Answer to Question #138898 in Electricity and Magnetism for Suraj kishore pandit

Question

A glass of relative permitivity 4 is kept in an external electric field of magnitude 10^2 Vm^-1. Calculate the polarization vector, molecular/ atomic polarisability and the reflective index of the glass.

Answer 1

Polarization vector, $P=\chi\epsilon_{o}E$

$\epsilon=1+\chi$

$\chi=\epsilon-1=4-1=3$

Therefore, $P=\chi\epsilon_{o}E=3\times8.85\times10^{-12}\times100=26.55\times10^{-10}$ $\frac{FV}{m^{2}}$

Molecular/atomic polarisability, $P$ $=N\alpha\epsilon_{o}E$

$\alpha=\frac{P}{N\epsilon_{o}E}= \frac{26.55\times10^{-10}}{6.02\times10^{23}\times8.85\times10^{-12}\times100}$ =$5.0\times10^{-24}m^{3}$

the refractive index, $\eta$ $=\sqrt\epsilon\mu = \sqrt(4\times0.99)=1.99$