This document discusses conductors and dielectrics. It defines conductors as materials that allow free movement of charges, like metals. The key properties of conductors are that the electric field inside is zero, the charge density inside is zero, and free charges exist only on the surface. These properties influence how conductors behave in external electric fields, inducing opposite charges on surfaces. The document also discusses equipotential surfaces, Poisson's and Laplace's equations, and provides examples of calculating electric fields and charge distributions for various conductor configurations.

1. Conductors and
Dielectrics
Prepared by:
Anuj Watal (130010111051)
Guided by:
Prof. Shailesh Khant

2. Conductors
A conductor (typically, a metal like Cu, Ag etc. or ionic conductors like HCl or
NaCl dissolved in water) allows free movement of charges. They have low resis-
tivity 10−8Ωm as compared to typical insulators like quartz, glass etc. which have
resistivity of the order of 1017Ωm. However, the property that really distinguishes
a metal from insulators or semi-conductors is the fact their temperature coefficient
of resistivity is positive while that of semi-conductors is negative.
• The electric field inside a conductor is zero. In an equilibrium situa- tion,
there cannot be an electric field inside a conductor as this would cause
charges (electrons or ions) to move around. In the presence of an external
field, there is charge separation inside a conductors with opposite charges
accumulating on the surface. This creates an internal electric field which
cancels the effect of the external field in such a way that the net electric
field inside the conductor volume is zero.
+
+
+
+
+
+
+
+
2
− E
−
−
−
−
−
−
−
int
Eext
• Charge density inside a conductor is zero. This follows from Gauss’slaw
∇ ·E = ρ/s0
As E˙ = 0,the charge density ρ = 0.

3. (This does not suggest that there is no charge inside, only that the positive
and negative charges cancel inside a conductor.)
• Free charges exist only on the surface of a conductor. Since there is no
net charge inside, free charges, if any, have to be on the surface.
• At the surface of a conductor, the electric field is normal to the surface.
If this were not so, the charges on the surface would move along the surface
because of the tangential component of the field, disturbing equilibrium.
E=0
Induced Charges in a conductor:
The above properties of a conductor influence the behaviour of a conductor placed
in an electric field. Consider, for instance, what happens when a charge +q is
brought near an uncharged conductor. The conductor is placed in the electric field
of the point charge. The field inside the conductor should, however, be zero. his
is achieved by a charge separation within the conductor which creates its own
electric field which will exactly compensate the field due to the charge +q. The
separated charges must necessarily reside on the surface.
Another way of looking at what is
happening is to think of the free
charges in the conductor being at-
tracted towards (or repelled from)
the external charge. Thus thesurface
nal charge is oppositely charged. To
keep the charge neutrality, the sur-
face away from the external charge
is similarly charged.
+q
++
+
+
+
of the conductor towards the exter- +
+
++
−
− −
−
−
−
−
−−
3

4. Example 1 :
A charge Qis located in the cavity inside a conducting shell. In addition, a charge
2Q is distributed in the conducting shell. Find the distribution of charge in the
shell. What is the electric field in the region outside the shell.
+Q
4
+
+
+
+
+
+
+
+
+
+
+ +
+
+
+
+
+
+
++
Take a gaussian surface entirely within the conducting shell, completely enclosing
the cavity. Everywhere on the gaussian surface E˙= 0. The flux and therefore, the
charge enclosed is zero within the gaussian surface. As the cavity contains a
charge Q, the surface of the cavity must have charge −Q. As the conductor has
distributed charge 2Q, the charge on the outside surface is 3Q.
The principle illustrated in the above problem is known as Faraday’s Cage. If a
hollow conducting box is kept in an electric field, the charges in the cavity are
redistributed in such a way that the electric field inside the cavity is zero. This is
used to provide an enclosure for sensitive electronic equpment which must be
kept free of external electronic disturbance.
Example 2 :
Calculate the electric field outside a conductor carrying a surface charge density
σ.

5. ++ +
+ +
dSE
+
+  ++
+ +
r +
L
+ +
+ +
+
+
+
+
+
+
+
+
+ ++
+ +++
Take a gaussian pillbox in the shape of a cylinder of height h with h/2 inside
and h/2 outside the conductor. Lat the cross sectional area be dS normal to the
surface. The electric field is normal to the surface. As the field inside is zer and
there is no tangential component of the field at the surface, the flux goes out only
through the outer cap of the cylider. The charge enclosed is σdS and the flux is
EdS. The electric field is normal to the surface. applyinng Gauss’s law
σ
s0
E = nˆ
.
Exercise :
Two parallel, infinite plates made of material of perfect conductor, carry charges
Q1 and Q2. The plates have finite thickness. Show that the charge densities on the
two adjecent inside surfaces are equal and opposite while that on the two outside
surfaces are equal. (Hint : Field inside the plates due to four charged surfaces
must be zero.)
Poisson’s and Laplace’s Equations
Differential form of Gauss’s law,
∇˙·E˙ =
ρ
s0
5
Using E˙ = −∇V,
∇˙E˙ = −∇ ·(∇V ) = −∇2
V

