Question

Sea water at frequency $\nu =4\times {10}^{8}Hz$ has permittivity $ϵ\approx 80{ϵ}_0$, permeability $\mu \approx {\mu }_0$ and resistivity $\rho =0.25\Omega$. Imagine a parallel plate capacitor immersed in sea water and driven by an alternating voltage source $V\left(t\right)=V_{0}sin(2\pi \nu t$). The ratio of amplitude of the conduction current density to the displacement current density is:

• A. $2/3$
• B. $4/9$
• C. $9/4$
• D. $2$

Answer 1

The correct option is B $4/9$

Let the distance between the plates=$d$

The electric field between the plates is given as:

$E=\frac{V\left(t\right)}{d}$

$V\left(t\right)=V_{0}sin2\pi \nu t$

$\Rightarrow E=\frac{V_0}{d}\left[sin2\pi \nu t\right]$

The conduction current density

$J_c=\frac{E}{\rho }=\frac{V_0}{\rho d}\left[sin2\pi \nu t\right]$

$\Rightarrow J_c=(J_0{)}_{c}sin2\pi \nu t$

Where $(J_o{)}_c$ is maximum conduction current density.

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

$\Rightarrow ϵ\frac{d}{dt}\left[\frac{V_0}{d}\right(sin\left(2\pi \nu t\right)\left)\right]$
$=\frac{2\pi \nu ϵ}{d}\left(V_{0}cos2\pi \nu t\right)$

$J_d=\left(J_0{)}_{d}cos\right(2\pi \nu t)$

The ratio of amplitude of two current densities:
$\frac{(J_0{)}_d}{(J_0{)}_c}=\frac{2\pi \nu }{V_0}\left(ϵ\right)\frac{V_0}{d}$

$\Rightarrow 2\pi \nu ϵ\rho =2\pi (4\times {10}^8)\left(80{ϵ}_0\right)\left(0.25\right)$

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

$⟹\frac{4}{9}$

Option $\left(B\right)$ is correct.