E CONTENT


E CONTENT LINK


.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 poten


tial 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, like the gravitational
force, 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
Physics
52
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 Fext just enough to
counter the repulsive electric force FE (i.e, Fext= –FE).
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
W
RP =
P R
d
F r ext
=
P R
d
F r E (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
∆ = − = U U U W P R RP (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., –WRP.)
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.
Electrostatic Potential
and Capacitance
53
(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 α to potential energy at
every point, since this will not change the potential energy
difference:
( ) ( ) U U U U P R P R + − + = − 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)
W U U U ∞ ∞ P P P = − = (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
P – VR U U P R
q
  −
  =
  (2.4)
where V
P and VR 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 electricity 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 (electrolyte) sandwiched
between disks of metal
(electrodes).
Physics
54
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
δq, obtain the work done δW in bringing it from
infinity to the point and determine the ratio
δW/δq. 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 Pon the path, the
electrostatic force on a unit positive charge is
2
0
1
ˆ
4 '
Q
ε r
×

π
r (2.5)
where r ˆis the unit vector along OP. Work done
against this force from r′ to r′ + ∆r′ is
2
Q 4 '0
W r
ε r
∆ = − ∆ ′
π (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,
2
4 ' 0 4 4 0 0
r
Q Q Q r
W dr
ε r ε ε r r


= − = = ′
π ′ π π (2.7)
This, by definition is the potential at P due to the charge Q
0
( )
4
Q
V r
ε r
=
π (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.
Electrostatic Potential
and Capacitance
55
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 2 ) varies with r.
Example 2.1
(a) Calculate the potential at a point P due to a charge of 4 × 10–7C
located 9 cm away.
(b) Hence obtain the work done in bringing a charge of 2 × 10–9 C
from infinity to the point P. Does the answer depend on the path
along which the charge is brought?
Solution
(a)
7
9 2 –2
0
1 4 10 C
9 10 Nm C
4 0.09m
Q
V
ε r

×
= = × ×
π
= 4 × 104 V
(b) W qV = = × × × 2 10 C 4 10 V 9 4
= 8 × 10–5 J
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/4πε0) m-1] (blue curve) and field with r [in units
of (Q/4πε0) m-2] (black curve) for a point charge Q.
Physics
56
the field falls off, at large distance, not as
1/r 2 (typical of field due to a single charge)
but as 1/r3. 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
1
4
q q
V
ε r r
 
= −  
π   (2.9)
where r
1 and r2 are the distances of the
point P from q and –q, respectively.
Now, by geometry,
2 2 2
r r a ar 1 = + − 2 cosθ
2 2 2
r r a ar 2 = + + 2 cosθ (2.10)
We take r much greater than a ( r >> a ) and retain terms only upto
the first order in a/r
2
2 2
1 2
2 cos
r r 1 a a
r r
  θ
= − +  
 
2 2 cos
r 1 a
r
  θ
≅ −  
  (2.11)
Similarly,
2 2
2
2 cos
r r 1 a
r
  θ
≅ +  
  (2.12)
Using the Binomial theorem and retaining terms upto the first order
in a/r ; we obtain,
1/2
1
1 1 2 cos 1
1 1 cos a a
r r r r r
θ
θ
   
≅ − ≅ +    
    [2.13(a)]


