Chapter 24 Electrostatic Energy and Capacitance

241

Electrostatic Energy and Capacitance

Conceptual Problems

*1 •

Determine the Concept The capacitance of a parallel-plate capacitor is a function of the surface area of its plates, the separation of these plates, and the electrical properties of the matter between them. The capacitance is, therefore, independent of the voltage across the capacitor. (c)iscorrect.

2 •

Determine the Concept The capacitance of a parallel-plate capacitor is a function of the surface area of its plates, the separation of these plates, and the electrical properties of the matter between them. The capacitance is, therefore, independent of the charge of the capacitor. (c)iscorrect.

3 •

Determine the Concept True.The energy density of an electrostatic field is given by 2

0 2 1

e

E

u

=

∈

.

4 •

Picture the Problem The energy stored in the electric field of a parallel-plate capacitor is related to the potential difference across the capacitor by

_U

=

₂1

_QV

.

Relate the potential energy stored in the electric field of the capacitor to the potential difference across the

capacitor:

QV

U

=

₂1

. doubles doubling

Hence, . to al proportion directly

is constant,

With Q U V V U

*5 ••

Picture the Problem The energy stored in a capacitor is given by

U

=

₂1

QV

_{and the} capacitance of a parallel-plate capacitor by

C

=

∈

₀

A

d

.

We can combine these relationships, using the definition of capacitance and the condition that the potential difference across the capacitor is constant, to express U as a function of d.

Express the energy stored in the capacitor:

Use the definition of capacitance to express the charge of the capacitor:

CV

Q

=

Substitute to obtain:

_U

=

₂1

_CV

2 Express the capacitance of a

parallel-plate capacitor in terms of the separation d of its plates:

d

A

C

=

∈

0

where A is the area of one plate. Substitute to obtain:

d

AV

U

2

2 0

∈

=

Because

d

U

∝

1

, doubling the separation of the plates will reduce the energy stored in the capacitor to

1/2 its previous value: _{(d)iscorrect.} 6 ••

Picture the Problem Let V represent the initial potential difference between the plates, U the energy stored in the capacitor initially, d the initial separation of the plates, and V ′, U

′, and d ′ these physical quantities when the plate separation has been doubled. We can use

U

=

₂1

QV

_{to relate the energy stored in the capacitor to the potential difference} across it and V = Ed to relate the potential difference to the separation of the plates. Express the energy stored in the

capacitor before the doubling of the separation of the plates:

QV

U

=

₂1

Express the energy stored in the capacitor after the doubling of the separation of the plates:

QV'

U'

=

₂1

because the charge on the plates does not change.

Express the ratio of U′ to U:

V

V'

U

U'

₌

Express the potential differences across the capacitor plates before and after the plate separation in terms of the electric field E between the plates:

Ed V = and

Ed' V' =

charge does not change during the separation process.

Substitute to obtain:

d

d'

Ed

Ed'

U

U'

₌

₌

For d ′ = 2d:

2

2

'

₌

₌

d

d

U

U

and (b)iscorrect

7 •

Determine the Concept Both statements are true. The total charge stored by two capacitors in parallel is the sum of the charges on the capacitors and the equivalent capacitance is the sum of the individual capacitances. Two capacitors in series have the same charge and their equivalent capacitance is found by taking the reciprocal of the sum of the reciprocals of the individual capacitances.

8 ••

(a) False. Capacitors connected in series carry the same charge.

(b) False. The voltage across the capacitor whose capacitance is C0 is Q/C0 and that across the second capacitor is Q/2C0.

(c) False. The energy stored by the capacitor whose capacitance is C0 is ₀ 2

2

C

Q

and the energy stored by the second capacitor is

4

0

.

2

C

Q

(d) True

9 •

Determine the ConceptTrue. The capacitance of a parallel-plate capacitor filled with a dielectric of constant κ is given by

d

A

C

=

κ

∈

0

or C∝κ.

*10 ••

Picture the Problem We can treat the configuration in (a) as two capacitors in parallel and the configuration in (b) as two capacitors in series. Finding the equivalent

capacitance of each configuration and examining their ratio will allow us to decide whether (a) or (b) has the greater capacitance. In both cases, we’ll let C1 be the capacitance of the dielectric-filled capacitor and C2 be the capacitance of the air capacitor.

Express C1 and C2: d A d A d A C 2 0 2 1 0 1 1 0 1 ∈ = ∈ = ∈ =

κ

κ

κ

and d A d A d A C 2 0 2 1 0 2 2 0 2 ∈ = ∈ = ∈ =

Substitute for C1 and C2 and

simplify to obtain:

2

2

2

(

1

)

0 0

0

+

∈

=

∈

+

∈

=

κ

κ

d

A

d

A

d

A

C

a

In configuration (b) we have:

2 1

1

1

1

C

C

C

_b

=

+

⇒ _{1 2} 2 1

C

C

C

C

C

_b

+

=

Express C1 and C2:

d

A

d

A

d

A

C

0 2 1 0 1 1 0 1

2

∈

=

∈

=

∈

=

and

d

A

d

A

d

A

C

0 2 1 0 2 2 0 2

2

∈

=

∈

=

∈

=

κ

κ

κ

Substitute for C1 and C2 and simplify to obtain:

(

)

