Question

A charge q is located at the centre of a cube. The electric flux through any face is?
A. $\dfrac{\pi q}{6(4 \pi \epsilon_0)}$
B. $\dfrac{q}{6(4 \pi \epsilon_0)}$
C. $\dfrac{2 \pi q}{6(4 \pi \epsilon_0)}$
D. $\dfrac{4 \pi q}{6(4 \pi \epsilon_0)}$

Answer 1

Hint: By Gauss's law, we have an easy expression to find out the total flux escaping for entering a surface. The surface and our case is the surface area of the cube. For all its six faces the electric field will pass in a symmetrical manner through it giving equal flux through all faces.
Formula used:
The flux of electric field passing through a closed surface is:
$\phi = E.A = \dfrac{q}{\epsilon_0}$,
where, q is the total charge inside the closed surface.

Complete answer:
In our case we have a cube as a closed surface at the centre of which a charge of q has been placed.

By Gauss's law, the total flux escaping out (or entering in) the sphere is given as:
$\phi = \dfrac{q}{\epsilon_0}$.
As this net flux is spread out uniformly in all directions, it can be said that an equal amount of flux must pass through all the faces of the cube. As a cube has 6 faces, we can write that flux through a single face is just:
$\phi_1= \dfrac{q}{6 \epsilon_0}$.
Now, as we look at the options, none of them appear to be in this form but consider option (D);
$\dfrac{4 \pi q}{6 (4 \pi \epsilon_0)}$.
Cancelling $4 \pi$ in the numerator and the denominator gives us the same form as we had found out by using Gauss's law simply.

So, the correct answer is “Option D”.

Additional Information:
For a point charge, the electric field lines tend to spread out equally in all directions and form a sphere as an equipotential surface i.e., locus of points having the same potential form a sphere. To the surface of this sphere, electric field lines as perpendicular.

Note:
One might go for another path here as the flux is also given by the product of electric field and the area through which it is passing i.e., the area of one face of the cube here. But, we have a much simpler and elegant option in the form of Gauss's law and besides that, the electric field of a point charge gives spherical symmetry so electric field intensity is not the same at all points through a face.