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

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

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

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

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

answer is B.

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Detailed Solution

In the diagram balance all the forces

For equilibrium, consider forces along DA and equate the resultant to zero

$\begin{array}{l} \frac{1}{4 π ε_{0}} \frac{Q \times Q}{(D A)^{2}} + \frac{1}{4 π ε_{0}} \frac{Q \times Q}{(C A)^{2}} \cos 45^{\circ} - \frac{1}{4 π ε_{0}} \frac{Q \times q}{(E A)^{2}} \cos 45^{\circ} = 0 \\ \Rightarrow \frac{Q}{a^{2}} + \frac{Q}{2 a^{2}} \times \frac{1}{\sqrt{2}} - \frac{q}{a^{2} / 2} \times \frac{1}{\sqrt{2}} = 0 \\ \Rightarrow Q [1 + \frac{1}{2 \sqrt{2}}] = q \sqrt{2} \\ \Rightarrow q = \frac{Q}{\sqrt{2}} [\frac{2 \sqrt{2} + 1}{2 \sqrt{2}}] = \frac{Q}{4} (1 + 2 \sqrt{2}) . \end{array}$