Total charge - Q is uniformly spread along length of a ring of radius R. A small test charge +q of mass m is kept at the centre of the ring and is given a gentle push along the axis of the ring.
(a) Show that the particle executes a simple harmonic oscillation.
(b) Obtain its time period.

Explanation

(a) The force acting on a point charge q at point P at distance x <<<< R from the centre of ring of radius R on its axis by (- dQ) charge at A on ring is,
`dF = (k(-dQ)q)/(sqrt(R^(2)+x^(2)))^(2) =-k.((dQ)q)/(R^(2) + x^(2))`
As shown in figure, rfFsin0 are same in magnitude but opposite in direction. Hence, they cancelled the effect of each other. Hence, F will be the sum of only `dFcostheta` components which are parallel to axis and towards centre O.
`F = ointdFcos theta`
`=oint -k((dQ)q)/(R^(2) + x^(2)) xx x/sqrt(R^(2) + x^(2))`
`therefore F = -kQq x/(R^(2) + x^(2))^(3//2)(-Q)`
Here, `x^(2) lt lt lt lt lt R^(2)` hence, `R^(2) + x^(2) = R^(2)`
`F = -(kQq)/R^(3)`x.........(1)
`therefore F prop (-x)`........(2)
The above relation shows that the oscillations of +q about axis of ring is simple harmonic motion. If force constant is K., then
`k. = -F/x = (Qq)/(4piepsilon_(0)R^(3))`..........(3)
From eqns. (1)
(b) Here, time period of +q charge of mass M is
`T = 2pisqrt(m/k.)`
`therefore T = 2pi sqrt(m/((Qq)/(4piepsilon_(0)R^(3)))`
`therefore T = 2pisqrt((4piepsilon_(0)mR^(3))/(Qq))`........(4)
Equation (2) and (4) show the asked results.