Question

A hollow charged conductor has a tiny hole cut into its surface.Show that the electric field in the hole is (s/2e0) ˆn , where ˆn is theunit vector in the outward normal direction, and s is the surfacecharge density near the hole.

Answer 1

The electric field inside the cavity in a conductor is zero.

Let, the electric field just outside the conductor is E, the charge is q and the charge density is σ.

The net charge is given by,

q=σ×ds(1)

From Gauss’s law, the electric flux is,

ϕ=E⋅ds ϕ= q ε 0

Therefore,

E⋅ds= q ε 0 (2)

Where, ε 0 is the permittivity of free space.

From equation (1) and (2), we get

E⋅ds= σ×ds ε 0 E= σ ε 0 n ^ (3)

The electric field just outside the conductor is σ ε 0 n ^ .

This field is the electric field due to superposition of the field due to cavity and the field due to rest of the charged conductor.

These fields are equal in magnitude and opposite in direction inside the conductor. Similarly equal in magnitude and same in direction outside the conductor.

Let E ′ be the magnitude of electric field inside and outside the conductor.

The total electric field is given as the sum of the electric fields inside and outside the conductor.

E= E ′ + E ′ E=2 E ′ E ′ = E 2

Substitute the value of E in above expression.

E ′ = σ 2 ε 0 n ^

Hence, the electric field in the hole is σ 2 ε 0 n ^ .