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(B) For the system to be in equilibrium,the net force on every charge must be zero. Let the side of the square be $a$. The distance of the centre from

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

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(B) For the system to be in equilibrium,the net force on every charge must be zero. Let the side of the square be $a$. The distance of the centre from each corner is $r = \frac{a}{\sqrt{2}}$.Consider the force on a charge $-Q$ at one corner. The forces acting on it are:$1$. Forces from the other three $-Q$ charges at the corners.$2$. Force from the charge $q$ at the centre.The resultant force due to the two adjacent corners is $F_{adj} = \sqrt{(\frac{kQ^2}{a^2})^2 + (\frac{kQ^2}{a^2})^2} = \frac{\sqrt{2}kQ^2}{a^2}$.The force due to the diagonally opposite corner is $F_{diag} = \frac{kQ^2}{(a\sqrt{2})^2} = \frac{kQ^2}{2a^2}$.The total force from the three corners is $F_{corners} = \frac{kQ^2}{a^2}(\sqrt{2} + \frac{1}{2})$.For equilibrium,this must be balanced by the force from the central charge $q$: $F_{centre} = \frac{kQq}{(a/\sqrt{2})^2} = \frac{2kQq}{a^2}$.Equating the magnitudes: $\frac{kQ^2}{a^2}(\sqrt{2} + \frac{1}{2}) = \frac{2kQq}{a^2}$.Solving for $q$: $q = \frac{Q}{2}(\sqrt{2} + \frac{1}{2}) = \frac{Q}{4}(2\sqrt{2} + 1)$.Since the corner charges are negative,the central charge $q$ must be positive to provide an attractive force.

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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