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(D) To find the electric flux through the square,we can use Gauss's Law by constructing a Gaussian surface.Consider the square as one face of a cube w

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$\frac{q}{6 \epsilon_0}$

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(D) To find the electric flux through the square,we can use Gauss's Law by constructing a Gaussian surface.Consider the square as one face of a cube with side length $a$.Since the charge $q$ is placed at a distance $\frac{a}{2}$ directly above the centre of the square,it is located exactly at the centre of this imaginary cube.According to Gauss's Law,the total electric flux through the entire closed surface of the cube is $\Phi_{total} = \frac{q}{\epsilon_0}$.Since the cube has $6$ identical faces and the charge is at the centre,the flux through each face is equal.Therefore,the electric flux through the square (one face of the cube) is $\Phi = \frac{\Phi_{total}}{6} = \frac{q}{6 \epsilon_0}$.

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

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