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(A) The potential energy $U$ of a system of point charges is given by $U = \sum \frac{k q_i q_j}{r_{ij}}$.For the given system,the potential energy $U

Vedclass · Accepted Answer

$\frac{1}{4 \pi \varepsilon_0} \frac{3q^2}{L}$

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(A) The potential energy $U$ of a system of point charges is given by $U = \sum \frac{k q_i q_j}{r_{ij}}$.For the given system,the potential energy $U$ is:$U = \frac{k(q)(q)}{L} + \frac{k(q)(-2q)}{L} + \frac{k(q)(-2q)}{L}$$U = \frac{k q^2}{L} - \frac{2k q^2}{L} - \frac{2k q^2}{L} = -\frac{3k q^2}{L}$Where $k = \frac{1}{4 \pi \varepsilon_0}$.When the charges are moved far apart (to infinity),the final potential energy $U_{\infty} = 0$.The work done by an external force is given by $W_{	ext{ext}} = U_{	ext{final}} - U_{	ext{initial}} = U_{\infty} - U$.$W_{	ext{ext}} = 0 - \left(-\frac{3k q^2}{L}ight) = \frac{3k q^2}{L} = \frac{1}{4 \pi \varepsilon_0} \frac{3q^2}{L}$.

Three point charges $q, q$ and $-2q$ are placed at the corners of an equilateral triangle of side $L$. Calculate the work done by an external force in moving all the charges far apart without acceleration.

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