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Lost1ne

- 47

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## 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?