Question

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.

Answer 1

The force on thepoint charge `Q` due to the element `dq` of the ring is
`F=(1)/(4piepsilon_(0))(dQ)/(r^(2))` along `AB`
For every element of the ring ,there is symmetrcially situated diametrically oppsite element ,the components of froces along the axis will add up while those perpendicular to it will cancel each other .Hence, net force on the charge `-Q` is `-ve` sign shows that this force will be towards the centre of ring.
`F= intdFcos theta=costhetaintdF`
`=(x)/(F) int(1)/(4piepsilon_(0))[(-Qdq)/(r^(2))]`
`F=-(1)/(4piepsilon_(0)(Qr)/(r^(3)))intdq=-(1)/(4piepsilon_(0))(Qqx)/((a^(2)+x^(2))^((3)/(2))....(1)`
(as `r=(a^(2)+x^(2))^((1)/(2))` and `r=(a^(2)+x^(2))^((1)/(2))`)
As the resorting froce is not linerar ,so the motion will be oscillatory .However if `x lt lt a` then
`F=-(1)/(4piepsilon_(0))(Qq)/(a^(3))x=-kx`
with `k=(Qq)/(4 pi epsilon_(0)a^(3))`
i.e, 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))`