Electromagnetism (B U study material)

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UNIT I

Lesson – 1

ELECTROMAGNETIC THEORY AND OPTICAL PHYSICS

1.0 Aims and Objectives
1.1 (a) Electric potential at a point due to an electric dipole:
(b) Electric field at any point due to an electric dipole
1.2 Dielectric polarization
1.3 Relative permitivity (dielectric constant) and displacement vector
1.4 External field of a dielectric medium
1.5 Relation between electric displacement, electric field and polarisation (D, E and P)
1.6 Molecular field in a dielectric (Clausius-Mossotti relation)
1.7 Polarizaiton of polar-molecules (Debye’s formula)
1.8 Electrostatic energy and energy density in free space and in dielectric
1.9 Let us sum up
1.10 Check your progress
1.11 Lesson-end activities
1.12 Points for discussion
1.13 References

1.0 Aims and Objectives

This lesson deals with potential and fields due to electric dipole. The relation

between electric susceptibility, polarization, displacement will be obtained. The molecular

field, derivation of claussion mossotti relation for non-polar molecules, Debge formula for

polar molecules are explained in detail. The derivation of electrostatic energy and energy

density who has has discussed.

1.1(a) ELECTRIC POTENTIAL AT A POINT DUE TO AN ELETRIC DIPOLE:

Two charges –q at A and +q at B separated by a small distance 2d constitutes an electric

dipole and its dipole moment is p (Fig 1.1a).

Let r1 and r2 be the distances of the point P from +q and –q charges respectively. Let P

be the point at a distance r from the midpoint of the dipole O and q be the angle between PO

and the axis of the dipole OB.


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Fig 1.1(a)

1 q
Potential at P due to charge (+q) = 4pe 0 r1

1 æ q ö
Potential at P due to charge (-q) = ç-
ç r ÷
÷
4pe 0 è 2 ø

1 q 1 q
Total Potential at P due to dipole is, V = -
4pe 0 r1 4pe 0 r2

q æ1 1ö
V= ç - ÷
çr r ÷ (1)
4pe 0 è 1 2 ø

Applying cosine law,

2
r1 = r 2 + d 2 - 2rd cos q
2 æ cos q d 2 ö
r1 = r 2 ç1 - 2d
ç + 2 ÷
÷
è r r ø

Since d is very smaller than r, d2 /r2 can be neglected.

1/ 2
æ cos q ö
r1 = r ç1 - 2d ÷
è r ø


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-1 / 2
1 1 æ 2d ö
or = ç1 - cos q ÷
r1 r è r ø

Using the binomial theorem and neglecting higher powers

1 1æ d ö
= ç1 + cosq ÷ (2)
r1 r è r ø

Similarly,

r22 = r 2 + d 2 - 2rd cos (180 - q )
r22 = r 2 + d 2 + 2rd cos q

1/ 2
æ 2d ö
r2 = r ç1 + cos q ÷ (Since d2 / r2 is negligible)
è r ø

-1 / 2
1 1 æ 2d ö
or 2
= ç1 + cos q ÷
r rè r ø

Using the Binomial theorem and neglecting higher powers,

1 1æ d ö
= ç1 - cos q ÷ (3)
r2 r è r ø

Substituting equation (2) and (3) in equation (1) and simplifying

q 1æ d d ö
V= ç1 + cos q - 1 + cos q ÷
4pe 0 r è r r ø

q 2d cos q 1 p . cos q
V= = (4)
4pe 0 .r 2 4pe 0 r2

Special cases:

1. When the point P lies on the axial line of the dipole on the side of +q, then q = 0.

V = p / (4pe0 r2 )

2. When the point P lies on the axial line of the dipole on the side of –q, then q = 180.

V = -p / (4pe0 r2 )


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3. When the point P lies on the equatorial line of the dipole, then, q = 90,

V = 0.

1.1(b) Electric field at any point due to an Electric dipole

AB is an electric dipole of dipole moment p. O is its midpoint. R is a point with polar

coordinate (r, q) (Fig 1.1(b))

Fig. 1.1(b) Electric field at any point due to an Electric dipole

OR = r, LROX = q

The dipole moment p is resolved into two components along the horizontal and

vertical directions. The point R is on the axial line of a dipole of moment p cosq and on the

equatorial line of the dipole of moment p sinq.

Electric field at R along the axial line at a distance r from O

1 2p 1 2 p cos q
E1 = 3
= (along RA) ---(1)
4pe 0 r 4pe 0 r3

Electric field at R along the equatorial line at a distance r from O is

1 p 1 p sin q
E2 = 3
= (along RB) ---(2)
4pe 0 r 4pe 0 r 3


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Magnitude of the resultant electric field is,

E = Ö (E1 2 + E2 2 ) = p / (4pe0 r3 ) . Ö (4 cos2 q + sin2 q)

E = p / (4pe 0 r3 ) . Ö (1 + 3 cos2 q) ---(3)

If f is the angle between E1 and resultant, then

p sin q 4pe 0 r 3 1
tan f = 3
= tan q ---(4)
2 p cos q 4pe 0 r 2

This gives the direction of the resultant electric field.