1/2
2
1 1 2 cos 1
1 1 cos a a
r r r r r
θ
θ
   
≅ + ≅ −    
    [2.13(b)]
Using Eqs. (2.9) and (2.13) and p = 2qa, we get
2 2
0 0
2 cos cos
4 4
q a p
V
r r
θ θ
ε ε
= =
π π (2.14)
Now, p cos θ = p r ˆ
FIGURE 2.5 Quantities involved in the calculation
of potential due to a dipole.
Electrostatic Potential
and Capacitance
57
where r ˆ is the unit vector along the position vector OP.
The electric potential of a dipole is then given by
2
0
1 ˆ
4
V
ε r
=
π
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 (θ = 0, π ) is given by
2
0
1
4
p
V
ε r
= ±
π
(2.16)
(Positive sign for θ = 0, negative sign for θ = π.) The potential in the
equatorial plane (θ = 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 θ 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 2, 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 2 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 q1, q2,…, qn with position vectors r1, r2,…,
r
nrelative to some origin (Fig. 2.6). The potential V1at P due to the charge
q1 is
1
1
0 1P
1
4
q
V
ε r
=
π
where r
1P is the distance between q1 and P.
Similarly, the potential V2 at P due to q2 and
V3
due to q3 are given by
2
2
0 2P
1
4
q
V
ε r
=
π
,
3
3
0 3P
1
4
q
V
ε r
=
π
where r
2P and r3P are the distances of P from
charges q2 and q3, 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
1 + V2 + ... + Vn (2.17)
FIGURE 2.6 Potential at a point due to a
system of charges is the sum of potentials
due to individual charges.
Physics
58
EXAMPLE 2.2
1 2
0 1P 2P P
1
......
4
n
q n
q q
ε r r r
 
= + + +  
π   (2.18)
If we have a continuous charge distribution characterised by a charge
density ρ (r), we divide it, as before, into small volume elements each of
size v and carrying a charge ρ∆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
0
1
4
q
V
ε r
=
π ( ) 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
0
1
4
q
V
ε R
=
π [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
–8 –8
–2 –2
0
1 3 10 2 10
0
4 ε 10 (15 ) 10
× ×
− =
π × − ×
 
 
  x x
where x is in cm. That is,
3 2
0
x x 15
− =

which gives x = 9 cm.
If x lies on the extended line OA, the required condition is
3 2
0
x x 15
− =

Electrostatic Potential
and Capacitance
59
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 VP VQ; VB VA.
(b) Give the sign of the potential energy difference of a small negative
charge between the points Q and P; A and B.
(c) Give the sign of the work done by the field in moving a small
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
1
V
r

, V
P > VQ. Thus, (VP VQ) is positive. Also VB is less negative
than V
A . Thus, VB > VA or (VB VA) 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.)A > (P.E.)B and hence sign of potential energy
differences is positive.
(c) In moving a small positive charge from Q to P, work has to be
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:

Physics
60
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):
1
4
o
q
V
ε r
=
π
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 x -axis, the equipotential
surfaces are planes normal to the x -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.
Electrostatic Potential
and Capacitance
61
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 + δV, where δV is the change in V in the
direction of the electric field E. Let P be a point on the
surface B. δl 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|δ l.
This work equals the potential difference
VA
VB
.
Thus,
|E|δ l = V (V +δV)= –δV
i.e., |E|= V
l
δ
δ
(2.20)
Since δV is negative, δV = – V|. we can rewrite
Eq (2.20) as
V V
l l
δ δ
δ δ
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 q1and q2 with position vector
r
1 and r2 relative to some origin. Let us calculate the work done
(externally) in building up this configuration. This means that we consider
the charges q1 and q2 initially at infinity and determine the work done by
an external agency to bring the charges to the given locations. Suppose,
first the charge q1 is brought from infinity to the point r1. There is no
external field against which work needs to be done, so work done in
bringing q1 from infinity to r1 is zero. This charge produces a potential in
space given by
1
1
0 1P
1
4
q
V
ε r
=
π
where r
1P is the distance of a point P in space from the location of q1.
From the definition of potential, work done in bringing charge q2 from
infinity to the point r2 is q2 times the potential at r2 due to q1:
work done on q2 = 1 2
0 12
1
4
q q
πε r
FIGURE 2.12 From the
potential to the field.
Physics
62
where r
12 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 q1and q2 is
1 2
0 12
1
4
q q
U
ε r
=
π
(2.22)
Obviously, if q2 was brought first to its present location and
q1 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 q1and q2. If q1q2 > 0, potential
energy is positive. This is as expected, since for like charges (q1q2 > 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 (q1 q2 < 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 q1, q2 and q3 located at r1, r2, r3, respectively. To bring q1 first
from infinity to r1, no work is required. Next we bring q2 from infinity to
r
2. As before, work done in this step is
1 2
2 1 2
0 12
1
( )
4
q q
q V
ε r
=
π
r (2.23)
The charges q1 and q2 produce a potential, which at any point P is
given by
1 2
1,2
0 1P 2P
1
4
q q
V
ε r r
 
