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*non uniformity*of the electric field in the following question is throwing me off. If the electric field were uniform I'd have no problem.

I assume I would use the following equation to solve for each of the surfaces:A cube has lengthL= 0.300 m. It is placed with one corner at the origin. The electric field is not uniform, but is given by E = (-5.00 N/C*m)xi + (3.00 N/C*m)zk.

a) Find the electric flux through each of the 6 cube faces.

b) Find the total charge inside the cube.

[tex]\Phi = \int \vec{E} \cdot d \vec{A}[/tex]

I'm having a difficult time picturing what's going on. How do I know what surfaces the E field is passing through given just a vector? As far as I can tell there are four surfaces that are not parallel to the E field direction so four surfaces should have a flux value, right? The answer to the question a says there's only two surfaces with flux values not equal to zero. The answer to b says there's a charge in the cube. If there's a charge in the cube why doesn't its E field pass through all surfaces of the box it's enclosed in?