6. so that
∇2
V = −
ρ
s0
This is Poisson equation. In cartesian form,
∂2V ∂2V ∂2V ρ
∂x2 +
∂y2 +
∂z2 = −
s0
For field free region, the equation becomes Laplace’s equation
∇2
V = 0
Equipotential surface
Equipotential surfaces are defined as surfaces over which the potential is constant
V(r˙)= constant
At each point on the surface, the electric field is perpendicular to the surface since
the electric field, being the gradient of potential, does not have component along
a surface of constant potential.
• We have seen that any charge on a conductor must reside on its surface.
These charges would move along the surface if there were a tangential com-
ponent of the electric field. The electric field must therefore be along the
normal to the surface of a conductor. The conductor surface is, therefore,
an equipotential surface.
• Electric field lines are perpendicular to equipotential surfaces (or curves)
and point in the direction from higher potential to lower potential.
• In the region where the electric field is strong, the equipotentials are closely
packed as the gradient is large.
6

7. −0.5 kV
3 kV
2 kV
1 kV
P  x
−1 kV
 0
−2 kV
The electric field strength at the point P may be found by finding the slope of the
potential at the point P. If ∆x is the distance between two equipotential curves
close to P,
E = −
∆V
∆x
where ∆V is the difference between the two equipotential curves near P.
Example 3:
Determine the equipotential surface for a point charge.
Solution : Let the point charge qbe located at the origin. The equation to the
equipotential surface is given by
V(x, y, z) =
1 q
4πs0
,
x 2 + y2 + z2
7
= V0 = constant

8. Equipotential surfaces (magenta)
and field lines (blue) for a positive
charge.
Thus the surfaces are concentric spheres with the origin (the location of the charge)
as the centre and radii given by
R =
q
4πs0φ0
The equipotential surfaces of an electric dipole is shown below.
Electric Field and Equipotential lines
for an electric dipole
8

9. Example 4 :
Determine the equipotential surface of an infinite line charge carrying a positive
charge density λ.
Solution :
Let the line charge be along the z- axis. The potential due to a line charge at a
point P is given by
λ
2πs0
V (r) = − lnr
where r is the distance of the point P from the line charge. Since the line charge
along the z-axis, r =
,
x 2 + y2 so that
V(r) = −
λ
4πs0
ln(x2
+ y2
)
The surface V = constant = V0 is givenby
ln(x2
+ y2
) = −
4πs0V0
λ
i.e.
x2
+ y2
= e−
4πs0V0
λ
which represent cylinders with axis along the z-axis with radii
r = e−
2πs0V0
λ

1
9
++++++++++++++++++++++++++++++
z−axis

2
As V0 increases, radius becomes
smaller. Thus the cylinders are
packed closer around the axis, show-
ing that the field is stronger near the
axis.

10. Example 5 :
Consider a charged sphere of radius R containing charge q, completely enclosed
by a spherical cavity of inner radius a and outer radius b.Calculate the charge
density on all surfaces and potential everywhere.
Solution :
R
ab
+q
−q
+q
As field inside the conductor is zero,
by taking a Gaussian surface com-
pletely in the region a < r < b, we
must have net charge enclosed by
such a surface to be zero. To com-
pensate for the charge q that exists
on the surface of the inner sphere,
the charge on the inside surface of
the shell must be −q. Since the shell
is charge neutral, a charge +q must,
therefore, appear on the outside sur- Gaussian surface
face of the shell.
For r > b, the field is
E˙ =
4πs0 r2
1 q
rˆ
The corresponding potential is
V(r) = −
¸ r
∞ 4πs0 r2
1 q
dr =
1 q
4πs0 r
At r = b,the potential is
V(b) =
1 q
4πs0 b
Since the field between a and b is zero, this is also the potential at all points from
r = bto r = a.
1 q
4πs0 b
V(a ≤ r ≤ b)=
For R < r < a, the potential is given by
V(r) =
1 q
4πs0 b
−
¸ R
a
1 q
4πs0 r2
=
1 . q q
+ −
4πs0 b R a
10
q.

