Question

Sea water at frequency $\nu =4x{10}^8$ Hz has permittivity $\epsilon \approx 80{\epsilon }_0$, permeability $\mu \approx {\mu }_0$ and resistivity $\rho =0.25\Omega m$. Imagine a parallel plate capacitor immersed in sea water and driven by an alternating voltage source V(t) = $V_0\sin \left(2\pi \nu t\right)$. The of amplitude of the displacement current density to the conduction current density is

• A. $\frac{2}{3}$
• B. $\frac{4}{9}$
• C. $\frac{9}{4}$
• D. 2

Answer 1

The correct option is B $\frac{4}{9}$

Suppose distance between the parallel plates is $D$ and applied voltage $V_{\left(t\right)}=V_{0}2\pi vt$.

thus electric field

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

Now using Ohm's law

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

$\frac{V_0}{\varphi d}\sin \left(2\pi vt\right)=J_0^c\sin 2\pi vt$

Here $J_0^c=\frac{V_0}{pd}$

Now the displacement current density is given as

$Jd=\in \frac{\delta E}{dt}=\frac{\in \delta }{dt}$ $\left[\frac{V_0}{dt}\sin \right(2\pi vt\left)\right]$

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

$\Rightarrow =J_0^d\cos \left(2\pi vt\right)$

Where $J_0^d=\frac{2\pi V\in V_0}{d}$

$\Rightarrow \frac{J_0^d}{J_0^c}=\frac{2\pi v\in V_0}{d}.\frac{pd}{V_0}=2\pi v\in \rho$

$=2\pi \times 80{\in }_{0}v\times 0.25=4\pi {\in }_{0}v\times 10$

$=\frac{10v}{9\times {10}^9}=\frac{4}{9}$