Charges $-q,\, q,\,q$ are placed at the vertices $A$, $B$, $C$ respectively of an equilateral triangle of side $'a'$ as shown in the figure. If charge $-q$ is released keeping remaining two charges fixed, then the kinetic energy of charge $(-q)$ at the instant when it passes through the mid point $M$ of side $BC$ is
Medium
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$ \mathrm{U}_{\mathrm{i}}=-\frac{\mathrm{kq}^{2}}{\mathrm{a}}, \mathrm{U}_{\mathrm{f}} =-\frac{2 \mathrm{kq}^{2}}{\mathrm{a}}-\frac{2 \mathrm{kq}^{2}}{\mathrm{a}}+\frac{\mathrm{kq}^{2}}{\mathrm{a}} $
$=-\frac{3 \mathrm{kq}^{2}}{\mathrm{a}} $
$ \mathrm{W} =-\left(\mathrm{U}_{\mathrm{f}}-\mathrm{U}_{\mathrm{i}}\right) $
$ = - \left[ {\frac{{ - 3{\rm{k}}{{\rm{q}}^2}}}{{\rm{a}}} + \frac{{{\rm{k}}{{\rm{q}}^2}}}{{\rm{a}}}} \right] = \frac{{2{\rm{k}}{{\rm{q}}^2}}}{{\rm{a}}} = \frac{{{{\rm{q}}^2}}}{{2{ \in _0}\pi {\rm{a}}}}$
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