⎟

⎠

⎞

⎜

⎝

⎛

+

∈

=

+

∈

⎟

⎠

⎞

⎜

⎝

⎛

∈

⎟

⎠

⎞

⎜

⎝

⎛ ∈

=

∈

+

∈

⎟

⎠

⎞

⎜

⎝

⎛

∈

⎟

⎠

⎞

⎜

⎝

⎛ ∈

=

1

2

1

2

2

2

2

2

2

2

0 0 0 0 0 0 0 0

κ

κ

κ

κ

κ

κ

d

A

d

A

d

A

d

A

d

A

d

A

d

A

d

A

C

_b

Divide Cb by Ca:

(

)

(

)

2 0 0

1

4

1

2

1

2

+

=

+

∈

⎟

⎠

⎞

⎜

⎝

⎛

+

∈

=

κ

κ

κ

κ

κ

d

A

d

A

C

C

a b Because

(

1

)

1

4

2

<

+

κ

κ

11 •

(a) False. The capacitance of a parallel-plate capacitor is defined to be the ratio of the charge on the capacitor to the potential difference across it.

(b) False. The capacitance of a parallel-plate capacitor depends on the area of its plates A, their separation d, and the dielectric constant κ of the material between the plates

according to

C

=

κ

∈

0

A

d

.

(c) False. As in part (b), the capacitance of a parallel-plate capacitor depends on the area of its plates A, their separation d, and the dielectric constant κ of the material between the plates according to

C

=

κ

∈

0

A

d

.

12 ••

Picture the Problem We can use the expression 2 2 1

_CV

U

=

to express the ratio of the energy stored in the single capacitor and in the identical-capacitors-in-series combination. Express the energy stored in

capacitors when they are connected to the 100-V battery:

2 eq 2 1_{C V} U =

Express the equivalent capacitance of the two identical capacitors connected in series:

C

C

C

C

₂1

2 eq

2

=

=

Substitute to obtain:

( )

4 2

1 2 2 1 2

1

_C

_V

_CV

U

=

=

Express the energy stored in one capacitor when it is connected to the 100-V battery:

2 2 1 0

CV

U

=

Express the ratio of U to U0:

2

1

2 2 1

2 4 1 0

=

=

CV

CV

U

U

or 0 2 1

_U

U

=

and (d)iscorrect

Estimation and Approximation

13 ••

Picture the Problem The outer diameter of a "typical" coaxial cable is about

cylindrical capacitor to estimate the capacitance per unit length of a coaxial cable. The capacitance of a cylindrical

dielectric-filled capacitor is given

by:

_⎟⎟

⎠

⎞

⎜⎜

⎝

⎛

∈

=

1 2 0

ln

2

R

R

L

C

πκ

where L is the length of the capacitor, R1 is the radius of the inner conductor, and R2 is the radius of the second (outer) conductor. Divide both sides by L to obtain an

expression for the capacitance per

unit length of the cable:

_⎟⎟

⎠

⎞

⎜⎜

⎝

⎛

=

⎟⎟

⎠

⎞

⎜⎜

⎝

⎛

∈

=

1 2 1

2 0

ln

2

ln

2

R

R

k

R

R

L

C

πκ

κ

If κ = 3:

(

)

0

.

104

nF/m

mm

5

.

0

mm

5

.

2

ln

C

/

m

N

10

99

.

8

2

3

2 2 9

=

⎟⎟

⎠

⎞

⎜⎜

⎝

⎛

⋅

×

=

L

C

If κ = 5:

(

)

0

.

173

nF/m

mm

5

.

0

mm

5

.

2

ln

C

/

m

N

10

99

.

8

2

5

2 2 9

=

⎟⎟

⎠

⎞

⎜⎜

⎝

⎛

⋅

×

=

L

C

A reasonable range of values for C/L,

corresponding to 3 ≤κ≤ 5, is: 0.104nF/m≤ ≤0.173nF/m

L C

*14 ••

Picture the Problem The energy stored in a capacitor is given by ₂1 2

.

CV

U

=

Relate the energy stored in a capacitor to its capacitance and the potential difference across it:

2 2 1

CV

U

=

Solve for C:

2

2

V

U

C

=

The potential difference across the spark gap is related to the width of the gap d and the electric field E in the gap:

Substitute for V in the expression for

C to obtain: 2 2

2

d

E

U

C

=

Substitute numerical values and evaluate C:

(

)

(

)

(

)

F

2

.

22

m

001

.

0

V/m

10

3

J

100

2

2 2

6

µ

=

×

=

C

15 ••

Picture the Problem Because ∆R << RE we can treat the atmosphere as a flat slab with an area equal to the surface area of the earth. Then the energy stored in the atmosphere can be estimated from U = uV, where u is the energy density of the atmosphere and V is its volume.

Express the electric energy stored in the atmosphere in terms of its energy density and volume:

uV U =

Because ∆R << RE = 6370 km, we

can consider the volume:

_R

_R

R

A

V

∆

=

∆

=

2 E

earth the of area surface

4

π

Express the energy density of the Earth’s atmosphere in terms of the average magnitude of its electric field:

2 0 2 1

_E

u

=

∈

Substitute for V and u to obtain:

(

)(

)

k

R

E

R

R

R

E

U

2

4

2 2 E 2

E 2

0 2

1

∆

=

∆

=

∈

π

Substitute numerical values and evaluate U:

(

) (

) (

)

(

)

J

10

03

.

