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Chapter 26. Capacitance Physics, 6th Edition

Chapter 26. Capacitance
Capacitance

26-1. What is the maximum charge that can be placed on a metal sphere 30 mm in diameter and

surrounded by air?

kQ Er 2 (3 x 106 N/C)(0.015 m) 2
E = 2 = 3 x 106 N/C; Q= = ; Q = 75.0 nC
r k (9 x 109 N ⋅ m 2 /C2 )

26-2. How much charge can be placed on a metal sphere of radius 40 mm if it is immersed in

transformer oil whose dielectric strength is 16 MV/m?

kQ Er 2 (16 x 106 N/C)(0.040 m) 2
E= = 16 x 106 N/C; Q= = ; Q = 2.48 µC
r2 k (9 x 109 N ⋅ m 2 /C 2 )

26-3. What would be the radius of a metal sphere in air such that it could theoretically hold a

charge of one coulomb?

kQ (9 x 109 N ⋅ m 2 /C 2 )(1 C)
E= ; r= ; r = 54.8 m
r2 3 x 106 N/C

26-4. A 28-µF parallel-plate capacitor is connected to a 120-V source of potential difference.

How much charge will be stored on this capacitor?

Q = CV = (28 µF)(120 V); Q = 3.36 mC

26-5. A potential difference of 110 V is applied across the plates of a parallel-plate capacitor. If

the total charge on each plate is 1200 µC, what is the capacitance?

Q 1200 µ C
C= = ; C = 10.9 µF
V 110 V

103


26-6. Find the capacitance of a parallel-plate capacitor is 1600 µC of charge is on each plate

when the potential difference is 80 V.

Q 1600 µ C
C= = ; C = 20.0 µF
V 80 V

26-7. What potential difference is required to store a charge of 800 µC on a 40-µF capacitor?

Q 800 µ C
V= = ; V = 20.0 V
C 40 µ F

26-8. Write an equation for the potential at the surface of a sphere of radius r in terms of the

permittivity of the surrounding medium. Show that the capacitance of such a sphere is

given by C = 4πεr.

kQ Q CV
V= = ; Q = CV; V= ; C = 4πεοr
r 4πε 0 r 4πε 0 r

*26-9. A spherical capacitor has a radius of 50 mm and is surrounded by a medium whose

permittivity is 3 x 10-11 C2/N m2. How much charge can be transferred to this sphere by a

potential difference of 400 V?

We must replace εο with ε for permittivity of surrounding medium, then

kQ Q CV
V= = ; Q = CV ; V= ; C = 4πε r
r 4πε r 4πε r

C 4πε r
C = 4πε r ; Q= =
V V

4π (3 x 10-11C2 /N ⋅ m 2 )(0.05 m)
Q= ; Q = 4.71 x 10-14 C
400 V

104


Calculating Capacitance

26-10. A 5-µF capacitor has a plate separation of 0.3 mm of air. What will be the charge on

each plate for a potential difference of 400 V? What is the area of each plate?

Q = CV = (5 µ F)(400 V); Q = 2000 µC

26-11. The plates of a certain capacitor are 3 mm apart and have an area of 0.04 m2. What is the

capacitance if air is the dielectric?

E (8.85 x 10-12 C 2 /N ⋅ m 2 )(0.04 m 2 )
C = ε0 = ; C = 118 pF
A 0.003 m

26-12. A capacitor has plates of area 0.034 m2 and a separation of 2 mm in air. The potential

difference between the plates is 200 V. What is the capacitance, and what is the electric

field intensity between the plates? How much charge is on each plate?

E (8.85 x 10-12 C 2 /N ⋅ m 2 )(0.034 m 2 )
C = ε0 = ; C = 150 pF
A 0.002 m

V 400 V
E= = ; E = 2.00 x 105 N/C
d 0.002 m

Q = CV = (150 pF)(400 V); Q = 30.1 µC

26-13. A capacitor of plate area 0.06 m2 and plate separation 4 mm has a potential difference of

300 V when air is the dielectric. What is the capacitance for dielectrics of air (K = 1)

and mica (K = 5)?

E (1)(8.85 x 10-12 C2 /N ⋅ m 2 )(0.06 m 2 )
C = Kε 0 = ; C = 133 pF
A 0.004 m

E (5)(8.85 x 10-12 C2 /N ⋅ m 2 )(0.06 m 2 )
C = Kε0 = ; C = 664 pF
A 0.004 m

105


26-14. What is the electric field intensity for mica and for air in Problem 26-13?

