Seawater at a frequency $f=9 \times 10^{2} \mathrm{~Hz}$, has permittivity $\varepsilon=80 \varepsilon_{0}$ and resistivity $\rho=0.25 \Omega \mathrm{m}$. Imagine a parallel plate capacitor is immersed in seawater and is driven by an alternating voltage source $\mathrm{V}(\mathrm{t})=\mathrm{V}_{0} \sin (2 \pi \mathrm{ft})$. Then the conduction current density becomes $10^{\mathrm{x}}$ times the displacement current density after time $\mathrm{t}=\frac{1}{800} \mathrm{~s}$. The value of $\mathrm{x}$ is
(Given : $\frac{1}{4 \pi \varepsilon_{0}}=9 \times 10^{9} \mathrm{Nm}^{2} \mathrm{C}^{-2}$ )
$\mathrm{J}_{\mathrm{c}}=\frac{\mathrm{E}}{\rho}=\frac{\mathrm{V}}{\rho \mathrm{d}}$
$\mathrm{J}_{\mathrm{d}}=\frac{1}{\mathrm{~A}} \frac{\mathrm{dq}}{\mathrm{dt}}$
$=\frac{\mathrm{C}}{\mathrm{A}} \frac{\mathrm{dV}_{\mathrm{c}}}{\mathrm{dt}}$
$=\frac{\in}{d} \frac{\mathrm{dV}_{\mathrm{c}}}{\mathrm{dt}}$
$\Rightarrow \frac{\mathrm{V}_{0} \sin 2 \pi \mathrm{ft}}{\rho \mathrm{d}}=10^{\mathrm{x}} \times \frac{80 \varepsilon_{0}}{\mathrm{~d}} \mathrm{~V}_{0}(2 \pi \mathrm{f}) \cos 2 \pi \mathrm{ft}$
$\tan \left(2 \pi \times \frac{900}{800}\right)=10 x \times \frac{40}{9 \times 10^{9}} \times 900$
$=x=6$
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