A parallel-plate capacitor with circular plates of the radius $\mathit{R}$is being charged. Show that, the magnitude of the current density of the displacement current is

role="math" localid="1663149858775" ${\mathit{J}}_{\mathbf{d}}\mathbf{=}{\mathit{\epsilon }}_{\mathbf{0}}\left(\frac{dE}{dt}\right)\mathit{f}\mathit{o}\mathit{r}\mathbf{}\mathit{r}\mathbf{⩽}\mathit{R}$.

A parallel plate capacitor with circular a radius $R$.

By using the relation between the current density and the displacement current, and the expression of flux in terms of area and electric field, find the current density as a function of the electric field. In electromagnetism, displacement current density are the quantity appearing in Maxwell's equations that is defined in terms of the rate of change of D, the electric displacement field.

Formulae are as follows:

$J_d=\frac{i_d}{A}$

where,$J_d$ is the displacement current density, $i$is the current and, $A$is the area.

The displacement current density is uniform and normal to the area. Its magnitude is given by,

$J_d=\frac{i_d}{A}\dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots .\left(1\right)$

Where displacement current can be written as,

$i_d={\epsilon }_0\frac{d{\varphi }_E}{dt}$

Whereis the electric flux, and it is${\varphi }_E=AE.$

So,

$i_d=A{\epsilon }_0\frac{dE}{dt}$

By substituting the above equation in equation (1),

$$\begin{gathered} J_d=\frac{1}{A}A{\epsilon }_0\frac{dE}{dt} \\ ={\epsilon }_0\frac{dE}{dt} \end{gathered}$$

Therefore, the magnitude of the current density of the displacement current is $J_d={\epsilon }_0\frac{dE}{dt}$.