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(B) Let the side of the square be $a$. The distance of the center from each corner is $r = \frac{a}{\sqrt{2}}$.For the system to be in equilibrium, th

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

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(B) Let the side of the square be $a$. The distance of the center from each corner is $r = \frac{a}{\sqrt{2}}$.For the system to be in equilibrium, the net force on any charge at the corner must be zero.Consider a charge $-Q$ at one corner. The forces acting on it are:$1$. Repulsive forces from the other three charges $-Q$ at the corners.Let $F = \frac{1}{4 \pi \epsilon_0} \frac{Q^2}{a^2}$.The resultant of the two forces along the sides is $F_{res} = \sqrt{F^2 + F^2} = \sqrt{2} F$.The force from the diagonal charge is $F_{diag} = \frac{1}{4 \pi \epsilon_0} \frac{Q^2}{(\sqrt{2}a)^2} = \frac{F}{2}$.Total repulsive force $F_{total} = \sqrt{2} F + \frac{F}{2} = F(\sqrt{2} + \frac{1}{2})$.$2$. Attractive force from the central charge $q$: $F_q = \frac{1}{4 \pi \epsilon_0} \frac{q Q}{r^2} = \frac{1}{4 \pi \epsilon_0} \frac{q Q}{(a/\sqrt{2})^2} = \frac{2 q Q}{4 \pi \epsilon_0 a^2}$.For equilibrium, $F_q = F_{total} \implies \frac{2 q Q}{4 \pi \epsilon_0 a^2} = \frac{Q^2}{4 \pi \epsilon_0 a^2} (\sqrt{2} + \frac{1}{2})$.$2q = Q(\sqrt{2} + 0.5) = Q(\frac{2\sqrt{2}+1}{2})$.$q = \frac{Q}{4}(1 + 2\sqrt{2})$.Since the force must be attractive, $q$ must have a sign opposite to $-Q$, so $q$ is positive.

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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