JEE Main · 2020 · 08 Jan 2020 · Shift 1 · Electrostatics

Gauss's law in general reads $\oint \vec{E}\cdot d\vec{A}=q_{enc}/{\epsilon }_0$, and it collapses to the simple form $|\vec{E}|=q_{enc}/({\epsilon }_0|A|)$ only under two conditions. The field must be perpendicular to the Gaussian surface everywhere, that is the surface must be an equipotential, so that $\vec{E}\cdot d\vec{A}$ reduces to $|\vec{E}|dA$; and the magnitude $|\vec{E}|$ must be constant over the surface so it can be pulled outside the integral. Both requirements together are what turn the surface integral into the product $|\vec{E}|A$.