Exams › JEE Advanced › Physics
A uniformly charged ring of radius 3a carrying total charge q lies in the xy-plane centred at the origin. A point charge q (mass m) moves toward the ring along the z-axis with speed u at the location z = 4a. The minimum value of u for which the charge can just reach (cross) the origin is:
- sqrt(2/m) * (2/15 * q²/(4*pi*epsilon0*a))^(1/2)
- sqrt(2/m) * (1/15 * q²/(4*pi*epsilon0*a))^(1/2)
- sqrt(2/m) * (4/15 * q²/(4*pi*epsilon0*a))^(1/2)
- sqrt(2/m) * (1/5 * q²/(4*pi*epsilon0*a))^(1/2)
Correct answer: sqrt(2/m) * (2/15 * q²/(4*pi*epsilon0*a))^(1/2)
Solution
Axial potential of the ring: V(z) = kq/sqrt((3a)² + z²), with k = 1/(4*pi*epsilon0). At z = 4a: V = kq/sqrt(9a²+16a²) = kq/(5a). At z = 0: V = kq/(3a). The potential is highest at the centre, so for the moving charge to just reach the origin its kinetic energy there is zero. Energy conservation: (1/2)mu² + qV(4a) = qV(0). So (1/2)mu² = q[V(0) - V(4a)] = kq²(1/(3a) - 1/(5a)) = kq²*(2/(15a)). Thus u² = (2/m)*kq²*(2/15a), giving u = sqrt(2/m) * (2/15 * q²/(4*pi*epsilon0*a))^(1/2).
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