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In the diagram shown , the charge +Q is fixed another charge +2q and mass M is projected from a distance R from the fixed charge . Minimum separation between the charges if the velocity becomes $\;\large\frac{1}{\sqrt{3}}\;$ times of the projected velocity , at this moment


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Answer : (a) $\;\large\frac{\sqrt{3}R}{2}$
Explanation :
Angular momentum of 2q charge about Q is constant . Therefore
$m\;v\;R\;sin 30^{0}=\large\frac{m\;v}{\sqrt{3}}\;r_{min} \;sin \theta$
When the two charges will be at minimum distance their relative velocity along the line joining them is zero . So $\theta=90^{0}$
$r_{min}=\large\frac{\sqrt{3} R}{2}\;.$
answered Feb 12, 2014 by yamini.v

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