A hollow charged conductor has a tiny hole cut into its surface. Show that the electric field in the hole is (σ/2ε0) nˆ , where nˆ is the unit vector in the outward normal direction, and σ is the surface charge density near the hole.

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

According to Gauss’s law, flux, φ = E.ds = q/ε₀

⇒E.ds=(σ x ds)/ε₀

Therefore E= (σ/2ε₀) ñ 

Therefore, the electric field just outside the conductor is (σ/2ε₀) ñ . This field is a superposition of field due to the cavity É  and the field due to the rest of the charged conductor É .These fields are equal and opposite inside the conductor and equal in magnitude and direction outside the conductor. Therefore  É +É = E

⇒ É =E/2 = (σ/2ε₀) ñ 

Hence, the field due to the rest of the conductor is (σ/ε₀) ñ .