Exams › JEE Advanced › Physics
A charge -q is placed on the axis of a uniformly charged ring of charge +Q and radius r, at an axial distance of 2*sqrt(2)*r from the centre of the ring. The charge -q is released from rest. What is the kinetic energy of the charge -q when it reaches the centre of the ring?
- qQ / (4*pi*epsilon0*r)
- qQ / (12*pi*epsilon0*r)
- qQ / (6*pi*epsilon0*r)
- qQ / (2*pi*epsilon0*r)
Correct answer: qQ / (6*pi*epsilon0*r)
Solution
The charge -q gains kinetic energy equal to the decrease in potential energy. V_initial = kQ/sqrt(r² + 8r²) = kQ/(3r). V_final = kQ/r. Change in PE of charge -q: delta_PE = (-q)*(V_final - V_initial) = (-q)*(kQ/r - kQ/(3r)) = (-q)*(2kQ/(3r)). KE = -delta_PE = q*2kQ/(3r) = 2kqQ/(3r) = 2qQ/(12*pi*epsilon0*r) = qQ/(6*pi*epsilon0*r).
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