Use Gauss's law and symmetry

The charge $Q$ is placed at a distance $a/2$ above the center of a square of side $a$.

This suggests imagining a cube of side $a$ built on that square as one face.

Since the charge is at height $a/2$ above the square, it lies exactly at the center of the cube.

Apply Gauss's law to the cube

For a closed cube enclosing charge $Q$, $\Phi_{\text{total}} = \frac{Q}{\varepsilon_0}$

Distribute flux equally among 6 faces

Because the charge is at the center of the cube, all 6 faces are equivalent by symmetry.

Therefore, flux through each face is $\Phi_{\text{one face}} = \frac{1}{6}\cdot \frac{Q}{\varepsilon_0} = \frac{Q}{6\varepsilon_0}$

Identify the required square surface

The given square surface is one face of this cube.

Hence the electric flux through the square surface is $\boxed{\frac{Q}{6\varepsilon_0}}$

Check options

• A: $\dfrac{Q}{\varepsilon_0}$
• B: $\dfrac{Q}{2\varepsilon_0}$
• C: $\dfrac{Q}{3\varepsilon_0}$
• D: $\dfrac{Q}{6\varepsilon_0}$

So the correct option is D.