Question

Two thin wire rings each having a radius R are placed at a distance d apart with their axes coinciding. The charges on the two rings are + q and – q. The potential difference between the centers of the two rings is .... 

(A) 0 

(B) [q/(2π∈₀)] [(1/R) – {1/√(R² + d²)}] 

(C) [q/(4π∈₀)] [(1/R) – {1/√(R² + d²)}] 

(D) [(qR)/(4π∈₀d²)]

Answer 1

The correct option (B) [q/(2π∈₀)] [(1/R) – {1/√(R² + d²)}]

Explanation:

from given data, diagram can be drawn as shown.

V_A = Potential due to charge + q on ring A + potential due to – q on ring B

∴ V_A = k[(q/R) + {(– a)/d₁}] & d₁ = √(R² + d²)

∴ V_A = [q/(4π∈₀)] [(1/R) – {1/√(R² + d²)}]  (1)

similarly V_B = [q/(4π∈₀)] [{(– 1) / R} + {1/√(R² + d²)}]    (2)

Potential difference = V_A – V_B

∴ from (1) & (2), [q/(4π∈₀)] [(1/R) – {1/√(R² + d²)} + (1/R) – {1/√(R² + d²)}]

= [2q/(4π∈₀)] [(1/R) – {1/√(R² + d²)}]

= [q/(2π∈₀)] [(1/R) – {1/√(R² + d²)}]