Q66P - Fundamentals Of Physics

Question

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

Step-by-Step Solution

Answer

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

Listing the given quantities:

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

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

Understanding the concepts of displacement current:

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)

Calculations of the greatest displacement current through the area:

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 $t=\text{6 μs}$  to  $t=\text{7 μs}$.

$\begin{array}{rcl}\frac{dE}{dt}& =& \frac{E_2-E_1}{t_2-t_1}\\ & =& \frac{(4-2){\displaystyle }{\displaystyle \raisebox{1ex}{${\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  $3.5\times {10}^{-5}\text{ A}$.