Question

A hollow charged conductor has a tiny hole cut into its surface. Show that the electric field in the hole is σ/2ε_o n cap , where n cap is the unit vector in the outward normal direction and σ is the surface charge density near the hole.

Answer 1

Let us consider a conductor with a cavity or a hole. Electric field inside the cavity is zero.
Let E is the electric field just outside the conductor, q is the electric charge, σ is the charge density and ϵ0 is the permittivity of free space.
Charge q= σ x ds

Therefore, the electric field just outside the conductor is σ/2ε_o  n cap. This field is a superposition of field due to the cavity E` and the field due to the rest of the

charged conductor E`. These fields are equal and opposite inside the conductor and equal in magnitude and direction outside the conductor.