Question

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

• A. periodic for all values of $z_0$ satisfying $0<z_0<\infty$
• B. simple harmonic for all values of $z_{0}satisfying0<z_0\le R$
  
• C. approximately simple harmonic provided $z_0<<R$
  
• D. such that P crosses O and continues to move along the
  negative z-axis towards z=- $\infty$

Answer 1

Answer: (A), (C)

Let Q be the charge on the ring, the negative charge - q is
released from point P (0,0, $z_0).$ The electric field at P due to
the charged ring will be along positive z-axis and its
magnitude will be
$E=\frac{1}{4\pi {\epsilon }_0}\frac{Q{z}_0}{(R^2+z_0^2{)}^{3/2}}$
E = 0 at eentre of the ring because $z_0=0$
Force on charge at P will be towards centre as shown, and its
magnitude is
$F_e=qE=\frac{1}{4\pi {\epsilon }_0}.\frac{Qq}{(R^2+z_0^2{)}^{3/2}}.z_0...(i)$
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 $z_0$
lying between 0 and $\infty$
Secondly, if $z_0<<R,(R^2+z_0^2{)}^{3/2}=R^3$
$F_e=\frac{1}{4\pi {\epsilon }_0}.\frac{Qq}{R^3}.z_0[FromEq.(i)]$
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.)