Energy and Capacitance

(d ) The charges on the three capacitors show n in Diagram 1 are: Q C Q e

n

Qright

V

2.01 10

13

2.01 10 14 C 1.60 10-19 C

Cright

V

F 100 10-3 V

14

C

1.26 105

6.00 F 36.0 V

Yes. Q4 Q2 Qright

2.01 10

216 C

Qtotal as it should.

(e) The charge on each capacitor in the series combination in the rightm ost branch of the original circuit (Figure P16.38) is 0

Ceq

(f)

Ceq

(g)

V8

A

where d

d

d1

d2

d3

3.33 F Q8 C8

216 C 8.00 F

27.0 V

N ote that

V8

V24

16.39

The circuit m ay be red uced in steps as show n above. Using the Figure 3, Then, in Figure 2,

and

V

bc

Qac V

V

ac

ab

4.00 F 24.0 V

96.0 C

Qac Cab

96.0 C 6.00 F

V

24.0 V 16.0 V 8.00 V

ab

16.0 V

V

36.0 V as it should .

67

Electrical Energy and Capacitance

Finally, using Figure 1,

Energy stored

and

16.40

Q4

4.00 F

From Q C

Q10

Q1 C1

120 10

Q42 2C4

4

V

V

1.00 F 16.0 V

ab

6

2 4.00 10

16.0 C

2

C 6

68

F

1.80 10

3

J

1.80 mJ ,

6.00 F

32.0 C

bc

V , the initial charge of each capacitor is 10.0 F 12.0 V

120 C and C2

After the capacitors are connected in parallel, the potential d ifference across each is V 3.00 V , and the total charge of Q Q10 Qx 120 C is d ivid ed betw een the tw o capacitors as

Q10

Qx

90.0 C 3.00 V

(a) From Q C

V , Q25

Q40

and

30.0 C and

Q Q10 120 C 30.0 C 90.0 C Qx V

Thus, Cx

16.41

10.0 F 3.00 V

30.0 F

25.0 F 50.0 V

40.0 F 50.0 V

1.25 103 C

2.00 103 C

1.25 mC

2.00 mC

(b) When the tw o capacitors are connected in parallel, the equiv alent capacitance is Ceq C1 C2 25.0 F+40.0 F 65.0 F . Since the negative plate of one w as connected to the positive plate of the other, the total charge stored in the parallel combination is

Q Q40 Q25

2.00 103 C 1.25 103 C 750 C

The potential d ifference across each capacitor of the parallel com bination is V

Q Ceq

750 C 65.0 F

11.5 V

Electrical Energy and Capacitance

and the final charge stored in each capacitor is

Q25 and

16.42

C1

V

25.0 F 11.5 V

Q40

Q Q25

288 C

750 C 288 C

462 C

(a) The original circuit red uces to a single equivalent capacitor in the steps show n below .

Cs

1 C1

C p1 Cs Cp2

1 C2

1

1 1 5.00 F 10.0 F

C3 Cs

C2 C2

2 3.33 F

2 10.0 F

1

2.00 F 8.66 F

20.0 F

1

Ceq

1 C p1

1 Cp2

3.33 F

1 8.66 F

1 20.0 F

1

6.04 F

69

Electrical Energy and Capacitance

70

(b) The total charge stored betw een points a and b is

Qtotal

Ceq

V

6.04 F 60.0 V

ab

362 C

Then, looking at the third figure, observe that the charges of the series capacitors of that figure are Qp1 Qp 2 Qtotal 362 C . Thus, the potential d ifference across the upper parallel com bination show n in the second figure is V

Q p1 p1

362 C 8.66 F

C p1

41.8 V

Finally, the charge on C3 is

Q3

16.43

From Q C

Q1

C3

V

p1

2.00 F 41.8 V

83.6 C

V , the initial charge of each capacitor is 1.00 F 10.0 V

10.0 C and Q2

2.00 F 0

0

After the capacitors are connected in parallel, the potential d ifference across one is the sam e as that across the other. This gives V

Q1 1.00 F

Q2 or Q2 2.00 F

From conservation of charge, Q1 Q2 Equation [1], this becomes

Q1 2 Q1 10.0 C , giving

Finally, from Equation [1],

[1]

2 Q1 Q1 Q2 10.0 C . Then, substituting from

Q1

10 3

C

Q2

20 3

C

Electrical Energy and Capacitance

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 1 5.00

Cs

1 7.00

1

2.92 F

Then recognize that this capacitor, the 4.00 F capacitor, and the 6.00 F capacitor are all connected in parallel betw een points a and b. Thus, the equivalent capacitance betw een points a and b is

