Question

The electric field between the plates of charged parallel plate capacitor decreases from 12V/m to X V/m. If the number of atoms per unit volume is N x 1028 atoms m-3, the electronic polarisability of the material is
X = 6.00 N = 13.84

Answer 1

Electronic polarisability , $\alpha_{e}$

$\alpha_{e} = \frac {\epsilon_{0}(\epsilon_{r}-1)} {N}$

we know that,

$X_{e} = \frac { \epsilon_{r}}{\epsilon_{0}}-1$

$6= \frac { 12}{\epsilon_{0}}-1$

$\epsilon_{0} = 1.7142$

Answer : $\alpha_{e} = \frac {\epsilon_{0}(\epsilon_{r}-1)} {N}$

$\alpha_{e} = \frac {1.7142(12-1)} {13.84\times 1028} = 1.3253\times10^{-3} F-m^{2}$