A positive charge Q is distributed uniformly over a circular ring of radius R . A particle having a mass m and a negative charge q is placed on its axis at a distance x from the centre. Determine the force on the particle. Assuming x R , find the time period of oscillation of the particle if it is released from this position.

Step-by-step solution

The electric field at a distance x on the axis of a uniformly charged ring of radius R and charge Q is given by: E = 1 4 ₀ Qx (R^2+x^2)^ 3/2 The force on a particle of negative charge q is: F = qE = 1 4 ₀ Qqx (R^2+x^2)^ 3/2 Since the charge on the particle is negative and the ring is positively charged, the force is attractive and directed towards the centre of the ring. For x R , we can approximate (R^2+x^2)^ 3/2 R^3 . The restoring force becomes: F = ( Qq 4 ₀ R^3 ) x This represents simple harmonic motion with fo