9

/C

m

N

10

99

.

8

2

km

1

V/m

200

km

6370

10

2 2 9

2 2

×

=

⋅

×

=

U

16 ••

Picture the Problem We’ll approximate the balloon by a sphere of radius R = 3 m and use the expression for the capacitance of an isolated spherical conductor.

Relate the capacitance of an isolated

Substitute numerical values and

evaluate C:

8.99

10

N

m

/C

0

.

334

nF

m

3

2 2

9

⋅

=

×

=

C

Electrostatic Potential Energy

17 •

The electrostatic potential energy of this system of three point charges is the work needed to bring the charges from an infinite separation to the final positions shown in the diagram.

Express the work required to assemble this system of charges:

⎟

⎟

⎠

⎞

⎜

⎜

⎝

⎛

+

+

=

+

+

=

3 , 2

3 2 3 , 1

3 1 2 , 1

2 1

3 , 2

3 2 3

, 1

3 1 2

, 1

2 1

r

q

q

r

q

q

r

q

q

k

r

q

kq

r

q

kq

r

q

kq

U

Find the distances r1,2, r1,3, and r2,3:

r

1,2

=

3

m,

r

2,3

=

3

m,

and

r

1,3

=

6

m

(a) Evaluate U for q1 = q2 = q3 = 2 µC:

(

)

(

)(

) (

)(

) (

)(

)

mJ 0 . 30

m 3

C 2 C 2 m

6 C 2 C 2 m

3 C 2 C 2 /C m N 10 99 .

8 9 2 2

=

⎥ ⎦ ⎤ +

⎢ ⎣ ⎡

+ ⋅

×

=

µ

µ

µ

µ

µ

µ

U

(b) Evaluate U for q1 = q2 = 2 µC and q3 = −2 µC:

(

)

(

)(

) (

)(

) (

)(

)

mJ 99 . 5

m 3

C 2 C 2 m

6 C 2 C 2 m

3 C 2 C 2 /C m N 10 99 .

8 9 2 2

− =

⎥ ⎦ ⎤ −

+ ⎢

⎣

⎡ −

+ ⋅

×

=

µ

µ

µ

µ

µ

µ

U

(c) Evaluate U for q1 = q3 = 2 µC and q2 = −2 µC:

(

)

(

)(

) (

)(

) (

)(

)

mJ 0 . 18

m 3

C 2 C 2 m

6 C 2 C 2 m

3 C 2 C 2 /C m N 10 99 .

8 9 2 2

− =

⎥ ⎦ ⎤ −

+ ⎢

⎣ ⎡

+ −

⋅ ×

=

µ

µ

µ

µ

µ

µ

18 •

Picture the Problem The electrostatic potential energy of this system of three point charges is the work needed to bring the charges from an infinite separation to the final positions shown in the diagram.

Express the work required to assemble this system of charges:

⎟

⎟

⎠

⎞

⎜

⎜

⎝

⎛

+

+

=

+

+

=

3 , 2

3 2 3 , 1

3 1 2 , 1

2 1

3 , 2

3 2 3

, 1

3 1 2

, 1

2 1

r

q

q

r

q

q

r

q

q

k

r

q

kq

r

q

kq

r

q

kq

U

Find the distances r1,2, r1,3, and r2,3:

r

1,2

=

r

2,3

=

r

1,3

=

2

.

5

m

(a) Evaluate U for q1 = q2 = q3 = 4.2 µC:

(

)

(

)(

) (

)(

)

(

)(

)

J

190

.

0

m

5

.

2

C

2

.

4

C

2

.

4

m

5

.

2

C

2

.

4

C

2

.

4

m

5

.

2

C

2

.

4

C

2

.

4

/C

m

N

10

99

.

8

9 2 2

=

⎥

⎦

⎤

+

+

⎢

⎣

⎡

⋅

×

=

µ

µ

µ

µ

µ

µ

U

(b) Evaluate U for q1 = q2 = 4.2 µC and q3 = −4.2 µC:

(

)

(

)(

) (

)(

)

(

)(

)

mJ

4

.

63

m

5

.

2

C

2

.

4

C

2

.

4

m

5

.

2

C

2

.

4

C

2

.

4

m

5

.

2

C

2

.

4

C

2

.

4

/C

m

N

10

988

.

8

9 2 2

−

=

⎥

⎦

⎤

−

+

−

+

⎢

⎣

⎡

⋅

×

=

µ

µ

µ

µ

µ

µ

U

(

)

(

)(

) (

)(

)

(

)(

)

mJ

4

.

63

m

5

.

2

C

2

.

4

C

2

.

4

m

5

.

2

C

2

.

4

C

2

.

4

m

5

.

2

C

2

.

4

C

2

.

4

/C

m

N

10

99

.

8

9 2 2

−

=

⎥

⎦

⎤

−

+

−

+

⎢

⎣

⎡

−

−

⋅

×

=

µ

µ

µ

µ

µ

µ

U

*19 •

Picture the Problem The potential of an isolated spherical conductor is given by r

kQ

V = ,where Q is its charge and r its radius, and its electrostatic potential energy by

U

=

1₂

QV

_{. We can combine these relationships to find the sphere’s electrostatic} potential energy.

