Question

Seawater at a frequency $f=9\times {10}^2\text{Hz}$ has permittivity $ϵ=80{ϵ}_o$ and resistivity $\rho =0.25\Omega \text{m}$. Imagine a parallel plate capacitor is immersed in seawater and is driven by an alternating voltage source $V\left(t\right)=V_o\sin \left(2\pi ft\right)$. Then, the conduction current density becomes ${10}^x$ times the displacement current density after time $t=\frac{1}{800}\text{s}$. The value of $x$ is ______.

$(\frac{1}{4\pi {ϵ}_o}=9\times {10}^9{\text{Nm}}^2{\text{/C}}^2)$

Answer 1

Given:
$f=9\times {10}^2\text{Hz}$
$ϵ=80{ϵ}_o$
$\rho =0.25\Omega \text{m}$
$V\left(t\right)=V_o\sin \left(2\pi ft\right)$
$t=\frac{1}{800}\text{s}$

Conduction current density,

$J_c=\frac{E}{\rho }=\frac{V}{\rho d}=\frac{V_o\sin \left(2\pi ft\right)}{\rho d}...\left(1\right)$

Displacement current density,

$J_d=ϵ\frac{d\varphi }{Adt}=\frac{ϵ}{d}\frac{dV}{dt}$

$\Rightarrow J_d=\frac{ϵ}{d}\left[2\pi V_{o}f\cos \right(2\pi ft\left)\right]...\left(2\right)$

As per question,

$J_c={10}^x{J}_d$

$\Rightarrow \frac{V_o\sin \left(2\pi ft\right)}{\rho d}={10}^x\times \frac{ϵ}{d}\left[2\pi V_{o}f\cos \right(2\pi ft\left)\right]$

$\left[\text{From 1 and 2}\right]$

$\Rightarrow {10}^x=\frac{\sin \left(2\pi ft\right)}{\rho \times 80{ϵ}_o\left[2\pi f\cos \right(2\pi ft\left)\right]}$

$\Rightarrow {10}^x=\frac{9\times {10}^9\times \tan (2\pi \times 9\times {10}^2\times \frac{1}{800})}{0.25\times 40\times 9\times {10}^2}$

$\Rightarrow {10}^x={10}^6$

$\Rightarrow x=6$