JEE Main · 2025 · 24 Jan 2025 · Shift 1 · Electrostatics

Place the point charge $q$ at a perpendicular distance $a/2$ from the plane of the square loop of side $a$, directly in front of its center — this makes the loop one face of an imaginary cube of side $a$ centered on the charge.

By Gauss's law, the total flux through the closed cubical surface is
${\varphi }_{cube}=\frac{q}{{\epsilon }_0}$

By symmetry this splits equally among the 6 faces, so the flux through the entire square face (the loop) is
${\varphi }_{face}=\frac{1}{6}\cdot \frac{q}{{\epsilon }_0}$

Drawing the two diagonals and the two lines joining midpoints of opposite sides divides the square into 8 congruent triangles, each subtending an equal solid angle at the charge, so each carries an equal share of the face flux:
${\varphi }_{triangle}=\frac{1}{8}{\varphi }_{face}=\frac{1}{48}\cdot \frac{q}{{\epsilon }_0}$

The shaded region consists of 5 of these 8 triangles:
${\varphi }_{shaded}=5\times \frac{1}{48}\cdot \frac{q}{{\epsilon }_0}=\frac{5}{48}\cdot \frac{q}{{\epsilon }_0}$

With $q=1$ C, this is $\frac{5}{48}\cdot \frac{1}{{\epsilon }_0}$ Nm²/C. Comparing with $\frac{5}{p}\cdot \frac{1}{{\epsilon }_0}$,

$p=48$