Electric flux charge through a cube

1. Oct 9, 2011

ZEli

1. The problem statement, all variables and given/known data
The electric flux through each face of a 4.50m x 4.50m x 4.50m cube is 130Nm^2/C .
How much charge is inside the cube?

2. Relevant equations
Electric Flux = E*A = Qin/ε

3. The attempt at a solution
I know electric flux is given, 130Nm^2/C and I'm trying to solve for Qin. So, I plug those numbers into the equation and get
130Nm^2/C = Qin/ε
I end up with Qin = 1.1505 * 10^-9 C, but when I type this into the answer, it doesn't work!
I don't see what I'm doing wrong!:(

2. Oct 9, 2011

WJSwanson

You're looking for the total flux through the faces of the cube. If the flux was only through one face, you'd be correct, but there are six faces of the cube that have the same flux through it.

3. Oct 9, 2011

ZEli

Ohhhhh, I see. So since there are 6 faces of the cube, I would multiply the given flux by 6 and then multiply by ε.

4. Oct 9, 2011

WJSwanson

Yep. Your net flux through an entire object is just the sum of the net fluxes through the component surfaces. This holds true for any Gaussian surface.

5. Oct 9, 2011

WJSwanson

As it turns out, in fact, that combination of properties (that the net flux through a surface is dependent only on the enclosed charge and that any Gaussian surface that encloses exactly the same amount of charge will exhibit exactly the same flux) will prove incredibly useful in your future studies! It has some very helpful applications in electrostatics.