2.1 INTRODUCTION

In Chapters 6 and 8 (Class XI), the notion of potential energy was

introduced. When an external force does work in taking a body from a

point to another against a force like spring force or gravitational force,

that work gets stored as potential energy of the body. When the external

force is removed, the body moves, gaining kinetic energy and losing

an equal amount of potential energy. The sum of kinetic and

potential energies is thus conserved. Forces of this kind are called

conservative forces. Spring force and gravitational force are examples of

conservative forces.

Coulomb force between two (stationary) charges is also a conservative

force. This is not surprising, since both have inverse-square dependence

on distance and differ mainly in the proportionality constants – the

masses in the gravitational law are replaced by charges in Coulomb’s

law. Thus, like the potential energy of a mass in a gravitational

field, we can define electrostatic potential energy of a charge in an

electrostatic field.

Consider an electrostatic field E due to some charge configuration.

First, for simplicity, consider the field E due to a charge Q placed at the

origin. Now, imagine that we bring a test charge q from a point R to a

point P against the repulsive force on it due to the charge Q. With reference

Chapter Two

ELECTROSTATIC

POTENTIAL AND

CAPACITANCE

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Physics

to Fig. 2.1, this will happen if Q and q are both positive

or both negative. For definiteness, let us take Q, q > 0.

Two remarks may be made here. First, we assume

that the test charge q is so small that it does not disturb

the original configuration, namely the charge Q at the

origin (or else, we keep Q fixed at the origin by some

unspecified force). Second, in bringing the charge q from

R to P, we apply an external force F

ext

just enough to

counter the repulsive electric force F

(i.e, F

ext

= –F

This means there is no net force on or acceleration of

the charge q when it is brought from R to P, i.e., it is

brought with infinitesimally slow constant speed. In

this situation, work done by the external force is the negative of the work

done by the electric force, and gets fully stored in the form of potential

energy of the charge q. If the external force is removed on reaching P, the

electric force will take the charge away from Q – the stored energy (potential

energy) at P is used to provide kinetic energy to the charge q in such a

way that the sum of the kinetic and potential energies is conserved.

Thus, work done by external forces in moving a charge q from R to P is

F r

ext

∫

−

∫

F r

(2.1)

This work done is against electrostatic repulsive force and gets stored

as potential energy.

At every point in electric field, a particle with charge q possesses a

certain electrostatic potential energy, this work done increases its potential

energy by an amount equal to potential energy difference between points

R and P.

Thus, potential energy difference

P R RP

U U U W

∆ = − =

(2.2)

(Note here that this displacement is in an opposite sense to the electric

force and hence work done by electric field is negative, i.e., –W

Therefore, we can define electric potential energy difference between

two points as the work required to be done by an external force in moving

(without accelerating) charge q from one point to another for electric field

of any arbitrary charge configuration.

Two important comments may be made at this stage:

(i) The right side of Eq. (2.2) depends only on the initial and final positions

of the charge. It means that the work done by an electrostatic field in

moving a charge from one point to another depends only on the initial

and the final points and is independent of the path taken to go from

one point to the other. This is the fundamental characteristic of a

conservative force. The concept of the potential energy would not be

meaningful if the work depended on the path. The path-independence

of work done by an electrostatic field can be proved using the

Coulomb’s law. We omit this proof here.

FIGURE 2.1 A test charge q (> 0) is

moved from the point R to the

point P against the repulsive

force on it by the charge Q (> 0)

placed at the origin.

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Electrostatic Potential

and Capacitance

(ii) Equation (2.2) defines potential energy difference in terms

of the physically meaningful quantity work. Clearly,

potential energy so defined is undetermined to within an

additive constant.What this means is that the actual value

of potential energy is not physically significant; it is only

the difference of potential energy that is significant. We can

always add an arbitrary constant

to potential energy at

every point, since this will not change the potential energy

difference:

( ) ( )

P R P R

U U U U

α α

+ − + = −

Put it differently, there is a freedom in choosing the point

where potential energy is zero. A convenient choice is to have

electrostatic potential energy zero at infinity. With this choice,

if we take the point R at infinity, we get from Eq. (2.2)

P P P

W U U U

∞ ∞

= − =

(2.3)

Since the point P is arbitrary, Eq. (2.3) provides us with a

definition of potential energy of a charge q at any point.

Potential energy of charge q at a point (in the presence of field

due to any charge configuration) is the work done by the

external force (equal and opposite to the electric force) in

bringing the charge q from infinity to that point.

2.2 ELECTROSTATIC POTENTIAL

Consider any general static charge configuration. We define

potential energy of a test charge q in terms of the work done

on the charge q. This work is obviously proportional to q, since

the force at any point is qE, where E is the electric field at that

point due to the given charge configuration. It is, therefore,

convenient to divide the work by the amount of charge q, so

that the resulting quantity is independent of q. In other words,

work done per unit test charge is characteristic of the electric

field associated with the charge configuration. This leads to

the idea of electrostatic potential V due to a given charge

configuration. From Eq. (2.1), we get:

Work done by external force in bringing a unit positive

charge from point R to P

= V

– V

−













U U

P R

(2.4)

where V

and V

are the electrostatic potentials at P and R, respectively.

Note, as before, that it is not the actual value of potential but the potential

difference that is physically significant. If, as before, we choose the

potential to be zero at infinity, Eq. (2.4) implies:

Work done by an external force in bringing a unit positive charge

from infinity to a point = electrostatic potential (V) at that point.

COUNT ALESSANDRO VOLTA (1745 –1827)

Count Alessandro Volta

(1745 – 1827) Italian

physicist, professor at

Pavia. Volta established

that the animal electri-

city observed by Luigi

Galvani, 1737–1798, in

experiments with frog

muscle tissue placed in

contact with dissimilar

metals, was not due to

any exceptional property

of animal tissues but

was also generated

whenever any wet body

was sandwiched between

dissimilar metals. This

led him to develop the

first voltaic pile, or

battery, consisting of a

large stack of moist disks

of cardboard (electro-

lyte) sandwiched

between disks of metal

(electrodes).

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In other words, the electrostatic potential (V)

at any point in a region with electrostatic field is

the work done in bringing a unit positive

charge (without acceleration) from infinity to

that point.

The qualifying remarks made earlier regarding

potential energy also apply to the definition of

potential. To obtain the work done per unit test

charge, we should take an infinitesimal test charge

q, obtain the work done

W in bringing it from

infinity to the point and determine the ratio

q. Also, the external force at every point of the

path is to be equal and opposite to the electrostatic

force on the test charge at that point.

2.3 POTENTIAL DUE TO A POINT CHARGE

Consider a point charge Q at the origin (Fig. 2.3). For definiteness, take Q

to be positive. We wish to determine the potential at any point P with

position vector r from the origin. For that we must

calculate the work done in bringing a unit positive

test charge from infinity to the point P. For Q > 0,

the work done against the repulsive force on the

test charge is positive. Since work done is

independent of the path, we choose a convenient

path – along the radial direction from infinity to

the point P.

At some intermediate point P′ on the path, the

electrostatic force on a unit positive charge is

4 '

′

(2.5)

where

′

is the unit vector along OP′. Work done

against this force from r′ to r′ + ∆r′ is

4 '

W r

∆ = − ∆

′

(2.6)

The negative sign appears because for ∆r′

< 0, ∆W is positive. Total

work done (W) by the external force is obtained by integrating Eq. (2.6)

from r

′

= ∞ to r

′

= r,

= −

′

′ =

′

∞

∫

4 4 4

0 0

π π π

ε ε ε

(2.7)

This, by definition is the potential at P due to the charge Q

( )

V r

(2.8)

FIGURE 2.2 Work done on a test charge q

by the electrostatic field due to any given

charge configuration is independent

of the path, and depends only on

its initial and final positions.

FIGURE 2.3 Work done in bringing a unit

positive test charge from infinity to the

point P, against the repulsive force of

charge Q (Q > 0), is the potential at P due to

the charge Q.

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Electrostatic Potential

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EXAMPLE 2.1

Equation (2.8) is true for any

sign of the charge Q, though we

considered Q > 0 in its derivation.

For Q < 0, V < 0, i.e., work done (by

the external force) per unit positive

test charge in bringing it from

infinity to the point is negative. This

is equivalent to saying that work

done by the electrostatic force in

bringing the unit positive charge

form infinity to the point P is

positive. [This is as it should be,

since for Q < 0, the force on a unit

positive test charge is attractive, so

that the electrostatic force and the

displacement (from infinity to P) are

in the same direction.] Finally, we

note that Eq. (2.8) is consistent with

the choice that potential at infinity

be zero.

Figure (2.4) shows how the electrostatic potential (

∝

1/r) and the

electrostatic field (

∝

1/r

) varies with r.

Example 2.1

(a) Calculate the potential at a point P due to a charge of 4 × 10

–7

located 9 cm away.

(b) Hence obtain the work done in bringing a charge of 2 × 10

–9

from infinity to the point P. Does the answer depend on the path

along which the charge is brought?

Solution

(a)

= 4 × 10

(b) W = qV = 2 × 10

–9

C × 4 × 10

= 8 × 10

–5

No, work done will be path independent. Any arbitrary infinitesimal

path can be resolved into two perpendicular displacements: One along

r and another perpendicular to r. The work done corresponding to

the later will be zero.

2.4 POTENTIAL DUE TO AN ELECTRIC DIPOLE

As we learnt in the last chapter, an electric dipole consists of two charges

q and –q separated by a (small) distance 2a. Its total charge is zero. It is

characterised by a dipole moment vector p whose magnitude is q × 2a

and which points in the direction from –q to q (Fig. 2.5). We also saw that

the electric field of a dipole at a point with position vector r depends not

just on the magnitude r, but also on the angle between

r and p. Further,

FIGURE 2.4 Variation of potential V with r [in units of

(Q/4π

) m

-1

] (blue curve) and field with r [in units

of (Q/4π

) m

-2

] (black curve) for a point charge Q.