Express the electrostatic potential energy of the isolated spherical conductor as a function of its charge Q and potential V:

QV

U

=

₂1

Express the potential of the spherical

conductor:

r

kQ

V

=

Solve for Q to obtain:

k

rV

Q

=

Substitute to obtain:

k rV V k rV U

2

2 2

1 ⎟ =

⎠ ⎞ ⎜ ⎝ ⎛ =

Substitute numerical values and evaluate U:

(

)(

)

(

)

J

2

.

22

/C

m

N

10

8.99

2

kV

2

m

0.1

2 2 9

2

µ

=

⋅

×

=

U

20 ••

Express the work required to assemble this system of charges:

⎟

⎟

⎠

⎞

+

+

+

⎜

⎜

⎝

⎛

+

+

=

+

+

+

+

+

=

4 , 3 4 3 4 , 2 4 2 3 , 2 3 2 4 , 1 4 1 3 , 1 3 1 2 , 1 2 1 4 , 3 4 3 4 , 2 4 2 3 , 2 3 2 4 , 1 4 1 3 , 1 3 1 2 , 1 2 1

r

q

q

r

q

q

r

q

q

r

q

q

r

q

q

r

q

q

k

r

q

kq

r

q

kq

r

q

kq

r

q

kq

r

q

kq

r

q

kq

U

Find the distances r1,2, r1,3, r1,4, r2,3,

r2,4, and r3,4,:

m

4

4 , 1 4 , 3 3 , 2 2 ,

1

=

r

=

r

=

r

=

r

and

m

2

4

4 , 2 3 ,

1

=

r

=

r

(a) Evaluate U for q1 = q2 = q3 = q4 = −2 µC:

(

)

(

)(

) (

)(

) (

)(

)

(

)(

) (

)(

) (

)(

)

mJ

7

.

48

m

4

C

2

C

2

m

2

4

C

2

C

2

m

4

C

2

C

2

m

4

C

2

C

2

m

2

4

C

2

C

2

m

4

C

2

C

2

/C

m

N

10

99

.

8

9 2 2

=

⎥

⎦

⎤

−

−

+

−

−

+

−

−

+

−

−

+

−

−

+

⎢

⎣

⎡

−

−

⋅

×

=

µ

µ

µ

µ

µ

µ

µ

µ

µ

µ

µ

µ

U

(b) Evaluate U for q1 = q2 = q3 = 2 µC and q4 = −2 µC:

(

)

(

)(

) (

)(

) (

)(

)

(

)(

) (

)(

) (

)(

)

0

m

4

C

2

C

2

m

2

4

C

2

C

2

m

4

C

2

C

2

m

4

C

2

C

2

m

2

4

C

2

C

2

m

4

C

2

C

2

/C

m

N

10

99

.

8

9 2 2

=

⎥

⎦

⎤

−

+

−

+

+

−

+

+

⎢

⎣

⎡

⋅

×

=

µ

µ

µ

µ

µ

µ

µ

µ

µ

µ

µ

µ

U

(c) Let q1 = q2 = 2 µC and q3 = q4 = −2 µC:

(

)

(

)(

) (

)(

) (

)(

)

(

)(

) (

)(

) (

)(

)

mJ

7

.

12

m

4

C

2

C

2

m

2

4

C

2

C

2

m

4

C

2

C

2

m

4

C

2

C

2

m

2

4

C

2

C

2

m

4

C

2

C

2

/C

m

N

10

99

.

8

9 2 2

−

=

⎥

⎦

⎤

−

−

+

−

+

−

+

−

+

−

+

⎢

⎣

⎡

⋅

×

=

µ

µ

µ

µ

µ

µ

µ

µ

µ

µ

µ

µ

U

(

)

(

)(

) (

)(

) (

)(

)

(

)(

) (

)(

) (

)(

)

mJ

2

.

23

m

4

C

2

C

2

m

2

4

C

2

C

2

m

4

C

2

C

2

m

4

C

2

C

2

m

2

4

C

2

C

2

m

4

C

2

C

2

/C

m

N

10

99

.

8

9 2 2

−

=

⎥

⎦

⎤

−

+

−

−

+

−

+

−

+

+

⎢

⎣

⎡

−

⋅

×

=

µ

µ

µ

µ

µ

µ

µ

µ

µ

µ

µ

µ

U

21 ••

Picture the Problem The diagram shows the four charges fixed at the corners of the square and the fifth charge that is released from rest at the origin. We can use conservation of energy to relate the initial potential energy of the fifth particle to its kinetic energy when it is at a great distance from the origin and the electrostatic potential at the origin to express Ui. Use conservation of energy to relate the initial potential energy of the particle to its kinetic energy when it is at a great distance from the origin:

0

= ∆ +

∆K U

or, because Ki = Uf = 0,

0

i f −U = K

Express the initial potential energy of the particle to its charge and the electrostatic potential at the origin:

( )

0

i qV U =

Substitute for Kf and Ui to obtain:

( )

0

0

2

2

1

_mv

−

_qV

=

Solve for v:

( )

m qV v= 2 0

Express the electrostatic potential at

the origin:

( )

a

kq

a

kq

a

kq

a

kq

a

kq

V

2

6

2

6

2

3

2

2

2

0

=

+

−

+

+

=

Substitute and simplify to obtain:

ma

k

q

a

kq

m

q

v

6

2

2

6

2

₌

Capacitance

*22 •

Picture the Problem The charge on the spherical conductor is related to its radius and potential according to V = kQ/r and we can use the definition of capacitance to find the capacitance of the sphere.

