Prove that the displacement current in a parallel-plate capacitor of capacitance$\mathit{C}$can be written as${\mathit{i}}_{\mathbf{d}}\mathbf{=}\mathit{C}\mathbf{(}\mathbf{dV}\mathbf{/}\mathbf{dt}\mathbf{)}$, where$\mathit{V}$is the potential difference between the plates.

The expression for the fictitious displacement current due to changing electric field is as follows:

${\mathit{i}}_{\mathbf{d}}\mathbf{=}{\mathit{\epsilon }}_{\mathbf{0}}\frac{{\mathbf{d\varphi }}_{\mathbf{E}}}{\mathbf{dt}}$

Here, $\frac{d{\varphi }_E}{dt}$is the change in electric field, and ${\epsilon }_0$is the permittivity of free space.

Write the expression for the electric flux.

${\varphi }_E=AE$

Here,$A$ is the area of the plate and $E$is the magnitude of the electric field.

Write the expression for the magnitude ofelectric field.

$E=\frac{V}{d}$

Here, V is the potential difference between the plates, and d is the distance of separation between the plates.

Substitute the above value in $\varphi =AE$.

${\varphi }_E=\frac{AV}{d}$

Substitute the above value in $i_d={\epsilon }_0\frac{d{\varphi }_E}{dt}$.

$$\begin{gathered} i_d={\epsilon }_0\frac{d\left(\frac{AV}{d}\right)}{dt} \\ =\frac{A{\epsilon }_0}{d}\frac{dV}{dt} \end{gathered}$$

It is known that the value of capacitance is $C=\frac{A{\epsilon }_0}{d}$, so substitute it in the above expression.

$i_d=C\left(\frac{dV}{dt}\right)$

Thus, the given expression has been proved.