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A point charge q is placed at a vertical height d from a corner of a square plate of side d placed horizontally as shown. The electric flux through the square plate is

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$\frac{q}{ε_{0}}$

$\frac{q}{6 ε_{0}}$

Zero

$\frac{q}{24 ε_{0}}$

answer is D.

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Detailed Solution

Apply Gauss’s law, $ϕ_{T o t a l} = \frac{q}{\in_{0}}$
$ϕ = \frac{ϕ_{T o t a l}}{6} = \frac{1}{6} = \frac{q}{24 \in_{0}}$

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