(a) Relate the potential V of the spherical conductor to the charge on it and to its radius:

r

kQ

V

=

Solve for and evaluate Q:

(

)(

)

nC

2

.

22

/C

m

N

10

8.99

kV

2

m

0.1

2 2

9

⋅

=

×

=

=

k

rV

Q

(b) Use the definition of capacitance to relate the capacitance of the sphere to its charge and potential:

pF

11.1

kV

2

nC

22.2

₌

=

=

V

Q

C

(c) It doesn't.Thecapacitanceofasphereisafunction of itsradius.

23 •

Picture the ProblemWe can use its definition to find the capacitance of this capacitor. Use the definition of capacitance to

obtain:

400

V

75

.

0

nF

C

30

₌

=

=

µ

V

Q

C

24 ••

Picture the Problem Let the separation of the spheres be d and their radii be R. Outside the two spheres the electric field is approximately the field due to point charges of +Q and −Q, each located at the centers of spheres, separated by distance d. We can derive an expression for the potential at the surface of each sphere and then use the potential difference between the spheres and the definition of capacitance and to find the capacitance of the two-sphere system.

The capacitance of the two-sphere

system is given by:

V

Q

C

∆

=

The potential at any point outside the two spheres is:

( ) ( )

2

1

r

Q

k

r

Q

k

V

=

+

+

−

where r1 and r2 are the distances from the given point to the centers of the spheres. For a point on the surface of the

sphere with charge +Q:

δ

+ =

=R r d

r1 and 2 where

δ

<R

Substitute to obtain:

( ) ( )

δ

+

−

+

+

=

+

d

Q

k

R

Q

k

V

_Q

For δ << d:

d

kQ

R

kQ

V

_+Q

=

−

and

d

kQ

R

kQ

V

−Q

=

−

+

The potential difference between the spheres is:

⎟ ⎠ ⎞ ⎜

⎝ ⎛ − =

⎟ ⎠ ⎞ ⎜

⎝

⎛− ₊

− − =

− =

∆ ₋

d R kQ

d kQ R

kQ d

kQ R kQ

V V

V Q Q

1 1 2

Substitute for ∆V in the expression for C to obtain:

d

R

R

d

R

d

R

kQ

Q

C

−

∈

=

⎟

⎠

⎞

⎜

⎝

⎛ −

∈

=

⎟

⎠

⎞

⎜

⎝

⎛ −

=

1

2

1

1

2

1

1

2

0

0

π

π

For d very large: _{C R}

0

2 ∈

=

π

The Storage of Electrical Energy

25 •

Picture the Problem Of the three equivalent expressions for the energy stored in a charged capacitor, the one that relates U to C and V is

_U

=

₂1

_CV

2_.

(a) Express the energy stored in the capacitor as a function of C and V:

Substitute numerical values and evaluate U:

(

3 F

)(

100V

)

2 15.0mJ

2

1 =

=

µ

U

(b) Express the additional energy required as the difference between the energy stored in the capacitor at 200 V and the energy stored at 100 V:

(

)

(

)

(

)(

)

mJ 0 . 45

mJ 0 . 15 V 200 F 3

V 100 V

200

2 2

1

=

− =

− =

∆

µ

U U

U

26 •

Picture the Problem Of the three equivalent expressions for the energy stored in a charged capacitor, the one that relates U to Q and C is

C

Q

U

2

2

1

=

.

(a) Express the energy stored in the

capacitor as a function of C and Q:

_C

Q

U

2

2

1

=

Substitute numerical values and evaluate U:

(

)

J

800

.

0

F

10

C

4

2

1

2

_µ

µ

µ

₌

=

U

(b) Express the energy remaining

when half the charge is removed:

( )

(

)

J

0.200

F

10

C

2

2

1

2

2

1

µ

µ

µ

₌

=

Q

U

27 •

Picture the Problem Of the three equivalent expressions for the energy stored in a charged capacitor, the one that relates U to Q and C is

C

Q

U

2

2

1

=

.

(a) Express the energy stored in the

capacitor as a function of C and Q:

C

Q

U

2

2

1

=

Substitute numerical values and

evaluate U:

(

)

(

)

J

625

.

0

pF

20

C

5

2

1

C

5

2

=

=

µ

µ

U

(b) Express the additional energy required as the difference between the energy stored in the capacitor when its charge is 5 µC and when its charge is 10 µC:

(

)

(

)

(

)

J

.88

1

J

0.625

J

2.50

J

625

.

0

pF

20

C

10

2

1

C

5

C

10

2

=

−

=

−

=

−

=

∆

µ

µ

µ

U

*28 •

Picture the Problem The energy per unit volume in an electric field varies with the square of the electric field according to

u

=

∈

0

E

2

2

.

Express the energy per unit volume in an electric field:

2 0 2 1

E

u

=

∈

Substitute numerical values and evaluate u:

(

)

(

)

3

2 2

2 12 2

1

J/m 8 . 39

MV/m 3 m /N C 10 85 . 8

=

⋅ ×

= −

u

29 •

Picture the Problem Knowing the potential difference between the plates, we can use E = V/d to find the electric field between them. The energy per unit volume is given by

2 0 2 1

_E

u

=

∈

and we can find the capacitance of the parallel-plate capacitor using

.

