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\[{z_o} > 0\]. Then the motion of P is
A. Periodic for all values of zo satisfying $0 < {z_o} < \infty $ .
B. Simple harmonic for all values of zo satisfying $0 < {z_o} \leqslant R$
C. Approximately simple harmonic provided \[\;{z_o} < < R\]
D. Such that P crosses O and continues to move along the negative z-axis towards $z = - \infty $

Answer 1

Hint: : The charge will be evenly distributed across the ring since it is uniformly charged. Field intensity will be produced with each charge. We will use differentiation of the equation in this example to determine the maximum electric field intensity. Following that, we will change the value of x in the first equation.

Formula used:
The electric field due to the ring is given by $E = \dfrac{{kQx}}{{{{({r^2} + {x^2})}^{3/2}}}}$ where \[k=\dfrac{1}{4\pi {{\varepsilon }_{o}}}\]
and the force becomes $F = qE = \dfrac{{kqQx}}{{{{({r^2} + {x^2})}^{3/2}}}}$
The positively charged ring is in the x y plane having let charge =+Q. Let the charge on negatively charged particles is –Q present above the plane that is z axis. The point P is at a distance x from the plane and radius of the ring be r. The direction of net force is towards the center of the ring.

Complete answer:
If the position of the particle is reversed, that means if the negatively charged particle is placed on the negative z axis then also the net force direction remains the same. Hence it is periodic in nature. Hence option (1) is correct. But option (4) is incorrect as the direction of force is not outward but inside.
When,
\[x = {\text{ }}r\], then $F = \dfrac{{kqQx}}{{{{({r^2} + {r^2})}^{3/2}}}}$
$F = \dfrac{{kqQx}}{{{{(2{r^2})}^{3/2}}}}$
$F = \dfrac{{kqQx}}{{{{(2)}^{3/2}}{r^2}}}$
As F is inversely proportional to square of radius, hence it is not simple harmonic motion.

Option3: if $x = {z_0}$
Then $F = \dfrac{{kqQ{z_0}}}{{{{({z_0}^2 + {r^2})}^{3/2}}}}$
As $z_o$ is very small than r
Then $F = \dfrac{{kqQ{z_0}}}{{{{({r^2})}^{3/2}}}}$
$F = \dfrac{{kqQ{z_0}}}{{({r^3})}}$

Here F is directly proportional to ${z_0}$ so option C. is correct. It is SHM.

Note: A periodic motion is a motion which repeats itself after a particular time period. Simple harmonic motion is also a periodic motion but it is oscillating in nature and the displacement is always in the direction opposite to the restoring force. But this is not the case in any periodic motion. Motion of the clock is periodic motion and the motion of the pendulum is periodic but oscillating hence is simple harmonic motion. If F is directly proportional to –x then it is called simple harmonic motion.