Charge enclosed in uniform electric field

[yeahhyeahyeah]

1b]1. Homework Statement

A cube oriented so that its corner lies on the origin and it extends positive 1m in the x, y, and z direction. An electric field of 3.00 i + 2.5 j passes through this space. What is the charge enclosed by the cube?

The Attempt at a Solution

Gauss' law is
integr(E . dA) = Q/epsilon

since the surface is of a cube, you can add the dot product of E and dA for every face except for the one lying in the x-y plane because there is no 'k' component of the E field.
The dot products of the faces cancel out though
So the the total charge enclosed is 0
is that right?

 

[LowlyPion]

If the electric field is uniform across the space the cube is in, then you are right that there is no charge inside, because for there to be a charge inside there would have to be a net non-zero flux through all of the faces of the cube.

 

[thomate1]

Yes, that is correct. Since the cube is oriented in such a way that one of its faces is parallel to the x-y plane, the electric field passing through that face (and the corresponding dot product) will be zero. Therefore, the total charge enclosed by the cube will also be zero. This is because Gauss' law states that the integral of the electric field over a closed surface is equal to the total charge enclosed by that surface divided by the permittivity of the medium. In this case, since the integral is zero, the total charge enclosed must also be zero.