Chapter 4 CAPACITANCE & DIELECTRICS

Recommended Problems:

1,3,59,16,17,18,19,21,22,23,25,27,29,31,33,47,49,54,61,62.

CAPACITANCE Capacitor is a device consists of two conductors with an insulator

between them. It is used to store charges and consequently

electrostatic energy

The conductors usually have equal but opposite charges so that

the net charge of any capacitor is zero.

When we said that a capacitor has a charge Q, we mean that one

of the conductors has a charge Q and the other has a charge -Q.

The capacitance C of a capacitor is defined as the ratio of the

magnitude of the charge Q on either conductors to the magnitude

of the potential difference between them, denoted hereafter by V,

i.e.,

V

QC The capacitance is always

a positive quantity.

The SI unit of the capacitance C is Coulomb per volt (C/V). This

unit is referred as Farad (F), that is 1F = 1 C/V

Calculation of Capacitance

The capacitor consists of two parallel metal

plates of equal area A separated by a distance

d and immersed in vacuum.

If the separation d is small

compared to the size of the

plates we can assume that any

point inside the capacitor is very

closed to a conductor (just

outside a conductor).

I- The Parallel Plates Capacitor

E=/o E=0

d

E=0

Note that the e.field outside the

capacitor is zero (why).

The value of the electric field inside the capacitor is, therefore,

uniform and equal to

o

E

plate.either on density charge theiswith AQ

The P.D between the two plates is

lE dV

dEdlEV

Since E is constant and directed from the positive to the negative

plate we can write

dA

QdV

oo

V

QCNow or

oAQd

QC

d

AC o

We will see later that if the medium between the plates is not

vacuum this constant should be multiplied by a factor.

In circuits the capacitors and the batteries are represented as

shown.

Capacitor Symbol Baterry Symbol

This means that the capacitance depends only on the geometry of

the capacitor; directly proportional to the area of the plates and

inversely proportional to their separation with the permittivity o

stands for the proportionality constant.

d

AC o

• Test Your Understanding (1)

The capacitance of a parallel-plate capacitor can be increased by:

a) increasing the charge Q b) increasing the voltage V

c) increasing the plates

separation d d) increasing the plates area A

We can charge a capacitor by connecting it

across the terminal of a battery.

In doing so electrons transfer from the plate

that is connected to the positive terminal of

the battery to the plate that is connected to

the negative terminal of the battery.

This process continues for a short time

until the potential difference across the

capacitor becomes equal to the

potential difference of the battery.

-Q +Q

II- The Spherical Capacitor

A spherical capacitor consists of two

concentric conducting spherical shells of radii

a and b. The inner shell has a charge +Q

while the outer shell has a charge -Q.

+Q

-Q

a

b

To find the capacitance of such a capacitor we have to calculate

the potential difference between the two shells

b

a

dV rE

The e.field between the two shells can be found using Gauss’ law

rE ˆ4 2

or

Q

e. Field between a and b

Note that the e.field outside the capacitor is zero.

b

a r

drQV

2o4

b

ar

Q

1

4 o

ab

QV

11

4or

o

ab

abC

o4

From which we have

A single isolated sphere can be considered as a capacitor where

the outer shell is assumed to be of infinite radius. If we let b in

the last equation, the capacitance of an isolated sphere of radius

a is found to be

aC o4

II- The Cylindrical Capacitor

This capacitor consists of a conducting

cylinder of radius a and length L concentric

with a conducting cylindrical shell of radius b.

The inner cylinder charged with charge Q and

the outer shell with charge -Q.

We will assume that the length of the capacitor L

is much greater than the radius b such that we can

neglect the fringing of the electric field at the ends.

The electric field in the region between the two conductors can be

calculated using Gauss’ law as

Lr

QE

o2

-Q

Q

The potential difference is then found to be

b

a

b

a

EdrdV rE

b

a r

dr

L

Q

o2 bar

L

Qln

2 o

The magnitude of the potential difference is then

a

b

L

QV ln

2 o

a

b

L

Q

Q

V

QC

ln2 o

ab

LC

ln2 o

C1

C2

Combinations of Capacitors

There are two basic combinations of capacitors: The parallel, and

the series combinations.

I- Parallel Combination

If two capacitors are connected as

shown, we say that the two capacitors

are connected in parallel.

In such a combination, the right plates

are connected, by a conducting wire,

together to form an equipotential surface

(They have the same potential). The

other two plates, the left plates, form a

another equipotential surface.

21eq VVV

where Veq stands for the potential difference across the equivalent

capacitor.

The charge stored by this equivalent capacitor Qeq is equal to the

sum of the charges stored by each capacitor, i.e.,

21eq QQQ

2211eqeq VCVCVC 21eq CCC

Capacitors are said to be connected in parallel if the potential

across each one is the same and equal to the potential

across an equivalent capacitor.