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the field falls off, at large distance, not as

1/r

(typical of field due to a single charge)

but as 1/r

. We, now, determine the electric

potential due to a dipole and contrast it

with the potential due to a single charge.

As before, we take the origin at the

centre of the dipole. Now we know that the

electric field obeys the superposition

principle. Since potential is related to the

work done by the field, electrostatic

potential also follows the superposition

principle. Thus, the potential due to the

dipole is the sum of potentials due to the

charges q and –q

= −













0 1 2

(2.9)

where r

and r

are the distances of the

point P from q and –q, respectively.

Now, by geometry,

2 2 2

r r a ar

= + −

cos

2 2 2

r r a ar

= + +

cos

(2.10)

We take r much greater than a (

) and retain terms only upto

the first order in a/r

r r

2 2

= − +













cos

≅ −













2 cos

(2.11)

Similarly,

r r

2 2

≅ +













cos

(2.12)

Using the Binomial theorem and retaining terms upto the first order

in a/r ; we obtain,

1 1

2 1

1 2

r r

≅ −













≅ +













−

cos

[2.13(a)]

1 1

2 1

1 2

r r

≅ +













≅ −













−

cos

[2.13(b)]

Using Eqs. (2.9) and (2.13) and p = 2qa, we get

q a

= =

4 4

π π

θ θ

2 cos cos

(2.14)

Now, p cos

p.r

FIGURE 2.5 Quantities involved in the calculation

of potential due to a dipole.

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Electrostatic Potential

and Capacitance

where

is the unit vector along the position vector OP.

The electric potential of a dipole is then given by

p.r

; (r >> a) (2.15)

Equation (2.15) is, as indicated, approximately true only for distances

large compared to the size of the dipole, so that higher order terms in

a/r are negligible. For a point dipole p at the origin, Eq. (2.15) is, however,

exact.

From Eq. (2.15), potential on the dipole axis (

= 0, π ) is given by

= ±

(2.16)

(Positive sign for

= 0, negative sign for

= π.) The potential in the

equatorial plane (

= π/2) is zero.

The important contrasting features of electric potential of a dipole

from that due to a single charge are clear from Eqs. (2.8) and (2.15):

(i) The potential due to a dipole depends not just on r but also on the

angle between the position vector r and the dipole moment vector p.

(It is, however, axially symmetric about p. That is, if you rotate the

position vector r about p, keeping

fixed, the points corresponding

to P on the cone so generated will have the same potential as at P.)

(ii) The electric dipole potential falls off, at large distance, as 1/r

, not as

1/r, characteristic of the potential due to a single charge. (You can

refer to the Fig. 2.5 for graphs of 1/r

versus r and 1/r versus r,

drawn there in another context.)

2.5 POTENTIAL DUE TO A SYSTEM OF CHARGES

Consider a system of charges q

, q

,…, q

with position vectors r

, r

,…,

relative to some origin (Fig. 2.6). The potential V

at P due to the charge

0 1P

where r

is the distance between q

and P.

Similarly, the potential V

at P due to q

and

due to q

are given by

0 2P

0 3P

where r

and r

are the distances of P from

charges q

and q

, respectively; and so on for the

potential due to other charges. By the

superposition principle, the potential V at P due

to the total charge configuration is the algebraic

sum of the potentials due to the individual

charges

V = V

+ V

+ ... + V

(2.17)

FIGURE 2.6 Potential at a point due to a

system of charges is the sum of potentials

due to individual charges.

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EXAMPLE

2.2

= + + +













nP P P

......

(2.18)

If we have a continuous charge distribution characterised by a charge

density

(r), we divide it, as before, into small volume elements each of

size ∆v and carrying a charge

∆v. We then determine the potential due

to each volume element and sum (strictly speaking , integrate) over all

such contributions, and thus determine the potential due to the entire

distribution.

We have seen in Chapter 1 that for a uniformly charged spherical shell,

the electric field outside the shell is as if the entire charge is concentrated

at the centre. Thus, the potential outside the shell is given by

( )

r R

≥

[2.19(a)]

where q is the total charge on the shell and R its radius. The electric field

inside the shell is zero. This implies (Section 2.6) that potential is constant

inside the shell (as no work is done in moving a charge inside the shell),

and, therefore, equals its value at the surface, which is

[2.19(b)]

Example 2.2 Two charges 3 × 10

–8

C and –2 × 10

–8

C are located

15 cm apart. At what point on the line joining the two charges is the

electric potential zero? Take the potential at infinity to be zero.

Solution Let us take the origin O at the location of the positive charge.

The line joining the two charges is taken to be the x-axis; the negative

charge is taken to be on the right side of the origin (Fig. 2.7).

FIGURE 2.7

Let P be the required point on the x-axis where the potential is zero.

If x is the x-coordinate of P, obviously x must be positive. (There is no

possibility of potentials due to the two charges adding up to zero for

x < 0.) If x lies between O and A, we have

3 10

2 10

15 10

−

− ×













–

( )x x

where x is in cm. That is,

3 2

15x x

− =

−

which gives x = 9 cm.

If x lies on the extended line OA, the required condition is

3 2

15x x

− =

−

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Electrostatic Potential

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EXAMPLE 2.2

which gives

x = 45 cm

Thus, electric potential is zero at 9 cm and 45 cm away from the

positive charge on the side of the negative charge. Note that the

formula for potential used in the calculation required choosing

potential to be zero at infinity.

Example 2.3 Figures 2.8 (a) and (b) show the field lines of a positive

and negative point charge respectively.

FIGURE 2.8

(a) Give the signs of the potential difference V

– V

; V

– V

(b) Give the sign of the potential energy difference of a small negative

charge between the points Q and P; A and B.

positive charge from Q to P.

(d) Give the sign of the work done by the external agency in moving

a small negative charge from B to A.

(e) Does the kinetic energy of a small negative charge increase or

decrease in going from B to A?

Solution

(a) As

∝

, V

> V

. Thus, (V

– V

) is positive. Also V

is less negative

than V

. Thus, V

> V

or (V

– V

) is positive.

(b) A small negative charge will be attracted towards positive charge.

The negative charge moves from higher potential energy to lower

potential energy. Therefore the sign of potential energy difference

of a small negative charge between Q and P is positive.

Similarly, (P.E.)

> (P.E.)

and hence

sign of potential energy

differences is positive.

done by an external agency against the electric field. Therefore,

work done by the field is negative.

(d) In moving a small negative charge from B to A work has to be

done by the external agency. It is positive.

(e) Due to force of repulsion on the negative charge, velocity decreases

and hence the kinetic energy decreases in going from B to A.

EXAMPLE 2.3

Electric potential, equipotential surfaces:

http://video.mit.edu/watch/4-electrostatic-potential-elctric-energy-ev-conservative-field-

equipotential-sufaces-12584/

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Physics

FIGURE 2.10 Equipotential surfaces for a uniform electric field.

2.6 EQUIPOTENTIAL SURFACES

An equipotential surface is a surface with a constant value of potential

at all points on the surface. For a single charge q, the potential is given

by Eq. (2.8):

This shows that V is a constant if r is constant. Thus, equipotential

surfaces of a single point charge are concentric spherical surfaces centred

at the charge.

Now the electric field lines for a single charge q are radial lines starting

from or ending at the charge, depending on whether q is positive or negative.

Clearly, the electric field at every point is normal to the equipotential surface

passing through that point. This is true in general: for any charge

configuration, equipotential surface through a point is normal to the

electric field at that point. The proof of this statement is simple.

If the field were not normal to the equipotential surface, it would

have non-zero component along the surface. To move a unit test charge

against the direction of the component of the field, work would have to

be done. But this is in contradiction to the definition of an equipotential

surface: there is no potential difference between any two points on the

surface and no work is required to move a test charge on the surface.

The electric field must, therefore, be normal to the equipotential surface

at every point. Equipotential surfaces offer an alternative visual picture

in addition to the picture of electric field lines around a charge

configuration.

FIGURE 2.9 For a

single charge q

(a) equipotential

surfaces are

spherical surfaces

centred at the

charge, and

(b) electric field

lines are radial,

starting from the

charge if q > 0.

For a uniform electric field E, say, along the

-axis, the equipotential

surfaces are planes normal to the

-axis, i.e., planes parallel to the y-z

plane (Fig. 2.10). Equipotential surfaces for (a) a dipole and (b) two

identical positive charges are shown in Fig. 2.11.

FIGURE 2.11 Some equipotential surfaces for (a) a dipole,

(b) two identical positive charges.

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Electrostatic Potential

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2.6.1 Relation between field and potential

Consider two closely spaced equipotential surfaces A and B (Fig. 2.12)

with potential values V and V +

V, where

V is the change in V in the

direction of the electric field E. Let P be a point on the

surface B.

l is the perpendicular distance of the

surface A from P. Imagine that a unit positive charge

is moved along this perpendicular from the surface B

to surface A against the electric field. The work done

in this process is |E|

This work equals the potential difference

–V

Thus,

|E|

l = V – (V +

V)=

–

i.e., |E|=

−

(2.20)

Since

V is negative,

V = – |

V|. we can rewrite

Eq (2.20) as

E = − = +

(2.21)

We thus arrive at two important conclusions concerning the relation

between electric field and potential:

(i) Electric field is in the direction in which the potential decreases

steepest.

(ii) Its magnitude is given by the change in the magnitude of potential

per unit displacement normal to the equipotential surface at the point.

2.7 POTENTIAL ENERGY OF A SYSTEM OF CHARGES

Consider first the simple case of two charges q

and q

with position vector

and r

relative to some origin. Let us calculate the work done

(externally) in building up this configuration. This means that we consider

the charges q

and q

initially at infinity and determine the work done by

an external agency to bring the charges to the given locations. Suppose,

first the charge q

is brought from infinity to the point r

. There is no

external field against which work needs to be done, so work done in

bringing q

from infinity to r

is zero. This charge produces a potential in

space given by

where r

is the distance of a point P in space from the location of q

From the definition of potential, work done in bringing charge q

from

infinity to the point r

is q

times the potential at r

due to q

work done on q

1 2

q q

FIGURE 2.12 From the

potential to the field.