V0 V 300 V
E0 = E0 = 0 = ; E0 = 7.50 x 104 V/m
d d 0.004 m

E0 E0 7.50 x 104
K= ; E= = ; E = 1.50 x 104 V/m
E K 5

26-15. Find the capacitance of a parallel-plate capacitor if the area of each plate is 0.08 m2, the

separation of the plates is 4 mm, and the dielectric is (a) air, (b) paraffined paper (K = 2)?

E (1)(8.85 x 10-12 C2 /N ⋅ m 2 )(0.08 m 2 )
C = Kε 0 = ; C = 177 pF
A 0.004 m

E (2)(8.85 x 10-12C 2 /N ⋅ m 2 )(0.08 m 2 )
C = Kε0 = ; C = 354 pF
A 0.004 m

26-16. Two parallel plates of a capacitor are 4.0 mm apart and the plate area is 0.03 m2. Glass

(K = 7.5) is the dielectric, and the plate voltage is 800 V. What is the charge on each

plate, and what is the electric field intensity between the plates?

E (7.5)(8.85 x 10-12 C2 /N ⋅ m 2 )(0.03 m 2 )
C = Kε 0 = ; C = 498 pF
A 0.004 m

Q = CV = (498 x 10-12 F)(800 V); Q = 398 nC

*26-17. A parallel-plate capacitor with a capacitance of 2.0 nF is to be constructed with mica

(K = 5) as the dielectric, and it must be able to withstand a maximum potential

difference of 3000 V. The dielectric strength of mica is 200 MV/m. What is the

minimum area the plates of the capacitor can have?

V 3000 V
Q = CV = (2 x 10-9 F)(3000 V); Q = 6 µC ; d= =
E 200 x 106 V/m

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*16-17. (Cont.) d = 1.50 x 10-5 m; C = 2 nF; K=5

A Cd (2 x 10-9 F)(1.5 x 10-5 m)
C = Kε 0 ; A= = ; A = 6.78 x 10-4 m2
d K ε 0 5(8.85 x 10-12 C2 /N ⋅ m 2 )

Capacitors in Series and in Parallel

26-18. Find the equivalent capacitance of a 6-µF capacitor and a 12-µF capacitor connected (a)

in series, (b) in parallel.

C1C2 (6 µ F)(12 µ F)
Capacitors in series: Ce = = ; Ce = 4.00 µF
C1 + C2 6 µ F + 12 µ F

Capacitors in parallel: Ce = C1 + C2 = 6 µF + 12 µF; Ce = 18 µF

26-19. Find the effective capacitance of a 6-µF capacitor and a 15-µF capacitor connected (a)

in series, (b) in parallel?

C1C2 (6 µ F)(15 µ F)
Capacitors in series: Ce = = ; Ce = 4.29 µF
C1 + C2 6 µ F + 15 µ F

Capacitors in parallel: Ce = C1 + C2 = 6 µF + 12 µF; Ce = 21.0 µF

26-20. What is the equivalent capacitance for capacitors of 4, 7, and 12 µF connected (a) in

series, (b) in parallel?

1 1 1 1 1 1 1
Series: = + + = + + ; Ce = 2.10 µF
Ce C1 C2 C3 4 µ F 7 µ F 12 µ F

Parallel: Ce = ΣCi = 4 µF + 7 µC + 12 µC; Ce = 23.0 µC

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26-21. Find the equivalent capacitance for capacitors of 2, 6, and 8 µF connected (a) in series,

(b) in parallel?

1 1 1 1 1 1 1
Series: = + + = + + ; Ce = 1.26 µF
Ce C1 C2 C3 2 µ F 6 µ F 8 µ F

Parallel: Ce = ΣCi = 2 µF + 6 µC + 8 µC; Ce = 16.0 µC

26-22. A 20-µF and a 60-µF capacitor are connected in parallel. Then the pair are connected

in series with a 40-µF capacitor. What is the equivalent capacitance?
20 µF
C’ = 20 µF + 60 µF = 80 µF;
40 µF
C ' C40 (80 µ F)(40µ F)
Ce = = ; Ce = 26.7 µF
C '+ C40 (80µ F + 40µ F
60 µF

*26-23. If a potential difference of 80 V is placed across the group of capacitors in Problem

26-22, what is the charge on the 40-µF capacitor? What is the charge on the 20-µF

capacitor? ( First find total charge, then find charge and voltage on each capacitor.)

