Question

The electric field in a region is given by $\overrightarrow{E}=\frac{3}{5}{E}_0\hat{j}$ with $E_0=2\times {10}^{3}N{C}^{-1}$ Find the flux of this field through a rectangular surface of area $0.2m^2$ parallel to the $Y-Z$ plane.

• A. $320N{m}^2{C}^{-1}$
• B. $240N{m}^2{C}^{-1}$
• C. $400N{m}^2{C}^{-1}$
• D. none of these

Answer 1

Answer: (D)

none of these

The electric flux , $\varphi =\overrightarrow{E}.\hat{n}dS$

Since the surface is parallel to $Y-Z$ plane so the normal unit vector will be along x-axis i.e $\hat{n}=\hat{i}$

Thus, $\varphi =\left(3/5\right)E_0\hat{j}.\hat{i}\left(0.2\right)=0$ (as $\hat{j}.\hat{i}=0$)

Thus, option D will correct.