11. If the outer surface is grounded, the potential on the shell becomes zero. There is
no charge on the outer surface. However the inner surface must have a charge −q
to keep the field in the shell zero,
R
V (0) = V (R) = −
¸
E˙·d˙l =
a
1
4πs0 R a
.
1 1
.
−
Exercise :
Determine the equipotential surface of an infinite plane with charge density σ.
Laplace’s Equation
Let us look at Laplace’s equation in one dimension. It becomes
d2V
dx2
= 0
which has the solution
V = mx + c
The solution shows two important characteristics of the solution of Laplace’s
equation, which are not immediately obvious in higher dimensions. The first
property is the potential at a point can be expressed asaverage of potentials at
neighbouring points. For instance,
1
V (x) = (V (x + x0) + V(x − x0))
2
This also illustrates the second property of the solutions, viz., the solution has no
local minimum or maximum. If it did, it would not be possible to express the
function as average of values at neigbouring points.
To see this consider a function
a
f (x, y) = (x2
+ y2
)
4
in two dimensions, which does not satisfy Laplace’s equation as
∂x2
2 2
∇2
V =
∂ f
+
∂ f
∂y2 = a
11
The function has a positive curvature everywhere and there exists a local minimum
at x = 0,y = 0.The function looks like thefollowing.

12. 0
2
4
6
8
10
12
14
16
18
-3 -2 -1 0 1 2
-2
3-3
-1
0 1
01..015523..235545..455567..675589..8955110111..0155112311..2355114511..4555116711..675518
x
2 3
y
V(x,y)
Consider, on the other hand, a function V(x, y) that satisfies Laplace’s equation
a
V (x, y) = (x2
− y2
)
12
4
The function has no minimum or maximum and looks like the following. It has a
saddle point at x = 0,y = 0.

13. -10
-8
-6
-4
-2
0
2
4
6
8
10
-3 -2 -1 0 1 2
-2
3-3
-1
0 1
--87--..9855--65--..7655--43--..5455--21--..3255-00-.015512..125534..345556..565578..78559
x
13
2 3
y
V(x,y)
An interesting consequence of Laplace equation is Earnshaw Theorem which
states that a charge cannot be held in stable equilibrium only by electrostatic
forces.
For instance, suppose we position a charge Q exactly at the centre of a cube which
has a positive charge q at each of its eight corners. We would expect the charge to
be in equilibrium as it is being pulled equally in all directions. However, this will
not be a stable equilibrium because at the centre, there being no charge density,
Laplace equation is obeyed. Thus there cannot be a minimum of the potential V
and hence of potential energy QV of the charge at the centre.
Consider again the case of cavity in a conductor. If the interior of the cavity does
not contain any charge, Laplace equation is obeyed. Thus the potential has no
minimum or maximum inside the cavity. Further, since the boundary of the cavity
is an equipotential, the potential inside the cavity is also constant.
Uniqueness Theorem :
This theorem states that the solution of Laplace’s equation is uniquely deter-
mined by the values of potential on the boundaries.
Suppose V1 and V2 are two potentials which satisfy Laplace’s equation in some
region with identical coundary conditions, i,e V1(boundary) = V2(boundary).
Consider a function V3 = V1 −V2. This satisfies Laplaces equation with thecondi-

14. tion V3(boundary) = 0. However, as V3 does not have a minimum or a maximum
in the region, its value has to be the same value as its value at the boundary, i.e.
V3 is constant. Hence V1 = V2.
Laplace’s Equations in 3-dimensions
We will consider the solutions of Laplace’s equations in problems with spherical
geometry having azimuthal symmetry. The equation to be solved is
∇2
V =
1
r2 ∂r
∂
.
r2
+
∂V
.
1
∂r r2 sin θ∂θ
∂
. 2∂V
.
1 ∂ V
sin θ + = 0
∂θ r2 sin θ∂φ2
where we have explicitly written down the Laplacian operator in spherical polar
coordinates.
For problems with azimuthal symmetry, ∂V/∂φ = 0so that we have
∂
.
r 2 ∂V
∂r ∂r
.
+
.
1 ∂
sinθ ∂θ ∂θ
.∂V
sin θ = 0
The equation above is conveniently solved by a technique called separation of
variables where we write the function V (r, θ) as a product of two functions, one
R(r) which is a function of radial variable r only and the other a function Θ(θ)
which is a function of the angle variable θ alone. Writing V (r, θ) = R(r)Θ(θ)
and dividing throghout by RΘ, weget
r21 ∂
.
∂R
.
1
R ∂r ∂r Θsin θ ∂θ
∂
.
+ sinθ
∂Θ
.
∂θ
= 0
Since the two terms on the left depend on two independent variables, this equation
can be satisfied only if each of the term equals to constants of opposite sign. We
write
r21 ∂
.
∂R
.
= l(l + 1)
1
Θsin θ ∂θ
R ∂r
∂
.
sinθ
∂r
∂Θ
.
∂θ
14
= −l(l + 1)
We will not attempt to solve these equations but merely quote the results. The
solution of the angular equation is in terms of what are known as Legendre Poly-
nomials. Pl(cos θ).These are polynomials of degree lin cosine of angle θ. The
first few polynomials are as follows :