QT = CeV = (26.7 µ F)(80 V); QT = 2133 µ C ; Q40 = 2133 µC

Note: 2133 µC are on each of the combination C’ and the 40-µF capacitor. To find the

charge across the 20 µF, we need to know the voltage across C’:

2133 µ C
VC ' = = 26.7 V; This is V across each of 20 and 60-µF capacitors
80 µ F

Thus, charge on 20-µF capacitor is: Q20 = (20 µF)(26.7 V); Q20 = 533 µC

Note that V40 = 2133 µC/26.7 µC or 53.3 V. Also 53.3 V + 26.7 V = 80 V !

Also, the charge on the 60-µC capacitor is 1600 µC and 1600 µC + 533 µC = QT

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*26-24. Find the equivalent capacitance of a circuit in which a 6-µF capacitor is connected in

series with two parallel capacitors whose capacitances are 5 and 4 µF.

C’ = 5 µF + 4 µF = 9 µF; 5 µF
6 µF
C ' C40 (9 µ F)(6µ F)
Ce = = ; Ce = 3.60 µF
C '+ C40 9µ F + 6µ F
4 µF

*26-25. What is the equivalent capacitance for the circuit drawn in Fig. 26-15?

C6C3 (6µ F)(3µ F)
C'= = ; C’ = 2.40 µF 200 V
C6 + C3 6 µ F + 3µ F 3 µF
4 µF
6 µF
Ce = 2 µF + 4 µF; Ce = 6.00 µF

*26-26. What is the charge on the 4-µF capacitor in Fig. 26-15? What is the voltage across the

6-µF capacitor?

Total charge QT = CeV = (6.00 µF)(200 V); QT = 1200 µC

Q4 = C4V4 = (4 µ C)(200 V); Q4 = 800 µC

The rest of the charge is across EACH of other capacitors:

Q3 = Q6 = 1200 µC – 800 µC; Q6 = 400 µC

Q6 400 µ C
V6 = = ; V6 = 66.7 V
C6 6 µF

You should show that V3 = 133.3 V, so that V3 + V6 = 200 V.

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*26-27. A 6-µF and a 3-µF capacitor are connected in series with a 24-V battery. If they are

connected in series, what are the charge and voltage across each capacitor?

C6C3 (6 µ F)(3µ F) 24 V
C= = ; Ce = 2.00 µF 3 µF
C6 + C3 6 µ F + 3µ F
6 µF

QT = CeV = (2 µF)(24 V) = 48 µF; Q3 =Q6 = QT = 48 µC

Q3 48 µ C
V3 = = ; V3 = 16.0 V, Q3 = 48.0 µC
C3 3 µF

Q6 48 µ C
V6 = = ; V6 = 8.00 V, Q6 = 48.0 µC
C6 6 µF

*26-28. If the 6 and 3-µF capacitors of Problem 26-27 are reconnected in parallel with a 24-V

battery, what are the charge and voltage across each capacitor?
24 V
Ce = 3 µF + 6 µF; Ce = 9 µF 3 µF 6 µF

QT = (9 µF)(24 V) = 216 µC; V3 = V6 = 24 V

Q3 = (3 µF)(24 V); Q6 = (6 µF)(24 V); Q3 = 72 µC; Q6 = 144 µC

*26-29. Compute the equivalent capacitance for the entire circuit shown in Fig. 26-16. What is
3 µF
the total charge on the equivalent capacitance? 4 µF

C2,3 = 2 µF + 3 µF = 5 µF
12 V
1 1 1 1 2 µF
= + + ; Ce = 1.74 µF 8 µF
Ce 4 µ F 5 µ F 8 µ F
1.74 µF
QT = CeV = 20.9 µC

12 V 8 µF 12 V
4 µF 5 µF
Also: Q4 = Q5 = Q8 = 20.9 µC

110


*26-30. What are the charge and voltage across each capacitor of Fig. 26-16? (See Prob. 26-29.)

Q4 20.9 µ C Q 20.9 µ C
Q4 = Q5 = Q8 = 20.9 µC; V4 = = = 5.22 V; V8 = 8 = = 2.61 V
C4 4 µF C8 8 µF

Q5 20.9 µ C
V5 = V3 = V2 = = = 4.17 V ; Q3 = C3V3 = (3 µF)(4.17 V) = 12.5 µC
C5 5 µF

Q2 = (2 µF)(4.17 V) = 8.35 µC; Note that Q2 + Q3 = Q5 = QT = 20.9 µC

Answer summary: Q2 = 8.35 µC; Q3 = 12.5 C, Q4 = Q8 = 20.9 C

V2 = V3 = 4.17 V; V4 = 5.22 V; V8 = 2.61 V

The Energy of a Charged Capacitor

26-31. What is the potential energy stored in the electric field of a 200-µF capacitor when it is

charged to a voltage of 2400 V?

