Question 29

# A hollow charged conductor has a tiny hole cut into its surface. Show that the σ/2ε0 n̂ , where n̂ 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 ε 0 is the permittivity of free space.

Charge q = σ × ds

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

⇒ E. ds = σ × ds / ε0

∴ E= σ/ 2ε0 n̂

Therefore, the electric field just outside the conductor is σ/ 2ε0 n̂. 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.

∴  E+ E = E

⇒ E` = E/2  = σ/2ε0

Hence, the field due to the rest of the conductor is σ/2ε0 n̂.