Question

Sea water at frequency $\upsilon = 4 \times 10^{8}\, Hz$ has permittivity $\varepsilon \approx 80 \,\varepsilon_{0}$, 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)

$V\left(t\right) = V_{0}\, sin \, 2\pi\, \upsilon t$
Let distance between the plates $= d$.
Electric field $= \frac{V\left(t\right)}{d} = \frac{V_{0}}{d} \, sin \, 2\pi \, \upsilon t$
Conduction current density
$J_{c} = \frac{E}{\rho} = \frac{V_{0}}{\rho d} \, sin \, 2\pi \, \upsilon t = J_{o_c} \, sin \, 2\pi \, \upsilon t$
where $J_{o_c}$ is maximum conduction current density. Displacement current density,
$J_{d} = \varepsilon \frac{d}{dt}\left[\frac{V_{o}}{d}sin \, \left(2\pi \, \upsilon t\right)\right]$
$ = \frac{2\pi\upsilon\varepsilon}{d}V_{o}\, cos \, 2\pi\upsilon t$
$J_{d} = J_{o_d} cos \, \left(2\pi\upsilon t\right)$
$\frac{J_{o_d}}{J_{o_c}} = \frac{2\pi\upsilon}{V_{o}}\varepsilon \frac{V_{o}}{d}$
$ = 2\pi\upsilon\varepsilon\rho $
$= 2\pi\left(4 \times 10^{8}\right)\,\left(80\,\varepsilon_{0}\right)\,\left(0.25\right)$
$= \frac{Jo_{c}}{Jo_{d}} = \frac{9 \times10^{9}}{4 \times 10^{9}} = \frac{9}{4}$
$\left(\frac{1}{4\pi\varepsilon_{0}} = 9 \times10^{9}\,N\,m^{2}/ C^{2}\right)$