15. P0(cos θ) = 1
P1(cos θ) = cos θ
1
P2(cos θ) =
P3(cos θ) =
2
1
2
(3 cos2
θ − 1)
(5 cos3
θ −3cos θ)
-0.5
0
0.5
1
-1
-1 -0.5
Pl(cos)
0 0.5 1
cos 
P0
P1
P2
P3
P4
V(r, θ)=
The solution of radial equation is consists of a power series in r and 1/r. The
complete solution is
∞
.
i=0
.
l
l
A r +
Bl
rl+1
.
lP (cosθ) (A)
We will illustrate the use of these solution by an example.
Example 6 :
Consider an uncharged conducting sphere in a uniform electric field and deter-
mine the potential at all points in space.
Solution :
The sphere, being a conductor, is an
equipotential. Let the potential be
zero. Far from the sphere, the field
is uniform. Let the field strength be
E0 and be in z-direction,
The boundary conditions are :
V = 0at r = R
V = −E0z = −E0 cos θ for r R. −
−
− − − −−
−
+
+ + +
+
+
Using Eqn. (A) and substituting the first boundary condition, we get a relationship
between Al and Bl
Al Rl
+
Bl
Rl+1
15
= 0

16. Thus Bl = −AlR2l+1. Thus, wehave
∞ .
V (r, θ) =
.
Al rl
+
i=0
rl+1
R2l+1.
Pl(cosθ)
For r R, we may neglect the second term in bracket and get
∞.
Alrl
Pl(cos θ) = −E0r cosθ
i=0
On comparing both sides, we get l = 1 which gives A1 = −E0. Substituting these
we get
R3
r2
V (r, θ) = −E0(r − )cos θ
The induced charge density is
∂V
σ = −s0 |r +R = 3s0E0 cosθ
16
∂r
It can be seen that the charge density is positive in the upper hemisphere and
negative in the lower hemisphere.
Dielectrics
A conductor is characterized by existence of free electrons. These are electrons
in the outermost shells of atoms (the valence electrons) which get detatched from
the parent atoms during the formation of metallic bonds and move freely in the
entire medium in such way that the conductor becomes an equipotential volume.
In contrast, in dielectrics (insulators), the outer electrons remain bound to the
atoms or molecules to which they belong. Both conductors and dielectric, on the
whole, are charge neutral. However, in case of dielectrics, the charge neutrality is
satisfied over much smaller regions (e.g. at molecular level).
2.9.1 Polar and non-polar molecules :
A dielectric consists of molecules which remain locally charge neutral. The
molecules may be polar or non-polar. In non-polar molecules, the charge cen-
tres of positive and negative charges coincide so that the net dipole moment of
each molecule is zero. Carbon dioxide molecule is an example of a non-polar
molecule.

17. +6e+8e
Oxygen atom
+8e
Oxygen atomCarbon atom
+8e
17
+e
+e
Hydrogen atom
Hydrogen atom
In a polar molecules, the arrange-
ment of atoms is such that the
molecule has a permanent dipole
moment because of charge separa-
tion. Water molecule is an example
of a polar molecule.
Oxygen atom
When a non-polar molecule is put in an electric field, the electric forces cause a
small separation of the charges. The molecule thereby acquires an induced dipole
moment.
A polar molecule, which has a dipole moment in the absence of the electric field,
gets its dipole moment aligned in the direction of the field. In addition, the mag-
nitude of the dipole moment may also increase because of increased separation of
the charges.

18. A non−polar molecule
in an Electric Field
E=0
E
+
− +
E=0
E
A polar molecule in an
Electric Field
Dielectric in an Electric Field
A dielectric consists of molecules which may (polar) or may not (non-polar) have
permanent dipole moment. Even in the former case, the dipoles in a dielectric
are randomly oriented because dipole energies are at best comparable to thermal
energy.
+
+
+
+
+
+
+ +
+
+
18
+
+
+
+
+ +
+
+
+ + +
++
+ +
+
+
Randomly oriented dipole in a dielectric (E=0)
Polarised Dipoles in an electric field
When a dielectric is placed in an electric field the dipoles get partially aligned in
the direction of the field. The charge separation is opposed by a restoring force

