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 .

Question

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 .

Philoid · Accepted Answer

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 =