Ceq

4.00 F 2.92 F+6.00 F

Q2 2C

1 C 2

2

1 4.50 10 2

6

2

Energy stored

16.46

(a) The equivalent capacitance of a series com bination of C1 and C2 is 1 18.0 F

1 36.0 F

2 1 36.0 F

F 12.0 V

4

16.45

1 Ceq

V

12.9 F

or

Ceq

3.24 10

J

12.0 F

When this series com bination is connected to a 12.0-V battery, the total stored energy is Total energy stored

1 Ceq 2

V

2

1 12.0 10 2

6

F 12.0 V

2

8.64 10

4

J

(b) The charge stored on each of the tw o capacitors in the series com bination is

Q1

Q2

Qtotal

Ceq

V

12.0 F 12.0 V

144 C 1.44 10

4

and the energy stored in each of the ind ivid ual capacitors is

and

Energy stored in C1

Q12 2C1

Energy stored in C2

Q22 2C2

1.44 10

4

6

2 18.0 10 1.44 10

4

2 36.0 10

2

C

5.76 10

F C

6

4

J

2

F

2.88 10

4

J

C

71

Electrical Energy and Capacitance

72

Energy stored in C1 Energy stored in C2 5.76 10 4 J 2.88 10 4 J 8.64 10 4 J , w hich is the sam e as the total stored energy found in part (a). This m ust be true if the computed equivalent capacitance is truly equivalent to the original combination. (c) If C1 and C2 had been connected in parallel rather than in series, the equivalent capacitance w ould have been Ceq C1 C2 18.0 F 36.0 F 54.0 F . If the total 1 2 Ceq V in this parallel com bination is to be the sam e as w as 2 stored in the original series com bination, it is necessary that

energy stored

V

2 Total energy stored

2 8.64 10

Ceq

54.0 10

6

4

J

5.66 V

F

Since the tw o capacitors in parallel have the same potential d iffer ence across them , 1 2 C V the energy stored in the ind ivid ual capacitors is d irectly proportional 2 to their capacitances. The larger capacitor, C2 , stores the most energy in this case. 16.47

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

1

Qi2 2Ci

1 Ci 2

V

1 3.00 F 6.00 V 2

2 i

2

54.0 J

(b) When the capacitor is d isconnected from the battery, the stored charge becom es isolated w ith no w ay off the plates. Thus, the charge rem ains constant at the value Qi as long as the capacitor rem ains d isconnected . Since the ca pacitance of a parallel plate capacitor is C e0 A d , w hen the d istance d separating the plates is d oubled , the capacitance is d ecreased by a factor of 2 C f

Ci 2 1.50 F . The stored energy

(w ith Q unchanged ) becom es

Energy stored

2

Qi2 2C f

Qi2 2 Ci 2

2

Qi2 2C f

2 Energy stored

1

108 J

(c) When the capacitor is reconnected to the battery, the potential d ifference betw een the plates is reestablished at the original value of V V i 6.00 V , w hile the capacitance rem ains at C f

Ci 2 1.50 F . The energy stored und er these

cond itions is Energy stored

3

1 Cf 2

V

2 i

1 1.50 F 6.00 V 2

2

27.0 J

Electrical Energy and Capacitance

16.48

73

The energy transferred to the w ater is

W

1 1 Q 100 2

50.0 C 1.00 108 V

V

2.50 107 J

200

Thus, if m is the m ass of w ater boiled aw ay,

W

m c

T

2.50 107 J m

m

giving

16.49

Lv becom es

4186

2.50 107 J 2.55 J kg

J

100 C 30.0 C

kg C

2.26 106 J kg

9.79 kg

(a) N ote that the charge on the plates rem ains constant at the original value, Q0 , as the d ielectric is inserted . Thus, the change in the potential d ifference, V Q C , is d ue to a change in capacitance alone. The ratio of the final and initial capacitances is Cf Ci

A d A d

0 0

and

Cf

Q0

V

Ci

Q0

V

f i

V V

Thus, the d ielectric constant of the inserted m aterial is

i f

85.0 V 25.0 V

3.40

3.40 , and the m aterial is

probably nylon (see Table 16.1). (b) If the d ielectric only partially filled the space betw een the plates, leaving the rem aining space air-filled , the equivalent d ielectric constant w ould be som ew here betw een 1.00 (air) and 3.40 . The resulting potential d ifference w ould then lie som ew here betw een V i 85.0 V and V f 25.0 V .

