Electrostatics Ques 184
- 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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Answer:
Correct Answer: 184.$v_{\min }=\sqrt{\frac{q \lambda}{2 \varepsilon_{0} m}}$
Solution:
Formula:
Electrostatic Potential Energy
- 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
$$ v_{\min }=\sqrt{\frac{q \lambda}{2 \varepsilon_{0} m}} $$
BETA
