**An atom is **said to be polarized if it possesses and effective dipole moment, that is, if there is a separation between the centres of negative and positive charge distribution. It is important to realize that a polarization p=0 does not mean that the material does contain dipole moment , but only the vector sum of all dipole moments zero . This will, always be the case if the dipole moments vectors are randomly distributed with respect to their directions. Let us consider a simple plate capacitor or condenser with a homogenously polarized material inside its plate's .More generally; this describes an isotropic dielectric slab of material in a homogenous electrical field. We have the subsequent idealized condition. To calculate the polarization p for the polarized dielectric, we need to sum of all the dipoles in the medium and divide by the volume Ad. _{However}, the polarized medium can be represented in terms of surface charge +QP and -QP, which are separated by the thickness distance d. We can view this arrangement as one big dipole moment P_{total }

-QP to +QP. Thus P_{total =} Qp.d

Since the polarization is defined as the total dipole moment per unit volume, the magnitude of p is

P= T_{otal} /volume = Qp.d /Ad = QP / A

But QP /A is the surface polarization charge density sigma p, so

P=sigma p

Polarization is a vector and equation only gives its magnitude. For the rectangular slab, the direction of P is normal to surface. For +σP, it comes out from the surface and for -σ, it is directed into the surface.

The polarization P induced in a dielectric medium when it is placed in an electric field depends on the field itself. To express the dependence of P on the field E, we define a quantity called the electric susceptibility ? e by

P=? e.σ0.E

An effect P due to a cause E and the quantity ? e relates the effect to its cause. Electric susceptibility, ? e is a material quantity that measures the extent of polarization in the material per unit field. Further, polarizibility is defined by

I_{nduced}=alpha

So, P=N i_{nduced }=N .alpha E

Where N is the density of dipoles. Then ? e and alpha are related by

? e=N. alpha/epsilen0

It is important to recognize the difference between free and polarization charges. The charge

Q_{o }and Q, before and after the dielectric insertion are free charges that arrive on the plates from the battery. The polarization charges +QP to -QP on the other hand, are bound to the atoms. They cannot move within the dielectric or on its surface. The field E before the dielectric was inserted is given by

E=V/d=Q_{0}/c_{o }d

After the insertion of the dielectric, this field remains the same v/d, but the free charges on the plates are different. The free surface charge on the plates in now Q in addition, there are bound polarization charges on the dielectric surfaces next to the plates. It is apparent that the flow of current during the insertion of the dielectric, is due to the additional free Q-Q _{0} needed on capacitor plates to neutralize the opposite polarity polarization charges U_{p} appearing on the dielectric surfaces, Q-U_{p} must be the same as before, Q_{0} , so that the field, does not change inside the dielectric that is,

Q-U_{p}=Q_{0}

Or Q=Q_{0}+Q_{P}

Dividing by A, defining sigma=Q/A as the free surface charge density on the plates with the dielectric inserted, we obtain

σ=ε0.E+σP

Since σP=P and P=zie .ε .E, we can eliminate σ P to obtain

σ=ε0 (1+zie.e) E

From the definition of the relative permittivity ?, we have

ε.r=Q/Q_{0}=σ/σ.0

So substituting for σ@, we obtain

ε.r=1+?

In terms of polarizability, this is

ε.r=1+N.@/ε.0

There are essentially four types of polarization Mechanisms.

(1) Electronic polarization

(2) Ionic polarization

(3) Orientation Polarization

(4) Interface polarization