Electrical Energy and Capacitance

16.50

74

(a) The capacitance of the capacitor w hile air-filled is

C0

12

8.85 10

A

0

d

C2 N m 2

25.0 10 2

1.50 10

4

m2

1.48 10

m

12

F 1.48 pF

The original charge stored on the plates is

Q0

C0

V

1.48 10-12 F 2.50 102 V

0

370 10

12

C

370 pC

Since d istilled w ater is an insulator, introd ucing it betw een the isolated capacitor plates d oes not allow the charge to change. Thus, the final charge is Q f 370 pC . (b) After im m ersion d istilled w ater

Cf

C0

80 1.48 pF

See Table 16.1 , the new capacitance is

80

118 pF

V

and the new potential d ifference is

Qf f

370 pC 118 pF

Cf

3.14 V

(c) The energy stored in a capacitor is: Energy stored Q2 2C . Thus, the change in the stored energy d ue to im m ersion in the d istilled w ater is

E

Q 2f 2C f

Q02 2Ci

4.57 10

16.51

Q02 2 8

J

1 Cf

370 10

1 Ci

45.7 10

9

C

2.1 8.85 10

A

2

1 118 10

J

12

1 F 1.48 10

12

F

45.7 nJ

2.1 , so the capacitance is

C2 N m2 175 10

4

m2

0.040 0 10-3 m

d

C 8.13 10

12

C

2

(a) The d ielectric constant for Teflon ® is

0

12

9

F

8.13 nF

(b) For Teflon ®, the d ielectric strength is Emax is

60.0 106 V m , so the m axim um voltage

60.0 106 V m 0.040 0 10-3 m

Vmax

Emax d

Vmax

2.40 103 V

2.40 kV

Electrical Energy and Capacitance

16.52

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

0

C

w L

0

d

d

w here A is the surface area of one sid e of a foil strip. Thus, the required length is

C d 0w

L

16.53

(a) V

8

3.70 8.85 10

12

1.00 10 12 kg 1100 kg m3

m

A 4 r

2

9.09 10

4

23

3V 4

2

C N m

16

m 2

7.00 10

1.04 m

m

m3

3V 4

13

, and the surface area is

3 9.09 10

4

2

3

16

m3

23

4.54 10

4

12

C 2 N m 2 4.54 10

Q C

V

2.01 10

13

9

10

m2

2.01 10

m

F 100 10-3 V

2.01 10

14

C

and the num ber of electronic charges is n

Q e

2.01 10 14 C 1.60 10-19 C

1.26 105

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

or

m2

d

100 10

16.54

10

A

0

5.00 8.85 10

(c)

F 0.025 0 10

4 r3 , the rad ius is r 3

Since V

(b) C

9.50 10

Ceq

C1 C2

0

d

A

C3

where A

A1 d

0

A1

A2 d

0

A2

A3

A3 d

0

0

A1

A2 d

A3

13

F

75

Electrical Energy and Capacitance

16.55

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

or

16.56

1 Ceq

1 C1

Ceq

0

1 C2 A

d

1 C3

where d

d1 0 A

d1

d2 0 A

d2

d3 0 A

d1

d 2 d3 0 A

d3

(a) Please refer to the solution of Problem 16.37 w here the follow ing results w ere obtained :

Ceq

3.33 F

Q3

Q6 180 C

Q2

Q4 120 C

The total energy stored in the full circuit is then Energy stored

total

1 1 2 Ceq V 3.33 10 6 F 90.0 V 2 2 1.35 10 2 J 13.5 10 3 J 13.5 mJ

2

(b) The energy stored in each ind ivid ual capacitor is

For 2.00 F :

For 3.00 F :

For 4.00 F :

For 6.00 F :

Energy stored

Energy stored

Energy stored

Energy stored

2

3

4

6

Q22 2C2 Q32 2C3 Q42 2C4 Q62 2C6

120 10

6

2 2.00 10 180 10

6

2 3.00 10 120 10

6

2 4.00 10 180 10

6

2 6.00 10

C 6

F

F

C 6

C 6

3.60 10

3

J

3.60 mJ

5.40 10

3

J

5.40 mJ

1.80 10

3

J

1.80 mJ

2.70 10

3

J

2.70 mJ

2

C 6

2

2

F 2

F

(c) The total energy stored in the ind ivid ual capacitors is

Energy stored= 3.60 5.40 1.80 2.70 mJ

13.5 mJ

Energy stored

total

Thus, the sum s of the energies stored in the ind ivid ual capacitors equals t he total energy stored by the system .

