Question

A parallel plate capacitor has two parallel plates each of area $A$ separated by a distance $d$. The space between plates is filled with a material having dielectric constant $K$. The material is not a perfect insulator but has a resistivity $\rho$. The capacitor is initially charged to $Q_0$ using a battery. The battery is removed. The capacitor discharges gradually by conduction through dielectric. The displacement current density at any instant is :

• A. $\frac{-Q_0}{pk{\epsilon }_{0}A}{e}^{-t/pk{\epsilon }_{0}A}$
• B. $\frac{-Q_0}{pk{\epsilon }_0}{e}^{-t/pk{\epsilon }_0}$
• C. $\frac{-Q_0}{pk{\epsilon }_{0}A}{e}^{-t/pk{\epsilon }_0{A}^2}$
• D. $\frac{-Q_0}{pk{\epsilon }_{0}A}{e}^{-t/pk{\epsilon }_0}$

Answer 1

The correct option is D $\frac{-Q_0}{pk{\epsilon }_{0}A}{e}^{-t/pk{\epsilon }_0}$

In the above diagram capacitor is having a di electric material of dielectric constant l also it is not a perfect insulator so current can pass through it. It can be represented separately as a combination of perfect capacitor and a resistor
So capacitance$=C=\frac{{\epsilon }_{0}A}{d-t+\frac{t}{k}}$
as $t=d$(completely filled)
$C=\frac{K{\epsilon }_{0}A}{d}$
Resistance (figure:2)
$R=\rho \frac{l}{A}=\rho \frac{d}{A}$
$R=\rho \frac{d}{A}$
Let the potentail difference across the capacitor just after removing battery is $v_0$ (figure:3)
So at time $t=0,{max}^m$ current will flow through resistance. Let it be $i_0$
So, $i_0=\frac{v_0}{R}$ But $v_0=\frac{Q_0}{C}$
$\Rightarrow i_0=\frac{Q_0}{RC}$
The capacitor is discharging so at any instant t, to find current we apply
$i=i_0{e}^{-\frac{t}{RC}}$
$\Rightarrow i=\frac{Q_0}{RC}{e}^{-\frac{t}{RC}}$
$R\times C=\rho \frac{d}{A}\times \frac{K{\epsilon }_{0}A}{d}=\rho K{\epsilon }_0$
$i=\frac{Q_0}{\rho K{\epsilon }_0}{e}^{-\frac{t}{\rho K{\epsilon }_0}}$
For current density $J=i/A$
$\Rightarrow J=\frac{Q_0}{\rho K{\epsilon }_{0}A}{e}^{-\frac{t}{\rho K{\epsilon }_0}}$