19. due to attaraction between the charges until the forces are balanced. Since the
dipoles are partially aligned, there is a net dipole moment of the dielectric which
opposes the electric field. However, unlike in the case of the conductors, the net
field is not zero. The opposing dipolar field reduces the electric field inside the
dielectric.
Dielectric Polarization
Electric polarization is defined as the dipole moment per unit volume in a di-
electric medium. Since the distribution of dipole moment in the medium is not
uniform, the polarization P˙ is a function of position. If p˙(r˙) is the sum of the
dipole moment vectors in a volume element dτ located at the position ˙r,
p˙(r˙)= P˙(r˙)dτ
It can be checked that the dimension of P˙is same as that of electric field divided
by permittivity s0. Thus the source of polarization field is also electric charge,
except that the charges involved in producing polarization are bound charges.
Denoting the local bound charge density by ρb, one can write
∇˙·P˙ = −ρb
The equation above is obtained in a manner that is identical to the way we de-
rived the equation ∇˙ ·E˙ = ρ/s0. The absence of the factor so in the equation is
because of the dimensional difference between E˙ and P˙ while the minus sign
arises because the dipole moment vector (and hence the polarization) is defined to
be directed from negative to positive charge as against E˙ which is directed from
positive to negative charge. Clearly, if polarization is uniform, the volume den-
sity of bound charges is equal to zero. Even in this case, there are surface bound
charges given by the normal component of the polarization vector. Summarizing,
we have,
∇ ·P˙ = −ρb P˙ ·nˆ= σb
We will derive these relations shortly.
Free and Bound Charges
The charge density of a medium consists of free charges, which represent a surplus
or deficit of electrons in the medium, and bound charges. The term free charge is
19

20. used to denote any charge other than that due to polarization effect. For instance,
the valence charges in a metal or charges of ions embedded in a dielectric are
considered as free charges.
The total charge density of a medium is a sum of free and bound charges
ρ = ρf + ρb
Gauss’s Law takes the form
∇˙·E˙ =
ρ
=
ρf + ρb
s0 s0
Potential due to a dielectric
Consider the dielectric to be built up of volume elements dτ . The dipole moment
of the volume element is P˙dτ .
The potential at a point S, whose po-
sition vector with respect to the vol-
ume element is ˙ris
dV =
4πso
1 P˙ ·rˆ
r2
dτ
P d
r
S
The potential due to the whole volume is
V =
1
4πs0
¸
volume
P˙·rˆ
r2
dτ =
1
4πs0
¸
volume
1
r
P˙ ·∇( )dτ
where, we have used
1
r
∇( ) =
rˆ
r2
19
Use the vector identity
∇˙·(A˙f(r)) = A˙·∇f (r) + f (r)∇˙ ·A˙
Substituting A˙ = P˙ and f (r) = 1/r,

21. ∇˙·(
P˙
r
) = P˙ ·∇( ) +
1 1
r r
∇˙·P˙
we get
V =
1
4πso
¸
vol
P˙
r
∇˙ ·( )dτ −
1
4πso
¸
1
vol r
∇˙·P˙dτ
The first integral can be converted to a surface integral using the divergence theo-
rem giving,
V =
1
4πso
¸
P˙
surface r
·dS˙−
1
4πs0
¸
1
vol r
∇˙·P˙dτ
The first term is the potential that one would expect for a surface charge density
σb where
σb = P˙ ·nˆ
where nˆis the unit vector along outward normal to the surface. The second term
is the potential due to a volume charge density ρb given by
ρb = −∇˙ ·P˙
The potential due to the dielectric is, therefore, given by
V =
1
4πso
¸
surf ace
σbdS
r
+
1
4πs0
¸
vol
ρbdτ
r
and the electric field
E˙ = −∇V
1
=
4πs0
¸
dS +
1
4πs0
¸
ρbrˆσbrˆ
surface r2
vol r2
29
dτ
Electric Displacement Vector D˙
The electri displacement vector D˙ is defined by
D˙ = s0E˙ + P˙
which has the same dimension as that of P˙. The equation satisfied by D˙ is

22. ∇˙ ·D˙ = s0∇˙ ·E˙ + ∇˙ ·P˙ = ρ − ρb = ρf
which is the differential form of Gauss’s law for a dielectric medium.
Integrating over the dielectric volume,
¸
∇˙·D˙dτ =
¸
volume volume
ρfdτ = Qf
where Qf is the free charge enclosed in the volume. The volume integral can be
converted to a surface integral using the divergence theorem, which gives
¸
surface
D˙ ·dS˙ = Qf
Thus the flux over the vector D˙over a closed surface is equal to the free charged
enclosed by the surface.
Example 5:
An uncharged spherical dielectric has polarization vector given by P˙= k˙r. Find
the electric field both outside and inside the dielectric.
Solution :
The dielectric has both bound surface charge and volume charge. The surface
charge density is σb = P˙·nˆ= kR where R is the radius of the sphere. The
volume charge density is
ρb = −∇ ·P˙ = −k∇ ·˙r= −3k
. The field inside the dielectric is given by Gauss’s law,
4πr2
E =
Q
s0
encl 34πr ρb
=
3s0
which gives
rρb
3s0
E = = −
kr
s0
22
The field outside is zero.
Example 6 :
Consider a spherical dielectric shell of inner radius a and outer radius b.The
space in the region between r = a and r = b is filled with a dielectric hasving
r
polarization P˙ = k rˆ. Determine the field inside and outside the shell.

