A point charge q is placed at a distance a/2 | vedclass

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(D) According to Gauss's Law,the total electric flux through a closed surface is $\frac{q}{\varepsilon_0}$.To calculate the flux through the square of

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$\frac{q}{6\varepsilon_0}$

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(D) According to Gauss's Law,the total electric flux through a closed surface is $\frac{q}{\varepsilon_0}$.To calculate the flux through the square of side $a$,we can imagine this square as one face of a cube of side $a$.Since the charge $q$ is placed at a distance $a/2$ above the centre of the square,it is exactly at the centre of this imaginary cube.By symmetry,the total flux $\frac{q}{\varepsilon_0}$ is distributed equally among the $6$ faces of the cube.Therefore,the electric flux through the given square is $\phi = \frac{1}{6} \left( \frac{q}{\varepsilon_0} \right) = \frac{q}{6\varepsilon_0}$.

$A$ point charge $q$ is placed at a distance $a/2$ directly above the centre of a square of side $a$. The electric flux through the square is

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