# Calculate flux through face of a cube

1. Sep 12, 2013

### mindarson

1. The problem statement, all variables and given/known data

A point charge q is located at the corner of a cube with edges of length
L. Find the electric flux through the shaded face at y = L.

2. Relevant equations

Gauss's Law; flux = ∫E$\bullet$n dA = qenc

3. The attempt at a solution

Here's what I'm thinking. In the problem statement, the charge q is located at a corner of the cube. I can't use Gauss's Law, due to lack of symmetry. However, I can draw a bigger cube instead, whose center is at the location of the charge. Then I have symmetry to use Gauss's law and say that the flux through the whole cube is q/ε. The face whose flux the problem wants me to calculate is now just 1/24 of the surface area of the whole (bigger) cube. Therefore I can say that the flux through the original face is just q/(24ε).

Is this correct, or am I just trying to be too clever by half, as they say?

Last edited: Sep 12, 2013
2. Sep 12, 2013

### Tsunoyukami

This is correct and in fact is the manner in which the problem is solved in the solution manual. Good job!

3. Sep 12, 2013

### mindarson

Nice! Seemed a bit too good to be true. :)