A thin fixed ring of radius 2m has a positive charge of `10^(-6)` C uniformly distributed over it. A particle of mass `0.9` g and having a negative charge of `10^(-7)C` is placed on the axis at at a distance 2 cm from centre of ring. Show that motion of negatively charged particle is approximately SHM. Calculate the time period of oscillation.

Explanation

Given, radius of ring `r=2m,q_(1)=10^(-6)C`,
`q_(2)=-10^(-7)C,m=0.9g=9xx10^(-4)"kg"`,
`l=2" cm"=0.02" m", T=?`
Force on particle is given by
`F=(q_(1)q_(2)l)/(4pi epsilon_(0)(l^(2)+r^(2))^(3//2))`
`=((9xx10^(9))(10^(-6))(10^(-7))l)/((4+0.004)^(3//2))`
`=(9xx10^(-4)l)/(8.001)=1.125xx10^(-4)l`
As, acceleration, `a=(F)/(m)=(1.125xx10^(-4)l)/(9xx10^(-4))`
`=0.125lrArra prop l`
Thus, the motion of particle is SHM.
Now, time period of oscillation of particle is given by
`T=2pisqrt((l)/(a))=2pisqrt((l)/(0.125l))`
`rArr" "T=17.7` s