= +  
π   (2.24)
Work done next in bringing q3 from infinity to the point r3 is q3times
V
1, 2 at r3
1 3 2 3
3 1,2 3
0 13 23
1
( )
4
q q q q
q V
ε r r
 
= +  
r π   (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)],
1 2 1 3 2 3
0 12 13 23
1
4
q q q q q q
U
ε r r r
 
= + +  
π   (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 q1 and q2 is
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.
Electrostatic Potential
and Capacitance
63
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 q0 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)
2
4 4 0 0
q q
q
ε ε d d
 
= − × = −  
  π π
(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 0 4 0
q q
q
ε d ε d
  + −
= + +  
  π π
2 0
1
1
4 2
q
ε d
−  
= −  
π  
(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 4 0 0 4 2 0
q q q
q
ε ε d d ε d
  + −
= − + +  
  π π π
2 0
1
2
4 2
q
ε d
−  
= −  
π  
Physics
64
EXAMPLE 2.4
Add the work done in steps (i), (ii), (iii) and (iv). The total work
required is
2 0
1 1
(0) (1) 1 2
4 2 2
q
ε d
−      
= + + − + −      
π      
( )
2 0
4 2
4
q
ε d

= −
π
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 q0 to the point E when
the four charges are at A, B, C and D is q0 × (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.
Electrostatic Potential
and Capacitance
65
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 qV = 1.6 ×
10–19J. This unit of energy is defined as 1 electron volt or 1eV, i.e.,
1 eV=1.6 × 10–19J. The units based on eV are most commonly used in
atomic, nuclear and particle physics, (1 keV = 103eV = 1.6 × 10–16J, 1 MeV
= 106eV = 1.6 × 10–13J, 1 GeV = 109eV = 1.6 × 10–10J and 1 TeV = 1012eV
= 1.6 × 10–7J). [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 q1
and q2 located at r1and r2, respectively, in an external field? First, we
calculate the work done in bringing the charge q1 from infinity to r1.
Work done in this step is q1 V(r1), using Eq. (2.27). Next, we consider the
work done in bringing q2to r2. In this step, work is done not only against
the external field E but also against the field due to q1.
Work done on q2 against the external field
= q2 V (r2)
Work done on q2 against the field due to q1
1 2
4 o 12
q q
ε r
=
π
where r
12 is the distance between q1 and q2. We have made use of Eqs.
(2.27) and (2.22). By the superposition principle for fields, we add up
the work done on q2 against the two fields (E and that due to q1):
Work done in bringing q2 to r2
1 2
2 2
12
( )
4
o
q q
q V
ε r
= +
π
r (2.28)
Thus,
Potential energy of the system
= the total work done in assembling the configuration
1 2
1 1 2 2
0 12
( ) ( )
4
q q
q V q V
ε r
= + +
π
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?
Physics
66
EXAMPLE 2.5
(c) Suppose that the same system of charges is now placed in an
external electric field E = A (1/r 2); A = 9 × 105 C m–2. What would
the electrostatic energy of the configuration be?
Solution
(a)
12
1 2 9
0
1 7 ( 2) 10
9 10
4 0.18
q q
U
ε r

