A charge $q$ is placed at the centre of a cube of side $l$. What is the electric flux passing through each face to the cube?

Hint:In order to solve this problem, we are going to apply the concept of Gauss’s law for electric flux.
According to Gauss’s law for electric flux we get:
$\phi = \dfrac{Q}
{{{\varepsilon _ \circ }}}$

Complete step by step answer:
-According to Gauss' theorem, the flux of the electric field E through any enclosed surface, also known as the Gaussian surface, is proportional to the net charge enclosed divided by the permittivity of free space.Gauss' law on the electrical field defines the static electrical field produced by the distribution of electrical charges.
-Gauss 's law is a general law applied on any closed floor. It is an important instrument since it enables the measurement of the volume of the enclosed charge to be carried out by mapping the area on the surface outside of the application of the charge. For geometries with sufficient symmetry, the measurement of the electrical field is simplified.
Electric flux, the property of an electrical field that can be thought of as the number of electric power lines (or electrical field lines) that intersect a given area. The negative flux is equal to the positive flux in magnitude, so that the net or cumulative electrical flux is zero.
We know that total flux passing through the cube is:
$\phi = \dfrac{q}{{{\varepsilon _ \circ }}}$
As the charge is symmetrically positioned on each face of the square, the electrical flux flowing through each face is proportional to the ${\dfrac{1}{6}^{th}}$ of the total flux.
Therefore, electric flux passing through each face
$\phi ' = \dfrac{\phi }{6} = \dfrac{q}{{6{\varepsilon _ \circ }}}$

Hence, the electric flux passing through each face to the cube is $\dfrac{q}{{6{\varepsilon _ \circ }}}$.

Note:Here we have to remember the formula for the electric flux. Also we have to be careful while writing the numerator and denominator.