(a) Show that the normal component of electrostatic field has a discontinuity from one side of a charged surface to another given by

(a) Electric field on one side of a charged body is E1 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,

$\vec{E}_{1}=-\frac{\sigma}{2 \epsilon_{0}} \hat{n}$  ..(i)

Where,

$\hat{n}=$ Unit vector normal to the surface at a point

σ = Surface charge density at that point

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

$\overrightarrow{E_{2}}=-\frac{\sigma}{2 \epsilon_{0}} \hat{n}$   ...(ii)

Electric field at any point due to the two surfaces,

$\overrightarrow{E_{2}}-\overrightarrow{E_{1}}=\frac{\sigma}{2 \epsilon_{0}} \hat{n}+\frac{\sigma}{2 \epsilon_{0}} \hat{n}=\frac{\sigma}{\epsilon_{0}} \hat{n}$

$\left(\overrightarrow{E_{2}}-\overrightarrow{E_{1}}\right) \cdot \hat{n}=\frac{\sigma}{\epsilon_{0}}$   ..(iii)

Since inside a closed conductor, $\overrightarrow{E_{1}}=0$

$\therefore \vec{E}=\overrightarrow{E_{2}}=-\frac{\sigma}{2 \epsilon_{0}} \hat{n}$

Therefore, the electric field just outside the conductor is $\frac{\sigma}{\epsilon_{0}} \hat{n}$.

(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 electrostatic field is continuous from one side of a charged surface to the other.