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where r

is the distance between points 1 and 2.

Since electrostatic force is conservative, this work gets

stored in the form of potential energy of the system. Thus,

the potential energy of a system of two charges q

and q

q q

1 2

(2.22)

Obviously, if q

was brought first to its present location and

brought later, the potential energy U would be the same.

More generally, the potential energy expression,

Eq. (2.22), is unaltered whatever way the charges are brought to the specified

locations, because of path-independence of work for electrostatic force.

Equation (2.22) is true for any sign of q

and q

. If

> 0, potential

energy is positive. This is as expected, since for like charges (q

> 0),

electrostatic force is repulsive and a positive amount of work is needed to

be done against this force to bring the charges from infinity to a finite

distance apart. For unlike charges (q

< 0), the electrostatic force is

attractive. In that case, a positive amount of work is needed against this

force to take the charges from the given location to infinity. In other words,

a negative amount of work is needed for the reverse path (from infinity to

the present locations), so the potential energy is negative.

Equation (2.22) is easily generalised for a system of any number of

point charges. Let us calculate the potential energy of a system of three

charges q

and q

located at r

, r

, respectively. To bring q

first

from infinity to r

, no work is required. Next we bring q

from infinity to

. As before, work done in this step is

1 2

2 1 2

0 12

( )

q q

q V

(2.23)

The charges q

and q

produce a potential, which at any point P is

given by

1 2

= +













P P

(2.24)

Work done next in bringing q

from infinity to the point r

is q

times

1, 2

at r

q V

q q

3 1 2 3

1 3

2 3

( )r = +













(2.25)

The total work done in assembling the charges

at the given locations is obtained by adding the work

done in different steps [Eq. (2.23) and Eq. (2.25)],

q q

= + +













1 2

1 3

2 3

(2.26)

Again, because of the conservative nature of the

electrostatic force (or equivalently, the path

independence of work done), the final expression for

U, Eq. (2.26), is independent of the manner in which

the configuration is assembled. The potential energy

FIGURE 2.13 Potential energy of a

system of charges q

and q

directly proportional to the product

of charges and inversely to the

distance between them.

FIGURE 2.14 Potential energy of a

system of three charges is given by

Eq. (2.26), with the notation given

in the figure.

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Electrostatic Potential

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EXAMPLE 2.4

is characteristic of the present state of configuration, and not the way

the state is achieved.

Example 2.4 Four charges are arranged at the corners of a square

ABCD of side d, as shown in Fig. 2.15.(a) Find the work required to

put together this arrangement. (b) A charge q

is brought to the centre

E of the square, the four charges being held fixed at its corners. How

much extra work is needed to do this?

FIGURE 2.15

Solution

(a) Since the work done depends on the final arrangement of the

charges, and not on how they are put together, we calculate work

needed for one way of putting the charges at A, B, C and D. Suppose,

first the charge +q is brought to A, and then the charges –q, +q, and

–q are brought to B, C and D, respectively. The total work needed can

be calculated in steps:

(i) Work needed to bring charge +q to A when no charge is present

elsewhere: this is zero.

(ii) Work needed to bring –q to B when +q is at A. This is given by

(charge at B) × (electrostatic potential at B due to charge +q at A)

= − ×













= −q

d4 4

π π

ε ε

(iii) Work needed to bring charge +q to C when +q is at A and –q is at

B. This is given by (charge at C) × (potential at C due to charges

at A and B)

= +

−













4 2

−













(iv) Work needed to bring –q to D when +q at A,–q at B, and +q at C.

This is given by (charge at D) × (potential at D due to charges at A,

B and C)

= −

−













4 2

−













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Physics

EXAMPLE 2.4

Add the work done in steps (i), (ii), (iii) and (iv). The total work

required is

−

+ + −













+ −

























0 1 1

( ) ( )

−

( )

4 2

The work done depends only on the arrangement of the charges, and

not how they are assembled. By definition, this is the total

electrostatic energy of the charges.

(Students may try calculating same work/energy by taking charges

in any other order they desire and convince themselves that the energy

will remain the same.)

(b) The extra work necessary to bring a charge q

to the point E when

the four charges are at A, B, C and D is q

× (electrostatic potential at

E due to the charges at A, B, C and D). The electrostatic potential at

E is clearly zero since potential due to A and C is cancelled by that

due to B and D. Hence, no work is required to bring any charge to

point E.

2.8 POTENTIAL

ENERGY IN AN EXTERNAL

FIELD

2.8.1 Potential energy of a single charge

In Section 2.7, the source of the electric field was specified – the charges

and their locations - and the potential energy of the system of those charges

was determined. In this section, we ask a related but a distinct question.

What is the potential energy of a charge q in a given field? This question

was, in fact, the starting point that led us to the notion of the electrostatic

potential (Sections 2.1 and 2.2). But here we address this question again

to clarify in what way it is different from the discussion in Section 2.7.

The main difference is that we are now concerned with the potential

energy of a charge (or charges) in an external field. The external field E is

not produced by the given charge(s) whose potential energy we wish to

calculate. E is produced by sources external to the given charge(s).The

external sources may be known, but often they are unknown or

unspecified; what is specified is the electric field E or the electrostatic

potential V due to the external sources. We assume that the charge q

does not significantly affect the sources producing the external field. This

is true if q is very small, or the external sources are held fixed by other

unspecified forces. Even if q is finite, its influence on the external sources

may still be ignored in the situation when very strong sources far away

at infinity produce a finite field E in the region of interest. Note again that

we are interested in determining the potential energy of a given charge q

(and later, a system of charges) in the external field; we are not interested

in the potential energy of the sources producing the external electric field.

The external electric field E and the corresponding external potential

V may vary from point to point. By definition, V at a point P is the work

done in bringing a unit positive charge from infinity to the point P.

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Electrostatic Potential

and Capacitance

EXAMPLE 2.5

(We continue to take potential at infinity to be zero.) Thus, work done in

bringing a charge q from infinity to the point P in the external field is qV.

This work is stored in the form of potential energy of q. If the point P has

position vector r relative to some origin, we can write:

Potential energy of q at r in an external field

= qV(r) (2.27)

where V(r) is the external potential at the point r.

Thus, if an electron with charge q = e = 1.6×10

–19

C is accelerated by

a potential difference of ∆V = 1 volt, it would gain energy of q∆V = 1.6 ×

–19

J. This unit of energy is defined as 1 electron volt or 1eV, i.e.,

1 eV=1.6 × 10

–19

J. The units based on eV are most commonly used in

atomic, nuclear and particle physics, (1 keV = 10

eV = 1.6 × 10

–16

J, 1 MeV

= 10

eV = 1.6 × 10

–13

J, 1 GeV = 10

eV = 1.6 × 10

–10

J and 1 TeV = 10

= 1.6 × 10

–7

J). [This has already been defined on Page 117, XI Physics

Part I, Table 6.1.]

2.8.2 Potential energy of a system of two charges in an

external field

Next, we ask: what is the potential energy of a system of two charges q

and q

located at r

and r

, respectively, in an external field? First, we

calculate the work done in bringing the charge q

from infinity to r

Work done in this step is q

V(r

), using Eq. (2.27). Next, we consider the

work done in bringing q

to r

. In this step, work is done not only against

the external field E but also against the field due to q

Work done on q

against the external field

= q

V (r

)

Work done on q

against the field due to q

1 2

q q

where r

is the distance between q

and q

. We have made use of Eqs.

(2.27) and (2.22). By the superposition principle for fields, we add up

the work done on q

against the two fields (E and that due to q

Work done in bringing q

to r

1 2

2 2

( )

q q

q V

= +

(2.28)

Thus,

Potential energy of the system

= the total work done in assembling the configuration

1 2

1 1 2 2

0 12

( ) ( )

q q

q V q V

= + +

r r

(2.29)

Example 2.5

(a) Determine the electrostatic potential energy of a system consisting

of two charges 7 µC and –2 µC (and with no external field) placed

at (–9 cm, 0, 0) and (9 cm, 0, 0) respectively.

(b) How much work is required to separate the two charges infinitely

away from each other?

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Physics

EXAMPLE 2.5

external electric field E = A (1/r

); A = 9 × 10

–1

. What would

the electrostatic energy of the configuration be?

Solution

(a)

1 2

1 7 ( 2) 10

9 10

4 0.18

q q

−

× − ×

= = × ×

= –0.7 J.

(b) W = U

– U

= 0 – U = 0 – (–0.7) = 0.7 J.

unchanged. In addition, there is the energy of interaction of the

two charges with the external electric field. We find,

( ) ( )

1 1 2 2

7 C 2 C

0.09m 0.09m

q V q V A A

µ − µ

+ = +r r

and the net electrostatic energy is

( ) ( )

1 2

1 1 2 2

0 12

7 C 2 C

0.7 J

4 0.09 m 0.09m

q q

q V q V A A

µ − µ

+ + = + −

r r

70 20 0.7 49.3 J

= − − =

2.8.3 Potential energy of a dipole in an external field

Consider a dipole with charges q

= +q and q

= –q placed in a uniform

electric field E, as shown in Fig. 2.16.

As seen in the last chapter, in a uniform electric field,

the dipole experiences no net force; but experiences a

torque

τ τ

τ given by

τ = τ =

τ = p × E (2.30)

which will tend to rotate it (unless p is parallel or

antiparallel to E). Suppose an external torque

ττ

ext

applied in such a manner that it just neutralises this

torque and rotates it in the plane of paper from angle

to angle

at an infinitesimal angular speed and without

angular acceleration. The amount of work done by the

external torque will be given by

(

)

cos cospE

θ θ

0 1

= −

(2.31)

This work is stored as the potential energy of the system. We can

then associate potential energy U(

) with an inclination

of the dipole.

