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

0.63

Explanation

`E_(ax is)=1/(4pi epsilon_0)(Qx)/((X^2+R^2)^(3//2))`
For `x< `E~~(Qx)/(4pi epsilon_0R^3)` , `F_(n et)` on negatively charged particle `F=(-Qqx)/((4pi epsilon_0)R^3`
`a=F/m=-(Qqx)/((4pi epsilon_0R^3)m)=-((Qq)/(4pi epsilon_0mR^3))(x)`
`veca=-omega^2x implies SHM T=(2pi)/omega=sqrt((4pi epsilon_0mR^3)/(Qq))implies r=0.628 sec~~063 sec`