Question

A square of side $L$ is in the plane of the paper. A uniform electric field $E$, also in the plane of the paper, is limited only to the lower half of the square surface as shown in the figure. The electric flux (in SI unit) associated with the surface is

• A. $2E{L}^2$
• B. $3E{L}^2$
• C. Zero
• D. $E{L}^2$

Answer 1

The correct option is C Zero


From the diagram, we can see that, area vector of the square surface is outward (along $+z-$ axis) and electric field is along $+x-$ axis.

Thus, the angle between positive normal to the square surface and electric field is ${90}^{\circ }$.

Hence, electric flux associated with the surface is

$\varphi =\overrightarrow{E}.\overrightarrow{A}=EA\cos {90}^{\circ }=0$

So, option $\left(c\right)$ is correct.