How Does Gauss' Law Apply to Electric Flux Through a Cube's Surface?

[Go Boom Now]

Homework Statement

A 10 nC point charge is at the center of a 2 m x 2 m x 2 m cube. What is the electric flux through the top surface of the cube?

Homework Equations

Electric Flux = E x A x cosθ
Electric Flux = ∫E x dA = Q(inside)/ε
E (point charge) = kq/r^2 where k = 1/(4pi ε)

The Attempt at a Solution

Aside from finding the area, volume, and anything else that might be considered obvious, I've no idea where to head from here. I'm not sure if the cube is completely closed or if the top surface isn't uniform. I'm hoping someone can help walk me through this problem.

 

[Dick]

It's an imaginary cube. So it's closed and doesn't have any properties that could be nonuniform. You can compute the total flux through the cube, right?? Any reason to believe that more passes through one face than any other?

 

[Go Boom Now]

I'm... not sure? The point charge is in the center of the cube itself, so... no?

 

[Dick]

I'm... not sure? The point charge is in the center of the cube itself, so... no?

Ok, so 1/6 of the total flux must pass through each face. Total flux is easy to compute.

 

[Go Boom Now]

I think I got it.

Total Flux of Cube = Q (charge) / ε = (10x10^-9) / (8.85x10^-12)

1129.94 / 6 = 188.32 N m^2 / C.

This is the answer I was given. Thanks for the help!