× − ×
= = × ×
π
= –0.7 J.
(b) W = U
2 U1 = 0 – U = 0 – (–0.7) = 0.7 J.
(c) The mutual interaction energy of the two charges remains
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 r r + = + µ − µ
and the net electrostatic energy is
1 1 2 2 ( ) ( ) 1 2
0 12
7 C 2 C
0.7 J
4 0.09m 0.09m
q q
q V q V A A
ε r
µ − µ
+ + = + −
π
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 q1 = +q and q2 = –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 given by
τ = p×E (2.30)
which will tend to rotate it (unless p is parallel or
antiparallel to E). Suppose an external torque τext is
applied in such a manner that it just neutralises this
torque and rotates it in the plane of paper from angle θ0
to angle θ1 at an infinitesimal angular speed and without
angular acceleration. The amount of work done by the
external torque will be given by
W d pE d ext( sin
θ θ
θ θ
1 1 τ θ θ θ θ
0 0
= ) = ∫ ∫
= − pE ( ) cos cos θ θ 0 1 (2.31)
This work is stored as the potential energy of the system. We can then
associate potential energy U(θ) with an inclination θ 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
θ
0= π / 2. (Αn explanation for it is provided towards the end of discussion.)
We can then write,
U pE pE (θ θ θ ) = − = = −     cos cos – cos π
  2 p E (2.32)
FIGURE 2.16 Potential energy of a
dipole in a uniform external field.
Electrostatic Potential
and Capacitance
67
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
( ) ( ) ( )
2
[ ] 1 2
4 2
q
U q V V
a
θ
ε
0
′ = − −
π ×
r r (2.33)
Here, r
1 and r2 denote the position vectors of +q and –q. Now, the
potential difference between positions r1 and r2 equals the work done
in bringing a unit positive charge against field from r2 to r1. The
displacement parallel to the force is 2a cosθ. Thus, [V(r1)–V (r2)] =
E × 2a cosθ . We thus obtain,
( )
2 2
cos
4 2 4 2
q q
U pE
a a
θ θ
ε ε
0 0
′ = − − = − −
π × π ×
p E (2.34)
We note that U(θ) differs from U(θ ) 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 θ0=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 (r1) – V (r2)]=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 106 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 × 1023 molecules,
total dipole moment of all the molecules, p = 6 × 1023 × 10–29 C m
= 6 × 10–6 C m
Initial potential energy, Ui = –pE cos θ = –6×10–6×106 cos 0° = –6 J
Final potential energy (when θ = 60°), Uf = –6 × 10–6 × 106 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
Physics
68
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,
Electrostatic Potential
and Capacitance
69
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
0
σ ˆ
ε
E n = (2.35)
where σ is the surface charge density and n ˆ 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 δ 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 ± EδS (positive for σ > 0,
negative for σ < 0), since over the small area δS, E
may be considered constant and E and δS are parallel
or antiparallel. The charge enclosed by the pill box
is σδS.
By Gauss’s law
EδS =
0
σ δ S
ε
E =
σ ε0
(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 σ. For σ > 0, electric field is
normal to the surface outward; for σ < 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.
Physics
70
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?
(c) Vehicles carrying inflammable materials usually have metallic
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.
Electrostatic Potential
and Capacitance
71
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.
(c) Reason similar to (b).
(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 (O2) and hydrogen (H2)
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
(H
2O) 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.
Physics
72
In an external electric field, the
positive and negative charges of a nonpolar 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 over all 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,
P E = χe (2.37)
where χe is a constant characteristic of the dielectric and is known as the
electric susceptibility of the dielectric medium.
It is possible to relate χe 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 E0 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.
Electrostatic Potential
and Capacitance
73
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 σ
p
and –σ
p
. 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
±σ
p
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 Q1 and Q2, and potentials
V1
and V
2. Usually, in practice, the two conductors have charges Q
and – Q, with potential difference V = V1 V2 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:
Q
C
V
= (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.
Physics
74
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 × 106 Vm–1. For a separation between
conductors of the order of 1 cm or so, this field corresponds to a potential
difference of 3 × 104 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 F,
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 (d2 << 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 σ = Q/A and plate 2 has a surface
charge density –σ. Using Eq. (1.33), the electric field
in different regions is:
Outer region I (region above the plate 1),
0 0
0
2 2
E σ σ
ε ε
= − = (2.39)
FIGURE 2.25 The parallel plate capacitor.
Electrostatic Potential
and Capacitance
75
Outer region II (region below the plate 2),
0 0
0
2 2
E 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
2 2 0 0 0 0
Q
E
A
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, σ will not
be strictly uniform on the entire plate. [E and σ are related by Eq. (2.35).]
However, for d2 << 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,
0
1 Qd
V E d
ε A
= = (2.42)
The capacitance C of the parallel plate capacitor is then
Q
C
V
= =
0A
d
ε
= (2.43)
which, as expected, depends only on the geometry of the system. For
typical values like A = 1 m2, d = 1 mm, we get
12 2 –1 –2 2
9
3
8.85 10 C N m 1m
8.85 10 F
10 m
C



