Exams › JEE Advanced › Physics

A particle of charge -q and mass m is released from rest on the axis of a fixed uniformly charged ring of total charge Q and radius R, at a distance sqrt(3)*R from the centre. Find the speed of the particle when it reaches the centre of the ring.

1. v = sqrt(Q*q / (2*pi*epsilon0*m*R))
2. v = sqrt(Q*q / (4*sqrt(3)*pi*epsilon0*m*R))
3. v = sqrt(Q*q / (8*pi*epsilon0*m*R))
4. v = sqrt(Q*q / (4*pi*epsilon0*m*R))

Correct answer: v = sqrt(Q*q / (4*pi*epsilon0*m*R))

Solution

The electric potential on the axis of the ring changes from k*Q/(2R) at the starting point to k*Q/R at the centre. The negative charge gains kinetic energy equal to the drop in its potential energy.

Related JEE Advanced Physics questions

• A particle with charge -q and mass m revolves in a circular path of radius r around an infinitely long charged wire with linear charge density +λ. What is the time period of its motion?
• What is the value of the electrostatic potential at point H?
• The electric potential V varies with distance x from a fixed point as follows: V = 5 V for 0 <= x < 10 m, then V decreases linearly from 5 V at x = 10 m to -10 V at x = 20 m. The electric field at x = 13 m is:
• Two capacitors are connected in a circuit with two batteries and a switch S. The system is in steady state with S open. When S is closed, which of the following statements correctly describes the energy exchange?
• In a capacitor bridge circuit, four capacitors C1, C2, C3, and C4 are connected in a bridge configuration with an EMF source E. Capacitor C1 is in the top-left arm, C2 in the top-right arm, C3 in the bottom-left arm, and C4 in the bottom-right arm. Points P and Q are the mid-points (junctions). Find the potential difference V_P - V_Q.
• An isolated charged spherical soap bubble of radius r has the pressure inside equal to atmospheric pressure outside. The surface tension of the soap solution is T. If the total charge on the bubble can be written as N*pi*r*sqrt(2*T/epsilon0), find the integer value of N.

⚔️ Practice JEE Advanced Physics free + battle 1v1 →