(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) hat n= (sigma)/(epsi_0)` where `hatn` 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 )/(epsi_0) hatn`. (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.]

Explanation

(a) Electric field on one side of a charged body is `E_(1)` and electric field on the other side of the same nody 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,
`vec(E)_(1) = -(sigma)/(2 in_(0)) hat(n)` ....(i)
Where `hat(n)` = Unit vactor 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,
`vec(E)_(2) = -(sigma)/(2 in_(0))hat(n)` ...(ii)
Electric field at anny due to the two surfaces,
`vec(E)_(2)-vec(E)_(1) = (sigma)/(2 in_(0)) hat(n)+(sigma)/(2 in_(0)) hat(n) = (sigma)/(in_(0))hat(n)`
`(vec(E)_(2)-vec(E)_(1)).hat(n) = (sigma)/(in_(0))` ...(iii)
Electric field at any point due to the two surfaces,
Since inside a closed conductor, `vec(E)_(1) = 0`,
`vec(E) = vec(E)_(2) = -(sigma)/(2 in_(0)) hat(n)`
Therefore, the electric field just outside the conductor is `(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.