Question

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.

Answer 1

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))`.