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# Determine the interaction energy of the point charges of the following setup.

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## 1 Answer

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As you know the interaction energy of an assembly of charges is given by
$\large\frac{1}{4 \pi \in_0} \sum _{i=j}^n \large\frac{q_{\gamma} q_j}{r_{ij}}$
$\therefore U= U_{12}+U_{13} +U_{14} +U_{23} +U_{24}+U_{34}$
$\quad=-\large\frac{Kq^2}{a} + \frac{kq^2}{(\sqrt 2 a)}-\frac{kq^2}{a}-\frac{kq^2}{a} + \frac{kq^2}{(\sqrt {2a})} -\frac{kq^2}{a}$
$\qquad= -\large\frac{2 K q^2}{a}$$[2 \sqrt 2 -1]$
answered Jun 11, 2014 by