JEE Advanced · 2011 · 10 Apr 2011 · Paper 1 · Electrostatics

Consider an electric field

E = E_{0} \overset{x}{^}

, where

E_{0}

is a constant. The flux through the shaded area (as shown in the figure) due to this field is

(A) $2 E_{0} a^{2}$
(B) $2 E_{0} a^{2}$
(C) $E_{0} a^{2}$ ✓ Correct
(D) $\frac{E _{0} a ^{2}}{2}$

Answer: (C)

Solution

The electric flux through a surface is

Φ = E \cdot A

, where

A

is the area vector (perpendicular to the surface). The shaded area appears to be a square of side

a

in a plane. The component of the electric field perpendicular to this surface determines the flux.

Since

E = E_{0} \overset{x}{^}

, the flux depends on the projected area perpendicular to the x-direction. For the geometry shown, the projected area is

a^{2}

, giving:

Φ = E_{0} \cdot a^{2}

JEE Advanced · 2011 · 10 Apr 2011 · Paper 1 · Single correct option

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