Question

Question #ccd75

Answer 1

Consider a Dielectric medium.

When placed in an external electric field `vec E`, the field polarizes the material (tiny little electric dipoles orienting in the direction of the field).

The process can take place in two ways.

If it is a polar Dielectric, i.e. the molecules have permanent dipole moment, then the electric field orients the dipoles in it's direction. The thermal agitation opposes this and soon an equilibrium is established so that a steady number of molecular dipoles are oriented in the direction of the field.

If however, the molecules have no net intrinsic dipole moment, the electric field acts on the electron cloud stretching it to induce dipole moments in molecules. Then a similar process of orientation of dipoles in the direction of the field takes place even though such a process is never complete because of the thermal agitation.

In either case, the material gets polarized, and the Polarization vector `vec P` is defined as,

`vec P = vec p/V` where `V` is the volume.

Thus polarization is dipole moment `vec p` in unit volume.

Now, from Gauss' law,

`nabla*vec E = rho/epsilon_0`

Where `rho` is total charge density which is sum of free charge density `rho_f` and bound charge density `rho_b`, the latter is due to Polarization only.

Now, `rho = rho_f + rho_b`

But, we have an important result from Electrostatics that,

`nabla*vec P = -rho_b`

Thus, Gauss' law in the presence of a dielectric becomes,

`nabla*vec E = (rho_f - nabla*vec P)/epsilon_0`

`implies nabla*(epsilon_0vec E) + nabla*vec P = rho_f`

`implies nabla*(epsilon_0vec E + vec P) = rho_f`

The electric displacement is then defined as,

`vec D = epsilon_0vec E + vec P` and then Gauss' law becomes,

`nabla*vec D = rho_f`

The displacement vector thus let's us write Gauss' law in terms of free charges only. It is only the free charges that we can control externally, the bound charges are a result of Polarization only.