Therefore, the potential difference across the two capacitors are

the same and equal to the potential difference across an

equivalent capacitor replacing the two capacitors

C1 C2 II- Series Combination

If two capacitors are connected as

shown, we say that the two capacitors

are connected in series.

When the battery is connected, electrons

transfer from the right plate of C1 to the

right plate of C2 through the battery.

Thus, the right plate of C1 acquires a positive charge while the left

plate of C2 acquires an equal negative charge.

As the other two plates, enclosed by the dashed line, form an

isolated conductor, electrons are attracted to the right end leaving

the left end with an excess positive charges.

This means that the battery induces a charge on the isolated

conductor.

It is clear here that the charges on each capacitor must be the

same and equal to the charge on an equivalent capacitor

replacing the two capacitors, that is,

21eq QQQ 21eqaband VVVV

2

2

1

1

eq

eq

C

Q

C

Q

C

Q

21eq

111

CCC

Capacitors are said to be connected in series if the charge on

each one is the same and equal to the charge on an

equivalent capacitor.

Example 26.1 Find the equivalent capacitance for the

combination shown. All capacitances are in microfarads.

1.0

3.0

4.0

6.0

2.0 8.0

Parallel

Parallel

4.0 4.0

8.0 8.0

Series

Series

2.0

4.0

Parallel

6.0

ENERGY STORED IN A CHARGED

CAPACITOR

As mentioned before, in charging a capacitor electrons are

transferred from one plate to the other building up a potential

difference across the capacitor.

This means that a work is required to

charge a capacitor and this work is stored

as a potential energy in the capacitor.

Consider a parallel plates capacitor that is

initially uncharged. Suppose that q is the

charge built up on the capacitor at some

instant during the charging process.

The potential difference across the capacitor at that instant is

-q +q

The work required to transfer a small charge dq is therefore

dqC

qvdqdW

C

Qdq

C

qW

Q

2

2

0

The total work required to charge a capacitor from uncharged

situation (q=0) to a final charge Q is, thus

This work will be stored in the capacitor as potential energy

QVCVC

QU

212

21

2

2

Cqv

• Test Your Understanding (2)

You charge a parallel-plate capacitor, remove it from the battery,

and prevent the wires connected to the plates from touching each

other. When you pull the plates apart to a larger separation, do the

following quantities increase, decrease, or stay the same?

a) C, b) Q, c) V, d) U, and e) E between the plates,

a) C decreases. b) Q stays the same.

c) V increases. d) U increases.

e) E stays the same.

• Test Your Understanding (3)

You charge a parallel-plate capacitor, pull the plates apart to a

larger separation while the battery remains connected to the

capacitor, do the following quantities increase, decrease, or stay

the same?

a) C, b) Q, c) V, d) U, and e) E between the plates,

a) C decreases. b) Q decreases.

d) V stays the same.c d) U decreases.

e) E decreases.

Example 26.5 Two capacitors C1=2F

and C2=1F (C1 > C2 ) are charged to the

same potential difference Vo=10V. The

charged capacitors are removed from the

battery and are then connected together as

shown.

C1

S1 S2

a b

C2

(a) Find the potential difference between points a and b after

closing the switches.

(b) Find the total energy stored before and after closing the

switches.

Vo +20C - 20C

+ 10C -10 C

Solution

(a) The charges on

each capacitor before

closing the switches are

CVCQ oi 2011

CVCQ oi 10and 22

After closing the switches, the net charge on the two capacitors is

CCCVQQQ oii 102121

The minus sign is because the two plates that are connected with

opposite polarities.

Since the two capacitors are connected together, this charge will

be distributed between the two capacitors provided that

ff VV 21

2

2

1

1

C

Q

C

Q ff

)1(10and 21 CQQQ ff

From Eqs. (1) and (2)

CQCQ ff 3

10103 22 CQ f

3

20and 1

Now from Eq. (2) we get

VVVV ffab3

1021

)2(2 21 ff QQ

(b) Before closing the switches

iii UUU 21

JVCCVCVC ooo 2

3002212

1222

1212

1

After closing the switches

fff UUU 21

2212

1222

1212

1abff VCCVCVC

JCC

CCVU of

6

100

21

2212

21

CAPACITORS WITH DIELECTRIC

A dielectric is an insulating material.

It is found that when a dielectric

material is inserted between the

plates of a capacitor, its capacitance

increases by a numerical factor

called the dielectric constant of the

material, that is

where C and Co are the

capacitance with and without

the dielectric respectively.

oCC

Since C is always greater than Co,

the dielectric constant must be

greater than unity.

