Electrostatics 7 Question 32
34. A circular ring of radius $R$ with uniform positive charge density $\lambda$ per unit length is located in the $y$-z plane with its centre at the origin $O$. A particle of mass $m$ and positive charge $q$ is projected from the point $P(R \sqrt{3}, 0,0)$ on the positive $x$-axis directly towards $O$, with an initial speed $v$. Find the smallest (non-zero) value of the speed $v$ such that the particle does not return to $P$.
(1993, 4M)
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Solution:
- Total charge in the ring is $Q=(2 \pi R) \lambda$
Potential due to a ring at a distance of $x$ from its centre on its axis is given by $V(x)=\frac{1}{4 \pi \varepsilon _0} \cdot \frac{Q}{\sqrt{R^{2}+x^{2}}}$
and at the centre is $V _{\text {centre }}=\frac{1}{4 \pi \varepsilon _0} \frac{Q}{R}$
Using the above formula
$$ \begin{aligned} V _p & =\frac{1}{4 \pi \varepsilon _0} \cdot \frac{2 \pi R \lambda}{\sqrt{R^{2}+3 R^{2}}}=\frac{\lambda}{4 \varepsilon _0} \\ V _o & =\frac{1}{4 \pi \varepsilon _0} \cdot \frac{2 \pi R \lambda}{R}=\frac{\lambda}{2 \varepsilon _0} \end{aligned} $$
$$ V _o>V _p $$
PD between points $O$ and $P$ is
$$ \begin{aligned} & & V & =V _o-V _p=\frac{\lambda}{2 \varepsilon _0}-\frac{\lambda}{4 \varepsilon _0}=\frac{\lambda}{4 \varepsilon _0} \\ & \therefore & \frac{1}{2} m v^{2} & \geq q V \text { or } v \geq \sqrt{\frac{2 q V}{m}} \\ & \text { or } & & \geq \sqrt{\frac{2 q \lambda}{4 \varepsilon _0 m}} \\ & \text { or } & & \geq \sqrt{\frac{q \lambda}{2 \varepsilon _0 m}} \end{aligned} $$
Therefore, minimum value of speed $v$ should be
$$ v _{\min }=\sqrt{\frac{q \lambda}{2 \varepsilon _0 m}} $$
