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A large sheet of charge having charge density ‘p’ coul/m² passes through a cube having length of side ‘l’. Then maximum electric flux over the surface of the cube is

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$\frac{\sqrt{2} {σl}^{2}}{\in_{o}}$

$\frac{{σl}^{2}}{\sqrt{2} \in_{o}}$

$\frac{2 {σl}^{2}}{\in_{o}}$

$\frac{2 \sqrt{2} {σl}^{2}}{\in_{o}}$

answer is A.

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Detailed Solution

Maximum area of the plane enclosed by the cube = area of the plane ADGF = $\sqrt{2} l \times l = \sqrt{2} l^{2}$

$∴$ maximum charge enclosed by the cube = $\sqrt{2} l^{2} \times σ coul .$

$∴ {(ϕ_{E})}_{\max} = Q_{\max} / \in_{0} = \frac{\sqrt{2} {σl}^{2}}{\in_{0}}$ .

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