0

A

d

C

=∈

(a) Express the electric field between the plates in terms of their separation and the potential difference between

them:

1

mm

100

kV/m

V

100

₌

=

=

d

V

E

(b) Express the energy per unit volume in an electric field:

2 0 2 1

_E

u

=

∈

Substitute numerical values and evaluate u:

(

)

(

)

3

2 2

2 12 2

1

mJ/m 3 . 44

kV/m 00 1 m /N C 10 85 . 8

=

⋅ ×

= −

u

(c) The total energy is given by:

(

)( )

(

)

J uAd uV U

µ

6 . 88

mm 1 m 2 mJ/m 3 .

44 3 2

= =

= =

(d) The capacitance of a parallel-plate capacitor is given by:

(

)( )

nF 7 . 17

mm 1

m 2 m /N C 10

8.85 12 2 2 2

0

=

⋅ ×

= ∈ =

− d

(e) The total energy is given by:

_U

=

1₂

_CV

2 Substitute numerical values and evaluate

U:

(

)(

)

). ( with agreement in

J, 5 . 88

V 100 nF

17.7 2

2 1

c U

µ

= =

30 ••

Picture the Problem The total energy stored in the electric field is the product of the energy density in the space between the spheres and the volume of this space.

(a) The total energy U stored in the electric field is given by:

uV

U =

where u is the energy density and V is the volume between the spheres.

The energy density of the field is: 2 0

2

1

E

u

=

∈

where E is the field between the spheres. The volume between the spheres is

approximately:

(

2 1

)

2 1

4

r

r

r

V

≈

π

−

Substitute for u and V to obtain:

(

2 1

)

2

1 2 0

2

E

r

r

r

U

=

π

∈

−

(1)

The magnitude of the electric field between the concentric spheres is the sum of the electric fields due to each charge distribution:

Q Q

E

E

E

=

+

₋

Because the two surfaces are so close together, the electric field between them is approximately the sum of the fields due to two plane charge distributions:

0 0 0 2

2∈ + ∈ = ∈

= Q −Q Q

E

σ

σ

σ

Substitute for σQ to obtain:

0 2 1

4

∈

≈

r

Q

E

π

Substitute for E in equation (1) and

simplify:

(

)

2 1

1 2 0 2

1 2 2 1 2 0 2 1 0

8

4

2

r

r

r

Q

r

r

r

r

Q

U

−

∈

=

−

⎟⎟

⎠

⎞

⎜⎜

⎝

⎛

∈

∈

=

π

Substitute numerical values and evaluate U:

(

) (

)

(

8

.

85

10

C

/

N

m

)

(

10

.

0

cm

)

56

.

0

nJ

8

cm

0

.

10

cm

5

.

10

nC

5

2 2

2 12 2

=

⋅

×

−

=

₋

π

U

(b) The capacitance of the

two-sphere system is given by:

V

Q

C

∆

=

where ∆V is the potential difference between the two spheres.

The electric potentials at the

surfaces of the spheres are: 0 1 1

4

r

Q

V

∈

=

π

and 2 _{0 2}

4

r

Q

V

∈

=

π

Substitute for ∆V and simplify to

obtain: 2 1

2 1 0 2

0 1

0

4

4

4

r

r

r

r

r

Q

r

Q

Q

C

−

∈

=

∈

−

∈

=

π

π

π

Substitute numerical values and evaluate C:

(

)

(

)(

)

0

.

234

nF

cm

0

.

10

cm

5

.

10

cm

5

.

10

cm

0

.

10

m

N

/

C

10

85

.

8

4

12 2 2

=

−

⋅

×

=

_π

−

C

Use ½ Q2/C to find the total energy stored in the electric field between the spheres:

(

)

nJ

4

.

53

nF

234

.

0

nC

5

2

1

2

₌

⎥

⎦

⎤

⎢

⎣

⎡

=

U

). ( in obtained

result exact our of 5% within is

) ( in result e approximat our

that Note

b

a

*31 ••

Picture the Problem We can relate the charge Q on the positive plate of the capacitor to the charge density of the plate σ using its definition. The charge density, in turn, is related to the electric field between the plates according to

σ

=

∈

₀

E

and the electric field can be found from E = ∆V/∆d. We can use

∆

U

=

1₂

Q

∆

V

_{in part (b) to find the increase} in the energy stored due to the movement of the plates.

(a) Express the charge Q on the positive plate of the capacitor in terms of the plate’s charge density σ and surface area A:

Relate σ to the electric field E between the plates of the capacitor:

E

0

∈

σ

=

Express E in terms of the change in V as the plates are separated a distance ∆d:

d

V

E

∆

∆

=

Substitute for σ and E to obtain:

d

V

A

EA

Q

∆

∆

=

=

∈

0

∈

0

Substitute numerical values and evaluate Q:

(

)(

)

11

.