The unit of potential difference is volt. The potential difference between two points is

1 volt if 1 joule of work is done in moving 1 Coulomb of charge from one point to another

against the electric force.

The electric potential in an electric field at a point is defined as the amount of work

done in moving a unit positive charge from infinity to that point against the electric forces.

Relation between electric field and potential

Let the small distance between A and B be dx. Work done in moving a unit positive

charge from A to B is dV = E.dx.

The work has to be done against the force of repulsion towards the charge +q. Hence,

dV = -E.dx

E = - (dV/dx)

The change of potential with distance is known as potential gradient, hence the
electric field is equal to the negative gradient of potential.
The negative sign indicates that the potential decreases in the direction of electric
field. The unit of electric intensity can also be expressed as Vm-1 .

1.2 DIELECTRIC POLARIZATION
In substances, called insulators or dielectrics, which do not have free electrons, or the
number of such electrons is too low, the electrons are tightly bound to the atom. When
potential difference is applied to insulators no electric current flows, even then their


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behaviour in fields is very important because the presence of the field may change the
behaviour of an insulator. The insulators whose behaviour gets modified in the electric field
are called as dielectrics. When the change in the behaviour of the dielectric is independent of
the direction of applied field, the dielectric is called as isotropic. On the other hand, if the
change in behaviour of the dielectric depends on the direction of applied field then the
dielectric is called as anisotropic.
If we consider a dielectric in an electric field it exerts a force on each charged particle.
The positive particles are pushed in one direction (direction of field) while the negative
particles in the opposite direction. As a result the positive and negative parts of each molecule
are displaced from their equilibrium positions in opposite directions. The overall effect is a
displacement of the entire positive charge in the dielectric relative to the negative charge. The
relative displacement of the charges is called polarization, and the dielectric is said to be
polarized.
The molecules of dielectrics are classified into two classes:

(1) polar molecules, and

(2) non-polar molecules.

The positive charge of the nucleus may be supposed to be concentrated at a point, say

the centre of gravity of positive charge. Similarly, the negative charge due to electrons in

orbit may be supposed to act at a point, known as centre of gravity of negative charge. When

two centres of gravity coincide in a molecule, the molecule as a whole possesses no resultant

charge and it is said to be non-polar, e.g., carbon tera-chloride. If two centres of gravity are

displaced from each other, the molecule as a whole possesses polarity and has permanent

electric moment and the molecule is said to be a polar molecule, e.g., CHCL3 etc. A polar

molecule in an electric field exhibits both permanent and induced dipole moments while non-

polar exhibits induced dipole moment only.

When either of these types of molecules is placed in an external field, the small

displacements of the orbital electrons will cause the distance between the centres of gravity to

change. As a result non-polar molecules become induced dipoles whereas polar molecules

will have both permanent and induced dipoles. The orientation of the induced dipoles or of


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permanent dipoles in an external electric field will be such as to align the dipole along the

field lines. Thus resultant dipole moment even in case of polar molecules gets modified.

Electric polarization effects in simple models of non-polar and polar dielectric materials are

shown in Fig 1.2

Fig 1.2 Electric polarization effect in simple modes of non-polar and polar
dielectric materials (a) Non-polar (b) Polar.

Definition of Dielectric Polarisation: In a homogeneous isotropic dielectric, the

polarisation (i.e., reorientation of electronic orbits) depends directly on the total electric fields

E in the medium. All the molecules are polarized to the same extent. The polarisation charges

within the body of the medium neutralize each other except at the surface of end faces


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(ABCD and EFGH Fig 1.2(a). The electric field due to induced charges is opposite to the

applied (external) electric field E0 . Thus resultant field E is less than the applied field E0.

Now the average dipole moment p induced in each molecule of the dielectric is proportional

to E provided E is not too great i.e.,

p µ E or p = aE ---(1)

where a is a constant called molecular polarizability and E is the intensity of field. If there
are n molecules per unit volume in a dielectric and p of each molecule be independent, we
have electric moment per unit volume = naE, and is represented by a vector P, given by
P = n a E ---(2)

Electric moment per unit volume is known as polarization.

Consider a rectangular block of dielectric material of length l [Fig 1.2a]. Let a be the
area of cross-section of the block perpendicular to the direction of the field (face ABCD or
EFGH). Let the surface density of charge appearing on these faces be sp due to polarization.
If l be the length of the block, then the charge induced on each face = sp a
Therefore total induced electric dipole moment

=spa.l………..(3)

A E

B F

+σP
-σP D
H

C G
E
Fig.1.2(a): Electric moment Calculation

Again the total induced dipole moment = P. a l
Because P is the polarization i.e., electric dipole moment per unit volume and volume of the

block is a l.

From equations (3) and (4) we have


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sp .al = P. a l

Or P =sp

Thus the polarization may also be defined as the surface density of charge appearing at faces

perpendicular to the direction of applied field.

Electrical Susceptibility:

It has been found experimentally that in a large number of dielectrics, polarization P

is proportional to the applied electric field E, atleast for field strengths that are not too large.

