Animated Solution for Physics - Electric Charges and Fields: A charge $q$ is placed at one corner of a cube as shown in figure. The flux of electrostatic field $\mathbf{E}$ through the shaded area is

Welcome to one of the most beautiful and classic problems in electrostatics! This question is a perfect test of your spatial visualization and your deep understanding of Gauss's Law.

At first glance, calculating the electric flux through a specific face of a cube when the charge is sitting right on its corner might seem like a nightmare of complex integration. But don't get intimidated! We are going to solve this elegantly using the sheer power of symmetry.

A Charge at the Corner

Imagine you are holding a perfect cube. Right at one of its corners, let's say the origin $(0,0,0)$, we place a point charge $q$.

Our goal is to find the electric flux passing through one of the faces that does not touch this charge. In our specific diagram, this is the shaded face located on the far right.

To tackle this, we need to recall the fundamental principle of Gauss's Law. Gauss's Law states that the total electric flux through any closed surface is equal to the enclosed charge divided by the permittivity of free space.

Building a Super-Cube

Here is the catch: Gauss's Law is most useful when we have a highly symmetric closed surface where the charge is exactly at the geometric center.

In our current setup, the charge is at the corner, not the center. So, how do we create symmetry? We build a larger system!

Visualize placing identical cubes around our original cube so that the charge $q$ becomes completely surrounded. To enclose the corner completely, you need exactly $8$ identical cubes. You place $4$ cubes to form a bottom layer, and another $4$ cubes to form a top layer.

Now, the charge $q$ sits perfectly at the center of this massive "super-cube".

Because the charge is at the exact center of the super-cube, the electric field radiates uniformly in all directions.

This means the total flux is shared equally among all $8$ constituent cubes. Therefore, the effective charge enclosed by our single, original cube is simply one-eighth of the total charge.

Consequently, the total electric flux passing through our single cube is:

Zero Flux Faces

Now, look closely at our single cube. It has $6$ faces in total. Does the flux ${\varphi }_{\text{cube}}$ pass through all of them?

Notice the three faces that meet exactly at the corner where the charge is located. These faces lie in the $XY$, $YZ$, and $ZX$ planes.

Because the electric field lines originate from the charge and travel radially outward, they will perfectly skim along the surfaces of these three adjacent faces.

Since the field lines are parallel to the area vectors of these faces, they never actually pierce through them. Therefore, the electric flux through these three adjacent faces is exactly zero!

Dividing the Remaining Flux

If no flux escapes through the three adjacent faces, then all of the flux entering the cube must exit through the remaining $3$ non-adjacent, outer faces.

Because our cube is perfectly symmetric with respect to the charge at its corner, the electric field treats these three outer faces equally. The total flux is divided perfectly into three equal parts.

To find the flux through just one of these faces—like our shaded face—we simply take one-third of the cube's total flux.

Let's substitute the value we found earlier and calculate the final answer.

And there we have it! By leveraging symmetry, we bypassed complex calculus and arrived at a beautifully simple result. The correct option is (d).