Seawater at a frequency $\mathrm{f}=9\times {10}^2 \mathrm{Hz}$ , has permittivity $\in =80{\in }_0$ and resistivity $\mathrm{\rho }=0.25\mathrm{\Omega m}$. Imagine a parallel plate capacitor is immersed in seawater and is driven by an alternating voltage source $\mathrm{V}\left(\mathrm{t}\right)={\mathrm{V}}_0\sin \left(2\mathrm{\pi ft}\right)$. Then the conduction current density becomes ${10}^{\mathrm{x}}$ times the displacement current density after time $\mathrm{t}=800\mathrm{s}$. The value of $\mathrm{x}$ is__________.

$\left(\text{ Given : }1/4\pi {\in }_0=9\times {10}^9{\mathrm{Nm}}^2{\mathrm{C}}^{-2}\right)$

Given $\mathrm{f}=9\times {10}^2 \mathrm{Hz}$

$\in ={\in }_0{\in }_{\mathrm{r}} ; \in =80{\in }_0 ; \mathrm{\rho }=0.25\mathrm{\Omega m}$

$\mathrm{V}\left(\mathrm{t}\right)={\mathrm{V}}_0\sin \left(2\mathrm{\pi ft}\right)$

$\text{ Displacement current, }{\mathrm{I}}_{\mathrm{d}}=\mathrm{dq}/\mathrm{dt}=(\mathrm{CdV}/\mathrm{dt})$

$\Rightarrow {\mathrm{I}}_{\mathrm{d}}=\frac{{\in }_0{\in }_{\mathrm{r}}\mathrm{A}}{\mathrm{d}}\frac{\mathrm{d}}{\mathrm{dt}}\left({\mathrm{V}}_0\sin \left(2\mathrm{\pi ft}\right)\right)$

$\Rightarrow {\mathrm{I}}_{\mathrm{d}}=\frac{{\in }_0{\in }_{\mathrm{r}}\mathrm{A}}{\mathrm{d}}{\mathrm{V}}_0\left(2\mathrm{\pi f}\right)\cos \left(2\mathrm{\pi ft}\right)----\left(1\right)$

Where d is the distance between plates & 

Conduction current ${\mathrm{I}}_{\mathrm{c}}=\mathrm{V}/\mathrm{R}$

$\Rightarrow {\mathrm{I}}_{\mathrm{c}}=\frac{{\mathrm{V}}_0\sin \left(2\mathrm{\pi ft}\right)}{\mathrm{\rho }\frac{\mathrm{d}}{\mathrm{A}}}=\frac{{\mathrm{AV}}_0\sin \left(2\mathrm{\pi ft}\right)}{\mathrm{\rho d}}----\left(2\right)$

Divide equation (1) and (2)

$\frac{{\mathrm{I}}_{\mathrm{d}}}{{\mathrm{I}}_{\mathrm{c}}}={\in }_0{\in }_{\mathrm{r}}2\mathrm{\pi f}\left(\mathrm{\rho }\right)\cot \left(2\mathrm{\pi ft}\right)$

$\frac{{\mathrm{I}}_{\mathrm{d}}}{{\mathrm{I}}_{\mathrm{c}}}=\frac{1}{4\mathrm{\pi }\times 9\times {10}^9}\times 80\times 2\mathrm{\pi }\times 9\times {10}^2\times (0.25)\times \cot \left(2\mathrm{\pi }\times 9\times {10}^2\times \frac{1}{800}\right)$

$\frac{{\mathrm{I}}_{\mathrm{d}}}{{\mathrm{I}}_{\mathrm{c}}}=\frac{{10}^3}{{10}^9}\left(\cot \left(\frac{9\mathrm{\pi }}{4}\right)\right)=\frac{{10}^3}{{10}^9}$

${\mathrm{I}}_{\mathrm{c}}={10}^6{\mathrm{I}}_{\mathrm{d}}$

$\text{ So }\mathrm{x}=6$