1

nC

cm

0.4

V

100

cm

500

m

/N

C

10

8.85

×

12 2

⋅

2 2

=

=

−

Q

(b) Express the change in the electrostatic energy in terms of the change in the potential difference:

V

Q

U

=

∆

∆

₂1

Substitute numerical values and evaluate ∆U:

(

11.1nC

)(

100V

)

0.553 J

2

1 =

µ

= ∆U

32 •••

Picture the Problem By symmetry, the electric field must be radial. In part (a) we can find Er both inside and outside the ball by choosing a spherical Gaussian surface first inside and then outside the surface of the ball and applying Gauss’s law.

(a) Relate the electrostatic energy density at a distance r from the center of the ball to the electric field due to the uniformly distributed charge Q:

2 0 2 1

e

E

u

=

∈

(1)

Relate the flux through the Gaussian surface to the electric field Eron the Gaussian surface at r < R:

(

)

0 inside 2

4

∈

π

r

Q

E

_r

=

(2)

Using the fact that the charge is uniformly distributed, express the ratio of the charge enclosed by the Gaussian surface to the total charge of the sphere:

3 3 3 3 4

3 3 4

ball surface Gaussian inside

R

r

R

r

V

V

Q

Q

=

=

=

π

π

Solve for Qinside to obtain:

3 3 inside

R

r

Q

Q

=

Substitute in equation (2):

( )

3 0

3 2

4

R

Qr

r

E

_r

∈

π

=

Solve for Er < R:

_r

R

kQ

R

Qr

E

_{r R 3 3}

0

4

=

=

<

_π

_∈

Substitute in equation (1) to obtain:

₍

₎

2 6

2 2 0

2 3 0 2 1 e

2R r Q k

r R kQ R

r u

∈

∈

=

⎟ ⎠ ⎞ ⎜ ⎝ ⎛ = <

Relate the flux through the Gaussian surface to the electric field Eron the Gaussian surface at r > R:

(

)

0 0 inside 2

4

∈

∈

π

r

Q

Q

E

_r

=

=

Solve for Er > R: 2

0 2

4

−

>

=

=

kQr

r

Q

E

_{r R}

∈

π

Substitute in equation (1) to obtain:

(

)

(

)

4 2 2 0 2 1

2 2 0

2 1 e

− −

=

=

>

r

Q

k

kQr

R

r

u

∈

∈

(b) Express the energy dU in a spherical shell of thickness dr and surface area 4π r2:

( )

r

dr

u

r

dU

_shell

=

4

π

2

For r < R:

₍

₎

dr r R kQ

dr r R

Q k r R r dU

4 6 2

2 6

2 2 0 2 shell

2 2 4

=

⎟⎟ ⎠ ⎞ ⎜⎜

⎝ ⎛ =

<

π

∈

For r > R:

(

)

(

)

dr r kQ

dr r Q k r

R r dU

2 2 2 1

4 2 2 0 2 1 2 shell 4

−

−

= =

>

π

∈

Integrate Ushell(r < R) from 0 to R:

(

)

R

kQ

dr

r

R

kQ

R

r

U

R

10

2

2 0

4 6 2 shell

=

=

<

∫

Integrate Ushell(r > R) from R to ∞:

(

)

R

kQ

dr

r

kQ

R

r

U

R

2

2 2

2 2 1

shell

>

=

∫

=

∞ −

Substitute in equation (3) to obtain:

R

kQ

R

kQ

R

kQ

U

5

3

2

10

2 2

2

=

+

=

sphere.

hin the

energy wit

field

the

includes

it

because

sphere

for the

greater

is

result

The

vanishes.

integral

first

the

so

zero,

is

shell

the

inside

field

The

Combinations of Capacitors

33 •

Picture the Problem We can apply the properties of capacitors connected in parallel to determine the number of 1.0-µF capacitors connected in parallel it would take to store a total charge of 1 mC with a potential difference of 10 V across each capacitor. Knowing that the capacitors are connected in parallel (parts (a) and (b)) we determine the potential difference across the combination. In part (c) we can use our knowledge of how potential differences add in a series circuit to find the potential difference across the combination and the definition of capacitance to find the charge on each capacitor.

(a) Express the number of

capacitors n in terms of the charge q on each and the total charge Q:

q

Q

n

=

Relate the charge q on one capacitor to its capacitance C and the potential difference across it:

CV

q

=

Substitute to obtain:

CV

Q

n

=

Substitute numerical values and

evaluate n:

( )(

1

F

10

V

)

100

mC

1

=

=

µ

(b) Because the capacitors are connected in parallel the potential difference across the combination is the same as the potential difference across each of them:

V 10

n combinatio

parallel =V = V

(c) With the capacitors connected in series, the potential difference across the combination will be the sum of the potential differences across the 100 capacitors:

(

)

kV 00 . 1

V 10 100 100

n combinatio series

= =

= V

V

Use the definition of capacitance to find the charge on each capacitor:

( )(

1

µ

F 10V

)

= 10.0

µ

C

=

=CV

q

34 •

Picture the Problem The capacitor array is shown in the diagram. We can find the equivalent capacitance of this combination by first finding the equivalent capacitance of the 3.0-µF and 6.0-µF capacitors in series and then the equivalent capacitance of this capacitor with the 8.0-µF capacitor in parallel.

