Question

A particle of mass m and charge `-Q` is constrained to move along the axis of a ring of radius a. The ring carries a uniform charge density `+lamda` along its length. Initially, the particle is in the centre of the ring where the force on it is zero. Show that the period of oscillation of the particle when it is displaced slightly from its equilibrium position is given by
`T=2pisqsqrt((2epsilon_0ma^2)/(lamdaQ))`

Answer 1

Correct Answer - A
Total charg on ring `=lamda(2pia)=q` (say)
Electric field at distance `x` from the centre of ring.
`E=1/(4piepsilon_0).(qx)/((a^2+x^2)^(3//2))=(lamddaax)/(2epsilon_0(a^2+x^2)^(3//2))`
Restoring force on `-Q` charge in this position would be
`F=-QE=-[(lambdaaQx)/(2epsilon_0(a^2+x^2)^(3//2))]`
For `xltlta`
`F=-[(lamdaaQ)/(2epsilon_0a^3)]x=-[(lamdaQ)/(2epsilon_0a^2)]x`
Comparing with `F=-kx`
`k=(lamdaQ)/(2epsilona_0^2)`
T=2pisqrt(m/k)=2pisqrt((2epsilon_0ma^2)/(lamdaQ)`