Material

Dielectric

constant

Dielectric

strength

(V/m)

Vacuum 1.0000 3 106

Air (dry) 1.0006 24 106

Paraffin 2.2 10 106

Polystyrene 2.6 24 106

Paper 3.7 15 106

Quartz 4.3 8 106

Oil 4 12 106

Glass 5 14 106

Rubber 6.7 12 106

Porcelain 6-8 5 106

Nylon 3.4 14 106

Water 80

+ - Eo

F-

F+

Let us now explain what happen when a dielectric material is

inserted between the plates of an isolated capacitor.

Suppose that the dielectric is a polar material (has a permanent

electric dipole).

The electric field of the capacitor Eo exerts a torque on the dipoles

of the material so that it tends to rotates these dipoles into the

direction of Eo.

When all dipoles aligned with the field Eo we said that the material

is polarized.

- +

- +

- +

- +

- +

+ - + - + -

+ - + - + -

+ - + - + -

+ - + - + -

+ - + - + - + - + - + -

Eo Eo

E’

As a result of this polarization, polarization charges are produced

at the faces of the dielectric with the +ve charges are near the –ve

plate and the –ve charges are near the +ve plate.(the charges

inside the dielectric cancel each other).

These charges create an electric field E ' opposite to Eo.

The net electric filed inside the conductor is therefore

EEE o

That is, a reduction is occurred in the electric field.

Since V=Ed, the potential difference across the capacitor

decreases, consequently, i.e.,

oE

E κ

VV oand

V

QC But And Q is constant o

o

CV

QC

Co

It should be noted that a limited value of the potential difference

can be applied across a capacitor.

If the potential exceeds this value, the insulating properties of the

dielectric will break down and form a conducting medium.

This limited value, called the breakdown potential, depends on the

dielectric strength, which is the maximum value of the electric field

that the dielectric can tolerate without breakdown.

Solution

Example 26.7 A parallel plate

capacitor is charged by a battery of

potential difference Vo .

Find the energy stored in the capacitor

before and after the slab is inserted.

Co

Vo

The battery charges the capacitor with a charge 0VCQ oo

After removing the battery, this charge will not be changed.

The energy stored in the capacitance before inserting the

dielectric is

o

2o2

oo21

o2C

QVCU

The battery is then removed and a

dielectric slab of dielectric constant is

inserted between the plates of the

capacitor, as shown.

Now, the energy stored after introducing the dielectric is

2C

2oQ

U o

oo

C

CV

2

2

oo UCV

2

20

The difference in the energy can be explained as the work done

by the capacitor.

• Test Your Understanding (4)

Two capacitors are connected in series as

shown in the figure. A dielectric material of

constant is inserted between the plates of

C1. Which of the following is correct?

a) Both C1 and C2 will increase. b) Both Q1 and Q2 will increase.

c) C1 will increase while C2 will

decrease. d) Q1 will increase while Q2 will

not change.

C1

C2

Example 26.9 A

parallel-plate capacitor has

a plate separation d and

plate area A. An uncharged

metallic slab of thickness a

is inserted midway between

the plates.

a) Find the capacitance of the device.

b) Show that the capacitance of the original capacitor

is unaffected by the insertion of the metallic slab if the

slab is infinitesimally thin.

Solution

The metal slab devide the capacitor into two capacitors. Hence,

the capacitor is equivalent to two capacitors in series, each having

a plate separation (d - a)/2

+ + + + + + + + + + + +

- - - - - - - - - - - - - - -

+ + + + + + + + + + + + + + + +

- - - - - - - - - - - - - - -

d a

ad

ACC o

12

1

b) Letting a→0 in the result of part (a)

d

A

ad

AC oo

a

0lim

21

111

CCC

2

with 21 ad

ACC o

Example 26.10 A parallel

plate capacitor has a plate

separation d has a

capacitance Co in the absence

of the dielectric. What is the

capacitance when a slab of

dielectric material of dielectric

constant and thickness is

inserted between the plates.

d31

+ + + + + + + + + + + +

- - - - - - - - - - - - - - -

d

d31

Solution

The capacitor is equivalent to two capacitors in series each with

31

d

AC o

and 32

2d

AC o

A

d

A

d

CCC oo

323111

21 d

AC

d

AC o

oo

12

3

Problem 27+28 Find the equivalent capacitance between a

and b for the combination shown. You are given C1 = 5F,

C2 = 10F and C3 = 2F. If Vab =60V find the charge on C3.

20

3.3

C1

Parallel

Series Series

C1

C2 C2

C2 C2

C3

Parallel

3.3 8.6

20

Series 6

a

b

60 V

360 C

360 C

360 C 42 V 42 V

84 C