Thus,

POOE

= ke0 E,

Where k is dimensionless constant called electrical susceptibility of the dielectric and e 0 the

permittivity of free space, has been included to make k dimensionless. Equation is true when

the polarization properties of medium are independent of direction of polarization which is

the case for liquids, gases, amorphous solids and cubic crystals.

1.3 RELATIVE PERMITTIVITY (DIELECTRIC CONSTANT) AND

DISPLACEMENT VECTOR

The capacity of a parallel plate capacitor increases if the gap between the plates is

filled with a dielectric. The increase in capacity of capacitor with a dielectric compared to

capacity of the same capacitor with vacuum between plates, measures a constant known as

dielectric constant, relative permittivity or specific inductive capacity (S.I.C.), i.e.,

capacity with dielectric present e
er = = ,
capacity with vacuum e0

where e and e 0 are absolute permittivities of the dielectric and vacuum.

Or e = er . eo

is absolute permittivity of dielectric.


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The dielectric constant is defined in terms of electric intensity as follows: When a

dielectric is polarized, the charges appearing on the surfaces perpendicular to the field are

opposite to the real charges present on the plates. Hence the field decreases in presence of

dielectric. Therefore relative permittivity is also defined as the ratio of two fields; that is

field in vacuum E
er = = 0
field in dielectric E

q 4pe 0 r 2 e 0
= =
q 4per 2 e

1 qr 1 qr
because E0 = . and E= .
4pe 0 r 3 4pe r 3

Thus the dielectric constant may defined as ratio of the absolute permitivity of the dielectric

to that of the free space.

Since the electric intensity depends on the medium, we define another electrical

vector which depends only on the magnitude of the charge and its distribution but

independent of the nature of medium. This vector is termed as displacement, D, and is given

by

1 qr
D= .
4p r 3

æ 1 qr ö
D = eç . ÷ = eE
è 4pe r 3 ø

1 qr
E= .
4pe r 3

vector D is analogous to E. Its unit is coul. /m2 .

While E is defined as force experienced by a positive charge in any medium, the product

D.dS gives the flux of normal displacement through surface dS and remains unaltered in any


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medium. The direction of the two vectors being the same in isotropic homogeneous

dielectric.

1.4 EXTERNAL FIELD OF A DIELECTRIC MEDIUM

In the presence of an electric field, a dielectric becomes polarized and aquires a dipole

moment P, per unit volume. The calculation of electric field intently produced by elementary

dipoles though out the space at an external point will be presented below.

Consider a finite piece of polarized dielectric. Let it be characterized at each point r

by a polarization P(r). The polarization gives rise to an electric field. The aim is to calculate

the potential and field at point r[denoted by Q(r’) in Fig.1.4] outside the dielectric. Let the

volume of the dielectric is dividd into small elementary volumes and

Fig.1.4 Dielectric body with external point of observations.

consider one such volume element dV of the dielectric.

The potential due to dielectric body occupying volume dV as observed

outside dV for single dipole potential can be expressed as,


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1 P(r ) .(r - r ')
f (r ') = ò (r - r ') 3
dV
4pe 0 V

P(r ).
(r - r ') dV
1 (r - r ')
f (r ') = ò 2
4pe 0 V
(r - r ')

P(r ) .(r - r ')
0
1
f (r ') = ò (r - r ') 2
dV
4pe 0 V

where (r-r’)0 represents unit vector directed from Q towards dV. It is obviously equal

to

(x'- x )i + ( y '- y ) j + (z '- z )k
{(x'- x ) + ( y'- y ) + (z '- z )
2 2 2
}
where x’, y’,z’ are the Cartesian coordinates of point r’ and (x,y,z)those at r. Further,

é ¶ ¶ ¶ ù
(r - r ')0
= - êi (r - r ') + j (r - r ') + k (r - r ')ú
ë ¶x ¶x ¶x û

and P (r ) = iPx ( x. y.z ) + jPy ( x. y.z ) + kPz ( x. y.z )

substituing these values in equation (1), we get

1 ì Px ¶ Py ¶ Pz ¶ ü
f (r ') = ò
íV . (r - r ')dV + ò (r - r ')dV +
. ò . (r - r ')dV ý (2)
4pe 0 î (r - r ') ¶x V (r - r ') ¶y V (r - r ') ¶z
2 2 2
þ

= f1 + f 2 + f 3
Now we shall consider the integrals separately. It is obvious that

1 ¶ (r - r ') ¶ 1
- 2
. =
(r - r ') ¶x ¶x (r - r ') 2

and dV = dxdydz


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dydz

x1 x2

Fig.1.4 (a) Integration with respect to ‘x’

Therefore

1 ¶ 1
f1 =
4pe 0 òòò Px
¶x (r - r ')
2
dxdydz

Let us first integrate with respect to x. If we consider integration along x-axis only, it would

be simply a contribution of a straight rod of material with area of cross-section dydz. If such

an elements cuts the surface of the body at x1 and x2 , coordinates of respective points, then x

integral would be

x2
¶ 1 é p X ( x, y , z ) ù é p X ( x, y , z ) ù x
1 ¶Px 2

ò1 ¶x (r - r ') 2 ê (r - r ') ú ê (r - r ') ú x (r - r ') 2 ¶x dx
x
Px dx = ê
ë
ú -ê
û x= x ë
ú -ò
û x= x 1
2 1

