(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)).n = sigma/epsilon_(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 `hatn` is from side 1 to side 2.) Hence, show that just outside a conductor, the electric field is `sigma hatn //epsilon_(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.]

Explanation

Suppose, xy is a charged surface of dielectric and electric field `vecE_(1)` and `vecE_(2)` are on its both side.

Let loop ABCD. Its length is `l` and width is negligible.
Line integration of closed path of ABCD ,
`intvecE.dvecl=vecE_(1).vecl-vecE_(2).vecl=0`
`= E_(1)l cos theta_(1)-E_(2) l cos theta_(2)=0`
`:. (E_(1)cos theta_(1)-E_(2)cos theta_(2))l =0`
`:. ( E_(1)-E_(2))l=0`
where `E_(1)` and `E_(2)` aretangenticalcomponent
`:. E_(1) = E_(2)`
`:. ` Hence, the tangential component of electrostatic field is continuous from one side of a charged surface to the other.