A particle of mass m having negative charge q | vedclass

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(A) For an elliptical orbit under a central force,the conservation of energy and angular momentum applies.At the points of closest approach $(r_1)$ an

Vedclass · Accepted Answer

$\sqrt{\frac{mr_1r_2Qq}{2\pi\varepsilon_0(r_1+r_2)}}$

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(A) For an elliptical orbit under a central force,the conservation of energy and angular momentum applies.At the points of closest approach $(r_1)$ and farthest distance $(r_2)$,the velocity is perpendicular to the radius vector.Conservation of angular momentum: $L = mv_1r_1 = mv_2r_2 \implies v_1 = \frac{L}{mr_1}$ and $v_2 = \frac{L}{mr_2}$.Conservation of energy: $\frac{1}{2}mv_1^2 - \frac{Qq}{4\pi\varepsilon_0r_1} = \frac{1}{2}mv_2^2 - \frac{Qq}{4\pi\varepsilon_0r_2}$.Substituting $v_1$ and $v_2$: $\frac{L^2}{2m}(\frac{1}{r_1^2} - \frac{1}{r_2^2}) = \frac{Qq}{4\pi\varepsilon_0}(\frac{1}{r_1} - \frac{1}{r_2})$.Using $a^2 - b^2 = (a-b)(a+b)$: $\frac{L^2}{2m} \frac{(r_2-r_1)(r_2+r_1)}{r_1^2r_2^2} = \frac{Qq}{4\pi\varepsilon_0} \frac{(r_2-r_1)}{r_1r_2}$.Solving for $L$: $L^2 = \frac{2mQq}{4\pi\varepsilon_0} \frac{r_1r_2}{r_1+r_2} = \frac{mr_1r_2Qq}{2\pi\varepsilon_0(r_1+r_2)}$.Thus,$L = \sqrt{\frac{mr_1r_2Qq}{2\pi\varepsilon_0(r_1+r_2)}}$.

$A$ particle of mass $m$ having negative charge $q$ moves along an ellipse around a fixed positive charge $Q$ such that its maximum and minimum distances from the fixed charge are $r_1$ and $r_2$ respectively. The angular momentum $L$ of this particle is:

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