Question

Sea water at frequency $\upsilon =4\times 10^{8}Hz$ 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 $V\left(t\right)=V_{0}sin\left(2\pi \upsilon t\right)$. 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

Answer: (C)

$V\left(t\right)=V_{0}sin2\pi \upsilon t$
Let distance between the plates $=d$.
Electric field $=\frac{V\left(t\right)}{d}=\frac{V_0}{d}sin2\pi \upsilon t$
Conduction current density
$J_c=\frac{E}{\rho }=\frac{V_0}{\rho d}sin2\pi \upsilon t=J_{o_c}sin2\pi \upsilon t$
where $J_{o_c}$ is maximum conduction current density. Displacement current density,
$J_d=\epsilon \frac{d}{dt}\left[\frac{V_o}{d}sin\left(2\pi \upsilon t\right)\right]$
$=\frac{2\pi \upsilon \epsilon }{d}{V}_{o}cos2\pi \upsilon t$
$J_d=J_{o_d}cos\left(2\pi \upsilon t\right)$
$\frac{J_{o_d}}{J_{o_c}}=\frac{2\pi \upsilon }{V_o}\epsilon \frac{V_o}{d}$
$=2\pi \upsilon \epsilon \rho$
$=2\pi \left(4\times 10^8\right)\left(80{\epsilon }_0\right)\left(0.25\right)$
$=\frac{J{o}_c}{J{o}_d}=\frac{9\times 10^9}{4\times 10^9}=\frac{9}{4}$
$\left(\frac{1}{4\pi {\epsilon }_0}=9\times 10^{9}N{m}^2/C^2\right)$