Question

A leaky parallel plate capacitor is filled completely with a material having dielectric constant $K=5$ and electric conductivity $\sigma =7.4\times {10}^{–12}{\Omega }^{–1}{\text{m}}^{–1}$. If the charge on the plate at the instant $t=0$ is $q=8.85\mu C$, then the leakage current at the instant $t=12\text{sec}$ is (approximately)
$\times {10}^{–1}\mu A$.

Answer 1

As, in case of discharging of a capacitor through a resistance
$q=q_0{e}^{-t/CR}$
$i=-\frac{dq}{dt}=\frac{q_0}{CR}{e}^{-t/CR}$
Here, $CR=\left(\frac{{ϵ}_{0}KA}{d}\right)\left(\rho \frac{d}{A}\right)=\frac{{ϵ}_{0}K}{\sigma }$
[as $\rho =1/\sigma$]
i.e., $CR=\frac{8.846\times {10}^{-12}\times 5}{7.4\times {10}^{-12}}=6$
So,
$i=\frac{8.85\times {10}^{-6}}{6}{e}^{-12/6}$
$i=\frac{8.85\times {10}^{-6}}{6\times 7.39}=0.20\mu A$
[As $e=2.718,e^2=7.39$]