A capacitor has two parallel plates with area 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 resistivity $\mathbf{\rho }$. The capacitor is initially charged with charge of magnitude${\mathbf{Q}}_{\mathbf{O}}$ on each plate that gradually discharges by conduction through the dielectric. (a) Calculate the conduction current density in the dielectric. (b) Show that at any instant the displacement current density${\mathbf{j}}_{\mathbf{c}}\left(t\right)$ in the dielectric is equal in magnitude to the conduction current density but opposite in direction, so the total current density is zero at every instant.

The term capacitor may be defined as the device which sued to store electric charge.

For a material when current passing through it, it equal to ratio of potential difference and its resistance.

$\begin{array}{r}i\left(t\right)=\frac{V\left(t\right)}{R}\\ V\left(t\right)=\frac{q\left(t\right)}{C}\\ C=K\frac{{\epsilon }_{0}A}{d}\end{array}$

Using above result

$\mathrm{i}\left(\mathrm{t}\right)=\frac{1}{\mathrm{R}}\left(\frac{\mathrm{d}}{{\mathrm{K\epsilon }}_{\mathrm{o}}\mathrm{A}}\right)\mathrm{q}\left(\mathrm{t}\right)$

Using ohm’s law

The resistance of dielectric current is$\mathrm{R}=\frac{\mathrm{\rho d}}{\mathrm{A}}$

Then

$$\begin{gathered} \mathrm{i}\left(\mathrm{t}\right)=\left(\frac{\mathrm{q}\left(\mathrm{t}\right)\mathrm{d}}{{\mathrm{K\epsilon }}_{\mathrm{o}}\mathrm{A}}\right)\left(\frac{\mathrm{A}}{\mathrm{\rho d}}\right) \\ \mathrm{i}\left(\mathrm{t}\right)=\frac{\mathrm{q}\left(\mathrm{t}\right)}{{\mathrm{K\epsilon }}_{\mathrm{o}}\mathrm{\rho }} \end{gathered}$$

Now we know that$\mathrm{i}\left(\mathrm{t}\right)=-\frac{\mathrm{dq}\left(\mathrm{t}\right)}{\mathrm{dt}}$

Negative sign denoted that charge on capacitor is decrease.

Than$\begin{array}{r}\frac{-\mathrm{dq}}{\mathrm{dt}}=\left(\frac{1}{\mathrm{K}\rho {\epsilon }_0}\right)\mathrm{q}\left(\mathrm{t}\right)\\ \frac{\mathrm{dq}}{\mathrm{q}}=-\frac{\mathrm{dt}}{\mathrm{K}\rho {\epsilon }_0}\end{array}$

Integration of both sides from$\mathrm{q}={\mathrm{Q}}_{\mathrm{o}}-\mathrm{q}$ and from

$\begin{array}{r}{\int }_{Q_0}^q \frac{dq}{q}=-\left(\frac{1}{K\rho {\epsilon }_0}\right){\int }_0^t dt\\ \ln \left(\frac{q}{Q_0}\right)=-\frac{t}{K\rho {\epsilon }_0}\\ q\left(t\right)=Q_0{e}^{t/K\rho {\epsilon }_0}\end{array}$

And the current is time derivative of this equation

$\begin{array}{r}\mathrm{i}\left(\mathrm{t}\right)=-\frac{\mathrm{dq}}{\mathrm{dt}}\\ =\left(\frac{{\mathrm{Q}}_0}{\mathrm{K}\rho {\epsilon }_0}\right){\mathrm{e}}^{-\mathrm{t}/K\rho {\epsilon }_0}\end{array}$

Now the conduction current density is

$\begin{array}{r}{J}_C=\frac{i\left(t\right)}{A}\\ J_C=\left(\frac{Q_0}{AK\rho {\epsilon }_0}\right)e^{-t/K\rho {\epsilon }_0}\end{array}$

Hence, the conduction current density is $j_C=\left(\frac{Q_o}{AK\rho {\epsilon }_o}\right)e^{-t/K\rho {\epsilon }_o}$.

Between the plates the electric field is calculate as

$E\left(t\right)=\frac{q\left(t\right)}{K{\epsilon }_{o}A}$

And the displacement current density is

$\begin{array}{r}{\mathrm{J}}_{\mathrm{D}}\left(\mathrm{t}\right)=\mathrm{K}{\epsilon }_0\frac{\mathrm{dE}}{\mathrm{dt}}\\ =\mathrm{K}{\epsilon }_0\frac{\mathrm{dq}\left(\mathrm{t}\right)/\mathrm{dt}}{\mathrm{K}{\epsilon }_0\mathrm{A}}\\ =-\frac{{\mathrm{i}}_{\mathrm{c}}\left(\mathrm{t}\right)}{\mathrm{A}}\\ =-{\mathrm{j}}_{\mathrm{C}}\left(\mathrm{t}\right)\end{array}$

Hence, the displacement current density in the dielectric is equal in magnitude to the conduction current density but opposite in direction, so the total current density is zero at every instant is hold by the result ${\mathrm{j}}_{\mathrm{D}}=-{\mathrm{j}}_{\mathrm{C}}$.