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(C) The work required to move a charge $q$ from a point with potential $V_0$ to infinity is given by $W = q(V_{\infty} - V_0)$. Since $V_{\infty} = 0$

Vedclass · Accepted Answer

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

Vedclass · Answer

(C) The work required to move a charge $q$ from a point with potential $V_0$ to infinity is given by $W = q(V_{\infty} - V_0)$. Since $V_{\infty} = 0$,the work required is $W = -qV_0$. Here,$q = -Q$,so $W = -(-Q)V_0 = QV_0$.The potential $V_0$ at the center of the square due to the four charges $Q$ at the corners is the sum of the potentials due to each charge.The distance from each corner to the center is $r = \frac{a}{\sqrt{2}}$.$V_0 = 4 	imes \left( \frac{1}{4\pi \varepsilon_0} \cdot \frac{Q}{a/\sqrt{2}} ight) = \frac{4\sqrt{2} Q}{4\pi \varepsilon_0 a} = \frac{\sqrt{2} Q}{\pi \varepsilon_0 a}$.Therefore,the work required is $W = (-Q) 	imes (-V_0) = Q 	imes \left( \frac{\sqrt{2} Q}{\pi \varepsilon_0 a} ight) = \frac{\sqrt{2} Q^2}{\pi \varepsilon_0 a}$.

Charges $Q$ are placed at each corner of a square of side $a$. How much work is required to remove a charge $-Q$ from the center of the square and move it to infinity?

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