Electric field on one side or a charged body Is Ei and electric field on the other side of the same body is E2. If infinite plane charged body has a uniform thickness, then electric field due to one surface of the charged body Is given by,

${\overline{E}}_1=-\frac{\sigma }{2{ϵ}_0}\hat{n} ...\left(\mathrm{i}\right)$

Where,

$\hat{\mathrm{n}}$ = Unit vector normal to the surface at a point a Surface charge
density at that point.

Electric field due to the other surface of the charged body,

$\overrightarrow{E_2}=-\frac{\sigma }{2{ϵ}_0}\hat{n} ...\left(\mathrm{ii}\right)$

Electric field at any point due to the two Surfaces,

$\begin{array}{l}\overline{E_2}-\overline{E_1}=\frac{\sigma }{2{ϵ}_0}\hat{n}+\frac{\sigma }{2{ϵ}_0}\hat{n}=\frac{\sigma }{{ϵ}_0}\hat{n}\\ (\overrightarrow{E_2}-\overrightarrow{E_1})\cdot n=\frac{\sigma }{{ϵ}_9}\end{array}$

Since inside a closed conductor, $\overrightarrow{E}=0$,

$\therefore \overrightarrow{E}={\overrightarrow{E}}_2=-\frac{\sigma }{2{ϵ}_0}\hat{n}$

Therefore, the electric field just outside the conductor is $\frac{\sigma }{{ϵ}_n}\hat{n}$.

(b) When a charged particle is moved from one point to the other on a closed-loop, the work done by the electrostatic field is zero. Hence, the tangential component of the electrostatic field is continuous from one side of a charged surface to the other.