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$(a)$ Show that the normal component of electrostatic field has a discontinuity from one side of a charged surface to another given by
$\left( E _{2}- E _{1}\right) \cdot \hat{ n }=\frac{\sigma}{\varepsilon_{0}}$
where $\hat{ n }$ is a unit vector normal to the surface at a point and $\sigma$ is the surface charge density at that point. (The direction of $\hat { n }$ is from side $1$ to side $2 .$ ) Hence, show that just outside a conductor, the electric field is $\sigma \hat{ n } / \varepsilon_{0}$
$(b)$ Show that the tangential component of electrostatic field is continuous from one side of a charged surface to another.
Solution
$(a)$ Electric field on one side of a charged body is $E_{1}$ and electric field on the other side of the same body. is $E_z$. If infinite plane charged body has a uniform thickness, then electric field due to one surface of the charged body is given by,
$\overline{E_{1}}=-\frac{\sigma}{2 \epsilon_{0}} \hat{n}$
Where,
$\hat{n}=$ Unit vector normal to the surface at a point
$\sigma=$ Surface charge density at that point Electric field due to the other surface of the charged body,
$\overrightarrow{E_{2}}=-\frac{\sigma}{2 \epsilon_{0}} \hat{n}$
Electric field at any point due to the two surfaces,
$\overrightarrow{E_{2}}-\overrightarrow{E_{1}}=\frac{\sigma}{2 \epsilon_{0}} \hat{n}+\frac{\sigma}{2 \epsilon_{0}} \hat{n}=\frac{\sigma}{\epsilon_{0}} \hat{n}$
$(\overrightarrow{E_{2}}-\overrightarrow{E_{1}}) \cdot \hat{n}=\frac{\sigma}{\epsilon_{0}}$
since inside a closed conductor, $\overline{E_{1}}=0$
$\therefore d \vec{E}=\overrightarrow{E_{2}}=-\frac{\sigma}{2 \epsilon_{0}} \hat{n}$
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.
