Answer:

Relation between polarization P, Susceptibility $\chi$ and $\in_{r}$:

Consider a parallel plate capacitor with plates between which an electric field $E_{0} $exists. If $\sigma$ is the change/unit area then using gauss law.

$E_{0} = \dfrac{\sigma}{ E_{0} }$ …………………………….(1)

Let the dielectric slab be placed between the two plates. Due to polarization, charges appear on the two faces of the slab, and establish yet another field within the dielectric. Let the field be $E_{i}$.As explained earlier, the field due to polarization will be in a direction apposite to the of $E_{0}$.

\therefore the resultant field E in the material can be written as,

$E= E_{0} - E_{i}$…………………………………………..…………(2)

If $\sigma_{p}$ is the charge/unit area on the slab surfaces, then by the following equation (1) we write

$E_{i} = \dfrac{\sigma_{p}}{ \epsilon_{0} }$ …………………………….(3)

\therefore from equation (1),(2),(3)

$E= \dfrac{\sigma}{ \epsilon_{0} } - \dfrac{\sigma_{p}}{ \epsilon_{0} }$  or  $\in_{0} E = \sigma - \sigma_{p}$…………………………….…(4)

Since magnitude of polarization P = charge/unit area,

$P = \sigma_{p}$

Also by gauss law we know that,

$D = \sigma$

Where D is the electric flux density

$\therefore$ equation (4) can be written as,

$\epsilon_{0} E = D – P$ or$ D = \epsilon_{0} E + P$………………………………………….(5)

We know by our studies of electrostatics,

$D = \epsilon_{0}\epsilon_{r} E$

Substituting the above in equation (5) we get

$\in_{0}\in_{r} E = \epsilon_{0}E + P$

OR

$ P = \epsilon_{0}(\in_{r} - 1)E$………………………………………………………………….(6)

OR

$P = \epsilon_{0}\chi E$,

Where$ \chi = (\epsilon_{r} - 1)$, is the dielectric susceptibility of the material.