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(A) The electric potential $V$ at the center of a square of side $a$ with charges $q$ at each of the $4$ corners is given by $V = 4 	imes \frac{Kq}{r

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$\frac{1}{2}$

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(A) The electric potential $V$ at the center of a square of side $a$ with charges $q$ at each of the $4$ corners is given by $V = 4 	imes \frac{Kq}{r}$,where $r$ is the distance from the center to any corner.For a square of side $a$,the diagonal is $a\sqrt{2}$,so the distance $r$ from the center to a corner is $r = \frac{a\sqrt{2}}{2} = \frac{a}{\sqrt{2}}$.For the first square,$a_1 = 1 \ m$,so $r_1 = \frac{1}{\sqrt{2}} \ m$. The potential is $V_1 = 4 	imes \frac{K 	imes 2}{1/\sqrt{2}} = 8\sqrt{2}K$.For the second square,$a_2 = 2 \ m$,so $r_2 = \frac{2}{\sqrt{2}} = \sqrt{2} \ m$. The potential is $V_2 = 4 	imes \frac{K 	imes 2}{\sqrt{2}} = \frac{8K}{\sqrt{2}} = 4\sqrt{2}K$.Therefore,the ratio is $\frac{V_2}{V_1} = \frac{4\sqrt{2}K}{8\sqrt{2}K} = \frac{1}{2}$.

Let $V_1$ be the potential at the center of a square of side $1 \ m$ when the charges at the $4$ corners are $2 \ C$ each. If the same charges are placed at the corners of a square of side $2 \ m$,then the potential at the center of this square is $V_2$. The value of $\frac{V_2}{V_1}$ is

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