Charges -q,, q,,q are placed at vertices A , B , C respectively of an equilateral triangle of side &

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2. Electric Potential and Capacitance

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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

A

$\frac{{{q^2}}}{{8\pi { \in _0}a}}$

B

$\frac{{{q^2}}}{{4\pi { \in _0}a}}$

C

$\frac{{{q^2}}}{{2\pi { \in _0}a}}$

D

$\frac{{{q^2}}}{{\pi { \in _0}a}}$

Solution

$ \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}}}}$

Standard 12

Physics

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