A parallel-plate capacitor with circular plates of radius $\mathit{R}$is being discharged. The displacement current through a central circular area, parallel to the plates and with radius $\raisebox{1ex}{$\mathbf{R}$}\!\left/ \!\raisebox{-1ex}{$\mathbf{2}$}\right.$, is $\mathbf{\text{2.0A}}$. What is the discharging current?

Discharging current, $I_d=\text{2.0A}$

Radius of the circular plates of the parallel plate capacitor is $R$.

Radius of the central circular plates through which the current flows is $\raisebox{1ex}{$R$}\!\left/ \!\raisebox{-1ex}{$2$}\right.$.

Use the formula of the displacement current through capacitor and the current for charging capacitor to find the displacement current.

Formulae:

The area is,

$A=\pi r^2$

The displacement current is,

$i_d={\epsilon }_{0}A\frac{dE}{dt}$

Here, ${\epsilon }_0$is the permittivity of free space, $E$is the electric field, and $t$is time.

Forcircular plates of the parallel plate capacitor of radius,

$A=\pi R^2$

And for central circular plates of radius $\raisebox{1ex}{$R$}\!\left/ \!\raisebox{-1ex}{$2$}\right.$. Thus, the area will be,

$A_c=\pi \frac{R^2}{4}$

For displacement current,

$i_d={\epsilon }_{0}A\frac{dE}{dt}$ ..... (1)

And for charging capacitor,

$i={\epsilon }_0{A}_c\frac{dE}{dt}$ ..... (2)

Take a ratio of equation (1) and (2), and you have

$$\begin{gathered} \frac{i_d}{i}=\frac{\pi R^2}{\pi \frac{R^2}{4}} \\ {i}_d=4i \end{gathered}$$

role="math" localid="1663091398478" $\begin{array}{rcl}{i}_d& =& 4\times \left(\text{2.0A}\right)\\ & =& 8.\text{0A}\end{array}$

Hence, the discharging current is $\begin{array}{rcl}& & 8.\text{0A}\end{array}$.