76

Electrical Energy and Capacitance

77

78

Electrical Energy and Capacitance

16.57

In the absence of a d ielectric, the capacitance of the parallel plate 0 A capacitor is C0 d With the d ielectric inserted , it fills one-third of the gap betw een the plates as show n in sketch (a) at the right. We m od el this situation as (a) consisting of a pair of capacitors, C1 and C2 , connected in series as show n in sketch (b) at the right. In reality, the low er plate of C1 and the upper plate of C2 are (b) one and the sam e, consisting of the low er surface of the d ielectric show n in sketch (a). The capacitances in the m od el of sketch (b) are given by: 0 A d 3

C1

3

0

A

and

d

0 A 2d 3

C2

3

0

A

2d

and the equivalent capacitance of the series com bination is 1 Ceq

and

16.58

d 3

2d 0

A

Ceq

3

0

1 A

3 2

2

For the series com bination:

Cp

C1

3

0

2 A

1

d 3

2 A

0

1 3

d 0

2 A

1 1 3 C0

C0

1

For the parallel com bination: C p

Thus, w e have

d

C2 Cs C1 C1 Cs

We w rite this result as :

1 Cs

Cs C1 C1 Cs

C1 C2 w hich gives 1 C1

1 C2

or

1 C2

1 Cs

C2 1 C1

[1]

Cp C1

C1 Cs Cs C1

and equating this to Equ ation [1] above gives

or

C p C1 C pCs

C12 C pC1 C pCs

and use the quad ratic form ula to obtain

C1

C12

CsC1

0

1 Cp 2

1 2 C p C p Cs 4

Cs C1

79

Electrical Energy and Capacitance

Then, Equation [1] gives

16.59

1 Cp 2

C2

1 2 C p C p Cs 4

The charge stored on the capacitor by the battery is

Q C

V

1

C 100 V

This is also the total charge stored in the parallel com bination w hen this charged capacitor is connected in parallel w ith an uncharged 10.0- F capacitor. Thus, if V the resulting voltage across the parallel com bination, Q C p

C 100 V

and 16.60

C

C 10.0 F 30.0 V or 70.0 V C

30.0 V 10.0 F 70.0 V

V

2

2

is

gives

30.0 V 10.0 F

4.29 F

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

V1

ke

q1 r1

(b) The potential at P d ue to the

V2

ke

q2 r2

1.0 10 6 C 0.50 m

8.99 109 N m 2 C 2

1.8 104 V

2.0- C charge located 0.50 m aw ay is 2.0 10 6 C 0.50 m

8.99 109 N m 2 C 2

(c) The total potential at point P is VP

V1 V2

3.6 10 4 V

1.8 3.6

104 V

1.8 104 V

(d ) The w ork required to m ove a charge q 3.0 C to point P from infinity is W

q V

q VP V

3.0 10

6

C

1.8 104 V 0

5.4 10

2

J

Electrical Energy and Capacitance

16.61

The stages for the red uction of this circuit are show n below .

Thus, Ceq

16.62

80

6.25 F

(a) Due to spherical sym m etry, the charge on each of the concentric spherical shells w ill be uniform ly d istributed over that shell. Insid e a spherical surface having a uniform charge d istribution, the electric field d ue to the charge on that surface is zero. Thus , in this region, the potential d ue to the charge on that surface is constant and equal to the potential at the surface. Outsid e a spherical surface having a uniform charge ke q d istribution, the potential d ue to the charge on that surface is given by V r w here r is the d istance from the center of that surface and q is the charge on that surface. In the region betw een a pair of concentric spherical shells, w ith the inner shell having charge Q and the outer shell having rad ius b and charge Q , the total electric potential is given by V

