Question

Seawater at a frequency $f=9\times {10}^{2}Hz$, has permittivity $\in =80{\in }_0$and resistivity $\rho =0.25\Omega m$. Imagine a parallel plate capacitor is immersed in seawater and is driven by an alternating voltage source $V\left(t\right)=V_0\sin \left(2\pi ft\right)$. Then the conduction current density becomes ${10}^x$ times the displacement current density after time $t=\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$800$}\right.\sec$. The value of $x$ is __________.

$\left(Given:\frac{1}{4\mathrm{\pi }{\in }_0}=9\times {10}^{9}N{m}^2{C}^{-2}\right)$

Answer 1

Step 1: Given data:

Frequency, $\left(f\right)=9\times {10}^{2}Hz$

Permittivity, $\in =80{\in }_0$

$\in ={\in }_0{\in }_r$

So, ${\in }_r=80$

Resistivity, $\rho =0.25\Omega m$

$V\left(t\right)=V_0\sin \left(2\pi ft\right)$

Where $V_0=$peak current, $V\left(t\right)=$alternating current, $f=$frequency, and $t=$time

The conduction current density becomes ${10}^x$ times the displacement current density after time $t=\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$800$}\right.\sec$

Step 2: Formula Used:

Displacement current, $I_d=\frac{dq}{dt}=\left(\frac{cdv}{dt}\right)$

$\in ={\in }_0{\in }_r$

So, ${\in }_r=80$

The generalized equation of capacitance of a parallel plate capacitor is given by

$c=\frac{\epsilon A}{d}$

$\in ={\in }_0{\in }_r$

$c=\frac{{\epsilon }_o{\epsilon }_{r}A}{d}$

$\epsilon$ is absolute permittivity

Step 3: Calculation of conduction current:

The displacement current can be given as-

$$\begin{gathered} \begin{array}{l}{I}_d=\frac{{ϵ}_0{ϵ}_{r}A}{d}\frac{d}{dt}\left(v_{0}sin\right(2\pi ft\left)\right)\end{array} \\ \begin{array}{l}{I}_d=\frac{{ϵ}_0{ϵ}_{r}A}{d}{V}_0\left(2\pi f\right)cos\left(2\pi ft\right)\dots \dots \dots \dots \dots \left(1\right)\end{array} \end{gathered}$$

Where $d$ is the distance between plates & conduction current $I_c=\frac{V}{R}$

The conduction current can be given as-

$\begin{array}{l}{I}_c=\frac{V_{0}sin\left(2\pi ft\right)}{\rho \frac{d}{A}}=\frac{A{V}_{0}sin\left(2\pi ft\right)}{\rho d}\dots \dots \dots \dots \dots \left(2\right)\end{array}$ As $R=\rho \frac{L}{A}=\rho \frac{d}{A}$

Step 4: Calculating the ratio of displacement current and conduction current:

From equation (1) and (2),

Divide (1) by (2)-

$\begin{array}{rcl}& & \begin{array}{l}\frac{I_d}{I_c}={ϵ}_0{ϵ}_{r}2\pi f\left(\rho \right)cot\left(2\pi ft\right)\end{array}\\ & & \begin{array}{c}\frac{I_d}{I_c}=\frac{1}{4\pi \times 9\times {10}^9}\times 80\times 2\pi \times 9\times {10}^2\times \left(0.25\right)\times cot\left(2\pi \times 9\times {10}^2\times \frac{1}{800}\right)\end{array}\\ & & \begin{array}{l}=\frac{{10}^3}{{10}^9}\left(cot\left(\frac{9\pi }{4}\right)\right)\end{array}\\ & & \begin{array}{c}=\frac{{10}^3}{{10}^9}\end{array}\\ & & \begin{array}{c}\frac{I_d}{I_c}=\frac{1}{{10}^6}\end{array}\\ I_c& =& {10}^6{I}_d\end{array}$

So, $x=6$