Calculate flux through face of a cube

[mindarson]

Homework Statement

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.

Homework Equations

Gauss's Law; flux = ∫E\bulletn dA = q_enc/ε

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?

 

[Tsunoyukami]

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

 

[mindarson]

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