91Ó°ÊÓ

Displacement current,$i_d=2.0\text{A}$

Circular plate radius,$R=0.10\text{m}$

${\mathrm{Î\mu }}_0=8.85×{10}^{-12}{\text{C}}^{\text{2}}{\text{/N.m}}^{\text{2}}$

In a parallel plate capacitor, the real current $\mathit{i}$,which charges the plates, changes the electric field $\stackrel{\mathbf{→}}{\mathbf{E}}$between the plates. The fictitious displacement current${\mathit{i}}_{\mathbf{d}}$between the plates is associated with that charging field,role="math" localid="1663155680805" $\stackrel{\mathbf{→}}{\mathbf{E}}$. Therefore, we consider${\mathbf{i}}_{\mathbf{d}}\mathbf{=}{\mathbf{Î\mu }}_{\mathbf{0}}\mathbf{A}\left(\mathbf{dE}\mathbf{/}\mathbf{dt}\right)$. In this problem, area A is the area over which a changing electric field is present. Also, it is given that,$\mathbf{r}\mathbf{>}\mathbf{R}$.

The formula is as follows:

$i_d={\mathrm{Î\mu }}_{0}A\left(dE/dt\right)$

where, $i_d$is the displacement current,$A$is the area and,$E$is the charging field.

To find the rate of electric field between plates with respect to the time,

$i_d={\mathrm{Î\mu }}_{0}A\left(dE/dt\right)$

$$\begin{gathered} \frac{dE}{dt}=\frac{i_d}{{\mathrm{Î\mu }}_{0}A} \\         =\frac{i_d}{{\mathrm{Î\mu }}_0\mathrm{Ï€}{R}^2} \end{gathered}$$

Putting values,

$$\begin{gathered} \frac{dE}{dt}=\frac{2.0}{\left(8.85×{10}^{-12}\right)\left(3.14\right){(0.10)}^2} \\          =7.2×{10}^{12}\text{V/m.s} \end{gathered}$$

Therefore, the rate of the electric field between the plates changing is $7.2×{10}^{12}\text{V/m.s}$.