JEE Main · 2023 · 01 Feb 2023 · Shift 1 · Electrostatics

Each infinite sheet of surface charge density $\sigma$ sets up a field of magnitude $\frac{\sigma }{2{ϵ}_0}$ on either side, pointing away from the sheet. Both sheets here carry $+\sigma$. Between them the two fields point towards each other and cancel, so ${\vec{E}}_{\mathrm{II}}=0$. To the left both fields point away to the left, adding to $\frac{\sigma }{{ϵ}_0}$ along $-\hat{n}$, and to the right both point away to the right, adding to $\frac{\sigma }{{ϵ}_0}$ along $+\hat{n}$. Hence ${\vec{E}}_{\mathrm{I}}=-\frac{\sigma }{{ϵ}_0}\hat{n}$, ${\vec{E}}_{\mathrm{II}}=0$ and ${\vec{E}}_{\mathrm{III}}=\frac{\sigma }{{ϵ}_0}\hat{n}$.