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

Explanation

The force on the point charge Q due to the element dq of the ring
`dF=(1)/(4piepsilon_(0))(dqQ)/(r^(2))` along AB.

As for every element of the ring there is symmetrically situated diametrically opposite element, the components of forces along the axis will add up while those perpendicular to it will cancel each other. Hence, net force on the charge `-Q` is
`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))(i)["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)`,
`F=-(1)/(4piepsilon_(0))(Qq)/(a^(3))x=-kx" with "k=(Qq)/(4piepsilon_(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))`
Substituting the given data, we have `T=2pisqrt((0.9xx10^(-3)xx1^(3))/(9xx10^(9)xx10^(-5)xx10^(-6)))=(2pi)/(10)=0.62sec`