Let Q be the charge on the ring, the negative charge -q is released from point ( 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 = 0 at centre of the ring because z0 = 0
Force on charge at P will be towards centre as shown, and its magnitude is
F0=qE= Qqz0/14πε0(R2+z20)3/2 ..........(i)
Similarly, when it crosses the origin, the force is again towards centre O.
Thus, the motion of the particle is peroidic for all values of z0 lying between 0 and ∞.
Secondly, if z0<<R,(R2+z02)3/2≈R3
F0≈Qqz0/14πε0⋅R3 [From Equation. ( i)]
i.e., the restoring force Fe∝−z0. Hence, the motion of the particle will be simple harmonic. (Here negative sign implies that the force is towards its mean position.)