23. Solution :
The charge densities are,
σouter−surface
b
k= P˙ ·nˆ=
b
σinner−surface
b
k= −P˙ ·nˆ= −
a
ρb = −∇ ·P˙ = −
k
r2
For r < a, no charge is included, hence the field is zero. For a < r < b, the
charges enclosed by a Gaussian surface are the surface bound charges on the inner
surface and the volume charge within the region. Thus
Qencl
r
a
b= 4πb2
σin
+
¸
ρ 4πr2
dr
a
r
a
= −4πa2 k
+
¸
(−k/r2
)4πr2
dr
= −4πka + 4πk(a − r) = −4πkr
Thus E = −(k/s0r)rˆ. For a Gaussian surface outside, the total charge enclosed
can be similarly calculated to be zero, so that field is zero.
Example 7 :Electric Field Due to Uniformly polarized sphere :
Since the polarization is uniform,
the bound charge density is zero.
Only on the surface, there are bound
charges. We have
σb = P˙ ·nˆ= P cosθ
where θ is the angle between the
direction of the external field (z-
direction) and a point on the sphere.
P
23
n^
z^

This is, once again, a problem with azimuthal symmetry with no charges inside
or outside the sphere. Hence Laplace’s equation is satisfied both in the interior of

24. the sphere and outside.
V(r, θ)=
∞
.
l=0
.
l
l
A r +
Bl
rl+1
.
lP (cosθ)
For r < R, the second term must vanish since the potential cannot become infinity
at the origin. Similarly, for r > R, the first tem must vanish as the potential must
be well defined at large distances.
For r < R,
V(r, θ) = Alrl
Pl(cos θ)
and, for r > R
V(r, θ)=
Bl
rl+1
Pl(cosθ)
At r = R, the potential is continuous. Hence,
Bl = AlR2l+1
At r = R, while the tangential component of the field is continuous, the normal
component has a discontinuity,
E above
n
below σ
0
− En =
s
nˆ
Using E˙ = −∇V,
∂Vabove
∂r
−
∂Vbelow
∂r
= −
σ
s0
Thus,
−
∞
.
l=0
.
B l
rl+2 − Alr l− 1
.
Pl(cos θ) |r=R= −
P cosθ
s0
Comparing both sides, we see that only l = 1 term is non-zero. We get,
2B1 P
R3
+ A1 =
s0
Using B1 = A1R3, we get A1 = P/3s0 and B1 = PR3/3s0 Finally, we get
V(r, θ) = r cos θ for r < R
=
P
3s0
P R3
3s0 r3
24
cos θ for r > R

25. The electric field inside the sphere is uniform and is equal to −∇V= −P/3s0zˆ.
Outside the sphere, the potential has the same form as that of a giant dipole with
dipole moment equal to volume of the sphere times the polarization vector, located
at the centre, because,
V =
P R3
3s0 r2
cosθ
=
3s0
3p/4πR3 R3
r2
cosθ
=
4πs0
1 pcos θ
r2
=
4πs0
1 p˙·rˆ
r2
Constitutive Relation
Electric displacement vector D˙ helps us to calculate fields in the presence of a
dielectric. This is possible only if a relationship between E˙ and D˙ is known.
For a weak to moderate field strength, the electric polarization P˙ is found to be
directly proportional to the external electric field E˙. We define Electric Suscepti-
bility χ through
P˙ = s0χE˙
so that
D˙ = s0E˙ + P˙
= so(1 + χ)E˙ = s0sr E˙ = sE˙
where κ ≡ sr = 1 + χ is called the relative permittivity or the dielectric constant
and s is the permittivity of the medium. Using differential form of Gauss’s law for
D˙, we get
∇˙·E˙ =
1
∇˙ ·D˙ =
ρf
s s
25
Thus the electric field produced in the medium has the same form as that in free
space, except that the field strength is reduced by a factor equal to the dielectric
constant κ.
Capacitance filled with Dielectric
If a material of dielectric constant κ is inserted between the plates of a capacitor,

