Show that the normal component of electrostatic field has a discontinuity from one side of a charged surface to another given by 
Where, 
is a unit vector normal to the surface at a point and σ is the surface charge density at that point. (The direction of 
is from side 1 to side 2.) Hence show that just outside a conductor, the electric field is 
.
Electric field on one side of the charged is body is E1 and the electric field on the other side of the same body be E2. If infinite plane charged body has uniform thickness then the electric field due to one surface of the body is given by,
= 
Where,
= a unit vector normal to the surface at a point
σ = surface charge density at that point.
ϵ0 = permittivity of free space.
Electric field due to other surface of the charged body,
= 
∴ Electric field due to both surfaces,
- 
= 
+ 
= 
Since, Inside a closed conductor electric field s zero,
∴ 
= 
= 
Hence, the electric field just outside the conductor = 