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The length of each side of a cubical closed surface is L metre. If charge 48C is situated at one of the corners of the cube, Find the flux passing through the cube. (In Volt-metre)

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$\frac{6}{ϵ_{0}}$

$\frac{3}{ϵ_{0}}$

$\frac{48}{ϵ_{0}}$

$\frac{8}{ϵ_{0}}$

answer is A.

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Detailed Solution

$ϕ = \frac{q}{8 ε_{0}} = \frac{48}{8 ε_{0}} = \frac{6}{ε_{0}}$

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