Question

The electric field in a region is given by $\overrightarrow{E}=\frac{2}{5}{E}_{\circ }\hat{i}+\frac{3}{5}{E}_{\circ }\hat{j}$ with $E_{\circ }=4.0\times {10}^{3}N/C$.The flux of this field through a rectangular surface are $0.4m^2$parallel to Y–Z plane is________$N{m}^2{C}^{-1}$.

Answer 1

Step 1: Given data

The electric field in the region, $\overrightarrow{E}=\frac{2}{5}{E}_{\circ }\hat{i}+\frac{3}{5}{E}_{\circ }\hat{j}$

Intensity $E_{\circ }=4.0\times {10}^{3}N/C$

Area=$0.4m^2$

Step 2: Formula Used

Gauss’ law can be used to compute the flux. The following expression represents the law,

$\varphi =\oint \overrightarrow{E}\cdot d\overrightarrow{A}----\left(1\right)$

Step 3: Compute the flux

Consider the following figure:

The plane is parallel to the yz-plane. Hence only $\frac{2}{5}{E}_{\circ }i$ passes perpendicular to the plane whereas$\frac{3}{5}{E}_{\circ }j$ goes parallel.

According to the figure, substitute the known values in the equation $\left(1\right)$,

$$\begin{gathered} \varphi =\frac{2}{5}{E}_{\circ }\times 0.4 \\ \varphi =\frac{{\displaystyle 2}}{{\displaystyle 5}}\times 4.0\times {10}^3\times 0.4 \\ \varphi =640N{m}^2{c}^{-1} \end{gathered}$$

Hence, the flux of this field through a rectangular surface will be $640N{m}^2{c}^{-1}$.