Question

Sea water at frequency $v=4\times {10}^{8}Hz$ has permittivity $ϵ\approx 80{ϵ}_0$,permeability $\mu \approx {\mu }_0$ and resistivity $\rho =0.25\Omega -m$. Imagine a parallelplate capacitor immersed in sea water and driven by an alternating voltage source $V\left(t\right)=V_0\sin \left(2\pi vt\right)$. What fraction of the conduction current density is the displacement current density?

Answer 1

Step 1: Find conduction current density.

Formula used: $J_c=\frac{E}{\rho }$

Given, voltage of the alternating source, $V\left(t\right)=V_0\sin \left(2\pi vt\right)$.

Let the distance between the plates be d.

Then the electric field $E=\frac{V}{d}=\frac{V_0}{d}\sin \left(2\pi vt\right)$.

The conduction current density is given by the Ohm's law, $J_c=\frac{E}{\rho }$

$\Rightarrow J_c=\frac{1}{\rho }\frac{V_0}{d}\sin \left(2\pi vt\right)$

$J_c=\frac{V_0}{\rho d}\sin \left(2\pi vt\right)$

$J_c=J_{0c}\sin \left(2\pi vt\right)$

where $J_{0c}=\frac{V_0}{\rho d}$ , is maximum conduction current density.

Step 2: Find displacement current density.
Formula used: $J_d=ϵ\frac{dE}{dt}$

The displacement current density,
$J_d=ϵ\frac{dE}{dt}$

$J_d=ϵ\frac{d}{dt}\left\{\frac{V_0}{d}\sin \right(2\pi vt\left)\right\}$

$J_d=\frac{ϵ2\pi v{V}_0}{d}\cos \left(2\pi vt\right)$

$J_d=J_{0d}\sin \left(2\pi vt\right)$

where $J_{0d}=\frac{2\pi vϵV_0}{d}$ = maximum displacement current density.

Step 3: Find fraction of conduction current density in the displacement current density.

Given, frequency of sea water, $v=4\times {10}^{8}Hz$
Resistivity, $\rho =0.25\Omega -m$

$\frac{J_{0d}}{J_{0c}}=\frac{2\pi vϵV_0}{d}.\frac{\rho d}{V_0}$

$\frac{J_{0d}}{J_{0c}}=2\pi vϵ\rho$

$\frac{J_{0d}}{J_{0c}}=2\pi (4\times {10}^8)\left(80{ϵ}_0\right)\left(0.25\right)$

$\frac{J_{0d}}{J_{0c}}=\left(4\pi {ϵ}_0\right)(4\times {10}^9)$

$\frac{J_{0d}}{J_{0c}}=\frac{4\times {10}^9}{9\times {10}^9}=\frac{4}{9}$

Final answer: $\frac{4}{9}$