JEE Main 2020 — Electrostatics Question with Solution

According to Gauss law of electrostatics, the electric flux through an enclosed surface is given by ${\varphi }_E=\frac{q_{\mathrm{enc}}}{{\epsilon }_0}$

From the formula of electric flux, $\varphi =\overrightarrow{E}·d\overrightarrow{s}$

Also, the relation between electric field and electric potential is given by $V=\overrightarrow{E}·d\overrightarrow{r}$

So, the formula $\left|\overrightarrow{E}\right|=\frac{q_{\mathrm{enc}}}{{\epsilon }_0\left|A\right|}$ is only applicable when the Gaussian surface is an equipotential surface and $\left|\overrightarrow{E}\right|$ is constant on the surface.