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

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

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v_{1} = \sqrt{\frac{Q q r_{1}}{(r_{1} + r_{2})}}; v_{2} = \sqrt{\frac{K Q r_{2}}{(r_{1} + r_{2})}}

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

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$v_{1} = \sqrt{\frac{Q q r_{2}}{m {r_{1}}^{2}}}; v_{2} = \sqrt{\frac{Q q r_{1}}{m {r_{2}}^{2}}}$

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Solution

The correct option is A
$v_{1} = \sqrt{\frac{Q q r_{2}}{2 π ε_{0} m r_{1} (r_{1} + r_{2})}}; v_{2} = \sqrt{\frac{Q q r_{1}}{2 π ε_{0} m r_{2} (r_{1} + r_{2})}}$

From energy conservation principle between $A$ and $B$ .
${K . E}_{A} + {P . E}_{A} = {K . E}_{B} + {P . E}_{B}$
Substituting the values from figure, we get

$\frac{1}{2} m {v_{1}}^{2} - \frac{Q q}{4 π ε_{0} r_{1}} = \frac{1}{2} m {v_{2}}^{2} - \frac{Q q}{4 π ε_{0} r_{2}} . . . . . (1)$

From angular momentum conservation between $A$ and $B$ :

$L_{A} = L_{B}$

$\Rightarrow m v_{1} r_{1} sin 90^{\circ} = m v_{2} r_{2} sin 90^{\circ}$

$\Rightarrow v_{1} r_{1} = v_{2} r_{2} . . . . . . . . . . . . . . (2)$

From $(1)$ and $(2)$ we have:

$\frac{1}{2} m {v_{1}}^{2} - \frac{Q q}{4 π ε_{0} r_{1}} = \frac{1}{2} m \frac{v_{1}^{2} r_{1}^{2}}{r_{2}^{2}} - \frac{Q q}{4 π ε_{0} r_{2}}$

$\Rightarrow \frac{1}{2} m v_{1}^{2} (\frac{r_{1}^{2}}{r_{2}^{2}} - 1) = \frac{Q q}{4 π ε_{0}} (\frac{1}{r_{2}} - \frac{1}{r_{1}})$

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

From $(2)$ , we get

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

Hence, option (a) is the correct answer.

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