× ×
= = × (2.44)
(You can check that if 1F= 1C V–1 = 1C (NC–1m)–1 = 1 C2 N–1m–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:
0
Cd
A
ε
= =
2
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 behavior 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 ±σ (with
σ = Q/A). When there is vacuum between the plates,
0
0
E σ
ε
=
Factors affecting capacitance, capacitors in action
Interactive Java tutorial

Physics
76
and the potential difference V0 is
V0
= E
0d
The capacitance C0 in this case is
0 0
0
Q A
C
V d
= = ε (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 σ
p
and –σ
p
. The electric field in the dielectric then corresponds
to the case when the net surface charge density on the plates is ±(σ σ
p
).
That is,
0
E σ σ P
ε

= (2.47)
so that the potential difference across the plates is
0
σ σPV E d dε
= = (2.48)
For linear dielectrics, we expect σ
p
to be proportional to E0, i.e., to σ.
Thus, (σ σ
p
) is proportional to σ and we can write
P
σ K
σ σ − = (2.49)
where K is a constant characteristic of the dielectric. Clearly, K > 1. We
then have
0 0
d Qd
V
K A K
σ
ε ε
= = (2.50)
The capacitance C, with dielectric between the plates, is then
Q 0KA
C
V d
ε
= = (2.51)
The product ε0K is called the permittivity of the medium and is
denoted by ε
ε = ε
0 K (2.52)
For vacuum K = 1 and ε = ε
0; ε0 is called the permittivity of the vacuum.
The dimensionless ratio
0
K ε
ε
= (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)
C 0
K
C
= (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
Electrostatic Potential
and Capacitance
77
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 σ
p
and the polarisation P.
We take without proof the result that
σP = P n ˆ
where n ˆ 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 n ˆ at the
right surface and opposite to n ˆ 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
0
ˆ
ˆ σ
ε

=
P n
E n
or (ε
0 E + P) n ˆ =σ
The quantity ε0 E + P is called the electric displacement and is denoted by D. It is a
vector quantity. Thus,
D = ε
0 E + P, D n ˆ = σ,
The significance of D is this : in vacuum, E is related to the free charge density σ.
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 σ, 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
0
0
P
D
K
E
σε
ε
σ σ
= =

Thus,
D = ε
0 K E
and P = D ε
0E = ε0 (K –1)E
This gives for the electric susceptibility χe defined in Eq. (2.37)
χe =ε0 (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
0 = V0/d be the electric field between the plates when
there is no dielectric and the potential difference is V0. If the dielectric
is now inserted, the electric field in the dielectric will be E = E
0/K.
The potential difference will then be
Physics
78
EXAMPLE 2.8
0
0
1 3
( ) ( )
4 4
E
V E d d
K
= +
0 0
1 3 3
( )
4 4 4
K
E d V
K K
+
= + =
The potential difference decreases by the factor (K + 3)/4K while the
free charge Q0 on the plates remains unchanged. The capacitance
thus increases
0 0
0
0
4 4
3 3
Q Q K K
C C
V K V K
= = =
+ +
2.14 COMBINATION OF CAPACITORS
We can combine several capacitors of capacitance C1, C2,…,Cn 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 C1 and C2 combined in series.
The left plate of C1 and the right plate of C2 are connected to two
terminals of a battery and have charges Q and –Q ,
respectively. It then follows that the right plate ofC1
has charge –Q and the left plate of C2 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 C1and C2. Charge
would flow until the net charge on both C1 and C2
is zero and there is no electric field in the conductor
connecting C1 and C2. 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 V1 and V2 across C1 and C2, respectively.
V = V
1 + V2 =
1 2
Q Q
C C
+ (2.55)
i.e.,
1 2
V 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
Q
C
V
= (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.
Electrostatic Potential
and Capacitance
79
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
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
Q1) on capacitor 1 and the plate charges (±Q2) on the
capacitor 2 are not necessarily the same:
Q1 = C1V, Q2 = C2V (2.61)
The equivalent capacitor is one with charge
Q = Q1 + Q2 (2.62)
and potential difference V.
Q = CV = C1V + C2V (2.63)
The effective capacitance C is, from Eq. (2.63),
C = C
1 + C2 (2.64)
The general formula for effective capacitance C for
parallel combination of n capacitors [Fig. 2.28 (b)]
follows similarly,
Q = Q1 + Q2 + ... + Qn (2.65)
i.e., CV = C
1V + C2V + ... CnV (2.66)
which gives
C = C
1 + C2 + ... Cn (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 charge 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
Physics
80
EXAMPLE 2.9
Solution
(a) In the given network, C1, C2 and C3 are connected in series. The
effective capacitance Cof these three capacitors is given by
1 2 3
1 1 1 1
C C C C
= + +

