StreakPeaked· Practice

ExamsJEE MainPhysics

A ring of radius R carries charge +q. A test charge -q0 is released from rest on the axis at a distance sqrt(3)*R from the centre. What kinetic energy does the test charge gain by the time it reaches the centre of the ring?

  1. (1/(4*pi*e0)) * (q*q0 / R)
  2. (1/(4*pi*e0)) * (q*q0 / (2*R))
  3. (1/(4*pi*e0)) * (q*q0 / (sqrt(3)*R))
  4. (1/(4*pi*e0)) * (q*q0 / (3*R))

Correct answer: (1/(4*pi*e0)) * (q*q0 / (2*R))

Solution

Potential on the axis at distance x: V(x) = k*q/sqrt(R² + x²). At x = sqrt(3)*R: sqrt(R² + 3R²) = 2R, so V_i = k*q/(2R). At centre x=0: V_c = k*q/R. The test charge -q0 is attracted; KE gained = q0*(V_c - V_i) = q0*k*q*(1/R - 1/(2R)) = k*q*q0/(2R).

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