Similar to other potential energies, there is a freedom in choosing the

angle where the potential energy U is taken to be zero. A natural choice

is to take

= π / 2. (Αn explanation for it is provided towards the end of

discussion.) We can then write,

(2.32)

FIGURE 2.16 Potential energy of a

dipole in a uniform external field.

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and Capacitance

EXAMPLE 2.6

This expression can alternately be understood also from Eq. (2.29).

We apply Eq. (2.29) to the present system of two charges +q and –q. The

potential energy expression then reads

( )

(

)

(

)

1 2

[ ]

4 2

U q V V

= − −′

π ×

r r

(2.33)

Here, r

and r

denote the position vectors of +q and –q. Now, the

potential difference between positions r

and r

equals the work done

in bringing a unit positive charge against field from r

to r

. The

displacement parallel to the force is 2a cos

. Thus, [V(r

)–V (r

)] =

–E × 2a cos

. We thus obtain,

( )

2 2

cos

4 2 4 2

θ θ

ε ε

0 0

= − − = − −′

π × π ×

p.E

q q

U pE

a a

(2.34)

We note that U

′

(

) differs from U(

) by a quantity which is just a constant

for a given dipole. Since a constant is insignificant for potential energy, we

can drop the second term in Eq. (2.34) and it then reduces to Eq. (2.32).

We can now understand why we took

=π/2. In this case, the work

done against the external field E in bringing +q and – q are equal and

opposite and cancel out, i.e., q [V (r

) – V (r

)]=0.

Example 2.6 A molecule of a substance has a permanent electric

dipole moment of magnitude 10

–29

C m. A mole of this substance is

polarised (at low temperature) by applying a strong electrostatic field

of magnitude 10

V m

–1

. The direction of the field is suddenly changed

by an angle of 60º. Estimate the heat released by the substance in

aligning its dipoles along the new direction of the field. For simplicity,

assume 100% polarisation of the sample.

Solution Here, dipole moment of each molecules = 10

–29

C m

As 1 mole of the substance contains 6 × 10

molecules,

total dipole moment of all the molecules, p = 6 × 10

× 10

–29

C m

= 6 × 10

–6

C m

Initial potential energy, U

= –pE cos

= –6×10

–6

×10

cos 0° = –6 J

Final potential energy (when

= 60°), U

= –6 × 10

–6

× 10

cos 60° = –3 J

Change in potential energy = –3 J – (–6J) = 3 J

So, there is loss in potential energy. This must be the energy released

by the substance in the form of heat in aligning its dipoles.

2.9 ELECTROSTATICS OF CONDUCTORS

Conductors and insulators were described briefly in Chapter 1.

Conductors contain mobile charge carriers. In metallic conductors, these

charge carriers are electrons. In a metal, the outer (valence) electrons

part away from their atoms and are free to move. These electrons are free

within the metal but not free to leave the metal. The free electrons form a

kind of ‘gas’; they collide with each other and with the ions, and move

randomly in different directions. In an external electric field, they drift

against the direction of the field. The positive ions made up of the nuclei

and the bound electrons remain held in their fixed positions. In electrolytic

conductors, the charge carriers are both positive and negative ions; but

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Physics

the situation in this case is more involved – the movement of the charge

carriers is affected both by the external electric field as also by the

so-called chemical forces (see Chapter 3). We shall restrict our discussion

to metallic solid conductors. Let us note important results regarding

electrostatics of conductors.

1. Inside a conductor, electrostatic field is zero

Consider a conductor, neutral or charged. There may also be an external

electrostatic field. In the static situation, when there is no current inside

or on the surface of the conductor, the electric field is zero everywhere

inside the conductor. This fact can be taken as the defining property of a

conductor. A conductor has free electrons. As long as electric field is not

zero, the free charge carriers would experience force and drift. In the

static situation, the free charges have so distributed themselves that the

electric field is zero everywhere inside. Electrostatic field is zero inside a

conductor.

2. At the surface of a charged conductor, electrostatic field

must be normal to the surface at every point

If E were not normal to the surface, it would have some non-zero

component along the surface. Free charges on the surface of the conductor

would then experience force and move. In the static situation, therefore,

E should have no tangential component. Thus electrostatic field at the

surface of a charged conductor must be normal to the surface at every

point. (For a conductor without any surface charge density, field is zero

even at the surface.) See result 5.

3. The interior of a conductor can have no excess charge in

the static situation

A neutral conductor has equal amounts of positive and negative charges

in every small volume or surface element. When the conductor is charged,

the excess charge can reside only on the surface in the static situation.

This follows from the Gauss’s law. Consider any arbitrary volume element

v inside a conductor. On the closed surface S bounding the volume

element v, electrostatic field is zero. Thus the total electric flux through S

is zero. Hence, by Gauss’s law, there is no net charge enclosed by S. But

the surface S can be made as small as you like, i.e., the volume v can be

made vanishingly small. This means there is no net charge at any point

inside the conductor, and any excess charge must reside at the surface.

4. Electrostatic potential is constant throughout the volume

of the conductor and has the same value (as inside) on

its surface

This follows from results 1 and 2 above. Since E = 0 inside the conductor

and has no tangential component on the surface, no work is done in

moving a small test charge within the conductor and on its surface. That

is, there is no potential difference between any two points inside or on

the surface of the conductor. Hence, the result. If the conductor is charged,

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electric field normal to the surface exists; this means potential will be

different for the surface and a point just outside the surface.

In a system of conductors of arbitrary size, shape and

charge configuration, each conductor is characterised by a constant

value of potential, but this constant may differ from one conductor to

the other.

5. Electric field at the surface of a charged conductor

E n

(2.35)

where σ is the surface charge density and

is a unit vector normal

to the surface in the outward direction.

To derive the result, choose a pill box (a short cylinder) as the Gaussian

surface about any point P on the surface, as shown in Fig. 2.17. The pill

box is partly inside and partly outside the surface of the conductor. It

has a small area of cross section

S and negligible height.

Just inside the surface, the electrostatic field is zero; just outside, the

field is normal to the surface with magnitude E. Thus,

the contribution to the total flux through the pill box

comes only from the outside (circular) cross-section

of the pill box. This equals ± E

S (positive for

> 0,

negative for

< 0), since over the small area

S, E

may be considered constant and E and

S are parallel

or antiparallel. The charge enclosed by the pill box

σδ

By Gauss’s law

S =

σ δ

E =

(2.36)

Including the fact that electric field is normal to the

surface, we get the vector relation, Eq. (2.35), which

is true for both signs of

. For

> 0, electric field is

normal to the surface outward; for

< 0, electric field

is normal to the surface inward.

6. Electrostatic shielding

Consider a conductor with a cavity, with no charges inside the cavity. A

remarkable result is that the electric field inside the cavity is zero, whatever

be the size and shape of the cavity and whatever be the charge on the

conductor and the external fields in which it might be placed. We have

proved a simple case of this result already: the electric field inside a charged

spherical shell is zero. The proof of the result for the shell makes use of

the spherical symmetry of the shell (see Chapter 1). But the vanishing of

electric field in the (charge-free) cavity of a conductor is, as mentioned

above, a very general result. A related result is that even if the conductor

FIGURE 2.17 The Gaussian surface

(a pill box) chosen to derive Eq. (2.35)

for electric field at the surface of a

charged conductor.

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Physics

EXAMPLE 2.7

FIGURE 2.18 The electric field inside a

cavity of any conductor is zero. All

charges reside only on the outer surface

of a conductor with cavity. (There are no

charges placed in the cavity.)

is charged or charges are induced on a neutral

conductor by an external field, all charges reside

only on the outer surface of a conductor with cavity.

The proofs of the results noted in Fig. 2.18 are

omitted here, but we note their important

implication. Whatever be the charge and field

configuration outside, any cavity in a conductor

remains shielded from outside electric influence: the

field inside the cavity is always zero. This is known

as electrostatic shielding. The effect can be made

use of in protecting sensitive instruments from

outside electrical influence. Figure 2.19 gives a

summary of the important electrostatic properties

of a conductor.

Example 2.7

(a) A comb run through one’s dry hair attracts small bits of paper.

Why?

What happens if the hair is wet or if it is a rainy day? (Remember,

a paper does not conduct electricity.)

(b) Ordinary rubber is an insulator. But special rubber tyres of

aircraft are made slightly conducting. Why is this necessary?

ropes touching the ground during motion. Why?

(d) A bird perches on a bare high power line, and nothing happens

to the bird. A man standing on the ground touches the same line

and gets a fatal shock. Why?

Solution

(a) This is because the comb gets charged by friction. The molecules

in the paper gets polarised by the charged comb, resulting in a

net force of attraction. If the hair is wet, or if it is rainy day, friction

between hair and the comb reduces. The comb does not get

charged and thus it will not attract small bits of paper.

FIGURE 2.19 Some important electrostatic properties of a conductor.

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Electrostatic Potential

and Capacitance

EXAMPLE 2.7

(b) To enable them to conduct charge (produced by friction) to the

ground; as too much of static electricity accumulated may result

in spark and result in fire.

(d) Current passes only when there is difference in potential.

2.10 DIELECTRICS AND POLARISATION

Dielectrics are non-conducting substances. In contrast to conductors,

they have no (or negligible number of) charge carriers. Recall from Section

2.9 what happens when a conductor is placed in an

external electric field. The free charge carriers move

and charge distribution in the conductor adjusts

itself in such a way that the electric field due to

induced charges opposes the external field within

the conductor. This happens until, in the static

situation, the two fields cancel each other and the

net electrostatic field in the conductor is zero. In a

dielectric, this free movement of charges is not

possible. It turns out that the external field induces

dipole moment by stretching or re-orienting

molecules of the dielectric. The collective effect of all

the molecular dipole moments is net charges on the

surface of the dielectric which produce a field that

opposes the external field. Unlike in a conductor,

however, the opposing field so induced does not

exactly cancel the external field. It only reduces it.

