Question

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 $(0,0,z_0)$ 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 $z_0$ satisfying $0<z_0\le R$
• C. Approximately simple harmonic, provided $z_0<<R$.
• D. Such that $P$ crosses $O$ & continues to move along the $–ve$ $z-$axis towards $z=–\infty$
  

Answer 1

Answer: (A), (C)

Approximately simple harmonic, provided $z_0<<R$.

The positively charge metal ring lies in x – y plane. Its centre is at the origin $O$.
Let $Q$ = Charge on the ring.

$\therefore$ Electric field at $P$
$E=\frac{1}{4\pi {ϵ}_0}\frac{Q{z}_0}{(R^2+z_0^2{)}^{1.5}}$ ........(i)
At centre of ring, $z_0=0$

$\therefore E=0$

Force on charge $P$ is $F_e=qE$

$F=\frac{1}{4\pi {ϵ}_0}\frac{Qq{z}_0}{(R^2+z_0^2{)}^{\frac{3}{2}}}$ .....(ii)

The force $F$ acts on $P$ towards centre $O$. As the particle reaches $O$, the force on $P$ becomes zero. When $P$ crosses $O$ and moves to negative side of z-axis, the force is again acts towards $O$. Thus the motion of $P$ is periodic for all values of $z_0$ satisfying $0<z_0<\infty$ .
Option (a) is correct.
Option (d) is incorrect under the above discussion.

From equation (ii). it is clear that F is not proportional to displacement $z_0$. Hence the motion is not simple harmonic.
(b) is incorrect.

Let $z_0\ll R,(z_0^2+R^2)\approx R^2$
$\therefore F=\frac{1}{4\pi {ϵ}_0}\frac{Qq{z}_0}{R^3}$
Force $\left(F\right)\propto –z_0$
The motion is simple harmonic if $z_0<<R$.
option (c) is correct.