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)` from the center of the ring. Show that the motion of the negatively charged particle is approximately simply harmonic. Calculate the time period of oscillation.

Explanation

The force on the point charge Q due to the element dq of the ring is `dF=(1)/(4piepsilon_(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 cancel each other. Hence, the net force on the charge is `-Q`, negative sign shows that this force will be toward the centre of ring.
`F=intdFcostheta=costhetaintdF=(x)/(r)int(1)/(4piepsilon_(0))[-(Qdq)/(r^(2))]`
So, `F=-(1)/(4piepsilon_(0))(Qx)/(r^(3))intdq=-(1)/(4piepsilon_(0))(Qqx)/((a^(2)+x^(2))^(3//2)).......(a)`
[as `r=(a^(2)+x^(2))^(1//2)andintdq=q`]
As the restoring force is not linear, the motion will be oscillatory. However, if `xltlta` so that `x^(2)ltlta^(2)`, then
`F=-(1)/(4piepsilon_(0))(Qq)/(a^(3))x=-kx` with `k=(Qq)/(4piepsilon_(0)a^(3))`
Thus, the restoring force will become linear and so the motion is simple harmonic with time period,
`T=(2pi)/(omega)=2pisqrt((m)/(k))=2pisqrt((4piepsilon_(0)ma^(3))/(qQ))`