The extent of the effect depends on the

nature of the dielectric. To understand the

effect, we need to look at the charge

distribution of a dielectric at the

molecular level.

The molecules of a substance may be

polar or non-polar. In a non-polar

molecule, the centres of positive and

negative charges coincide. The molecule

then has no permanent (or intrinsic) dipole

moment. Examples of non-polar molecules

are oxygen (O

) and hydrogen (H

)

molecules which, because of their

symmetry, have no dipole moment. On the

other hand, a polar molecule is one in which

the centres of positive and negative charges

are separated (even when there is no

external field). Such molecules have a

permanent dipole moment. An ionic

molecule such as HCl or a molecule of water

O) are examples of polar molecules.

FIGURE 2.20 Difference in behaviour

of a conductor and a dielectric

in an external electric field.

FIGURE 2.21 Some examples of polar

and non-polar molecules.

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Physics

In an external electric field, the

positive and negative charges of a non-

polar molecule are displaced in opposite

directions. The displacement stops when

the external force on the constituent

charges of the molecule is balanced by

the restoring force (due to internal fields

in the molecule). The non-polar molecule

thus develops an induced dipole moment.

The dielectric is said to be polarised by

the external field. We consider only the

simple situation when the induced dipole

moment is in the direction of the field and

is proportional to the field strength.

(Substances for which this assumption

is true are called linear isotropic

dielectrics.) The induced dipole moments

of different molecules add up giving a net

dipole moment of the dielectric in the

presence of the external field.

A dielectric with polar molecules also

develops a net dipole moment in an

external field, but for a different reason.

In the absence of any external field, the

different permanent dipoles are oriented

randomly due to thermal agitation; so

the total dipole moment is zero. When

an external field is applied, the individual dipole moments tend to align

with the field. When summed overall the molecules, there is then a net

dipole moment in the direction of the external field, i.e., the dielectric is

polarised. The extent of polarisation depends on the relative strength of

two mutually opposite factors: the dipole potential energy in the external

field tending to align the dipoles with the field and thermal energy tending

to disrupt the alignment. There may be, in addition, the ‘induced dipole

moment’ effect as for non-polar molecules, but generally the alignment

effect is more important for polar molecules.

Thus in either case, whether polar or non-polar, a dielectric develops

a net dipole moment in the presence of an external field. The dipole

moment per unit volume is called polarisation and is denoted by P. For

linear isotropic dielectrics,

ε χ

P E

(2.37)

where

is a constant characteristic of the dielectric and is known as the

electric susceptibility of the dielectric medium.

It is possible to relate

to the molecular properties of the substance,

but we shall not pursue that here.

The question is: how does the polarised dielectric modify the original

external field inside it? Let us consider, for simplicity, a rectangular

dielectric slab placed in a uniform external field E

parallel to two of its

faces. The field causes a uniform polarisation P of the dielectric. Thus

FIGURE 2.22 A dielectric develops a net dipole

moment in an external electric field. (a) Non-polar

molecules, (b) Polar molecules.

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Electrostatic Potential

and Capacitance

every volume element ∆v of the slab has a dipole moment

P ∆v in the direction of the field. The volume element ∆v is

macroscopically small but contains a very large number of

molecular dipoles. Anywhere inside the dielectric, the

volume element ∆v has no net charge (though it has net

dipole moment). This is, because, the positive charge of one

dipole sits close to the negative charge of the adjacent dipole.

However, at the surfaces of the dielectric normal to the

electric field, there is evidently a net charge density. As seen

in Fig 2.23, the positive ends of the dipoles remain

unneutralised at the right surface and the negative ends at

the left surface. The unbalanced charges are the induced

charges due to the external field.

Thus, the polarised dielectric is equivalent to two charged

surfaces with induced surface charge densities, say

and –

. Clearly, the field produced by these surface charges

opposes the external field. The total field in the dielectric

is, thereby, reduced from the case when no dielectric is

present. We should note that the surface charge density

arises from bound (not free charges) in the dielectric.

2.11 CAPACITORS AND CAPACITANCE

A capacitor is a system of two conductors separated by an insulator

(Fig. 2.24). The conductors have charges, say Q

and Q

, and potentials

and V

. Usually, in practice, the two conductors have charges Q

and – Q, with potential difference V = V

– V

between them. We shall

consider only this kind of charge configuration of the capacitor. (Even a

single conductor can be used as a capacitor by assuming the other at

infinity.) The conductors may be so charged by connecting them to the

two terminals of a battery. Q is called the charge of the capacitor, though

this, in fact, is the charge on one of the conductors – the total charge of

the capacitor is zero.

The electric field in the region between the

conductors is proportional to the charge Q. That

is, if the charge on the capacitor is, say doubled,

the electric field will also be doubled at every point.

(This follows from the direct proportionality

between field and charge implied by Coulomb’s

law and the superposition principle.) Now,

potential difference V is the work done per unit

positive charge in taking a small test charge from

the conductor 2 to 1 against the field.

Consequently, V is also proportional to Q, and the

ratio Q/V is a constant:

(2.38)

The constant C is called the capacitance of the capacitor. C is independent

of Q or V, as stated above. The capacitance C depends only on the

FIGURE 2.23 A uniformly

polarised dielectric amounts

to induced surface charge

density, but no volume

charge density.

FIGURE 2.24 A system of two conductors

separated by an insulator forms a capacitor.

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Physics

geometrical configuration (shape, size, separation) of the system of two

conductors. [As we shall see later, it also depends on the nature of the

insulator (dielectric) separating the two conductors.] The SI unit of

capacitance is 1 farad (=1 coulomb volt

-1

) or 1 F = 1 C V

–1

. A capacitor

with fixed capacitance is symbolically shown as ---||---, while the one with

variable capacitance is shown as .

Equation (2.38) shows that for large C, V is small for a given Q. This

means a capacitor with large capacitance can hold large amount of charge

Q at a relatively small V. This is of practical importance. High potential

difference implies strong electric field around the conductors. A strong

electric field can ionise the surrounding air and accelerate the charges so

produced to the oppositely charged plates, thereby neutralising the charge

on the capacitor plates, at least partly. In other words, the charge of the

capacitor leaks away due to the reduction in insulating power of the

intervening medium.

The maximum electric field that a dielectric medium can withstand

without break-down (of its insulating property) is called its dielectric

strength; for air it is about 3 × 10

–1

. For a separation between

conductors of the order of 1 cm or so, this field corresponds to a potential

difference of 3 × 10

V between the conductors. Thus, for a capacitor to

store a large amount of charge without leaking, its capacitance should

be high enough so that the potential difference and hence the electric

field do not exceed the break-down limits. Put differently, there is a limit

to the amount of charge that can be stored on a given capacitor without

significant leaking. In practice, a farad is a very big unit; the most common

units are its sub-multiples 1 µF = 10

–6

F, 1 nF = 10

–9

F, 1 pF = 10

–12

etc. Besides its use in storing charge, a capacitor is a key element of most

ac circuits with important functions, as described in Chapter 7.

2.12 THE PARALLEL PLATE CAPACITOR

A parallel plate capacitor consists of two large plane parallel conducting

plates separated by a small distance (Fig. 2.25). We first take the

intervening medium between the plates to be

vacuum. The effect of a dielectric medium between

the plates is discussed in the next section. Let A be

the area of each plate and d the separation between

them. The two plates have charges Q and –Q. Since

d is much smaller than the linear dimension of the

plates (d

<< A), we can use the result on electric

field by an infinite plane sheet of uniform surface

charge density (Section 1.15). Plate 1 has surface

charge density

= Q/A and plate 2 has a surface

charge density –

. Using Eq. (1.33), the electric field

in different regions is:

Outer region I (region above the plate 1),

0 0

2 2

σ σ

ε ε

= − =

(2.39)

FIGURE 2.25 The parallel plate capacitor.

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Electrostatic Potential

and Capacitance

Outer region II (region below the plate 2),

0 0

2 2

σ σ

ε ε

= − =

(2.40)

In the inner region between the plates 1 and 2, the electric fields due

to the two charged plates add up, giving

0 0 0 0

2 2

σ σ σ

ε ε ε ε

= + = =

(2.41)

The direction of electric field is from the positive to the negative plate.

Thus, the electric field is localised between the two plates and is

uniform throughout. For plates with finite area, this will not be true near

the outer boundaries of the plates. The field lines bend outward at the

edges — an effect called ‘fringing of the field’. By the same token,

will

not be strictly uniform on the entire plate. [E and

are related by Eq.

(2.35).] However, for d

<< A, these effects can be ignored in the regions

sufficiently far from the edges, and the field there is given by Eq. (2.41).

Now for uniform electric field, potential difference is simply the electric

field times the distance between the plates, that is,

V E d

= =

(2.42)

The capacitance C of the parallel plate capacitor is then

(2.43)

which, as expected, depends only on the geometry of the system. For

typical values like A = 1 m

, d = 1 mm, we get

12 2 –1 –2 2

8.85 10 C N m 1m

8.85 10 F

10 m

−

× ×

= = ×

(2.44)

(You can check that if 1F= 1C V

–1

= 1C (NC

–1

= 1 C

–1

This shows that 1F is too big a unit in practice, as remarked earlier.

Another way of seeing the ‘bigness’ of 1F is to calculate the area of the

plates needed to have C = 1F for a separation of, say 1 cm:

= =

9 2

12 2 –1 –2

1F 10 m

10 m

8.85 10 C N m

−

(2.45)

which is a plate about 30 km in length and breadth!

