A positively charged thin metal ring of radius R is fixed in the xy plane with its centre at the origin O. A negatively charged particle P is released from rest at the point $\left(0,0,z_0\right)$ Where $z_0>0$. Then the motion of P is 

Let Q be the charge on the ring, the negative charge-q is released from point P (0, 0, Z₀). The electric field at P due to the charged ring will be along positive z-axis and its magnitude will be

$E=\frac{1}{4\pi {\epsilon }_0}.\frac{Q{z}_0}{{\left(R^2+z_0^2\right)}^{3/2}}$

Therefore, force on charge P will be towards centre as shown, and its magnitude is

$F_e=qE=\frac{1}{4\pi {\epsilon }_0}.\frac{Qq}{{\left(R^2+z_0^2\right)}^{3/2}}.Z_0\mathrm{....}\left(1\right)$

Similarly, when it crosses the origin, the force is again towards centre O.

Thus the motion of the particle is periodic for all values of Z₀ lying between 0 and ∞.

Secondly if $Z_0<<R,{\left(R^2+Z_0^2\right)}^{3/2}\approx R^3$

$F_e=\frac{1}{4\pi {\epsilon }_0}.\frac{Qq}{R^3}.Z_0 \left[Fromequation1\right]$

i.e. the restoring force $F_e\propto -Z_0.$ Hence the motion of the particle will be simple harmonic. (Here negative sign implies that the force is towards its mean position).