# Electric flux through a cube

1. Aug 27, 2006

### erik-the-red

Question:

The cube in the figure (attachment) has sides of length $$L=10.0 {\rm cm}$$. The electric field is uniform, has a magnitude $$E=4.00 \times 10^{3} {\rm N}/{\rm C}$$, and is parallel to the xy-plane at an angle of $$36.9^\circ$$c measured from the $$+ x - {\rm axis}$$ toward the $$+ y - {\rm axis}$$.

The question asks for the electric fluxes through each of the faces and the sum.

I don't really understand why the electric flux through the top and bottom faces of the cube is zero. Is it because the angle between the face and the electric field is $$90^\circ$$?

The sum is zero because the electric field goes in through two of the faces and then leaves through two others, right?

My answer for the electric flux through face one was $$E \cdot L^2$$ or (4.00*10^(3))*(.10^2), but that was wrong. I thought the angle was 180?

Last edited: Aug 27, 2006
2. Aug 27, 2006

### Saketh

Yes. The flux is $$\oint \vec{E} \cdot d\vec{A}$$. The dot product can be rewritten with a cosine. Since the angle between the field and the surface is 90 degrees, the flux resolves to zero.

The electric field is piercing the surface at an angle. Draw a diagram. Since you know that $$\Phi = \oint \vec{E} \cdot d\vec{A}$$, evaluate that integral (which, in this simplified case, can be written as $$\Phi = EA\cos{\theta}$$). The angle is not 180.