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Chapter 26 Capacitance and

Dielectrics

We must do work, qV, to bring a point charge q from far away (V = 0 at infinity) to a

region where other charges are present. The work done is stored in electrostatic field

energy.

26.1 Definition of Capacitance

26.2 Calculating Capacitance

Measure of the capacity to store charge: V

QC =

Unit: farad (F): 1 F = 1 C / V; 1 µF = 10-6 F; 1 pF = 10-12 F

The ratio of charge Q to the potential depends on the size and the shape of the

conductor.

Capacitors

A device consisting of two conductors carrying equal but

opposite charges is called a capacitor.

Parallel Plate Capacitor

0

2εQ

AE = , 02

2εA

QE = , d

A

QEdV

0ε==

d

A

V

QC 0ε==

2

Cylindrical Capacitor

0

2ε

π QrLE = -->

02 επrL

QER =

=−=−= ∫ a

b

L

Qdr

rL

QVVV

a

b

ba ln22 00 πεεπ

==

a

bL

V

QC

ln

2 0πε

Spherical Capacitor

0

24ε

π QEr = ,

−=−= ∫ ba

Qdr

r

QV

a

b

1144 0

20 πεπε

ab

ab

V

QC

−== 04πε

Self-Capacitance

The potential of a spherical conductor of radius R carrying a charge Q is R

kQV = .

The self-capacitance of a spherical conductor is:

Rk

R

R

kQQ

V

QC 04πε====

26.3 Combination of Capacitors

Capacitors Connected in Parallel

Obtain V and Q to calculate C.

VVV == 21 & 111 /VQC =

3

( )VCCVCVCQQQ 21221121 +=+=+=

21/ CCVQC +==

Capacitors connected in parallel:

...4321 ++++= CCCCCeq

Capacitors connected in series

QQQ == 21 & 111 /VQC =

+=+=+=

212

2

1

121

11

CCQ

C

Q

C

QVVV

21

111

CCV

QC

+==

series --> sum voltage & parallel --> sum charges

Capacitors connected in series:

...11111

4321

++++=CCCCC

Example: Two capacitors are removed from the battery and

carefully connected from each other.

21 VV = & 1

11 V

QC =

2

2

1

1

C

Q

C

Q = & CQQ µ9621 =+

Capacitors in Series and in Parallel

321

111CCCC

++

=

Q = ? V = ?

213 QQQQ +== , 31 VVV += , 21 VV =

2

2

1

1

C

Q

C

Q = & QQQ =+ 21 --> QCC

CQ

21

11 +

= & QCC

CQ

21

22 +

=

C1

C2

C3

4

QC

QCCC

Q

C

QVVV

3213

3

1

131

11 ++

=+=+= -->

321

111

CCCV

QC

++

==

26.4 Energy Stored in a Charged

Capacitor

dqC

qVdqdU ==

22

0 21

21

2CVQV

C

Qdq

C

qdUU

Q

===== ∫∫

Electrostatic Field Energy (derived from energy

stored in a capacitor)

d

A

V

QC 0ε== , EdV =

( ) VEAdEEdd

ACVU

=

=== 20

20

202

21

21

21

21 εεε

Electrostatic Energy Density: 202

1E

V

Uue ε== (energy per unit volume)

Example: Calculate the energy stored in the conductor

carrying a charge Q.

Rr < : 0=E

Rr > : 2r

kQE =

( ) ( )drrr

QkdrrEdVudU e

24

22

022

0 421

421 πεπε

=

==

QVR

kQQdr

rQkdUU

RR 21

211

22

220 ==== ∫∫

∞∞

πε

+

+

+

-

-

- +

+Q R

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26.5 Capacitors and Dielectrics When the space between the two conductors of a capacitor is occupied by a dielectric,

the capacitance is increased by a factor κ ( 1>κ ) that is characteristic of the

dielectric.

If the dielectric field is 0E before the dielectric slab is inserted, after the dielectric

slab is inserted between the plates the field is

κ0E

E = --> the potential is κκ

00 VdEEdV ===

If 0

0 V

QC = , the capacitor is 0

0 /' C

V

Q

V

QC κ

κ=== .

The capacitance of a parallel-plate capacitor filled with a dielectric of constant κ is

d

A

d

A

d

A

dE

A

V

A

V

QC

εκεκ

εσ

σκσκ

σ ====== 0

0

00 / --> 0κεε = is called the

permittivity of the dielectric.

