JEE Main · 2026 · 23 Jan 2026 · Shift 1 · Electrostatics

By Gauss's law the flux through the cube equals the enclosed charge divided by ${\epsilon }_0$. The charge $2q$ sits at a vertex of the cube, shared equally by $8$ cubes, so it contributes $\frac{2q}{8}=\frac{q}{4}$. The charge $q$ lies at the centre of a face, shared by $2$ cubes, contributing $\frac{q}{2}$. The enclosed charge is $\frac{q}{4}+\frac{q}{2}=\frac{3q}{4}$, so the flux through the cube is $\frac{3q}{4{\epsilon }_0}$.