Express the equivalent capacitance for the 3.0-µF and 6.0-µF capacitors in series:

F

6

1

F

3

1

1

6

3

µ

µ

+

=

+

C

Solve for C3+6:

C

3+6

=

2

µ

F

Find the equivalent capacitance of a

2-µF capacitor in parallel with an

8-µF capacitor:

F 10 F 8 F 2

8

2+ =

µ

+

µ

=

µ

C

*35 •

Picture the Problem Because we’re interested in the equivalent capacitance across terminals a and c, we need to recognize that capacitors C1 and C3 are in series with each other and in parallel with capacitor C2.

Find the equivalent capacitance of

C1 and C3 in series: 13 1 3

1

1

1

Solve for C1+3:

3 1

3 1 3 1

C

C

C

C

C

+

=

+

Find the equivalent capacitance of

C1+3 and C2 in parallel: _{1 3}

3 1 2 3 1 2 eq

C C

C C C C

C C

+ + = +

= ₊

36 •

Picture the Problem Because the capacitors are connected in parallel we can add their capacitances to find the equivalent capacitance of the combination. Also, because they are in parallel, they have a common potential difference across them. We can use the definition of capacitance to find the charge on each capacitor.

(a) Find the equivalent capacitance of the two capacitors in parallel:

F 30 F 20 F 0 . 10

eq =

µ

+

µ

=

µ

C

(b) Because capacitors in parallel have a common potential difference across them:

V 00 . 6

20 10+ =

=V V

V

(c) Use the definition of capacitance to find the charge on each capacitor:

(

10 F

)( )

6V 60.0 C

10

10 =C V =

µ

=

µ

Q

and

(

20 F

)( )

6V 120 C

20

20 =C V =

µ

=

µ

Q

37 ••

Picture the Problem We can use the properties of capacitors in series to find the

equivalent capacitance and the charge on each capacitor. We can then apply the definition of capacitance to find the potential difference across each capacitor.

(a) Because the capacitors are connected in series they have equal charges:

V

C

Q

Q

₁₀

=

₂₀

=

_eq

Express the equivalent capacitance

of the two capacitors in series: 20 F 1 F 10

1 1

eq

µ

µ

+ =

C

Solve for Ceq to obtain:

(

)(

)

F

67

.

6

F

20

F

10

F

20

F

10

eq

µ

µ

µ

µ

µ

₌

+

=

C

Substitute to obtain:

(

_{6.67 F}

)( )

_{6V 40.0 C} 20

10 =Q =

µ

=

µ

(b) Apply the definition of capacitance to find the potential difference across each capacitor:

V

00

.

4

F

10

C

0

.

40

10 10

10

=

=

µ

=

µ

C

Q

V

and

V

00

.

2

F

20

C

0

.

40

20 20

20

=

=

µ

=

µ

C

Q

V

*38 ••

Picture the Problem We can use the properties of capacitors connected in series and in parallel to find the equivalent capacitances for various connection combinations.

(a) If their capacitanceis tobeamaximum, they must beconnectedin parallel.

Find the capacitance of each capacitor:

F

15

3

eq

=

C

=

µ

C

and

F

5

µ

=

C

(b) (1) Connect the three capacitors in series:

F 5

3 1

eq

µ

=

C and Ceq = 1.67

µ

F (2) Connect two in parallel, with the

third in series with that combination:

(

5

F

)

10

F

2

parallel in two

eq,

=

µ

=

µ

C

and

F 5

1 F 10

1 1

eq

µ

µ

+ =

C

Solve for Ceq:

(

)(

)

F

33

.

3

F

5

F

10

F

5

F

10

eq

_µ

_µ

µ

µ

µ

₌

+

=

C

(3) Connect two in series, with the third in parallel with that

combination:

F 5

2 1

series in two

eq,

µ

=

C

or

F

5

.

2

series in two

eq,

=

µ

C

Find the capacitance equivalent to 2.5 µF and 5 µF in parallel:

F 50 . 7 F 5 F 5 . 2

eq =

µ

+

µ

=

µ

C

39 ••

(a) Relate the equivalent

capacitance of the two capacitors in series to their individual

capacitances:

F

15

1

F

4

1

1

15

4

µ

µ

+

=

+

C

Solve for C4+15:

(

)(

)

F

16

.

3

F

15

F

4

F

15

F

4

15

4

_µ

_µ

µ

µ

µ

₌

+

=

+

C

Find the equivalent capacitance of C4+15 in parallel with the

12-µF capacitor:

F 2 . 15 F 12 F 16 . 3

eq =

µ

+

µ

=

µ

C

(b) Using the definition of capacitance, express and evaluate the charge stored on the 12-µF capacitor:

(

)(

)

mC 40 . 2

V 200 F 12

12 12 12 12

= =

= =

µ

V C V C Q

Because the capacitors in series

have the same charge:

(

)(

)

mC 632 . 0

V 200 F 16 . 3

15 4 15 4

= =

=

= ₊

µ

V C Q Q

(c) The total energy stored is given by:

2 eq total

2

1

V

C

U

=

Substitute numerical values and evaluate Utotal:

(

15

.

2

F

)(

200

V

)

0

.

304

J

2

1

2

total

=

µ

=

U

40 ••

Picture the Problem We can use the properties of capacitors in series to establish the results called for in this problem.

(a) Express the equivalent capacitance of two capacitors in series:

2 1

1 2 2 1 eq

1 1 1

C C

C C C C C

+ = + =

Solve for Ceq by taking the reciprocal of both sides of the equation to obtain:

2 1

2 1 eq

C C

C C C