Gauss's law and nonuniform electric field

1. Jul 4, 2012

boredbluejay

1. The problem statement, all variables and given/known data
A Gaussian surface is in the shape of a cube with edge length 1.40m. The electric field is E=[-4i+(6+3y)j]N/C.

I got an answer, but the solution manual stated that we treat the electric field as E=3yj+E0, where E0=-4i+6j, which does not contribute to the flux. Why is this? Please help!

2. Jul 4, 2012

ehild

The flux of a constant vector is zero as the contribution from one side of the cube is positive and negative from the opposite side.

ehild

3. Jul 4, 2012

boredbluejay

Ah, okay. I didn't know that. But when calculating the flux of individual faces, does the constant vector contribute?

4. Jul 4, 2012

ehild

You know how to get a flux of a vector field B(r) at an surface area A? you integrate the normal component of the vector B : Φ=∫BndA.
The normal component refers to the outward normal. Although there is nonzero flux from the constant vector on on side, there is the same with opposite sign on the opposite side.

ehild