Question

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

$(E_2-E_1)\dot{}\hat{n}\frac{\sigma }{{\epsilon }_0}$
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 $\sigma \hat{n}/{ϵ}_0$
(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.]

Answer 1

(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 $E_2$. If infinite plane charged body has a uniform thickness, then electric field due to one surface of the charged body is given by,

$\overline{E_1}=-\frac{\sigma }{2{\in }_0}\hat{n}$ ...(i)

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,

$\overline{E_2}=\frac{\sigma }{2{\in }_0}\hat{n}$ ...(ii)

Electric field at any point due to the two surfaces,

$\overline{E_2}-\overline{E_1}=\frac{\sigma }{2{\in }_0}\hat{n}+\frac{\sigma }{2{\in }_0}\hat{n}=\frac{\sigma }{{\in }_0}\hat{n}$

$(\overline{E_2}-\overline{E_1}).\hat{n}=\frac{\sigma }{{\in }_0}$ ...(iii)

Since inside a closed conductor, $\overline{E_1}=0$,

$\therefore \overline{E}=\overline{E_2}=\frac{\sigma }{{\in }_0}\hat{n}$

Therefore, the electric field just outside the conductor is $\frac{\sigma }{{\in }_0}\hat{n}$.

(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.