2.13 EFFECT OF DIELECTRIC ON CAPACITANCE

With the understanding of the behaviour of dielectrics in an external

field developed in Section 2.10, let us see how the capacitance of a parallel

plate capacitor is modified when a dielectric is present. As before, we

have two large plates, each of area A, separated by a distance d. The

charge on the plates is ±Q, corresponding to the charge density ±

(with

= Q/A). When there is vacuum between the plates,

Factors affecting capacitance, capacitors in action

Interactive Java tutorial

http://micro.magnet.fsu.edu/electromag/java/capacitance/

2020-21

Physics

and the potential difference V

= E

The capacitance C

in this case is

0 0

Q A

V d

= =

(2.46)

Consider next a dielectric inserted between the plates fully occupying

the intervening region. The dielectric is polarised by the field and, as

explained in Section 2.10, the effect is equivalent to two charged sheets

(at the surfaces of the dielectric normal to the field) with surface charge

densities

and –

. The electric field in the dielectric then corresponds

to the case when the net surface charge density on the plates is ±(

–

That is,

σ σ

−

(2.47)

so that the potential difference across the plates is

V E d d

σ σ

−

= =

(2.48)

For linear dielectrics, we expect

to be proportional to E

, i.e., to

Thus, (

–

)

is proportional to

and we can write

σ σ

− =

(2.49)

where K is a constant characteristic of the dielectric. Clearly, K > 1. We

then have

0 0

d Qd

K A K

ε ε

= =

(2.50)

The capacitance C, with dielectric between the plates, is then

V d

= =

(2.51)

The product

K is called the permittivity of the medium and is

denoted by

K (2.52)

For vacuum K = 1 and

;

is called the permittivity of the vacuum.

The dimensionless ratio

(2.53)

is called the dielectric constant of the substance. As remarked before,

from Eq. (2.49), it is clear that K is greater than 1. From Eqs. (2.46) and

(2. 51)

(2.54)

Thus, the dielectric constant of a substance is the factor (>1) by which

the capacitance increases from its vacuum value, when the dielectric is

inserted fully between the plates of a capacitor. Though we arrived at

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Electrostatic Potential

and Capacitance

EXAMPLE 2.8

Eq. (2.54) for the case of a parallel plate capacitor, it holds good for any

type of capacitor and can, in fact, be viewed in general as a definition of

the dielectric constant of a substance.

ELECTRIC DISPLACEMENT

We have introduced the notion of dielectric constant and arrived at Eq. (2.54), without

giving the explicit relation between the induced charge density

and the polarisation P.

We take without proof the result that

•

P n

where

is a unit vector along the outward normal to the surface. Above equation is

general, true for any shape of the dielectric. For the slab in Fig. 2.23, P is along

at the

right surface and opposite to

at the left surface. Thus at the right surface, induced

charge density is positive and at the left surface, it is negative, as guessed already in our

qualitative discussion before. Putting the equation for electric field in vector form

•

−

P n

E n

or (

E + P)

•

The quantity

E + P is called the electric displacement and is denoted by D. It is a

vector quantity. Thus,

D =

E + P, D

•

σ,

The significance of D is this : in vacuum, E is related to the free charge density

When a dielectric medium is present, the corresponding role is taken up by D. For a

dielectric medium, it is D not E that is directly related to free charge density

, as seen in

above equation. Since P is in the same direction as E, all the three vectors P, E and D are

parallel.

The ratio of the magnitudes of D and E is

σε

σ σ

= =

−

Thus,

D =

K E

and P = D –

E =

(K –1)E

This gives for the electric susceptibility

defined in Eq. (2.37)

= (K–1)

Example 2.8 A slab of material of dielectric constant K has the same

area as the plates of a parallel-plate capacitor but has a thickness

(3/4)d, where d is the separation of the plates. How is the capacitance

changed when the slab is inserted between the plates?

Solution Let E

= V

/d be the electric field between the plates when

there is no dielectric and the potential difference is V

. If the dielectric

is now inserted, the electric field in the dielectric will be E = E

/K.

The potential difference will then be

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Physics

EXAMPLE 2.8

1 3

( ) ( )

4 4

V E d d

= +

0 0

1 3 3

( )

4 4 4

E d V

K K

= + =

The potential difference decreases by the factor (K + 3)/4K while the

free charge Q

on the plates remains unchanged. The capacitance

thus increases

0 0

4 4

3 3

Q QK K

C C

V K V K

= = =

+ +

2.14 COMBINATION OF CAPACITORS

We can combine several capacitors of capacitance C

, C

,…, C

to obtain

a system with some effective capacitance C. The effective capacitance

depends on the way the individual capacitors are combined. Two simple

possibilities are discussed below.

2.14.1 Capacitors in series

Figure 2.26 shows capacitors C

and C

combined in series.

The left plate of C

and the right plate of C

are connected to two

terminals of a battery and have charges Q and –Q ,

respectively. It then follows that the right plate of

has charge –Q and the left plate of C

has charge Q.

If this was not so, the net charge on each capacitor

would not be zero. This would result in an electric

field in the conductor connecting C

and C

. Charge

would flow until the net charge on both C

and C

is zero and there is no electric field in the conductor

connecting C

and C

Thus, in the series

combination, charges on the two plates (±Q) are the

same on each capacitor. The total potential drop V

across the combination is the sum of the potential

drops V

and V

across C

and C

respectively.

V = V

+ V

1 2

Q Q

C C

(2.55)

i.e.,

1 2

1 1

Q C C

= +

, (2.56)

Now we can regard the combination as an

effective capacitor with charge Q and potential

difference V. The effective capacitance of the

combination is

(2.57)

We compare Eq. (2.57) with Eq. (2.56), and

obtain

1 2

1 1 1

C C C

= +

(2.58)

FIGURE 2.26 Combination of two

capacitors in series.

FIGURE 2.27 Combination of n

capacitors in series.

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Electrostatic Potential

and Capacitance

EXAMPLE 2.9

The proof clearly goes through for any number of

capacitors arranged in a similar way. Equation (2.55),

for n capacitors arranged in series, generalises to

1 2 n

... ...

Q Q Q

V V V V

C C C

= + + + = + + +

(2.59)

Following the same steps as for the case of two

capacitors, we get the general formula for effective

capacitance of a series combination of n capacitors:

1 2 3 n

1 1 1 1 1

...

C C C C C

= + + + +

(2.60)

2.14.2 Capacitors in parallel

Figure 2.28 (a) shows two capacitors arranged in

parallel. In this case, the same potential difference is

applied across both the capacitors. But the plate charges

(±Q

) on capacitor 1 and the plate charges (±Q

) on the

capacitor 2 are not necessarily the same:

= C

V, Q

= C

V (2.61)

The equivalent capacitor is one with charge

Q = Q

+ Q

(2.62)

and potential difference V.

Q = CV = C

V + C

V (2.63)

The effective capacitance C is, from Eq. (2.63),

C = C

+ C

(2.64)

The general formula for effective capacitance C for

parallel combination of n capacitors [Fig. 2.28 (b)]

follows similarly,

Q = Q

+ Q

+ ... + Q

(2.65)

i.e., CV = C

V + C

V + ... C

V (2.66)

which gives

C = C

+ C

+ ... C

(2.67)

Example 2.9 A network of four 10 µF capacitors is connected to a 500 V

supply, as shown in Fig. 2.29. Determine (a) the equivalent capacitance

of the network and (b) the char

ge on each capacitor. (Note, the charge

on a capacitor is the charge on the plate with higher potential, equal

and opposite to the charge on the plate with lower potential.)

FIGURE 2.28 Parallel combination of

(a) two capacitors, (b) n capacitors.

FIGURE 2.29

2020-21

Physics

EXAMPLE

2.9

Solution

(a) In the given network, C

, C

and C

are connected in series. The

effective capacitance C′ of these three capacitors is given by

1 2 3

1 1 1 1

C C C C

= + +

′

For C

= C

= 10 µF, C

′

= (10/3) µF. The network has C

′

and C

connected in parallel. Thus, the equivalent capacitance C of the

network is

C = C

′

+ C

10+













F =13.3µ

(b) Clearly, from the figure, the charge on each of the capacitors, C

and C

is the same, say Q. Let the charge on C

be Q

′

. Now, since

the potential difference across AB is Q/C

, across BC is Q/C

, across

CD is Q/C

, we have

1 2 3

500 V

Q Q Q

C C C

+ + =

Also, Q

′

= 500 V.

This gives for the given value of the capacitances,

500 F 1.7 10 C

Q V

−

= × µ = ×

and

500 10 F 5.0 10 C

Q V

−

= × µ = ×′

2.15 ENERGY STORED IN A CAPACITOR

A capacitor, as we have seen above, is a system of two conductors with

charge Q and –Q. To determine the energy stored in this configuration,

consider initially two uncharged conductors 1 and 2. Imagine next a

process of transferring charge from conductor 2 to conductor 1 bit by

bit, so that at the end, conductor 1 gets charge Q. By

charge conservation, conductor 2 has charge –Q at

the end (Fig 2.30 ).

In transferring positive charge from conductor 2

to conductor 1, work will be done externally, since at

any stage conductor 1 is at a higher potential than

conductor 2. To calculate the total work done, we first

calculate the work done in a small step involving

transfer of an infinitesimal (i.e., vanishingly small)

amount of charge. Consider the intermediate situation

when the conductors 1 and 2 have charges Q

′

and

–Q

′

respectively. At this stage, the potential difference

′

between conductors 1 to 2 is Q

′

/C, where C is the

capacitance of the system. Next imagine that a small

charge

′

is transferred from conductor 2 to 1. Work

done in this step (

W), resulting in charge Q′ on

conductor 1 increasing to Q

′

, is given by

W V Q Q

δ δ δ

′

= =

′ ′ ′

(2.68)

FIGURE 2.30 (a) Work done in a small

step of building charge on conductor 1

from Q

′

to Q

′

+ δ Q

′

. (b) Total work done

in charging the capacitor may be

viewed as stored in the energy of

electric field between the plates.