P.E. = ½CV 2 = ½(200 x 10-6 F)(2400 V) 2 ; P.E. = 576 J

26-32. What is the energy stored on a 25-µF capacitor when the charge on each plate is 2400

µF? What is the voltage across the capacitor?

Q 2 (2400 x 10-6 C) 2
P.E. = = ; P.E. = 115 mJ
2C 2(25 x 10-6 F)

Q 2400 µ C
V= = ; V = 96.0 V
C 25 µ F

26-33 How much work is required to charge a capacitor to a potential difference of 30 kV if

800 µC are on each plate?

Work = P.E. = ½QV; Work = ½(800 x 10-6 C)(30 x 103 V)

Work = 12.0 J

111


*26-34. A parallel plate capacitor has a plate area of 4 cm2 and a separation of 2 mm. A

dielectric of constant K = 4.3 is placed between the plates, and the capacitor is

connected to a 100-V battery. How much energy is stored in the capacitor?

A (4.3)(8.85 x 10-12 C2 /N ⋅ m 2 )(4 x 10-4 m 2 )
C = Kε0 = ; C = 7.61 pF
d 0.002 m

P.E. = ½CV2 = ½(7.61 x 10-12 F)(100 V)2; P.E. = 3.81 x 10-8 J

Challenge Problems

26-35. What is the break-down voltage for a capacitor with a dielectric of glass (K = 7.5) if the

plate separation is 4 mm? The average dielectric strength is 118 MV/m.

V = Emd = (118 x 106 V)(0.004 m); V = 472 kV

26-36. A capacitor has a potential difference of 240 V, a plate area of 5 cm2 and a plate

separation of 3 mm. What is the capacitance and the electric field between the plates?

What is the charge on each plate?

A (8.85 x 10-12 C2 /N ⋅ m 2 )(5 x 10-4 m 2 )
C = ε0 = ; C = 1.48 pF
d 0.003 m
240 V
E= ; E = 8.00 x 104 V/m Q = (1.48 x 10-12 F)(240 V); Q = 0.355 nC
0.003 m

26-37. Suppose the capacitor of Problem 26-34 is disconnected from the 240-V battery and then

mica (K = 5) is inserted between the plates? What are the new voltage and electric

field? If the 240-battery is reconnected, what charge will be on the plates?

V0 E0 240 V 80, 000 V/m
K= = ; V= = 48.0 V ; E = E= = 1.60 x 104 V/m
V E 5 5

C = KCo= 5(1.48 pF) = 7.40 pF; Q = CV = (7.40 pF)(240 V) = 1.78 nC

112


26-38. A 6-µF capacitor is charged with a 24-V battery and then disconnected. When a

dielectric is inserted, the voltage drops to 6 V. What is the total charge on the capacitor

after the battery has been reconnected?

Vo 24 V
K= = = 4; C = KCo = 4(6 µ F) = 24µ F;
V 6V

Q = CV = (24 µF)(24 V); Q = 576 µC

26-39. A capacitor is formed from 30 parallel plates 20 x 20 cm. If each plate is separated by 2

mm of dry air, what is the total capacitance?

Thirty stacked plates means that there are 29 spaces, which make for 29 capacitors:

A (29)(8.85 x 10-12 C 2 /N ⋅ m 2 )(0.2 m) 2
C = 29ε 0 = ; C = 5.13 nF
d 0.002 m

*26-40. Four capacitors A, B, C, and D have capacitances of 12, 16, 20, and 26 µF, respectively.

