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. Find the force on the particle. Assuming x `ltlt`R, find the time period of oscillation of the particle if it is released from there.

Explanation

When the negative charge is shifted at a distance x from the centre of the ring along its axis then force acting on the point charge dur to ring :
`F_(E) = qE` (towards centre)
`= q[(KQx)/((R^(2) + x^(2))^(3//2))]`
if `R gt gt x` then
`R^(2) + x^(2) = R^(2)`
`F_(E) = (1)/(4pi epsilon_(0)) (Qqx)/(R^(3))` (Towards centre)
Since restoring force `F_(E) prop x`, therefore motion of charge the particle will be S.H.M.
Time period of SHM.
`T = 2pisqrt((m)/(k)) = 2pisqrt((m)/((Qq)/(4pi epsilon_(0)R^(3))))`
`= [(16pi^(3) epsilon_(0)mR^(3))/(Qq)]^(1//2)`