A thin fixed of radius 1 metre has a positive charge `1xx10^-5` coulomb uniformly distributed over it. A particle of mass 0.9 gm and having a negative charge of `1xx10^-6` coulomb is placed on the axis at a distance of 1cm from the centre of the ring. Show that the motion of the negatively charged particle is approaximately simple harmonic. Calculate the time period of oscillations.

Explanation

Let us first find the force on a – q charge placed at a distance x from centre of ring along its axis. Figure shows the respective situation.

In this case force on particle P is
`F_P=-qE=-q(KQx)/((x^2+R^(2))^(3//2))`
For small x, `x ltlt R`, we can neglect x, compared to R, we have
`F=(KqQx)/(R^(3))`
Acceleration of particle is `a=-(KqQ)/(mR^(3))x` ,
[Here we have x = 1 cm and R = 1 m hence `xltltR` can be used ]
This shows that particle P excutes SHM, now comparing this acceleration with `a=-omega^2x`
`omega=sqrt((KqQ)/(mR^3))`
Thus time period of SHM is T
`=(2pi)/(omega)=2pisqrt((mR)/(KqQ))=2pisqrt((0.9xx10^(-3)xx(1)^(3))/(9xx10^(9)xx10^-5xx10^-6))`
`pi/5` seconds