ì é P ù é P ù ü
1 ï x x 1 ¶Px ï
f1 = í
4pe 0 ï òò ê 2
ú dydz -
ê (r - r ') ú x = x
òò ê 2
ú dydz -
ê (r - r ') ú x = x
òòò (r - r ') 2
¶x
dxdydz ý
ï
î ë û2
ë û 1 þ

if n is unit outward normal at dA then

dy dz = n.dA2 = nxdA2

and dy dz = n.dA1 = nxdA1

since at x1 which is on the part S1 , the x-component of n, i.e., nx has negative value.


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Now the integral of Φ1 can be written as

1 ì
ï Px nx Px nx 1 ¶Px ü
ï
f1 =
4pe 0
í
ï
î
ò S2
dA +
(r - r ') 2 òS1 (r - r ')
dA1 -
V
ò dV ý
(r - r ') ¶x ï þ
Since the two integrals over S1 and S2 cover whole boundary S, we have

1 Px n x 1 1 ¶ Px
f1 =
4 pe 0
ò ( r - r ' ) dA - 4 pe ò (r - r ' )
S
0
V ¶x
dV

similar integrals for Φ2 and Φ3 can be written and hence value of Φ will be

1 ì Px nx + Py n y + Pz nz
ï 1 æ ¶Px ¶Py ¶Pz ö ü ï
f (r ') = íSò dA - ç
ç ¶x + ¶y + ¶z ÷dV ý
ò ÷ ï
4pe 0 ï
î (r - r ') V (r - r ')
è ø þ

f (r ') =
1 P.n 1 (-divP )dV
or 4pe 0 ò
S (r - r ')
dA -
4pe 0 ò (r - r ')
V
(3)

Equation (3) obviously shows that the potential at point r’ due to dielectric body is the same as if it

were replaced by a system of bound charges in empty space. A part of these bound charges appears

on the dielectric surfaces as a surface density σ P ’ given by

σP ’=P.n

while the remaining bound charge appears throughout the volume V as a volume density rp’ given

as

' æ ¶P ¶Py ¶Pz ö
r P = -ç x +
ç ¶x + ÷
è ¶y ¶z ÷
ø

in the case of homogeneous and isotropic dielectrics, where the polarisation of each molecule is to

the same extent, the volume distribution of polarized charge is zero because the positive and

negative charges of dipoles neutralize each other. Hence the second term does not contribute

anything to the potential, i.e., -div P = 0. The potential is due to the surface distribution of charges

only. The intensity of electric field due to the polarized volume may be expressed as


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1 é s P (r - r ') r P (r - r ') ù
' '

E (r ') = êò dA + ò dV ú
4pe 0 ê S r - r ' 3 r - r'
3
ú
ë V
û

this gives the outer field due to polarisation.

The total bound charge of a dielectric body of volume V and complete boundary surfaces S

is zero as the total charge is

q total = s ¢dA + r ¢dV
¢ ò ò
S V

ò
= P.ndA - divPdV = 0
S
ò
V

since by divergence theorem of vector algebra,

ò P.ndA = ò div P dV
S V

Equation (3) is in M.K.S system units. If it is to be expressed in e.s.u then

1 P.n 1 1
f (r ') = ò dA - ò div P dV
e0 (r - r ') e0 (r - r ' ) (r - r ')

1.5 RELATION BETWEEN ELECTRIC DISPLACEMENT, ELECTRIC FIELD AND

POLARISATION (D, E and P)

(1) Electric Intensity: The electric intensity E at a point in the electric field is

numerically equal to the force experienced by a unit positive charge placed at that point. The

direction of E being the same as that of the field.

(2) Dielectric Polarisation: When a dielectric is placed between the two charged

metallic plates, the dielectric is polarized. The distorted atom is called as an electric dipole or

an electric doublet. The electric dipole moment per unit volume is called the dielectric

polarisation P. It is a vector parallel to the electric field.


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(3) Electric Displacement: When no electric field is present, Gauss’ law gives

q
ò E.dS = E0 A =
e0

q
E0 =
e0A
where E0 is the electric field and q is the free charge.

When dielectric is present, induced surface charges say-q’ appears, so that Gaussian

surface encloses (q-q’) charge . Gaussian’s law then becomes

q - q¢
ò E.dS = E A = e0

q q¢
E= -
e0A e0A

Now the field when dielectric is present is given by

E0 q
E= =
e r e re 0 A

q q q¢
= -
e re 0 A e 0 A e 0 A

q q q¢
= +
e 0 A e re 0 A e 0 A

q æ q ö q¢
= e0ç ÷
çe e A÷- A
A è r 0 ø

q
= e0E + P
A
As the quantity on left hand side is termed as electric displacement D, we write


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D = e0E + P

It can be noted that D is connected with free charge, q only. P is connected with

polarisation charges, q¢ only and E is connected with all charges that are actually present

whether free or polarised.