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Electrostatic Potential

and Capacitance

Since

′

can be made as small as we like, Eq. (2.68) can be written as

2 2

[( ) ]

W Q Q Q

δ δ

= + −

′ ′ ′

(2.69)

Equations (2.68) and (2.69) are identical because the term of second

order in

′

, i.e.,

′

/2C, is negligible, since

′

is arbitrarily small. The

total work done (W) is the sum of the small work (

W) over the very large

number of steps involved in building the charge Q

′

from zero to Q.

W W=

∑

sum over all steps

2 2

Q Q Q

sum over all steps

∑

′

−

′

[( ) ]

(2.70)

2 2 2

[{ 0} {(2 ) }

Q Q Q

δ δ δ

− + −

′ ′ ′

2 2

{(3 ) (2 ) } ...

Q Q

δ δ

+ − +

′ ′

2 2

{ ( ) }]

+ − −

Q Q Q'

(2.71)

[ 0]

2 2

C C

= − =

(2.72)

The same result can be obtained directly from Eq. (2.68) by integration

Q Q

′

∫

2 2

’

This is not surprising since integration is nothing but summation of

a large number of small terms.

We can write the final result, Eq. (2.72) in different ways

1 1

2 2 2

W CV QV

= = =

(2.73)

Since electrostatic force is conservative, this work is stored in the form

of potential energy of the system. For the same reason, the final result for

potential energy [Eq. (2.73)] is independent of the manner in which the

charge configuration of the capacitor is built up. When the capacitor

discharges, this stored-up energy is released. It is possible to view the

potential energy of the capacitor as ‘stored’ in the electric field between

the plates. To see this, consider for simplicity, a parallel plate capacitor

[of area A (of each plate) and separation d between the plates].

Energy stored in the capacitor

2 2

1 ( )

2 2

Q A d

C A

= ×

(2.74)

The surface charge density

is related to the electric field E between

the plates,

(2.75)

From Eqs. (2.74) and (2.75) , we get

Energy stored in the capacitor

U =

(

)

1/ 2

E A d

(2.76)

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Physics

EXAMPLE 2.10

Note that Ad is the volume of the region between the plates (where

electric field alone exists). If we define energy density as energy stored

per unit volume of space, Eq (2.76) shows that

Energy density of electric field,

u =(1/2)

(2.77)

Though we derived Eq. (2.77) for the case of a parallel plate capacitor,

the result on energy density of an electric field is, in fact, very general and

holds true for electric field due to any configuration of charges.

Example 2.10 (a) A 900 pF capacitor is charged by 100 V battery

[Fig. 2.31(a)]. How much electrostatic energy is stored by the capacitor?

(b) The capacitor is disconnected from the battery and connected to

another 900 pF capacitor [Fig. 2.31(b)]. What is the electrostatic energy

stored by the system?

FIGURE 2.31

Solution

(a) The charge on the capacitor is

Q = CV = 900 × 10

–12

F × 100 V = 9 × 10

–8

The energy stored by the capacitor is

= (1/2) CV

= (1/2) QV

= (1/2) × 9 × 10

–8

C × 100 V = 4.5 × 10

–6

(b) In the steady situation, the two capacitors have their positive

plates at the same potential, and their negative plates at the

same potential. Let the common potential difference be V′. The

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Electrostatic Potential

and Capacitance

EXAMPLE 2.10

SUMMARY

1. Electrostatic force is a conservative force. Work done by an external

force (equal and opposite to the electrostatic force) in bringing a charge

q from a point R to a point P is q(V

–V

), which is the difference in

potential energy of charge q between the final and initial points.

2. Potential at a point is the work done per unit charge (by an external

agency) in bringing a charge from infinity to that point. Potential at a

point is arbitrary to within an additive constant, since it is the potential

difference between two points which is physically significant. If potential

at infinity is chosen to be zero; potential at a point with position vector

r due to a point charge Q placed at the origin is given is given by

( )

3. The electrostatic potential at a point with position vector r due to a

point dipole of dipole moment p placed at the origin is

( )

p.r

The result is true also for a dipole (with charges –q and q separated by

2a) for r >> a.

4. For a charge configuration q

, q

, ..., q

with position vectors r

, ... r

, the potential at a point P is given by the superposition principle

1 2

0 1P 2P P

( ... )

q q

r r r

= + + +

where r

is the distance between q

and P, as and so on.

5. An equipotential surface is a surface over which potential has a constant

value. For a point charge, concentric spheres centred at a location of the

charge are equipotential surfaces. The electric field E at a point is

perpendicular to the equipotential surface through the point. E is in the

direction of the steepest decrease of potential.

charge on each capacitor is then Q

′

= CV

′

. By charge

conservation, Q

′

= Q/2. This implies V

′

= V/2. The total energy

of the system is

1 1

2 ' ' 2.25 10 J

2 4

Q V QV

−

= × = = ×

Thus in going from (a) to (b), though no charge is lost; the final

energy is only half the initial energy. Where has the remaining

energy gone?

There is a transient period before the system settles to the

situation (b). During this period, a transient current flows from

the first capacitor to the second. Energy is lost during this time

in the form of heat and electromagnetic radiation.

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Physics

6. Potential energy stored in a system of charges is the work done (by an

external agency) in assembling the charges at their locations. Potential

energy of two charges q

, q

at r

, r

is given by

1 2

0 12

q q

where r

is distance between q

and q

7. The potential energy of a charge q in an external potential V(r)

is qV(r).

The potential energy of a dipole moment p in a uniform electric field E is

–p

8. Electrostatics field E is zero in the interior of a conductor; just outside

the surface of a charged conductor, E is normal to the surface given by

E n

where

is the unit vector along the outward normal to the

surface and

is the surface charge density. Charges in a conductor can

reside only at its surface. Potential is constant within and on the surface

of a conductor. In a cavity within a conductor (with no charges), the

electric field is zero.

9. A capacitor is a system of two conductors separated by an insulator. Its

capacitance is defined by C = Q/V, where Q and –Q are the charges on the

two conductors and V is the potential difference between them. C is

determined purely geometrically, by the shapes, sizes and relative

positions of the two conductors. The unit of capacitance is farad:,

1 F = 1 C V

–1

. For a parallel plate capacitor (with vacuum between the

plates),

C =

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

10. If the medium between the plates of a capacitor is filled with an insulating

substance (dielectric), the electric field due to the charged plates induces

a net dipole moment in the dielectric. This effect, called polarisation,

gives rise to a field in the opposite direction. The net electric field inside

the dielectric and hence the potential difference between the plates is

thus reduced. Consequently, the capacitance C increases from its value

when there is no medium (vacuum),

C = KC

where K is the dielectric constant of the insulating substance.

11. For capacitors in the series combination, the total capacitance C is given by

1 2 3

1 1 1 1

...

C C C C

= + + +

In the parallel combination, the total capacitance C is:

C = C

+ C

+ ...

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Electrostatic Potential

and Capacitance

Physical quantity Symbol Dimensions Unit Remark

Potential or V [M

–3

–1

] V Potential difference is

physically significant

Capacitance C [M

–1

–2

–4

] F

Polarisation P [L

–2

AT] C m

-2

Dipole moment per unit

volume

Dielectric constant K [Dimensionless]

where C

, C

... are individual capacitances.

12. The energy U stored in a capacitor of capacitance C, with charge Q and

voltage V is

U QV CV

= = =

The electric energy density (energy per unit volume) in a region with

electric field is (1/2)

POINTS TO PONDER

1. Electrostatics deals with forces between charges at rest. But if there is a

force on a charge, how can it be at rest? Thus, when we are talking of

electrostatic force between charges, it should be understood that each

charge is being kept at rest by some unspecified force that opposes the

net Coulomb force on the charge.

2. A capacitor is so configured that it confines the electric field lines within

a small region of space. Thus, even though field may have considerable

strength, the potential difference between the two conductors of a

capacitor is small.

3. Electric field is discontinuous across the surface of a spherical charged

shell. It is zero inside and

outside. Electric potential is, however

continuous across the surface, equal to q/4π

R at the surface.

4. The torque p × E on a dipole causes it to oscillate about E. Only if there

is a dissipative mechanism, the oscillations are damped and the dipole

eventually aligns with E.

5. Potential due to a charge q at its own location is not defined – it is

infinite.

6. In the expression qV(r) for potential energy of a charge q, V(r) is the

potential due to external charges and not the potential due to q. As seen

in point 5, this expression will be ill-defined if V(r) includes potential

due to a charge q itself.

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Physics

EXERCISES

2.1 Two charges 5 × 10

–8

C and –3 × 10

–8

C are located 16 cm apart. At

what point(s) on the line joining the two charges is the electric

potential zero? Take the potential at infinity to be zero.

2.2 A regular hexagon of side 10 cm has a charge 5 µC at each of its

vertices. Calculate the potential at the centre of the hexagon.

2.3 Two charges 2 µC and –2 µC are placed at points A and B 6 cm

apart.

(a) Identify an equipotential surface of the system.

(b) What is the direction of the electric field at every point on this

surface?

2.4 A spherical conductor of radius 12 cm has a charge of 1.6 × 10

–7

distributed uniformly on its surface. What is the electric field

(a) inside the sphere

(b) just outside the sphere

2.5 A parallel plate capacitor with air between the plates has a

capacitance of 8 pF (1pF = 10

–12

F). What will be the capacitance if

the distance between the plates is reduced by half, and the space

between them is filled with a substance of dielectric constant 6?

2.6 Three capacitors each of capacitance 9 pF are connected in series.

(a) What is the total capacitance of the combination?

(b) What is the potential difference across each capacitor if the

combination is connected to a 120 V supply?

2.7 Three capacitors of capacitances 2 pF, 3 pF and 4 pF are connected

in parallel.

(a) What is the total capacitance of the combination?

(b) Determine the charge on each capacitor if the combination is

connected to a 100 V supply.

2.8 In a parallel plate capacitor with air between the plates, each plate

has an area of 6 × 10

–3

and the distance between the plates is 3 mm.

Calculate the capacitance of the capacitor. If this capacitor is

connected to a 100 V supply, what is the charge on each plate of the

capacitor?