Vacuum: κv = 1

the Dielectric: κ > 1

material: κm > 0

6

Energy Stored in The Presence of a Dielectric

The energy stored in a capacitor is:

VdqdU = --> dqC

qdU = --> ∫∫ =

QU

dqC

qdU

00

--> 22

21

2CV

C

QU ==

The energy of a capacitor with the dielectric is

( ) ( ) ( )AdEEdd

AEd

d

ACVU

==== 2222

21

21

21

21 εεε

20

2

21

21

EEue κεε ==

1. You lose electric force to separate the charge.

2. You enlarge the charging capacity as you know the dielectric will breakdown in a

high electric field. (If the same charge --> you lose some electric field,)

3. You increase the energy per unit volume.

Combination of Capacitors

Example: A parallel-plate capacitor has square plates of edge length 10 cm and a

separation of d = 4 mm. A dielectric slab of constant 2=κ has dimensions 10 cm X

10 cm X 4 mm. (a) What is the capacitance without the dielectric? (b) What is the

capacitance with the dielectric? (c) What is the capacitance if a dielectric slab with

dimensions 10 cm X 10 cm X 3 mm is inserted into the 4-mm gap?

(a) d

AC 0ε= , (b)

d

AC

ε=

(c) series connection:

4/3

1

4/

1

4/3

1

4/

11

000

d

A

d

A

d

A

d

AC κεεεε +=+=

Example: The parallel plates of a given capacitor are square with 2aA = and

separation distance d. If the plates are maintained at a constant potential V and a

Vacuum: κv = 1

the Dielectric: κ > 1

material: κm > 0

- + E0

E

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square of dielectric slab of constant κ , area 2aA = , thickness d is inserted between the capacitor plates to a distance x as shown in the following figure. Let 0σ

be the free charge density at the conductor-air surface. (a) Calculate the free charge

density κσ at the capacitor-dielectric surface. (b) What is the effective capacitance?

(c) What is the magnitude of the required force to prevent the dielectric slab from

sliding into the plates?

(a) In air: 0

0

εσ=E & dV

0

0

εσ= , in dielectric:

0εσK

E K= & dK

V K

0εσ=

� 0σσ KK =

(b) ( ) ( )( )xKa

d

a

d

xaa

d

axKCCC 1000

21 −+=−+=+= εεε

(c) 2

21

CVU = �

( )121

21

21 02222 −=+

−=+

−= Kd

aV

dx

dCV

dx

dCV

dx

dQV

dx

dCVF

ε

The first term is due to charge redistribution and the second is due to the additional

charges supplied by the constant voltage.

Example: A parallel plate capacitor with plates of area LW and separation t has the

region between its plates filled with wedges of two dielectric materials. Assume t is

much less than both W and L. (a) Please determine its capacitance.

The thickness of the k1 material decrease as a function of ( )

L

xLt − while that of the

k2 material is of L

tx

For a short stripe of dx, ( )

( ) LxLt

Wdx

d

AC

/01

1

11 −

== εκε,

( )Ltx

Wdx

d

AC

/02

1

12

εκε ==

The series connected capacitance 21

111

CCC+= ,

−+

=

L

xt

dxWC

121

0

111κκκ

ε

The total parallel connected capacitance

−

=

−+

= ∫2

1

12

0

0

121

0 ln11111 κ

κ

κκ

ε

κκκ

ε

t

LW

L

xt

dxWC

L

total

k2 k1 L W

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26.6 Electric Dipole in an Electric Field Inside the material � to make sure that the electric field lines are from the positive

charge to the negative charge

If the field is uniform, it can rotate the dipole.

( ) EpEdqEqrEqrrrrrrrrrr ×=×=−×+×= −+τ

θτ sinpEEp =×=rrv

The potential energy: ( ) θθθτ dpEdFdxdU sin−−=−=−=

� ( )if

f

i

pEdpEU θθθθθ

θ

coscossin −−== ∫

EpUrr ⋅−=

26.7 An Atomic Description of

Dielectrics

- +

-q

+q

E

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Example: A hydrogen atom consists of a proton nucleus of charge +e and an electron

of charge –e. The charge distribution of the atom is spherically symmetric, so the

atom is nonpolar. Consider a model in which the hydrogen atom consists of a positive

charge +e at the center of a uniformly charged spherical cloud of radius R and total

charge –e. Show that when such an atom is placed in a uniform external field Er

, the

induced dipole moment is proportional to Er

; that is, Eprr α= , where α is called

Qtotal = 0

Qtotal < 0 Qtotal > 0

bound charge

E0 E

bound charge

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the polarizability.

rR

e

r

Re

r

A

QE efrominside 3

002

33

0__ 44

3

434

πεεπ

ππ

ε−=

−

==−

304 R

erEerp

πεαα === --> 3

04 Rπεα =

Magnitude of The Bound Charge

0εσ b

bE = & 0

0 εσ fE =

bEEE

E −== 00

κ --> 0

11 EEb

−=κ

--> fb σκ

σ

−= 11

0

1 σσκ

σσσ ==−= fbfeffective � 0κσσ =f

+

+

+

+

-

-

-

-

+

+

+

+

-

-

-

-

-

-

+

+