Charges -q,, q,,q are placed at the vertices | vedclass

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Question

$ \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}}

Vedclass · Accepted Answer

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

Vedclass · Answer

$ \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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