7. A cavity inside a conductor is shielded from outside electrical influences.

It is worth noting that electrostatic shielding does not work the other

way round; that is, if you put charges inside the cavity, the exterior of

the conductor is not shielded from the fields by the inside charges.

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Electrostatic Potential

and Capacitance

2.9 Explain what would happen if in the capacitor given in Exercise

2.8, a 3 mm thick mica sheet (of dielectric constant = 6) were inserted

between the plates,

(a) while the voltage supply remained connected.

(b) after the supply was disconnected.

2.10 A 12pF capacitor is connected to a 50V battery. How much

electrostatic energy is stored in the capacitor?

2.11 A 600pF capacitor is charged by a 200V supply. It is then

disconnected from the supply and is connected to another

uncharged 600 pF capacitor. How much electrostatic energy is lost

in the process?

ADDITIONAL EXERCISES

2.12 A charge of 8 mC is located at the origin. Calculate the work done in

taking a small charge of –2 × 10

–9

C from a point P (0, 0, 3 cm) to a

point Q (0, 4 cm, 0), via a point R (0, 6 cm, 9 cm).

2.13 A cube of side b has a charge q at each of its vertices. Determine the

potential and electric field due to this charge array at the centre of

the cube.

2.14 Two tiny spheres carrying charges 1.5 µC and 2.5 µC are located 30 cm

apart. Find the potential and electric field:

(a) at the mid-point of the line joining the two charges, and

(b) at a point 10 cm from this midpoint in a plane normal to the

line and passing through the mid-point.

2.15 A spherical conducting shell of inner radius r

and outer radius r

has a charge Q.

(a) A charge q is placed at the centre of the shell. What is the

surface charge density on the inner and outer surfaces of the

shell?

(b) Is the electric field inside a cavity (with no charge) zero, even if

the shell is not spherical, but has any irregular shape? Explain.

2.16 (a) Show that the normal component of electrostatic field has a

discontinuity from one side of a charged surface to another

given by

2 1

( )

− •

E E

where

is a unit vector normal to the surface at a point and

is the surface charge density at that point. (The direction of

is from side 1 to side 2.) Hence, show that just outside a

conductor, the electric field is

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Physics

(b) Show that the tangential component of electrostatic field is

continuous from one side of a charged surface to another.

[Hint: For (a), use Gauss’s law. For, (b) use the fact that work

done by electrostatic field on a closed loop is zero.]

2.17 A long charged cylinder of linear charged density

is surrounded

by a hollow co-axial conducting cylinder. What is the electric field in

the space between the two cylinders?

2.18 In a hydrogen atom, the electron and proton are bound at a distance

of about 0.53 Å:

(a) Estimate the potential energy of the system in eV, taking the

zero of the potential energy at infinite separation of the electron

from proton.

(b) What is the minimum work required to free the electron, given

that its kinetic energy in the orbit is half the magnitude of

potential energy obtained in (a)?

energy is taken at 1.06 Å separation?

2.19 If one of the two electrons of a H

molecule is removed, we get a

hydrogen molecular ion H

. In the ground state of an H

, the two

protons are separated by roughly 1.5 Å, and the electron is roughly

1 Å from each proton. Determine the potential energy of the system.

Specify your choice of the zero of potential energy.

2.20 Two charged conducting spheres of radii a and b are connected to

each other by a wire. What is the ratio of electric fields at the surfaces

of the two spheres? Use the result obtained to explain why charge

density on the sharp and pointed ends of a conductor is higher

than on its flatter portions.

2.21 Two charges –q and +q are located at points (0, 0, –a) and (0, 0, a),

respectively.

(a) What is the electrostatic potential at the points (0, 0, z) and

(x, y, 0) ?

(b) Obtain the dependence of potential on the distance r of a point

from the origin when r/a >> 1.

point (5,0,0) to (–7,0,0) along the x-axis? Does the answer

change if the path of the test charge between the same points

is not along the x-axis?

2.22 Figure 2.32 shows a charge array known as an electric quadrupole.

For a point on the axis of the quadrupole, obtain the dependence

of potential on r for r/a >> 1, and contrast your results with that

due to an electric dipole, and an electric monopole (i.e., a single

charge).

FIGURE 2.32

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Electrostatic Potential

and Capacitance

2.23 An electrical technician requires a capacitance of 2 µF in a circuit

across a potential difference of 1 kV. A large number of 1 µF capacitors

are available to him each of which can withstand a potential

difference of not more than 400 V. Suggest a possible arrangement

that requires the minimum number of capacitors.

2.24 What is the area of the plates of a 2 F parallel plate capacitor, given

that the separation between the plates is 0.5 cm? [You will realise

from your answer why ordinary capacitors are in the range of µF or

less. However, electrolytic capacitors do have a much larger

capacitance (0.1 F) because of very minute separation between the

conductors.]

2.25 Obtain the equivalent capacitance of the network in Fig. 2.33. For a

300 V supply, determine the charge and voltage across each

capacitor.

FIGURE 2.33

2.26 The plates of a parallel plate capacitor have an area of 90 cm

each

and are separated by 2.5 mm. The capacitor is charged by

connecting it to a 400 V supply.

(a) How much electrostatic energy is stored by the capacitor?

(b) View this energy as stored in the electrostatic field between

the plates, and obtain the energy per unit volume u. Hence

arrive at a relation between u and the magnitude of electric

field E between the plates.

2.27 A 4 µ

F capacitor is charged by a 200 V supply. It is then disconnected

from the supply, and is connected to another uncharged 2 µF

capacitor. How much electrostatic energy of the first capacitor is

lost in the form of heat and electromagnetic radiation?

2.28 Show that the force on each plate of a parallel plate capacitor has a

magnitude equal to (½) QE, where Q is the charge on the capacitor,

and E is the magnitude of electric field between the plates. Explain

the origin of the factor ½.

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Physics

2.29 A spherical capacitor consists of two concentric spherical conductors,

held in position by suitable insulating supports (Fig. 2.34). Show

that the capacitance of a spherical capacitor is given by

0 1 2

1 2

–

r r

where r

and r

are the radii of outer and inner spheres,

respectively.

2.30 A spherical capacitor has an inner sphere of radius 12 cm and an

outer sphere of radius 13 cm. The outer sphere is earthed and the

inner sphere is given a charge of 2.5 µC. The space between the

concentric spheres is filled with a liquid of dielectric constant 32.

(a) Determine the capacitance of the capacitor.

(b) What is the potential of the inner sphere?

isolated sphere of radius 12 cm. Explain why the latter is much

smaller.

2.31 Answer carefully:

(a) Two large conducting spheres carrying charges Q

and Q

are

brought close to each other. Is the magnitude of electrostatic

force between them exactly given by Q

/4π

, where r is

the distance between their centres?

(b) If Coulomb’s law involved 1/r

dependence (instead of 1/r

would Gauss’s law be still true ?

electrostatic field configuration. Will it travel along the field

line passing through that point?

(d) What is the work done by the field of a nucleus in a complete

circular orbit of the electron? What if the orbit is elliptical?

FIGURE 2.34

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Electrostatic Potential

and Capacitance

(e) We know that electric field is discontinuous across the surface

of a charged conductor. Is electric potential also discontinuous

there?

(f) What meaning would you give to the capacitance of a single

conductor?

(g) Guess a possible reason why water has a much greater

dielectric constant (= 80) than say, mica (= 6).

2.32 A cylindrical capacitor has two co-axial cylinders of length 15 cm

and radii 1.5 cm and 1.4 cm. The outer cylinder is earthed and the

inner cylinder is given a charge of 3.5 µC. Determine the capacitance

of the system and the potential of the inner cylinder. Neglect end

effects (i.e., bending of field lines at the ends).

2.33 A parallel plate capacitor is to be designed with a voltage rating

1 kV, using a material of dielectric constant 3 and dielectric strength

about 10

–1

. (Dielectric strength is the maximum electric field a

material can tolerate without breakdown, i.e., without starting to

conduct electricity through partial ionisation.) For safety, we should

like the field never to exceed, say 10% of the dielectric strength.

What minimum area of the plates is required to have a capacitance

of 50 pF?

2.34 Describe schematically the equipotential surfaces corresponding to

(a) a constant electric field in the z-direction,

(b) a field that uniformly increases in magnitude but remains in a

constant (say, z) direction,

(d) a uniform grid consisting of long equally spaced parallel charged

wires in a plane.

2.35 A small sphere of radius r

and charge q

is enclosed by a spherical

shell of radius r

and charge q

. Show that if q

is positive, charge

will necessarily flow from the sphere to the shell (when the two are

connected by a wire) no matter what the charge q

on the shell is.

2.36 Answer the following:

(a) The top of the atmosphere is at about 400 kV with respect to

the surface of the earth, corresponding to an electric field that

decreases with altitude. Near the surface of the earth, the field

is about 100 Vm

–1

. Why then do we not get an electric shock as

we step out of our house into the open? (Assume the house to

be a steel cage so there is no field inside!)

(b) A man fixes outside his house one evening a two metre high

insulating slab carrying on its top a large aluminium sheet of

area 1m

. Will he get an electric shock if he touches the metal

sheet next morning?

conductivity of air is known to be 1800 A on an average over

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Physics

the globe. Why then does the atmosphere not discharge itself

completely in due course and become electrically neutral? In

other words, what keeps the atmosphere charged?

(d) What are the forms of energy into which the electrical energy

of the atmosphere is dissipated during a lightning?

(Hint: The earth has an electric field of about 100 Vm

–1

at its

surface in the downward direction, corresponding to a surface

charge density = –10

–9

C m

–2

. Due to the slight conductivity of

the atmosphere up to about 50 km (beyond which it is good

conductor), about + 1800 C is pumped every second into the

earth as a whole. The earth, however, does not get discharged

since thunderstorms and lightning occurring continually all

over the globe pump an equal amount of negative charge on

the earth.)

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