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Four charges equal to ‘-Q’ are placed at the four corners of a square and a charge ‘q’ is at its centre. If the system is in equilibrium the value of ‘q’ is

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$\frac{- Q}{4} (1 + 2 \sqrt{2})$

$\frac{- Q}{2} (1 + 2 \sqrt{2})$

$\frac{Q}{4} (1 + 2 \sqrt{2})$

$\frac{Q}{2} (1 + 2 \sqrt{2})$

answer is C.

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Detailed Solution

$U_{s y s t e m} = \sum \frac{1}{4 π ε_{0}} \frac{q_{1} q_{2}}{r} = 0$

$\frac{1}{4 π ε_{0}} [\frac{4 (- Q) (- Q)}{l} + \frac{4 (- Q) q}{l / \sqrt{2}} + \frac{2 (- Q) (- Q)}{\sqrt{2} l}] = 0$

$4 (- Q) + 4 q \sqrt{2} - \frac{2 Q}{\sqrt{2}} = 0$

$q = Q (\frac{4 + \sqrt{2}}{4 \sqrt{2}})$

$q = \frac{Q}{4} (2 \sqrt{2} + 1)$

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