# Can someone explain to me why a uniformly charged ring can produce a SHM [closed]

I am doing this homework question on Mastery Physics and I do not understand why this field would make a particle have Simple Harmonic Motion behaviors. Here is the question:

Imagine a small metal ball of mass m and negative charge $−q_0$. The ball is released from rest at the point $(0,0,d)$ and constrained to move along the $z$ axis, with no damping. If $0 < d ≪a$, what will be the ball's subsequent trajectory? • Can you clarify why you don't understand? Do you know how to apply Coulomb's law? Are you having trouble with the vector math? Are you confused about the equation of motion? How to recognize whether or not it supports SHO? Jan 9, 2016 at 21:32

To definitely be SHM you need to show that $F = -kz$ is true, so I'd start by trying to find the electric force $F$ as a function of $z$. Hyperphysics can help you out, although you may want to apply the approximation that $d << a$ to get SHM.
Electric field at a height $z$ on the axis of a ring carrying charge $Q$ uniformly distributed on it can be shown to be $${E(0,0,z)}=\frac{1}{4\pi\epsilon_0}\frac{Qz}{(a^2+z^2)^{3/2}} \hat{z}$$ if $d<<a$ the in the denominator $z$ can be neglected and force on the charge at a height $z$ will be $$F(0,0,z)=-\frac{1}{4\pi\epsilon_0}\frac{q_0Qz}{a^3}\hat{z}=-kz\hat{z}$$ Or $$m\ddot{z}+kz=0$$ Where $k=\frac{1}{4\pi\epsilon_0}\frac{q_0Q}{a^3}$ and frequency of oscillation will be $\omega=\sqrt{\frac{k}{m}}$.