26. the field E˙is reduced by a factor κ. The potential between the plates also reduces
by the same factor κ.
φ −→
φ
κ
Thus the capacitance
C =
Q
φ
+Q
−Q

increases by a factor κ.
Example:
A parallel plate capacitor with plate separation 3.54mm and area 2m2 is initially
charged to a potential difference of 1000 volts. The charging batteries are then
disconnected. A dielectric sheet with the same thickness as that of the separation
between the plates and having a dielectric constant of 2 is then inserted between
the capacitor plates. Determine (a) the capacitance , (b) potential difference across
the capacitor plates, (c) surface charge density (d) the electric field and (e) dis-
placement vector , before and after the insertion of the dielectric .
Solution :
(a) The capacitance before insertion of the dielectric is
A
d
C = s0 = 8.85×10−12 2
3.54 ×10−3
= 5 ×10−9
F
After the insertion the capacitance doubles and becomes 10−8 F.
(b)Potential difference between the plates before insertion is given to be 1000 V.
On introducing the dielectric it becomes half, i.e. 500 V.
(c)The charge on each capacitor plate was Q = CV = 5 × 10−6 coulomb, giving
a surface charge density of 2.5 × 10−6 C/m2. The free charge density remains the
same on introduction of the dielectric.
(d) The electric field strength E is given by
σ
E = = 2.8× 105
volts/meter
26
s0
The electric field strength is reduced to 1.4 × 105 volt/meter oninsertion.
(e) The displacement vector remains the same in both cases as the free charge
density is not altered. It is given by D = σ = 2.5 × 10−6 C/m2.
Example :
The parallel plates of a capacitor of plate dimensions a × b and separation d are

27. Forcex
x−axis
charged to a potential difference φ and battery is disconnected. A dielectric slab
of relative permittivity κ is inserted between the plates of the capacitor such that
the left hand edge of the slab is at a distance x from the left most edge of the
capacitor. Calculate (a) the capacitance and (b) the force on the dielectric.
y−axis
a
b
d
z−axis
Solution :
Since the battery is disconnected, the potential difference between the plates will
change while the charge remains the same. Since the capacitance of the part of
the capacitor occupied by the dielectric is increased by a factor κ, the effective
capacitance is due to two capacitances in parallel ,
b
C = s0 [x + (a −x)κ]
d
The energy stored in the capacitor is
U = =
Q2 Q2
d 1
2C 2 bs0 x + (a −x)κ
Let F be the force we need to apply in the x-direction to keep the dielectric in
place. For an infinitisimal increment dx of x, we have to do an amount of work
Fdx , which will increase the energy strored in the field by dU , so that
F =
dU
dx
27

28. the differentiation is to be done, keeping the charge Q constant. Thus
F =
dQ2 κ− 1
2bs0 [x + (a − x)κ]2
Since κ > 1, F is positive. This means the electric field pulls the dielectric inward
so that an external agency has to apply an outward force to keep the dielectric in
position. Since the initial potential difference φ is given by Q/C, one can express
the force in terms of this potential
F =
s0b
2d
φ2
(κ −1)
This is the force that the external agency has to apply to keep the left edge of the
dielectric at x. The force with which the capacitor pulls the dielectric in has the
same magnitude.
Example 22 :
In the above example, what would be the force if the battery remained connected
?
Solution :
If the battery remained connected Q does not remain the same, the potential φ
does. The battery must do work to keep the potential constant. It may be realised
that the force exerted on the dielectric in a particular position depends on the
charge distribution (of both free and bound charges) existing in that position and
the force is independent of whether the battery stays connected or is disconnected.
However, in order to calculate the force with battery remaining connected, one
must, explicitly take into account the work done by the battery in computing the
total energy of the system. The total energy U now has two parts, one the work
done by the external agency Fdx and the other the work done by the battery, viz.,
φdQ where dQ is the extra charge supplied by the battery to keep the potential
constant. Thus
U= Fdx + φdQ
which gives
F =
dU
dx
−φ
dQ
dx
Since φ is constant, we have
1
2
28
cφ2
U =
Q = Cφ

29. Using these
F = φ21 dC dC
− φ2
= − φ21 dC
2 dx dx 2 dx
(Note that if the work done by the battery were negnected, the direction of F will
be wrong, though, because we have used linear dielectrics, the magnitude,
accidentally, turns out to be correct !)
In the previous example, we have seen that
bs0
d
C = [x + (a − x)κ]
giving
dC bs0
dx d
= (1 −κ)
which is negative. Thus F is positive, as before,
F =
bs0φ2
d
(κ −1)
Example :
The space between the plates of a parallel plate capacitor is filled with two differ-
ent dielectrics, as shown. Find the effective capacitance.
 1
 2
d1
d2
29
Solution :

30. Take a Gaussian pill-box as shown.
Wehave
¸
D˙ ·d˙S = ρfree
= 0
as there are no free charges inside
the dielectric. Contribution to the in-
tegral comes only from the faces of
the pill-box parallel to the plates and
d˙S1 = d˙S2.Hence,
D1 = D2 = σ
d1
d2
 1
 2
dS
dS2
1
dz
Let φ1
where σ is the surface density of free
charges.
be the potential difference between the upper plate and the interface between the
dielectric and φ2 that between the interface and the lower plate. We have
φ = φ1 + φ2
= E1d1 + E2d2
=
D1
κ1s0
d1 +
D2
κ2s0
d2
= +
σd1 σd2
κ1s0 κ2s0
Thus the effective capacitance is given by
C =
Q
φ
=
σA
σ
.
d1 d2
.
=
s0 κ1
+ κ2
As0
+d1 d2
κ κ
=
1 2
C1C2
C1 + C2
30
where C1 and C2 are the capacitances for parallel plate capacitors with one type
of dielectric with separations d1 and d2 between the plates respectively.

