Question

A charge $q$ is placed at the corner of a cube of side $a$. The electric flux passing through the cube is :

• A. $\frac{q}{{a\epsilon }_0}$
• B. $\frac{q}{{\epsilon }_0{a}^2}$
• C. $\frac{q}{4\pi {\epsilon }_0{a}^2}$
• D. $\frac{q}{8{\epsilon }_0}$

Answer 1

The correct option is D $\frac{q}{8{\epsilon }_0}$

$\text{Step 1: Choosing Gaussian Surface [Refer Fig. 1 \& 2]}$

Consider the cube and the charge placed at its corner as shown in figure 1.

For application of Gauss Law, we have to find a gaussian surface inside which this charge$\left(q\right)$ lies symmetrically.

Therefore, Imagine 3D picture with the charge $q$ at the origin, and 8 cubes of side $a$ placed such that each having one vertex at the origin, touching the charge $q$ as shown in the figure 2.

Whereas the given smaller cube is one of the part of the larger cube, i.e. cube no. 7.

Combining these 8 cubes, we get a Larger cube of side 2a with charge$\left(q\right)$ placed at its center. Assuming it to be Gaussian Surface.

$\text{Note:}$ The given cube of side $a$ can not be taken as gaussian surface, as the charge$\left(q\right)$ lies on its boundary, therefore Gauss law will not be applicable.

$\text{Step 2: Applying Gauss Law}$

Now, By Gauss Law for the chosen gaussian surface:

Total flux through this larger cube $=\frac{\text{Charge Enclosed}}{{ϵ}_0}=\frac{q}{{ϵ}_0}$

$\text{Step 3: Finding Flux through given Cube using the symmetry}$

As the position of the charge is symmetric with respect to all 8 cubes.

Hence, the total flux through the larger cube will be divided equally among all the 8 cubes.

$\therefore$ Flux through smallar cube(no. 7) of side $a$ $=\frac{\text{Total Flux}}{8}=\frac{q}{8{ϵ}_0}$

Hence, Option A is correct.

$\text{Additional Info:}$ Further, this $q/\left(8{ϵ}_0\right)$ flux will be shared equally among the three opposite faces of the cube due to symmetry, Hence flux through each of these 3 faces $=\frac{1}{3}\times \frac{q}{8{ϵ}_0}=\frac{q}{24{ϵ}_0}$

And the flux through the 3 adjacent faces of the cube will be zero, as electric field will be parallel to these three faces.