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A cubical volume has uniform charge density. Electric potential at the centre of the cube is $V_{0} .$ If the potential at the centre of a cube of twice the side length and half the charge density of the previous cube is $V$ then, find $\frac{V}{V_{0}} .$

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answer is 2.

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Detailed Solution

$V α \frac{Q}{x}$ where $x$ is side length of the cube.
$Q$ = charge inside the cube
$= ρ x^{3}$
$∴ V α \frac{ρ x^{3}}{x} = ρ x^{2}$
$∴ V α ρ x^{2}$
$\frac{V}{V_{0}} = (\frac{ρ / 2}{ρ}) {(\frac{2 x}{x})}^{2} = 2$

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