Capacitors A and B are connected in parallel. The combination is then connected in

series with C and D. What is the effective capacitance? 12 µF 20 µF

C1 = 12 µF + 16 µF = 26 F; C2 = 28 µF + 26 µF = 46 µF

C1C2 (28 µ F)(46 µ F) 16 µF 26 µF
Ce = = ;
C1 + C2 28 µ F + 46 µ F
28 µF 46 µF
Ce = 17.4 µF

*26-41. Consider the circuit drawn in Fig. 32-17. What is the equivalent capacitance of the

circuit? What are the charge and voltage across the 2-µF capacitor? (Redraw Fig.)
2 µF
1 1 1
= + ; C6,2 = 1.50 µF
C6,2 2 µ F 6 µ F 4.5 µF 12 V

C6,2 = 1.5 µF + 4.5 µF = 6 µF Ce 6 µF
6µF

113


*26-41. (Cont.)
2 µF 4.5 µF 1.5 µF 4.5 µF 6 µF
12 V 12 V
12 V
6µF

QT = (6 µF)(12 V) = 72 µC; Q1.5 = Q2 = Q3 = (1.5 µF)(12 V) = 18 µC

Q2 18µ C
V2 = = = 9.00 V; Ce = 6.00 µF, V2 = 9.00 V; Q2 = 18 µC
C2 2 µ F

*26-42. Two identical 20-µF capacitors A and B are connected in parallel with a 12-V battery.

What is the charge on each capacitor if a sheet of porcelain (K = 6) is inserted between

the plates of capacitor B and the battery remains connected?
20 µF 120 µF 140 µF
C A = 20 µ F; CB = 6(20 µ F) = 120 µ F
12 V
12 V
Ce = 120 µF + 20 µF; Ce = 140 µF

QB = (120 µF)(12 V) = 1440 µC; QA = (20 µF)(12 V) = 240 µC

Show that BEFORE insertion of dielectric, the charge on EACH was 240 µC!

*26-43. Three capacitors A, B, and C have respective capacitances of 2, 4, and 6 µF. Compute

the equivalent capacitance if they are connected in series with an 800-V source. What

are the charge and voltage across the 4-µF capacitor? A B C

1 1 1 1 800 V
= + + ; Ce = 1.09 µF 2 µF 4 µF 6 µF
Ce 2 µ F 4 µ F 6 µ F

QT = CeV = (1.09 µF)(800 V); QT = 873 µC; Q2 = Q4 = Q6 = 873

µC

873 µ C
Also:; V4 = = 218 V; Q4 = 873 µC; V4 = 218 V
4 µF

114


*26-44. Suppose the capacitors of Problem 26-41 are reconnected in parallel with the 800-V

source. What is the equivalent capacitance? What are the charge and voltage across the

4-µF capacitor? 2 µF 4 µF 6 µF 12 µF
12 V
Ce = 2 µF + 4 µF + 6 µF; Ce = 12 µF 12 V

QT = CeV = (12 µF)(12 V) = 144 µC

VT = V2 = V3 = V4 = 12 V; Q4 = C4V4 = (4 µF)(12 V); Q4 = 48 µC; V4 = 12 V

*26-45. Show that the total capacitance of a multiple-plate capacitor containing N plates

separated by air is given by:

( N − 1)ε 0 A
C0 =
d

where A is the area of each plate and d is the separation of each plate.

For a multiplate capacitor, if there are a total of N plates, we have the equivalent of

(N – 1) capacitors each of area A and separation d. Hence, the above equation.

*26-46. The energy density, u, of a capacitor is defined as the energy (P.E.) per unit volume (Ad)

of the space between the plates. Using this definition and several formulas from this

chapter, derive the following relationship for finding the energy density, u:

1
u= ε 0E 2
2

where E is the electric field intensity between the plates?

A P.E.
P.E. = ½CV 2 ; V = Ed ; C = ε0 ; u=
d Ad

115


 A
½  ε0  ( E 2d 2 )
½CV 2
d u = ½εοE2
u= =  ;
Ad Ad

*26-47. A capacitor with a plate separation of 3.4 mm is connected to a 500-V battery. Use the

relation derived in Problem 26-44 to calculated the energy density between the plates?

V 500 V
E= = = 1.47 x 105 V/m ; u = ½εοE2
d 0.0034 m

u = ½(8.85 x 10-12 C2 /N ⋅ m 2 )(1.47 x 104 V/m)2 ; u = 95.7 mJ/m3

Critical Thinking Problems

26-48. A certain capacitor has a capacitance of 12 µF when its plates are separated by 0.3 mm

of vacant space. A 400-V battery charges the plates and is then disconnected from the

capacitor. (a) What is the potential difference across the plates if a sheet of Bakelite

(K = 7) is inserted between the plates? (b) What is the total charge on the plates?

(c) What is the capacitance with the dielectric inserted? (d) What is the permittivity of

Bakelite? (d) How much additional charge can be placed on the capacitor if the 400-V

battery is reconnected?

