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Let $\mathrm{V}_1$ be the potential at the center of the square of side $1 \mathrm{~m}$ when the charges at the 4 corners are $2 \mathrm{C}$ each. If the same charges are placed at the corners of a square of side $2 \mathrm{~m}$, then the potential at the centre of this square is $\mathrm{V}_2$. The value of $\frac{V_2}{V_1}$ is
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The correct answer is:
$\frac{1}{2}$
$\left(\mathrm{V}_{\text {centre }}\right)_1=\frac{\mathrm{K} \times 2}{\frac{1}{\sqrt{2}}} \times 4=\mathrm{V}_1$
$$
\left(\mathrm{V}_{\text {centre }}\right)_2=\frac{\mathrm{K} \times 2}{\frac{2}{\sqrt{2}}} \times 4=\mathrm{V}_2
$$
So, $\frac{\mathrm{V}_2}{\mathrm{~V}_1}=\frac{1}{2}$
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