The relationship between D, E and P can also be obtained as follows:

The differential form of Gauss law is

r r
Ñ.E = t (1)
e0

where rt is the total charge density. For dielectrics, rt must include charge densities of real

and polarized charges, that is,

rt = r + r¢,

where rt is volume density of real charge and r¢ that of polarized charges.

Therefore eq. (1) becomes

r r - r¢
Ñ.E =
e0

r r¢ r
Ñ.E = (2)
e0 e0
r
We have shown that r ¢ = -Ñ.P which when substituted in eq.(2) gives

Ñ ¢P r
Ñ ¢.E + =
e0 e0

r
Ñ.(e 0 E + P ) = P
r
Ñ.D = r

D = e0E + P (3)


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and defines electric displacement vector D. Eq. (3) shows that D is responsible for the partial

field due to real charges, and that D and E are additive. This result is called differential form

of Gauss’ theorem in a dielectric. The curl of the displacement vector, on the other hand, can

be written as
r r
Ñ ´ D = Ñ ´ (e 0 E + P ) (4)
r r
= e 0Ñ ´ E + Ñ ´ P (5)

in case of electric field of solenoidal type
r
Ñ ´ E =0

in case of polarisation of isotropic homogeneous dielectric
r
Ñ´ P = 0
r
so that Ñ´ D = 0

in such cases. But curl of displacement vector D does not vanish in the case of non-solenoidal

type of fields and anisotropic dielectric.

Further relation(4) can be put in the form,

D = e0E + P

æ P ö
= e0çE +
ç
÷
÷
è e0 ø

æ ke P ö
= e0çE + 0 ÷
ç
è e0 ÷ø

= e 0 E (1 + k ) (6)

= e 0e r E

where (1 + k ) = e r


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e
is the relative permitivity of medium defining dielectric constant by e r = , so that
e0

D=εE (7)

Now P is related to E by the relation

P = D- ε0 E = ε0 εr E - ε0 E

= ε0 (εr-1)E (8)

A relation can also be developed between r and r’ as follows:
r
r ¢ = -Ñ.P
r
= -Ñ.e 0 (e r - 1)E

r æ 1 ö
= -Ñ.e 0e r E ç1 -
ç e ÷
÷
è r ø

r æ 1 ö
( )
= Ñ.D ç1 -
ç e ÷
÷ by eq.(5)
è r ø

æ 1 ör
= -ç1 -
ç e ( )
÷ Ñ.D
÷
è r ø

æ 1 ö
= - r ç1 -
ç e ÷
÷
è r ø

As e r > 1, r¢ < -r that is bound charge density is always opposite in sign and less in

magnitude than free charge density.

1.6 MOLECULAR FIELD IN A DIELECTRIC (CLAUSIUS-MOSSOTTI RELATION)

The aim of this section is to examine the molecular nature of the dielectric and to

consider how the electric field & responsible for polarizing the molecule is related to the

macroscopic electric filed. The electric field which is responsible for polarizing a molecule of


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the dielectric is called the molecular field. This is the electric field at a molecular field or

local field or local field is produced by all external sources and by all polarized molecules in

the dielectric, except one molecule under consideration.

Fig.1.6(a) Fig.1.6(b)

The molecular field, Ein, may be calculated in the following way. Let us cut a

spherical cavity of radius r (such that its dimensions are very great as compared to the

molecular dimensions and very small as compared to the volume of the dielectric)

surrounding the point at which the molecular field is to be computed. The dielectric which is

now left, will be treated as a continuum. The cavity is put in its original position in the

dielectric (molecule by molecule) except the molecule where the molecular field is to be

computed. The molecules which are just replaced in the cavity are treated as individual

dipoles and not as a continuum.

Let the dielectric be placed in the uniform electric field between two parallel plates

(of a condenser) as shown in fig (1.6b). The dotted lines show the boundary of the dielectric.

Let the surface density of real charges on the capacitor plates be s. Again let the surface of

cavity has polarized charges of surface density sp .


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The field experienced by the molecule of the dielectric at the centre of the cavity C,

Ein , given by

Ein = E1 + E2 + E3 + E4

Where

(i) E1 is the field between two plates with no dielectric, so that E1 = s / e 0 ,

(ii) E2 is the field at C due to polarized charges on the plane surfaces of the

dielectric facing the capacitor plates and is given by E2 = sp / e0 .

(iii) E3 is the field at C due to polarized charges on the surface of cavity, to be

calculated.

(iv) E4 is the field at C due to permanent dipoles. But in present case for non-polar

isotropic dielectrics E4 = 0.

s sp
Thus, E in = - + E3 (1)
e0 e0

Evaluation of E3 : Consider a small elemental area ds on the surface of cavity of an

angular width dq and an angle q with the direction field E. The vector P shows the direction

of displacement at the centre of ds, Fig (1.6a). The normal component of displacement is

PN = P cos q.

