SATHEE: Electrostatics 7 Question 22

Electrostatics 7 Question 22

24. 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 $z_{0}$ satisfying $0 < z_{0} \leq R$

(d) such that $P$ crosses $O$ and continues to move along the negative $z$ -axis towards $z = - \infty$

Integer Answer Type Questions

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Solution:

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

$E = \frac{1}{4 π ε_{0}} \frac{Q z_{0}}{{(R^{2} + z_{0}^{2})}^{3 / 2}}$

$E = 0$ at centre of the ring because $z_{0} = 0$

Force on charge at $P$ will be towards centre as shown, and its magnitude is

$F_{e} = q E = \frac{1}{4 π ε_{0}} \cdot \frac{Q q}{{(R^{2} + z_{0}^{2})}^{3 / 2}} \cdot z_{0}$

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 $z_{0} « R, {(R^{2} + z_{0}^{2})}^{3 / 2} \approx R^{3}$

$F_{e} \approx \frac{1}{4 π ε_{0}} \cdot \frac{Q q}{R^{3}} \cdot z_{0} [From Eq. (i)]$

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.)