For C
1 = C2 = C3 = 10 µF, C= (10/3) µF. The network has Cand C4
connected in parallel. Thus, the equivalent capacitance C of the
network is
C = C+ C
4 =
10
10
3
 
  +
  µF =13.3µF
(b) Clearly, from the figure, the charge on each of the capacitors, C1,
C2
and C
3 is the same, say Q. Let the charge on C4 be Q. Now, since
the potential difference across AB is Q/C1, across BC is Q/C2, across
CD is Q/C3 , we have
1 2 3
Q Q Q 500 V
C C C
+ + =
.
Also, Q/C4 = 500 V.
This gives for the given value of the capacitances,
10 3
500 F 1.7 10 C
3
Q V = × µ = × and
Q V ′ = × µ = × 500 10 F 5.0 10 C 3
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 Qand
Qrespectively. At this stage, the potential difference
Vbetween conductors 1 to 2 is Q/C, where C is the
capacitance of the system. Next imagine that a small
charge δ Qis transferred from conductor 2 to 1. Work
done in this step (δ W), resulting in charge Qon
conductor 1 increasing to Q+ δ Q, is given by
Q
W V Q Q
C
δ δ δ = = ′ ′ ′ ′ (2.68)
FIGURE 2.30 (a) Work done in a small
step of building charge on conductor 1
from Qto 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.
Electrostatic Potential
and Capacitance
81
Since δQcan be made as small as we like, Eq. (2.68) can be written as
1 2 2
[( ) ]
2
W Q Q Q
C
δ δ = + − ′ ′ ′ (2.69)
Equations (2.68) and (2.69) are identical because the term of second
order in δ Q, i.e., δ Q2/2C, is negligible, since δ Qis arbitrarily small. The
total work done (W) is the sum of the small work (δ W) over the very large
number of steps involved in building the charge Qfrom zero to Q.
sum over all steps
W W = δ
=
2 2
sum over all steps
1
[( ) ]
2
Q Q Q
C
′ ′ ′ + − δ (2.70)
=
1 2 2 2
[{ 0} {(2 ) }
2
Q Q Q
C
δ δ δ ′ ′ ′ − + − + − + {(3 ) (2 ) } ... δ δ Q Q ′ ′ 2 2
2 2
+ − − { ( ) }] Q Q Q δ (2.71)
2
1 2
[ 0]
2 2
Q
Q
C C
= − = (2.72)
The same result can be obtained directly from Eq. (2.68) by integration
2 2
0 0
1
'
2 2
Q Q
Q Q Q
W Q
C C C
= = = ′ ′ δ
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
2
1 1 2
2 2 2
Q
W CV QV
C
= = = (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
0
1 ( )
2 2
Q A d
C A
σ
ε
= × (2.74)
The surface charge density σ is related to the electric field E between
the plates,
0
E σ
ε
= (2.75)
From Eqs. (2.74) and (2.75) , we get
Energy stored in the capacitor
U = (1/2)ε0E A d 2 × (2.76)
Physics
82
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)ε0E2 (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 C
The energy stored by the capacitor is
= (1/2) CV2 = (1/2) QV
= (1/2) × 9 × 10–8C × 100 V = 4.5 × 10–6 J
(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
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 6
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.
Electrostatic Potential
and Capacitance
83
2.16 VAN DE GRAAFF GENERATOR
This is a machine that can build up high voltages of the order of a few
million volts. The resulting large electric fields are used to accelerate
charged particles (electrons, protons, ions) to high energies needed for
experiments to probe the small scale structure of matter. The principle
underlying the machine is as follows.
Suppose we have a large spherical conducting shell of radius R, on
which we place a charge Q. This charge spreads itself uniformly all over
the sphere. As we have seen in Section 1.14, the field outside the sphere
is just that of a point charge Q at the centre; while the field inside the
sphere vanishes. So the potential outside is that of a point charge; and
inside it is constant, namely the value at the radius R. We thus have:
Potential inside conducting spherical shell of radius R carrying charge Q
= constant
0
1
4
Q R
ε
=
π
(2.78)
Now, as shown in Fig. 2.32, let us suppose that in some way we
introduce a small sphere of radius r, carrying some charge q, into the
large one, and place it at the centre. The potential due to this new charge
clearly has the following values at the radii indicated:
Potential due to small sphere of radius r carrying charge q
0
1
4
q r
ε
=
π
at surface of small sphere
0
1
4
q R
ε
=
π
at large shell of radius R. (2.79)
Taking both charges q and Q into account we have for the total
potential V and the potential difference the values
0
1
( )
4
Q q
V R
ε R R
 
