Four equal charges Q are placed at the four c | vedclass

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(C) The distance of each corner from the center $O$ of the square is $r = \frac{a}{\sqrt{2}}$.The electric potential $V_O$ at the center $O$ due to th

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$\frac{\sqrt{2} Q^2}{\pi \varepsilon_0 a}$

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(C) The distance of each corner from the center $O$ of the square is $r = \frac{a}{\sqrt{2}}$.The electric potential $V_O$ at the center $O$ due to the four charges $Q$ at the corners is:$V_O = 4 	imes \left( \frac{1}{4\pi \varepsilon_0} \cdot \frac{Q}{r} ight) = 4 	imes \left( \frac{Q}{4\pi \varepsilon_0 (a/\sqrt{2})} ight) = \frac{4\sqrt{2} Q}{4\pi \varepsilon_0 a} = \frac{\sqrt{2} Q}{\pi \varepsilon_0 a}$.The work done $W$ in moving a charge $q = -Q$ from the center to infinity is given by $W = q(V_{\infty} - V_O)$.Since the potential at infinity $V_{\infty} = 0$,we have:$W = (-Q)(0 - V_O) = Q V_O$.Substituting the value of $V_O$:$W = Q 	imes \left( \frac{\sqrt{2} Q}{\pi \varepsilon_0 a} ight) = \frac{\sqrt{2} Q^2}{\pi \varepsilon_0 a}$.

Four equal charges $Q$ are placed at the four corners of a square of side length $a$. The work done in removing a charge $-Q$ from its center to infinity is

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