31. Example :
A capacitor consists of an inner conducting sphere of radius R and an outer con-
ducting shell of radius 2R. The space between the spheres is filled with two dif-
ferent linear dielectrics, one with a dielectric constant κ from r = R to r = 1.5R
and the other with dielectric constant 2κ from r = 1.5R to r = 2R. The outer
shell has a charge −Q while the inner conductor has a charge +Q. Determine the
electric field for r > 0and find the effective capacitance.
Solution :
The electric field is radially symmet-
ric and may be obtained by apply-
ing Gauss’s law for the displacement
vector
¸
D˙ ·d˙S = 4πr2
D = Qfree
where Qfree is the free charge en-
closed within a sphere of radius r.
For r < R, the field is zero as the
free charges are only on the surface
of the inner cylinder.
+
+
+
+ +
+
+ +
+ −
−
−
−
−
−
−
−
−
−
−
R
1.5R
2R
For R <
r < 1.5R, the electric fieldis
E = =
D Q
κs0 4πs0κr2
and for 1.5R < r < 2R,
E = =
D Q
2κs0 8πs0κr2
30
For r > 2R, the field is zero. The fields are radial with the ineer sphere at a
higher potential. The potential difference is calculated by taking the taking the

32. line integral of the electric field along any radial line.
¸
R
2R
∆φ = E˙ ·d˙l =
¸
E dr
=
¸ 1.5R
R
Q
4πκs0r2
dr +
¸ 2R
Q
1.5R 8πκs0r2
dr
=
5Q
48πκs0
The effective capacitance is
C = =
Q 48πκs0
∆φ 5
D = s0κE = s0
.
1 +
x
d
.
E
As the insertion of dielectric does
not affect free charges, the displace-
ment vector D˙ is remains the same
as it would in the absence of the di-
electric. Thus D˙ = ˆıσ.

Example :
Aparallel plate capacitor has charge densities ±σ on its plates which are separated
by a disance d. The space between the capacitor plates is filled with a linear but
inhomogeneous dielectric. The dielectric constant varies with distance from the
positive plate linearly from a value 1 to a value 2 at the negative plate. Determine
the effective capacitance.
2
As the dielectric is linear,
1
d
x
0
distance from positive plate
dielectricconstant
++++++
____
Thusthe
electric field E˙ is given by
E˙ = ˆı
σa
s0(x + d)
31
The field close to x = d is given by E = σ/2s0, which shows that adjacent to
the negative plate there is a positive charge density σ/2. To find the effective

33. capacitance, we find the potential difference between the plates by integrating the
electric field
0 s0
φ =
¸
Edx =
σd
¸d d
0
dx
d + x s0
σd
= ln2
so that
C = =
Q Aσ
φ φ
=
As0
dln 2
The polarization P is given by
P = D − s0E =
σx
x + d
The volume density of bound charges, given by ∇ ·P˙ = −ρb is found as follows:
ρb = −σ
dx
d
.
x
.
σd= −
x + d (x + d)2
The bound charge density on the surface, given by nˆ·P˙= P , has a value σ/2 on
the dielectric adjacent to the negative plate (x = d). As the dielectric is charge
neutral, this requires a net volume charge of −σ/2 in the dielectric. This can be
verified by integrating over the volume charge density ρb given above.
Exercise :
A parallel plate capacitor of plate area S and separation d, contains a dielectric of
thickness d/2 and of dielectric constant 2, resting on the negatve plate.
d/2
d−
33
+
A potential difference of φ is maintained between the plates. Calculate the electric
field in the region between the plates and the density of bound charges on the
surface of the dielectric. [Ans. field in empty region = 4φ/3d, within dielectric
= 2φ/3d, bound charge density = 2s0φ/3d]

34. Exercise :
The permittivity of a medium filling the space between the plates of a spherical
capacitor with raddi a and b (b > a) is given by
s =
.
2s0a ≤ r ≤ (a+ b)/2
4s0 (a + b)/2 ≤ r ≤ b
Find the capacitance of the capacitor, distribution of surface bound charges and
.
1
the total bound charges in the dielectric. [Ans. C = 8πs0 −
1
a a + b 2b
34
1
. − 1
− ,
bound charges on dielectric surface with radii a, (a + b)/2 and b are respectively
−σ/2, 3σa2/(a + b)2 and 3σa2/4b2]