= +  
π  
0
1
( )
4
Q q
V r
ε R r
 
= +  
π  
0
1 1
( ) – ( ) –
4
q
V r V R
ε r R
 
=  
π   (2.80)
Assume now that q is positive. We see that,
independent of the amount of charge Q that may have
accumulated on the larger sphere and even if it is
positive, the inner sphere is always at a higher
potential: the difference V(r)–V(R) is positive. The
potential due to Q is constant upto radius R and so
cancels out in the difference!
This means that if we now connect the smaller and
larger sphere by a wire, the charge q on the former
FIGURE 2.32 Illustrating the principle
of the electrostatic generator.
Van de Graaff generator, principle and demonstration:


Physics
84
will immediately flow onto the matter, even
though the charge Q may be quite large. The
natural tendency is for positive charge to
move from higher to lower potential. Thus,
provided we are somehow able to introduce
the small charged sphere into the larger one,
we can in this way keep piling up larger and
larger amount of charge on the latter. The
potential (Eq. 2.78) at the outer sphere would
also keep rising, at least until we reach the
breakdown field of air.
This is the principle of the van de Graaff
generator. It is a machine capable of building
up potential difference of a few million volts,
and fields close to the breakdown field of air
which is about 3 × 106 V/m. A schematic
diagram of the van de Graaff generator is given
in Fig. 2.33. A large spherical conducting
shell (of few metres radius) is supported at a
height several meters above the ground on
an insulating column. A long narrow endless
belt insulating material, like rubber or silk, is wound around two pulleys –
one at ground level, one at the centre of the shell. This belt is kept
continuously moving by a motor driving the lower pulley. It continuously
carries positive charge, sprayed on to it by a brush at ground level, to the
top. There it transfers its positive charge to another conducting brush
connected to the large shell. Thus positive charge is transferred to the
shell, where it spreads out uniformly on the outer surface. In this way,
voltage differences of as much as 6 or 8 million volts (with respect to
ground) can be built up.
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 VP VR, 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
1
( )
4
o
Q
V
ε r
=
π
r
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
2
1 ˆ
( )
4
o
V
ε r
=
π
p r
r

