Flux Through a Cube's Face with our Point Charge at a Corner

[Lost1ne]

Homework Statement

A charge q is placed at one corner of a cube. What is the value of the flux of the charge's electric field through one of its faces?

Homework Equations

The flux surface integral of an electric field is equal to the value of the charge enclosed divided by the epsilon_naught constant.

The Attempt at a Solution

I've seen the creative solution to the problem where you place the charge q at the center of a cube of length 2s if the length of our first cube was s.

It follows that the total flux through the 2s cube is q/ε and that each of its 6 faces (with surface area 2s * 2s), by symmetry, have equal portions of this flux value. Now we can divide each of these faces on the 2s cube into four squares. By symmetry, each of these squares must have an equal portion of the flux value, and we realize that our 1s cube is composed out of 6 of these squares in one corner of our 2s cube. So we then conclude that the flux through one face of our 1s cube is (1/24)q/ε.

I still can't seem to accept this argument though. Let's take our 1s cube and have our charge q at one of its corners. 3 of the 1s cube's faces will have no flux. The other 3 faces will have flux, but by symmetry, these flux values are equal. So why may I not conclude from here that the flux through one of the faces is (1/3)q/ε?

Is taking a larger cube of side-length 2s logical? By Gauss's Law, any closed surface with the same and only the same charge value inside, q, will have the same net flux, q/ε. But by taking a larger geometrical object that still satisfies the previous sentence, aren't we dividing the net flux value into a larger number of "pieces", meaning that the flux through each of these pieces must then take a smaller value so that the net flux through the entire object has a constant value? So then, wouldn't the flux found through one of the 1s cube faces change if we speak of it as one-twenty-fourth of a larger cube instead of one-sixth of a smaller cube?

 

[Orodruin]

Placing the charge at the boundary of your volume, all of the flux will not go through your surface.

In other words, you need to be much more careful when there is charge on your gaussian surface.

 

[Orodruin]

Also note that if you place the charge just inside the corner, then you can no longer argue by symmetry that all sides must have the same flux.

 

[Lost1ne]

Placing the charge at the boundary of your volume, all of the flux will not go through your surface.

So then my interpretation of Gauss's Law or this problem must be incorrect.

My understanding coming in was that Gauss's Law couldn't care less where your charge is inside the surface. As long as that quantity of charge is enclosed by a surface, then that surface along with any other arbitrary, closed surface enclosing just that charge will produce the same net flux of q/ε.

I see that 3 of the 1s cube faces will have no flux. But my interpretation was that the net flux due to this electric field had to add to q/ε, thus, along with symmetry, meaning that the other faces would have a flux of (1/3)q/ε.

If my interpretation of Gauss's Law is correct, then is the issue because the charge is only partially enclosed by our 1s cube? (Would your edit on your first comment agree with this?)

2nd edit: Okay, I think things are beginning to click. So can I say that Gauss's Law assumes that, with the surface chosen, all the flux will go through the surface?

 

[Orodruin]

My understanding coming in was that Gauss's Law couldn't care less where your charge is inside the surface.

The keyword being inside the surface. If you place it in the corner, it is on the surface, not inside it.

I see that 3 of the 1s cube faces will have no flux.

This is true if you place your charge in the corner, ie, on the surface. If you place the charge just inside the corner it will no longer be true. Then most of the flux will go through the nearby faces.

I see that 3 of the 1s cube faces will have no flux. But my interpretation was that the net flux due to this electric field had to add to q/ε, thus, along with symmetry, meaning that the other faces would have a flux of (1/3)q/ε.

Since the charge is not enclosed by the surface, this is not true. Gauss law is true for enclosed charge.

If my interpretation of Gauss's Law is correct, then is the issue because the charge is only partially enclosed by our 1s cube? (Would your edit on your first comment agree with this?)

Well, it is a point charge so you need to be careful with those interpretations. But in essence yes. I suggest not using gaussian surfaces such that you have non-zero charge on the surface.

 

[Lost1ne]

Well, it is a point charge so you need to be careful with those interpretations. But in essence yes. I suggest not using gaussian surfaces such that you have non-zero charge on the surface.

Yeah, I think "on the surface" would be a better interpretation of what's going on.

So it's not that Gauss's Law assumes that your surface will have a non-zero flux on every part of your surface; it's that Gauss's law assumes that your surface encloses a charge.

So let me see if I have this straight. If the charge were exterior to the surface, the net flux by the charge's electric field is zero. If the charge is interior to the surface, the net flux by the charge's electric field is computed through Gauss's Law. And if the charge is on the surface, it's like encountering an indeterminate form in a math problem. We have to shift our perspective to calculate the flux, and we may make this process simpler by taking a cube of side length 2s and placing our charge at the cube's center. Then symmetry just turns things into a simple dividing problem (i.e, divide by 6 for the 6 faces of the 2s cube, divide by four for the four squares on each face).

 

[Orodruin]

That sounds much better.