A point charge $- q$ revolves around a fixed charge $+ Q$ in elliptical orbit. The minimum and maximum distance of $q$ from $Q$ are $r_{1}$ and $r_{2}$ , respectively. The mass of revolving particle is $m$ . $Q > q$ and assume no gravitational effects. Find the velocity of $q$ at positions when it is at $r_{1}$ and $r_{2}$ distance from $Q$ .

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None of above

$\sqrt{\frac{Q q r_{2}}{2 π ε_{0} m r_{1} (r_{1} + r_{2})}}, \sqrt{\frac{Q q r_{1}}{2 π ε_{0} m r_{2} (r_{1} + r_{2})}}$

$\frac{5}{8} \sqrt{\frac{3 Q q r_{2}}{π ε_{0} m r_{1} (r_{1} - r_{2})}}$

$\sqrt{\frac{9 Q q r_{2}}{7 π ε_{0} m r_{1} (r_{2} + r_{2})}} = \sqrt{\frac{2 Q q r_{1}}{π ε_{0} m_{2} (r_{1} - r_{2})}}$

answer is C.

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Detailed Solution

From conservation of mechanical energy and conservation of angular momentum about point $O$ ,

A point charge –q revolves around a fixed charge +Q in elliptical

we have

$\frac{1}{2} m v_{1}^{2} - \frac{1}{4 π ε_{0}}, \frac{Q q}{r_{1}} = \frac{1}{2} m v_{2}^{2} - \frac{1}{4 π ε_{0}}, \frac{Q q}{r_{1}}$
$\frac{1}{4 π ε_{0}} \cdot \frac{Q_{\cdot} q}{r_{2}} and m v_{1} r_{1} \sin 90 ° = m v_{2} r_{2} \sin 90 ° or v_{1} r_{1} = v_{2} \cdot r_{2}$
Solving these two equations, we have
$v_{1} = \sqrt{\frac{Q q r_{2}}{2 π ε_{0} m r_{1} (r_{1} + r_{2})}} and \sqrt{\frac{Q q r_{1}}{2 π ε_{0} m_{2} (r_{1} + r_{2})}} = v_{2}$