Q. The charges Q + q and +q are placed at the vertices of a right-angle isosceles triangle as shown below. The net electrostatic energy of the configuration is zero, it the value of Q is:

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Solution:

$U =K \left[\frac{q^{2}}{a} +\frac{Qq}{a} +\frac{Qq}{a\sqrt{2}}\right] = 0 $
$ \Rightarrow q=-Q \left[1+ \frac{1}{\sqrt{2}}\right] $
$\Rightarrow Q = \frac{-q\sqrt{2}}{\sqrt{2}+1} $

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