By definition of polarization, it is the surface charge per unit area. Such a charge on ds to

provide flux normal to ds is,

PN = P cos q ds

and the electric intensity at C due to this charge is given by,

P cos qds
=
4pe 0 r 2

where r is the radius of cavity. The field is directed along the radius CA. Resolving the

intensity along and perpendicular to the applied field, we have the components


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P cosqds
= cosq along the field
4pe 0 r 2

P cosqds
= sin q perpendicular to field
4pe 0 r 2

If the area ds be taken round through 2p radians about LM, it will describe a ring, the

surface area of which is given by,

= 2pr sin q .r.dq

= 2pr 2 sin qdq

Therefore intensity at C due to ring in the field direction is

P cos 2 qds
= 2
2pr 2 sin qdq
4pe 0 r

P
= 2
cos 2 q sin qdq
2e 0 r

while the normal components of intensity due to the ring cancel each other.

Integrating the intensity at C due to charges on the surface of cavity, we have

P p
E3 = ò cos 2 q sin qdq
2e 0 0

P 2
= ´
2e 0 3

P
=
3e 0

therefore the total intensity or Ein at C is given by eq.(1) as

s sp P
Ein = - + (2)
e 0 e 0 3e 0

The resultant field, E between the plates is


P a g e | 23

s sp
= -
e0 e0

P
Therefore Ein = E +
3e 0

Further we know that D = eE = e 0 E + P

P
So that E=
e -e0

Putting this value in eq.(2) for Ein, we get

P P
Ein = +
e - e 0 3e 0

P æ e + 2e 0 ö
= ç
ç e -e ÷
÷ (3)
3e 0 è 0 ø

if the number of the molecules per unit volume , a molecular polarizability then polarization,

P, is defined as electric moment per unit volume . That is

P = naE in

Putting this value in eq,(3), we get

P P æ e + 2e 0 ö
= ç ÷
na 3e 0 ç e -e ÷
è 0 ø

na æ e - e 0 ö
=ç ÷
3e 0 ç e + 2e 0
è
÷
ø

e e 0 -1
=
e e0 + 2

na e -1
= r
3e 0 e r + 2


P a g e | 24

where e r is the relative permittivity or dielectric constant. Equation (6) is well known

Claussius-Mossotti relation.

We know that e r = mg2 where mg is the refractive index of gas. Hence equation (6) can

be expressed in terms of refractive index. The relation in that case is

na m g2 - 1
= (4)
3e 0 m g2 - 2

called Lorentz formula and is valid only as long as e r is frequency independent.

Validity of Claussius-Mossotti Relation:

The number of molecules per unit volume is proportional to the density of gas. If

er -1
molecular polarizability a is taken to be a constant, the number n is proportional to .
er+2

Also the constant a is proportional to the cube of the radius of molecule. Hence finding e r

experimentally and calculating n from the density at definite temperature and pressure, the

value of the radius of the molecule of dielectric may be reckoned. The values, so obtained,

fairly agree with the values obtained by other methods and proves the validity of the relation.

It is true only in case of monatomic gases and weak solutions. The experimental and

theoretical values disagree in case of strong solutions and solids. It is due to the fact that in

these cases the interaction forces among molecules are sufficiently great, the account of

which has not been taken.

Molar Polarizability:

Multiplying Eq. (4) by M*/d on both sides, where M* is the molecular weight and d

the density of the dielectric, we have

er -1 M * a nM *
. = .
er +2 d 3e 0 d

But (M*n / d) = N and is called the Avogadro’s number. Therefore


P a g e | 25

Na e r - 1 M *
= .
3e 0 er +2 d

where Na is known as molar polarizability.

1.7 POLARIZAITON OF POLAR-MOLECULES,(DEBYE’S FORMULA):

Dependence of polarizability on temperature.

While discussing the polarization of non-polar molecules, it has been observed that

the molecular polarizability, a, does not depend on the temperature. But such is not the case

with polar molecules. If polar molecules are subjected to electric field then in addition to

induced polarization, the permanent polarization of molecules is also present. In absence of

electric field, the permanent dipoles point in random directions and the net polarization is

zero. On application of field, the permanent dipoles are forced to orient in the direction of the

field. But the orientations along the field direction are continuously disturbed by thermal

agitation. Therefore an equilibrium is attained and the dipoles make

Fig.1.7

zero to p angles with field direction. The polarization of such permanent molecules may

therefore be resolved in the direction of the field and the average value of Pp cos q may be

found (fig 1.7). Thus we have Pp the permanent polarization, and Pi the induced polarization.

Therefore total polarization is given by

P T = Pp + Pi


P a g e | 26

If Ein is the inner field acting on the dielectric molecules, the potential energy of

permanent dipoles per unit volume of moment Pp is given by

Ein . Pp = Ein . Pp cos q.

