# Flux through a cube

## Homework Statement

6 Gaussian cubes are shown below. The surfaces are located within space containing non-uniform electric
fields. The electric fields are produced by charge distributions located outside the cubes (no charges in
the cubes). Given for each case is the side length for the cube as well as the total electric flux through 5
out of 6 of the cube faces. Determine the electric field flux through the remaining sixth face. Rank the
electric flux through the remaining side from greatest positive to the greatest negative. Electric field flux
is a scalar quantity and not a vector. A negative value is possible and would be ranked lower than a
positive value (X = 200, Y = 0, Z = -200 would be ordered X=1, Y=2, Z=3)

2. Homework Equations

Q(enclosed)/E(knot)[/B]

## The Attempt at a Solution

I have a theory but I need a question answered first. Is the flux always equal to the charge enclosed divided by epsilon knot? Cause that would make this question easy to answer. However, if this is only true for a uniform electric field than that changes things. For instance, one of the cubes has a side length a = .5 m and the flux through 5 of the six sides is 40. Does that simply mean the flux through the sixth side must be -40? The reason I think that is because there is no charge enclosed, zero divided by anything equals zero.

Doc Al
Mentor
Is the flux always equal to the charge enclosed divided by epsilon knot?
Yes, that's Gauss' law. (Total flux through a closed surface.)

Cause that would make this question easy to answer.
Don't fight it.

For instance, one of the cubes has a side length a = .5 m and the flux through 5 of the six sides is 40. Does that simply mean the flux through the sixth side must be -40?
Yep.

The reason I think that is because there is no charge enclosed, zero divided by anything equals zero.
The total flux is zero, and the total flux equals the sum of the flux through each side of the cube.

The net flux if the charge is not enclosed should be zero. It shouldn't matter how the field is distributed, whether inside or outside the boundary. This fact allows for the simplified determination of flux through surfaces

Awesome okay thanks guys!