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

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$-(Q/ 4)(1+ 2\sqrt 2)$

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(A) Let the side of the square be $a$. The distance of the centre from each corner is $r = \frac{a}{\sqrt{2}}$.For the system to be in equilibrium,the net force on any one charge $-Q$ at a corner must be zero.The forces acting on a charge $-Q$ at one corner due to the other three charges at the corners are:$1$. Two forces of magnitude $F_1 = \frac{1}{4\pi\epsilon_0} \frac{Q^2}{a^2}$ along the sides of the square.$2$. One force of magnitude $F_2 = \frac{1}{4\pi\epsilon_0} \frac{Q^2}{(\sqrt{2}a)^2} = \frac{1}{4\pi\epsilon_0} \frac{Q^2}{2a^2}$ along the diagonal.The resultant of the two forces $F_1$ is $F_{1net} = \sqrt{F_1^2 + F_1^2} = \sqrt{2} F_1 = \frac{\sqrt{2}Q^2}{4\pi\epsilon_0 a^2}$.The total force due to the three charges is $F_{total} = F_{1net} + F_2 = \frac{1}{4\pi\epsilon_0} \left( \frac{\sqrt{2}Q^2}{a^2} + \frac{Q^2}{2a^2} \right) = \frac{Q^2}{4\pi\epsilon_0 a^2} \left( \sqrt{2} + \frac{1}{2} \right)$.For equilibrium,this must be balanced by the force due to charge $q$ at the centre: $F_q = \frac{1}{4\pi\epsilon_0} \frac{qQ}{r^2} = \frac{1}{4\pi\epsilon_0} \frac{qQ}{(a/\sqrt{2})^2} = \frac{2qQ}{4\pi\epsilon_0 a^2}$.Equating the magnitudes: $\frac{Q^2}{4\pi\epsilon_0 a^2} \left( \sqrt{2} + \frac{1}{2} \right) = \frac{2qQ}{4\pi\epsilon_0 a^2}$.$q = \frac{Q}{2} \left( \sqrt{2} + \frac{1}{2} \right) = \frac{Q}{4} (2\sqrt{2} + 1)$.Since the force must be attractive,$q$ must have the opposite sign of $Q$. Thus,$q = -\frac{Q}{4}(1+2\sqrt{2})$.

Four charges equal to $-Q$ are placed at the four corners of a square of side length $a$,and a charge $q$ is at its centre. If the system is in equilibrium,the value of $q$ is:

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