A thin fixed ring of radius a has a positive chage q unformly distributed over it.A particle of mass m having a megative 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 chaged particle is approximatly simple harmonic. Calculate the time period of oscillation.

Explanation

The force on the pointchaage Q due to the element dq of the ring is
`dF=(1)/(4 pi epsilon_(0)) (dqQ)/(r^2)along AB`
For every element of the ring, there is a diametrically opposite element . The components of forces along the axis will add up. while those perpendicular to it will. cancle each other. Hence, the net force will be toward the center of ring.
`F=int dF cos theta= cos theta int dF=x/r int (1)/(4 pi epsilon_(0)) [-(Qdq)/(r^2)]`
so, `F=-(1)/(4 pi epsilon_(0)) (Qx)/(r^(3)) int dq=-(1)/(4 pi epsilon_(0)) (Qqx)/((a^(2)+x^(2))^(3//2))`...(i)
[ as `r=(a^(2)+x^(2))^(1//2) and int dq=q]`
As the restoring force is not linear, the motion will be oscillatroy. However, if `xltlta` so that `x^(2)ltlta^(2)`, then
`F=(1)/(4 pi epsilon_(0)) (Qq)/(a^(3)) x = -kx` with `k=(Qq)/(4 pi epsilon_(0)a^(3))`
Thus the restroing force will become linear and so the motion is simple harmonic with time period.
`T=(2 pi)/(omega) 2pi sqrt((m)/(k)) -2pi sqrt((4 pi epsilon_(0)ma^(3))/(qQ))`.