Calculation of the average value of Pp cos q:

Following the statistical method of Boltzmann used in kinetic theory of gases, the

probability that a molecule is found in an energy state W, at temperature T is proportional to

e-W/KT, where K = R/N is the Boltzmann constant, R and N are gas constant and the Avogadro

number respectively. In present case, the potential energy of dipoles per unit volume is given

by - Ein . Pp. Hence probability of finding dipoles at an angle q with the field direction is

given by

Pq µ eEin . Pp /KT

Or Pq µ eEin . Pp cos q /KT

The number of dipoles at an orientation will be proportional to its probability. Hence

if Nq is the number of electric dipoles per unit volume at an angle q and Np/2 at q/2, we have

Nq µ eEin . Pp cos q /KT

Np/2 µ 1

(Nq / Np/2) = eEin . Pp cos q /KT

or Nq = Np/2 eEin . Pp cos q /KT

Differentiating, we obtain the number of dipoles in an angular width dq as

E in Pp cos q kT E in P p
dN q ® q + d q = - N p 2 e sin q d q
kT

The average moment per unit volume in the direction of field is given by

p

òP
0
p
cos q dN q

P p cos q = p

ò dN
0
q


P a g e | 27

Substituting the value of dNθ, we get

Pp2 Ein p

ò( )
E P cos q kT
N p 2 cosq sin qe dq in p

Pp cosq = KT 0

Pp Ein p

ò( )
E P cos q kT
N p 2 sin qe dq in p

KT 0

p

ò (cosq sin qe )dq Ein Pp cos q kT
Pp
0
Pp cosq = p

ò (sin qe )dq Ein Pp cos q kT

0

Pp Ein
u=
KT
p
cos q

Pp cosq =
Pp òu 0
cosq sin q dq
p
cos q
òu 0
sin q dq

PE in
t= = u cos q
KT

dt = u sin q dq

we get

u
Pp ò e t (u t )dt
-u
Pp cosq = u
t
ò e dt
-u

Pp [te t - e t ]- u
u

=
u [e t ]- u
u

Pp é u (e u + e - u ) e u - e - u ù
= - u
u ê e u + e -u
ë e - e -u ú û


P a g e | 28

æ 1ö
= Pp ç coth u - ÷
è uø

æ Pp E in KT ö
= Pp ç coth - ÷
ç KT Pp E in ÷
è ø

which is called Langevin equation

when the inner field, Ein , is small and temperature T is large

PpE in
i.e., áá 1
KT
then we can expand the exponentials in Langevin equation as

æ cosh u 1 ö
Pp cosq = Pp ç
ç sinh u - u ÷
÷
è ø

éæ u2 u4 ù
ç1 +
êç + .......... ú
êè 2! 4! 1ú
= Pp -
ê æ u3 ö uú
ê çu +
ç + ......... ÷
÷ ú
ë è 3! ø û

-1
éæ u 2 öæ u3 ö 1ù
= Pp êç1 +
ç + ....÷ç u +
÷ç + ....÷ - ú
÷
êè
ë 2! øè 3! ø uú
û

Pp u 2
= .
u 3

Pp u
=
3

Pp Ein
putting u=
KT
we get average polarisation of permanent molecules in the direction of the field as


P a g e | 29

Pp2 Ein
Pp cosq =
3KT

PT = Pi + P cos q

Pp2 Ein
PT = Pi +
3KT
Pi = a i E in

Pp2 Ein
PT = a i Ein +
3KT

æ Pp2 ö
= ça i +
ç
÷ Ein
÷
è 3KT ø

= a T E in

where a T is the total polarizability per unit volume is

Pp2
aT = ai +
3KT
Pp2
where a i is induced polarizability per unit volume and is permanent polarizability per
3KT

unit volume.

Study of Molecular Structure and Debye’s Relation:

The Claussius-Mossotti relation, modified for polar molecules, shows that the

constant a is temperature dependent. The modification was utilized by Debye to study

molecular structure. The modified Claussius-Mossotti relation or Debye’s relation is

e r - 1 aT
º
er +2 3e 0


P a g e | 30

2
er -1 1 é Pp ù
º ê + ai ú
er+2 3e 0 ë 3KT û

Pp2 1 ai
= .+
9e 0 K T 3e 0

e -1 Pp2 1 ai
y= r m= x= c=
er+2 9e 0 K T 3e 0

For non-polar molecules Pp = 0 and therefore m = 0, ie., the straight line is parallel to

er -1
x-axis. Therefore, if is plotted as ordinate against 1/T as abscissae, a straight line
er+2

parallel to x-axis is obtained.

er -1
The plot of against 1/T for CHCl3 , the chloroform, and CCl 4 , the carbon
er+2

tetrachloride, is shown in fig 1.6a. We interpret that:

(i) The line for CCl4 is parallel to x-axis which shows that its molecules are

non-polar. It implies that the chlorine atoms in CCl4 are symmetrically

placed with respect to carbon atom, so that the center of gravity of negative

charges coincides with that of the positive charges and no permanent

dipoles are present.

(ii) The straight line for CHCl3 is inclined and shows the presence of polar

molecules. The polarity appears due to the charge asymmetry of the

molecule. The permanent dipole moment per unit volume is calculated

from the slope of the line. The dipole moment is then calculated on

dividing it by the number of dipoles per unit volume n. Constant a i is

determined from the intercept on y- axis.


