Figure 32-41 gives the variation of an electric field that is perpendicular to a circular area of ${\mathbf{\text{2.0m}}}^{\mathbf{\text{2}}}$. During the time period shown, what is the greatest displacement current through the area?

Circular area,$A={\text{2.0m}}^{\text{2}}$

Figure 32.41, which shows the variation of an electric field.

Use the formula of the displacement current to find the greatest displacement current through the given area. For the rate of change of electric field $\left(\frac{\mathbf{d}\mathbf{E}}{\mathbf{d}\mathbf{t}}\right)$, you can consider the slope of the given graph.

Formula:

The displacement current is,

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

For displacement current, the greatest displacement current can have at greatest, $\frac{dE}{dt}$value.

From the given graph, the greatest slope $\left(\frac{dE}{dt}\right)$is from $t=\text{6μs}$ to $t=\text{7μs}$.

$\begin{array}{rcl}\frac{dE}{dt}& =& \frac{E_2-E_1}{t_2-t_1}\\ & =& \frac{(4-2){\displaystyle }{\displaystyle \raisebox{1ex}{${\displaystyle \text{N}}$}\!\left/ \!\raisebox{-1ex}{${\displaystyle \text{C}}$}\right.}}{(7-1)\text{μs}}\\ & =& 2\times {10}^{-6}\raisebox{1ex}{${\displaystyle \text{V}}$}\!\left/ \!\raisebox{-1ex}{$\text{m}$}\right.\end{array}$

Therefore, the displacement current will becomes,

$\begin{array}{rcl}i& =& {\epsilon }_{0}A\frac{dE}{dt}\\ & =& (8.85\times {10}^{-12}\frac{{\text{C}}^{\text{2}}}{{\text{Nm}}^{\text{2}}})(2.0{\text{m}}^{\text{2}})(2\times {10}^{-6}\frac{\text{V}}{\text{m}})\\ & =& 3.5\times {10}^{-5}\text{A}\end{array}$

Hence, the greatest displacement current through the area is $3.5\times {10}^{-5}\text{A}$.