The material filled between the plates of a parallel plate capacitor has resistivity 200 Ωm. The value of capacitance of the capacitor is 2pF. If a potential difference of 40 V is applied across the plates of the capacitor, then the value of maximum leakage current flowing out of the capacitor is (Given, the value of relative permittivity of material is 50.)

The given situation is shown below

$\begin{array}{l}\mathrm{Resistivity}, \mathrm{\rho }=200\mathrm{\Omega m}\text{, }\\ \mathrm{C}=2\mathrm{pF}=2\times {10}^{-12}\mathrm{F}\\ \mathrm{V}=40\mathrm{V}\\ \mathrm{K}\text{ or }{\mathrm{\epsilon }}_{\mathrm{r}}=50\\ {\mathrm{i}}_{\text{leakage }}=\text{ ? }\end{array}$

$\mathrm{C}=\frac{{\mathrm{K\epsilon }}_0\mathrm{A}}{\mathrm{d}}$ (K is the dielectric constant or relative permittivity) and $\mathrm{R}=\frac{\mathrm{\rho d}}{\mathrm{A}}$

Now, charge will be discharged through the resistance between the plates.

Now, time constant (τ) of discharging,

              $\begin{array}{l}\mathrm{\tau }=\mathrm{RC}=\frac{\mathrm{\rho d}}{\mathrm{A}}\times \frac{{\mathrm{K\epsilon }}_0\mathrm{A}}{\mathrm{d}}\\ \mathrm{\tau }={\mathrm{\rho K\epsilon }}_0\end{array}$

For a given R-C circuit, the discharged current is given by

              $\begin{array}{l}\mathrm{i}=\frac{\mathrm{Q}}{\mathrm{RC}}{\mathrm{e}}^{-\frac{\mathrm{t}}{\mathrm{RC}}}\\ \mathrm{i}=\frac{\mathrm{Q}}{{\mathrm{pK\epsilon }}_0}{\mathrm{e}}^{-\frac{\mathrm{t}}{{\mathrm{pK\epsilon }}_0}}\end{array}$

The above discharge current is the leakage current,

               ${\mathrm{i}}_{\text{leakage }}=\frac{\mathrm{Q}}{{\mathrm{\rho K\epsilon }}_0}{\mathrm{e}}^{-\frac{\mathrm{t}}{{\mathrm{\rho K\epsilon }}_0}}$

Maximum leakage current,

               $\begin{array}{l} {\left({\mathrm{i}}_0\right)}_{\text{leakage }}=\frac{\mathrm{Q}}{{\mathrm{\rho K\epsilon }}_0}=\frac{\mathrm{CV}}{{\mathrm{\rho K\epsilon }}_0}\\ =\frac{2\times {10}^{-12}\times 40}{200\times 50\times 8.85\times {10}^{-12}}\\ =903\mathrm{\mu A}=0.9\mathrm{mA}\\ \therefore {\left({\mathrm{i}}_0\right)}_{\text{leakage }}=0.9\mathrm{mA}\end{array}$