FIGURE 2.33 Principle of construction
of Van de Graaff generator.
Electrostatic Potential
and Capacitance
85
The result is true also for a dipole (with charges –q and q separated by
2a) for r >> a.
4. For a charge configuration q1, q2, ..., qn with position vectors r1,
r
2, ... rn, the potential at a point P is given by the superposition principle
1 2
0 1P 2P P
1
( ... )
4
n
q n
q q
V
ε r r r
= + + +
π
where r
1P is the distance between q1 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 centered 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.
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 q1, q2 at r1, r2 is given by
1 2
0 12
1
4
q q
U
ε r
=
π
where r
12 is distance between q1 and q2.
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.E.
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
0
σ ˆ
ε
E n = where n ˆ is the unit vector along the outward normal to the
surface and σ 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 =
0
A d
ε
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
C
0 when there is no medium (vacuum),
C = KC
0
where K is the dielectric constant of the insulating substance.
Physics
86
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
1 + C2 + C3 + ...
where C
1, C2, C3... are individual capacitances.
12. The energy U stored in a capacitor of capacitance C, with charge Q and
voltage V is
2
1 1 1 2
2 2 2
Q
U QV CV
C
= = =
The electric energy density (energy per unit volume) in a region with
electric field is (1/2)ε
0E2.
13. A Van de Graaff generator consists of a large spherical conducting shell
(a few metre in diameter). By means of a moving belt and suitable brushes,
charge is continuously transferred to the shell and potential difference
of the order of several million volts is built up, which can be used for
accelerating charged particles.
Physical quantity Symbol Dimensions Unit Remark
Potential φ or V [M1 L2 T–3 A–1] V Potential difference is
physically significant
Capacitance C [M–1 L–2 T–4 A2] F
Polarisation P [L–2 AT] C m-2 Dipole moment per unit
volume
Dielectric constant K [Dimensionless]
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
0
σ ˆ
ε n outside. Electric potential is, however
continuous across the surface, equal to q/4πε0R 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.
Electrostatic Potential
and Capacitance
87
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 × 10–7C
distributed uniformly on its surface. What is the electric field
(a) inside the sphere
(b) just outside the sphere
(c) at a point 18 cm from the centre of 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 m2 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?
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.
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.
Physics
88
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 r1 and outer radius r2
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
0
( ) ˆ σ
ε
E E − n =
where n ˆ is a unit vector normal to the surface at a point and
σ is the surface charge density at that point. (The direction of
n ˆ is from side 1 to side 2.) Hence show that just outside a
conductor, the electric field is σ n ˆ /ε
0.
(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 λ 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:
Electrostatic Potential
and Capacitance
89
(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)?
(c) What are the answers to (a) and (b) above if the zero of potential
energy is taken at 1.06 Å separation?
2.19 If one of the two electrons of a H
2 molecule is removed, we get a
hydrogen molecular ion H+ 2. In the ground state of an H+ 2, 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.
(c) How much work is done in moving a small test charge from the
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 2.34 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 2.34
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.]
Physics
90
2.25 Obtain the equivalent capacitance of the network in Fig. 2.35. For a
300 V supply, determine the charge and voltage across each capacitor.
FIGURE 2.35
2.26 The plates of a parallel plate capacitor have an area of 90 cm2 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 µF 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.
2.29 A spherical capacitor consists of two concentric spherical conductors,
held in position by suitable insulating supports (Fig. 2.36). Show
FIGURE 2.36
Electrostatic Potential
and Capacitance
91
that the capacitance of a spherical capacitor is given by
0 1 2
1 2
4

r r
C
r r
πε
=
where r
1 and r2 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?
(c) Compare the capacitance of this capacitor with that of an
isolated sphere of radius 12 cm. Explain why the latter is much
smaller.
2.31 Answer carefully:
(a) Two large conducting spheres carrying charges Q1 and Q2 are
brought close to each other. Is the magnitude of electrostatic
force between them exactly given by Q1 Q2/4πε0r 2, where r is
the distance between their centres?
(b) If Coulomb’s law involved 1/r3 dependence (instead of 1/r2),
would Gauss’s law be still true ?
(c) A small test charge is released at rest at a point in an
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?
(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 107 Vm–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,
Physics
92
(c) a single positive charge at the origin, and
(d) a uniform grid consisting of long equally spaced parallel charged
wires in a plane.
2.35 In a Van de Graaff type generator a spherical metal shell is to be a
15 × 106 V electrode. The dielectric strength of the gas surrounding
the electrode is 5 × 107 Vm–1. What is the minimum radius of the
spherical shell required? (You will learn from this exercise why one
cannot build an electrostatic generator using a very small shell
which requires a small charge to acquire a high potential.)
2.36 A small sphere of radius r1 and charge q1 is enclosed by a spherical
shell of radius r
2 and charge q2. Show that if q1 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 q2 on the shell is.
2.37 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 1m2. Will he get an electric shock if he touches the metal
sheet next morning?
(c) The discharging current in the atmosphere due to the small
conductivity of air is known to be 1800 A on an average over
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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