Q.
Consider an electric field E=$E_{0}\widehat{\mathbf{x}},$ where $E_{0}$ is a constant. The flux through the shaded area(as shown in the figure) due to this field is
A) $2E_{0}a^{2}$
B) $\sqrt{}2E_{0}a^{2}$
C) $E_{0}a^{2}$
D) $\frac{E_{0}a^{2}}{\sqrt{2}}$
Explanation
(c) Electric flux, $\phi = \mathbf{E.S}$
Or $\phi = ES\cos\theta$
Here, $\theta\ $is the angle between E and S. In this question$\theta = 45{^\circ}$, because S is perpendicular to surface.
$E = E_{0}$
$S = \left( \sqrt{2a} \right)(a) = \sqrt{2a^{2}}$
$\therefore\phi = \left( E_{0} \right)\left( \sqrt{2a^{2}} \right)\cos\ 45{^\circ} = E_{0}a^{2}$
∴ Correct option is (C).
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.