A charge Q is placed at a distance `a//2` above the centre of the square surface of edge a as shown in the figure

The electric flux through the square surface is :

Explanation

This problem can be solved by symmetry consideration and Gauss's law.

`*` We can enclose the charged particle by a cube of side 'a' and keep the particle at the center of the cube.
`*` The total flux passing through the close cube is `phi = q//epsilon_0`.
`*` All the six surfaces are symmetrical with respect to charge, hence they will have eqyal contribution of the flux. So, flux through any one face is `phi' = phi//6 = q//6epsilon_0`.