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 is very less than R, find the time period of oscillation of the particle if it is released from there.

Options

A. `[[16pi^(3)epsilon_(0)R^(3)m)/(Qq)]^(1//2)`
B. `[(8pi^(2)epsilon_(0)R^(3))/q]^(1/2)`
C. `[(2pi^(3)epsilonR^(3))/(3q)]^(1//2)`
D. None of these

Answer

A

Explanation

When the negative charge is shifted at a distance x from the center of the ring along its axis, then force acting on the point charge due to the ring.

F=qE(towards center)
`=q.(kOx)/(R^(2)+x^(2))^(3//2)`
If Rgtgtx, then `R^(2)+x^(2)~~R^(2)`
and `F=1/(4piepsilon_(0)).(Qqx)/(R^(3))` (towards centre)
`rArr a =F/m =1/(4piepsilon_(0).(Qqx)/(mR^(3)))`
Since, restoring force `F_(E) propto x`, therefore motion of charge particle will be SHM.
Time period of SHM,
`T=(2pi)/omega=[(16pi^(3)epsilon_(0)R^(3)m)/(Qq)]^(1/2)` `[therefore omega^(2)=(Qq)/(4piepsilon_(0)R^(3))]`