Question

A point charge $\mathrm{q}$ is placed at the corner of a cube with side $\mathrm{I}$, find flux through entire surface and flux through each face?

Answer 1

Step 1. Given data

From the given, the charge is placed at the corner of the cube as shown in the below figure.

We have to find the flux through the entire surface and flux through each face

Step 2. Concept to be used

Gauss's law states that electric flux through any surface enclosing charge is equal to $\frac{\mathrm{q}}{{\in }_0}$ where $\mathrm{q}$ is the net charge enclosed by the surface.

$\mathrm{\varphi }\overrightarrow{\mathrm{E}}·\mathrm{d}\overrightarrow{\mathrm{a}}=\frac{{\mathrm{q}}_{\mathrm{enc}}}{{\in }_0}$

Here, ${\mathrm{q}}_{\mathrm{enc}}$ is the net charge enclosed by the surface.

${\in }_0$is the permittivity of free space

Step 3. Find flux through the entire surface and flux through each face.

Let us consider another figure in which the whole charge is enclosed in the four such cubes.

According to the Gauss Law, the total flux from a charge $\mathrm{q}$ is $\frac{\mathrm{q}}{{\in }_0}$.
If the charge is on the corner of a cube, a few of the flux enters the cube and renounce through a few of its faces. But few of the flux doesn't enter the cube.
Let us consider, that the charge is on a corner of every 8 cubes. An equal amount of flux will disperse outwards symmetry via each of the cubes.
Charge for $1$ cube $=\frac{\mathrm{q}}{8}$
Now, three of its faces are touching each given cube. Therefore, the electric field vector and the area vector of that side will be perpendicular. Thus, the flux via those three sides will be zero. (Field and area would be perpendicular, resulting in a zero dot product)
The flux will come from the other three sides in equal amounts.
So,

Flux via one side $=\frac{\left({\displaystyle \frac{\mathrm{q}}{8}}\right)}{3}\times \frac{1}{{\in }_0}$

$=\frac{\mathrm{q}}{24{\in }_0}$

Therefore, the flux via the single face is, $\frac{\mathrm{q}}{24{\in }_0}$.