Vdue to

Vdue to

inner shell

outer shell

ke Q r

ke

Q b

ke Q

1 r

1 b

The potential d ifference betw een the tw o shells is therefore, V

V

r a

V

r b

ke Q

1 a

1 b

keQ

1 b

The capacitance of this device is given by

C

Q V

ab ke b a

1 b

keQ

b a ab

Electrical Energy and Capacitance

(b) When b a , then b a b . Thus, in the lim it as b above becom es ab ke b

C

16.63

4

0

From Q C C0

2 300 J

2W C

, the capacitance found

a

The energy stored in a charged capacitor is W

V

16.64

a ke

81

30.0 10-6 F

1 C 2

4.47 103 V

V

2

. H ence,

4.47 kV

V , the capacitance of the capacitor w ith air betw een the plates is Q0 V

150 C V

After the d ielectric is inserted , the potential d ifference is held to the original value, but the charge changes to Q Q0 200 C=350 C . Thus, the capacitance w ith the d ielectric slab in place is C

Q V

350 C V

The d ielectric constant of the d ielectric slab is th erefore C C0

16.65

350 C V

V 150 C

350 150

2.33

The charges initially stored on the capacitors are

and

Q1 C1

V

Q2

V

C2

6.0 F 250 V

i

i

2.0 F 250 V

1.5 103 C

5.0 102 C

When the capacitors are connected in parallel, w ith the negative plate of one connected to the positive plate of the other, the net stored charge is

Q Q1 Q2 1.5 103 C 5.0 102 C=1.0 103 C

Electrical Energy and Capacitance

The equivalent capacitance of the parallel com bination is Ceq

C1 C2

82

8.0 F . Thus, the

final potential d ifference across each of the capacitors is 1.0 103 C 125 V 8.0 F

Q Ceq

V

and the final charge on each capacitor is

and

16.66

Q1

C1

V

6.0 F 125 V

750 C

0.75 mC

Q2

C2

V

2.0 F 125 V

250 C

0.25 mC

The energy required to m elt the lead sam ple is W

m cPb

T

6.00 10

Lf 6

kg

128 J kg C 327.3 C 20.0 C

24.5 103 J kg

0.383 J

The energy stored in a capacitor is W

V

16.67

2W C

2 0.383 J 52.0 10-6 F

1 C 2

V

2

, so the required potential d ifference is

121 V

When excess charge resid es on a spherical surface that is far rem oved from any other charge, this excess charge is uniform ly d istributed over the spherical surface, an d the electric potential at the surface is the sam e as if all the excess charge w ere concentrated at the center of the spherical surface. In the given situation, w e have tw o charged spheres, initially isolated from each other, w ith charges and potentials of: Q1 6.00 C , V1 keQ1 R1 w here R1 12.0 cm , Q2 4.00 C , and V2 keQ2 R2 w ith R2 18.0 cm .

Electrical Energy and Capacitance

83

When these spheres are then connected by a long cond ucting thread , the charges are red istributed yielding charges of Q1 and Q2 respectively until the tw o surfaces com e to a com m on potential V1

kQ1 R1 V2

From conservation of charge: kQ1 R1

From equal potentials:

16.68

kQ2 R2 . When equilibrium is established , w e have:

Q1 Q2

Q1 Q2

kQ2 R2

Q1 Q2 R2 Q1 R1

Q2

2.00 C

or

2.00 C 2.50

Substituting Equation [2] into [1] gives:

Q1

Then, Equation [2] gives:

Q2 1.50 0.800 C

Q2 1.50Q1

[1]

[2]

0.800 C

1.20 C

The electric field betw een the plates is d irected dow nw ard w ith m agnitu d e V d

Ey

100 V 2.00 10-3 m

5.00 104 N m

Since the gravitational force experienced by the electron is negligible in com parison to the electrical force acting on it, the vertical acceleration is

ay

Fy me

1.60 10

qEy me

19

C

9.11 10

5.00 104 N m 31

kg

(a) At the closest approach to the bottom plate, vy from point O is found from v

y

0

v0 sin 2 ay

2 0

2 y

2 0y

v

8.78 1015 m s2

2 ay

0 . Thu s, the vertical displacem ent

y as

5.6 106 m s sin 45 2 8.78 1015 m s 2

2

0.89 mm

The m inim um d istance above the bottom plate is then d

D 2

y 1.00 mm 0.89 mm

0.11 mm

Electrical Energy and Capacitance

(b) The tim e for the electron to go from point O to the upper plate is found from 1 2 y v0 y t a y t as 2 1.00 10

3

m

5.6 106

m sin 45 t s

1 m 8.78 1015 2 t 2 2 s

Solving for t gives a positive solution of t 1.11 10 9 s . The horizontal d isplacem ent from point O at this tim e is x v0 xt

5.6 106 m s cos 45

1.11 10

9

s

4.4 mm

84