A thin disc of radius b = 2 a has a concentric hole of radius a in it (see figure). It carries uniform surface charge σ on it. If the electric field on its axis at a height h h < < a from its centre is given as C h then the value of C is

Step-by-step solution

∵ At the axial point of a uniformly charged disc electric field is given by, E = σ 2 ϵ 0 1 - c o s θ By superposition principle, when inner disc is removed , then, electric field due to remaining disc is, E = σ 2 ϵ 0   1 - c o s θ 2 - 1 - c o s θ 1 = σ 2 ϵ 0 c o s θ 1 - c o s θ 2 = σ 2 ϵ 0 h h 2 + a 2   - h h 2 + b 2 = σ 2 ϵ 0 h a 1 + h 2 a 2 - h b 1 + h 2 b 2   ∵ h   ≪ a and b. ∴ E = σ 2