A thin fixed ring of radius a has a positive charge q uniformly distributed over it.A particle of mass m having a negative charge Q, is placed on the axis at a distance of `x(xltlta)` form the center of the ring. Show that the motion of the negatively charged particle is approximately simple harmonic. Calculate the time period of oscillation.

Explanation

Let at some instant charged particle is at distance `x` from centre .
`E = (1)/(4 pi in_(0)) (Qx)/((R^(2) + x^(2))^((3)/(2)))`
Restoring force `F = -qE = -(1)/(4 pi in_(0)) (qQx)/((R^(2) + x^(2))^((3)/(2)))`
(`-ve` sign because force is directed towards centre)
Acceleration `a = (F)/(m) = -(Qq)/(4 pi in_(0) m) .(x )/((R^(2) + x^(2))^((3)/(2)))`
Since `(x lt lt R) ,(a lt lt R)`
`x^(2)` is negligible
`a = -(Qq)/(4 pi in_(0) m R^(3)) x`
`a alpha - x` (case of `S.H.M.`)
`T = 2 pi sqrt((4 pi in_(0) mR^(3))/(Qq))`
If `a` is comparable to `R` , particle will execute harmonic motion between
`x = -a `and `x = + a`