# Electric flux through a cube

1. Feb 3, 2009

### Oblivion77

1. The problem statement, all variables and given/known data

2. Relevant equations

E * A

3. The attempt at a solution

I need to find the electric flux through each face. I am a bit confused. I believe the flux through sides 3 and 1 are zero because those sides are parallel to the electric field. The whole flux through the cube should be 0, I am not sure how to find the electric flux through the other sides because the electric field has 2 components.

Last edited: Feb 3, 2009
2. Feb 3, 2009

3. Feb 3, 2009

### Oblivion77

Would the angle be Tan(3.27/5.20) for the electric field?

4. Feb 3, 2009

### Oblivion77

Ok, if I start to write out the area vectors I get.

A1 = -L$$^{2}$$j
A2 = L$$^{2}$$k
A3 = L$$^{2}$$j
A4 = -L$$^{2}$$k
A5 = L$$^{2}$$i
A6 = -L$$^{2}$$i

Now I need to multiple these by the electric field, what is the best way to do it?

5. Feb 4, 2009

anyone?

6. Feb 4, 2009

### Brian_C

Use the definition of flux. It is the dot product of the electric field and the outwardly directed area vector. Since the electric field is non-uniform, you will need to do a little calculus. You should notice that the x-coordinate is constant on two faces, and the z-coordinate is constant on two faces.

7. Feb 4, 2009

### Oblivion77

I am still confused, I haven't really done a problem before with a non-uniform electric field. My textbook doesn't really have any good examples of this.

8. Feb 4, 2009

### Oblivion77

Ahhh, I think I understand it now.