P a g e | 31

Fig.1.6(a)

The study has been extended to a great number of molecules of monatomic gases like

Argon and molecules like H2 which are made up of two similar atoms. They do not

exhibit charge asymmetry and are therefore non-polar. But for molecules made up of

different atoms the charge asymmetry comes in and the molecules act like permanent

dipoles. The examples are NH3 , H2 O etc.

1.8 ELECTROSTATIC ENERGY AND ENERGY DENSITY IN FREE SPACE AND

IN DIELECTRIC

We shall first calculate the potential energy of a group of n point charges. It will be
equal to the work done in assembling the charges, bringing in one at a time.
If in the field of a stationary charge q1 , another charge q2 is brought from infinity to a
distance r12 = r1 - r2 then work done is

1 q1 q 2
DW2 = (1)
4pe 0 r12
=If a third charge q3 is brought from infinity to a point distant r13 and r23 from charges q1 and
q2 respectively then work done is
1 q1 q 2 1 q1 q 2
DW = + (2)
4pe 0 r 13 4pe 0 r23
From eqs. (1) and (2), the total work done in assembling the three charges is


P a g e | 32

DW = DW2 + DW3

1 é q1 q 2 qq qq ù
= ê + 1 2 + 1 2ú
4pe 0 ë r12 r 13 r23 û
which can be extended for an assembly of n charges as

1 é q1 q 2 q1 q 3 q 2 q3 qi q j ù
= ê + + + .......... + + .....ú (3)
4pe 0 ê r12
ë r13 r23 rij ú
û
1 n n 1 qi q j
=
2
å å i =1 j =1
4pe 0 rij

where the prime on the second summation implies that the terms i = j are to be excluded from
1
the sum. The factor has been introduced because of the fact that each pair of charges
2
occurs twice in the summation.
Equation (3) then represents the total electrostatic energy U of the assembly of n charges and
if we put
n 1 qi
å j =1
4pe 0 q j
= ji

In eqn. (3), we get
1 n
U=
2
å i =1
q ij i

If the point charges have been assembled in a linear dielectric medium, instead of vacuum or
free space, then free space permittivity e 0 is to be replaced by absolute permittivity e = e 0 er ,
where e r is the relative permittivity or dielectric constant of the medium.
Arbitrary Charge Distribution:
The expression for the field energy, U can be expressed with much advantage in terms of
volume and surface integrals involving the useful quantities electric field E and displacement
D when there is arbitrary charge distribution.
Let us consider a system of arbitrary charge distribution with density of initial charge
distribution asp. Let us remove the charge to infinity so that at any time the charge density is
ar where a is a parameter with values is ranging from unity to zero. At a point, where
density is ar, the charge dq’ in a volume element ‘dv’ will be
dq’ = (ar) dv (1)
and the potential f’ = af (2)
because f µ r and f is the initial potential at that point.


P a g e | 33

As the charge is removed to infinity, charge density at every point in the distribution will fall,
say, from ar to (a - da) r so that decrease in the charge contained in the volume element dv
will be
dq” = (da) r dv (3)
And the energy withdrawn from the system in removing this charge from volume element dv
is
dU = f’ dq” (4)
= (af) (da) r dv (5)
On using Eqs. (2) and (3).
If the whole charge from the system is removed to infinity i.e., the charge density is
reduced to zero everywhere then the total energy withdrawn, equal energy of the system, will
be
1
U = ò a da ò r f dv
0
v

1
æa 2 ö
U =ç
ç 2 ÷ òv r f dv
÷
è ø0
1
= ò r f dv
2 v

We know that r = div D
1
So that field energy is U= f div D dv
2ò

Further
div (fD) = f div D + D grad f
1
U= ò div (fD ) - D. grad f dv
2 v
so that
1 r r
= ò Ñ. (fD) - D.Ñ f dv
2 v
r
But E = - Ñ f,
1 r 1
so that U= ò Ñ. (fD) dv + 2 ò D.E dv
2 v v


P a g e | 34

Changing the first volume integral into surface integral by Gauss divergence theorem, we get
1 1
U= ò fD dS + 2 ò E.D dv (6)
2 s v

as v can be any volume which includes all the charges in the system, we can choose the
bounding surface S at large distance from the charge distribution. At large distances
1
fa
r
1
Da
r2
dS a r 2
1
so that ò f D. dS = r
S

®0 when r®¥
Eq. (6) is then left with
1
V = E.D dv
2ò
(7)
v

1
predicting that the energy is distributed with a density ( E.D ) per unit volume. Hence the
2
energy density i.e., the ener gy per unit volume in an electrostatic field is
1
u= ( E .D ) (8)
2
Eq. (7) holds for linear dielectrics.
Form linear dielectric,
D=eE
1
So that energy density is u= e E2 (9)
2
In free space e = e0 (since e r = 1), we have energy density of electrostatic field in free
1
space as given by u= e0 E2 (10)
2
Eqs. (8) and (9) hold good for isotropic homogeneous dielectric in which D and E are
proportional to each other. But if the medium is anisotropic, the permittivity e has different
values in different directions and consequently e which relates D and E is a tensor e 0b. The
relation can be written as

Electromagnetism (B U study material)

Electromagnetism (B U study material)

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