Using a cube as a Gaussian surface

[anban]

Homework Statement

Suppose there is a uniformly charged cube with known side length. I then imagine a larger, closed cube surface surrounding it. This larger cube has double the side length and is symmetrical to the smaller cube.

Is is practical to use this Gaussian surface to compute the value of the E field at some arbitrary point outside the charged cube?

Homework Equations

Gauss' Law, \intE*da = q/\epsilon

The Attempt at a Solution

A classmate told me that this would be very difficult to do, and that no, using a cubical Gaussian surface would not work very well in a practical sense.

When I think of this problem, this is what I think:
Every charge on the surface of the inner cube will have an electric field pointing away from its center radially. Since the cube is symmetrical, though, many of these field lines will cancel each other out, leaving only field lines going outward from the inner cube perpendicular to each cube face. That would mean that the larger, imaginary cube would experience a flux out of each of its cube faces equal to the E field coming from each inner cube face. The sum of each face would then allow us to know the total flux leaving the Gaussian surface.

However, this would give us a discontinuous E field, one that would have no flux or E field coming off the edges. This would give us E field lines in a 3D cross shape.

I have no clue if this problem is very simple or very difficult. Any insights are very warmly welcome.

 

[anban]

One more idea for solving this:

If I simplify the the inner charged cube to just be a point charge surrounded by a cubical Gaussian surface, then computing the E field is very difficult. It is not simple because the electric field vector coming from the charge is often at an angle to the area vector of the imaginary surface. It would be a very difficult integral to execute.

 

[rude man]

I don't see how Gauss can help here.

I would orient the cube with its center at (0,0,0) and observation point along the x axis, then compute the potential by integration and finally take the negative gradient of that potential. At least that way you only have to integrate over 1/4th the cube volume.