Question

Three concentric spherical shells have radii a, b and c (a < b < c) and have surface charge densities σ, – σ and σ respectively. If V_A, V_B and V_C denote the Potentials of the three shells, then for c = a + b, we have 

(A) V_C = V_B = V_A 

(B) V_C = V_B ≠ V_A 

(C) V_C = V_B ≠ V_A 

(D) V_C = V_A ≠ V_B

Answer 1

The correct option (D) V_C = V_A ≠ V_B 

Explanation:

V = Potential for a concentric shell = [1/(4π∈₀)](q/r)  (1)

σ = surface charge density = (q/A) = [q/(4πr²)]  (2)

from (1), V_A = [1/(4π∈₀)] [(q_A/a) + (q_B/b) + (q_C/c)]

from (2), q = σ ∙ 4πr^2­­ also σ_A = σ, σ_B = – σ, σ_C = σ

∴ V_A = [1/(4π∈₀)] [{(4πσa²)/a} + {(– 4πσb²)/b} + {(4πσc²)/c}]

V_A = (1/∈₀)[σa – σb + σc] = (σ/∈₀)(a – b + c) (3)

similarly

V_B = [1/(4π∈₀)] [(q_A/b) + (q_B/b) + (q_C/c)]

= [1/(4π∈₀)] [{(4πσa²)/b} + {(– 4πσb²)/b} + {(4πσc²)/c}]

V_B = (σ/∈₀) [(a²/b) – b + c] (4)

&  V_C = (σ/∈₀) [(a²/c) – (b²/c) + c]   (3)

putting c = a + b

V_A = V_C > V_B

i.e. V_A = V_C ≠ V_B