The gap between the plates of a parallel plate capacitor is filled with glass of dielectric constant k = 6 and of specific resistivity 100 G$\mathrm{\Omega }$m. The capacitance of the capacitor is 4.0nF . When a voltage of 2.0kV is applied to the capacitor, the  leakage current of the capacitor will be

Let Q₀ be the initial charge on the capacitor and  V₀ its initial potential. Let A be the area of the plates and $\mathrm{\ell }$ be the distance between them.
Capacitance C of the capacitor is $\mathrm{C}=\frac{{\mathrm{k\epsilon }}_0\mathrm{A}}{\mathrm{\ell }}$
Resistance of the capacitor is $\mathrm{R}=\frac{\mathrm{\rho \ell }}{\mathrm{A}}=\frac{{\mathrm{\rho k\epsilon }}_0}{\mathrm{C}}$
If q be the charge at time t, then the leakage current 
i is  $\mathrm{i}=\frac{\mathrm{dq}}{\mathrm{dt}}=\frac{\mathrm{V}}{\mathrm{R}}$ where V is the volt age across the capacitor at time t. But q = CV
Hence $\mathrm{i}=\frac{\mathrm{dq}}{\mathrm{dt}}=\mathrm{C}\frac{\mathrm{dV}}{\mathrm{dt}}=\frac{\mathrm{V}}{\mathrm{R}}$
Hence $\frac{\mathrm{dV}}{\mathrm{V}}=-\frac{1}{\mathrm{CR}}\mathrm{dt}$The minus sign is because dV is negative.
Integrating between limits, $\mathrm{V}={\mathrm{V}}_0\exp \left(-\frac{\mathrm{t}}{\mathrm{CR}}\right)$
Hence , $\mathrm{i}=\frac{\mathrm{V}}{\mathrm{R}}=\frac{{\mathrm{V}}_0}{\mathrm{R}}\exp \left(-\frac{\mathrm{t}}{\mathrm{CR}}\right)$