(a) Show that the normal component of electrostatic field has a discontinuity from one side of a charged surface to another given by
(a) Show that the normal component of electrostatic field has a discontinuity from one side of a charged surface to another given by $\left(\overrightarrow{E_{2}}-\overrightarrow{E_{1}}\right) \cdot \hat{n}=\frac{\sigma}{\epsilon_{0}}$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 $\sigma \hat{n} / €_{0}$
(b) Show that the tangential component of electrostatic field is continuous from one side of a charged surface to another. [Hint: For (a), use Gauss’s law. For, (b) use the fact that work done by electrostatic field on a closed loop is zero.]
(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.