Question

determine the value of E in a material for which the electric susceptibility is 3.5 and p = 2.3 x 10 x 87 C/m^2

Answer 1

We know that the polarization density can be expressed as

$$P=\epsilon_0\chi E,\\\space\\ E=\frac{P}{\epsilon_0\chi}=\frac{2.3·10^{87}}{(8.85·10^{-12})·3.5}=7.42·10^{97}\text{ V/m}.$$

Quite a big number.