V0 400 V
E0 = = ; E0 = 1.33 x 106 V/m
d 0.0003 m

C = KC0 = (7)(12 µF) = 84 µF

V0 400 V 400 V 400 V
(a) K = ; V= ; V = 57.1 V
V 7

(b) Q0 = C0V0 = (12 µF)(400 V); Q0 = 4800 µC (disconnected) (c) 48 µF

(d) ε = Κεο = (7)(8.85 x 10-12 N m2/C2); ε = 6.20 x 10-11 N m2/C2

(e) Q = CV = (84 µF)(400 V); Q = 33.6 mC (400-V Battery reconnected)

116


The added charge is: Q – Q0 = 33.6 mC - 4.8 mC; ∆Q = 28.8 mC

*26-49. A medical defibrillator uses a capacitor to revive heart attack victims. Assume that a

65-µF capacitor in such a device is charged to 5000 V. What is the total energy stored?

If 25 percent of this energy passes through a victim in 3 ms, what power is delivered?

P.E. = ½CV2 = ½(65 x 10-6 F)(5000 V); P.E. = 162 mJ

0.25(0.162 J)
Power = = 13.5 W ; Power = 13.5 W
0.003 s

*26-50. Consider three capacitors of 10, 20, and 30 µF. Show how these might be connected to

produce the maximum and minimum equivalent capacitances and list the values. Draw

a diagram of a connection that would result in an equivalent capacitance of 27.5 µF.

Show a connection that will result in a combined capacitance of 22.0 µF.

The maximum is for a parallel connection: C = 10 µF + 20 µF + 30 µF; Cmax = 60 µF

1 1 1 1
Minimum is for series: = + + ; Cmin = 5.45 µF
Ce 20 µ F 20 µ F 30 µ F

1 1 1
Case 1: = + ; C1 = 7.5 µF 30 µF
C1 10 µ F 30 µ F 20 µF
10 µF
Ce = C1 + 20 µF = 12 µF + 20µF; Ce = 22.0 µF
1 1 1
Case 2: = + ; C1 = 12 µF 30 µF
C2 20 µ F 30 µ F 10 µF
20 µF
Ce = C1 + 20 µF = 12 µF + 20µF; Ce = 22.0 µF

117


*26-51. A 4-µF air capacitor is connected to a 500-V source of potential difference. The

capacitor is then disconnected from the source and a sheet of mica (K = 5) is inserted

between the plates? What is the new voltage on the capacitor? Now reconnect the

500-V battery and calculate the final charge on the capacitor? By what percentage did

the total energy on the capacitor increase due to the dielectric?

V0 500 V
(a) K = ; V= ; V = 100 V
V 5 500 V 500 V

(b) Q0 = C0V0 = (4 µF)(500 V); Q0 = 2000 µC

(c) (P.E.)0 = ½C0V2 = ½(4 x 10-6 F)(500 V)2; (P.E.)0 = 0.500 J; C = 5(4 µF) = 20 µF

(P.E.) = ½CV2 = ½(20 x 10-6 F)(500 V)2; (P.E.)0 = 2.50 J

2.50 J - 0.50 J
Percent increase = ; Percent increase = 400%
0.500 J

*26-52. A 3-µF capacitor and a 6-µF capacitor are connected in series with a 12-V battery. What

is the total stored energy of the system? What is the total energy if they are connected in

parallel? What is the total energy for each of these connections if mica (K = 5) is used

as a dielectric for each capacitor?
12 V 12 V
3 µF 6 µF 3 µF 6 µF
(3 µ F)(6 µ F)
Series: Ce = = 2.00 µ F
3 µF + 6 µF
3 µF 6 µF 3 µF 6 µF
-6 2
P.E. = ½(2 x 10 F)(12 V) = 0.144 mJ 12 V 12 V

Parallel: Ce = 3 µF + 6 µF = 9 µF

P.E. = ½(9 x 10-6 F)(12 V)2 = 0.648 mJ C3 = 5(3 µF) = 15 µF; C6 = 5(6 µF) = 30 µF

(15 µ F)(30 µ F)
Series: Ce = = 10.00 µ F ; Parallel: C = 15 µF + 30 µF = 45 µF
15 µ F + 30 µ F

118


P.E. = ½(10 x 10-6 F)(12 V)2 = 0.720 mJ; P.E. = ½(45 x 10-6 F)(12 V)2 = 3.24 mJ

119

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