Flux through one side of a cube

[JamesMay]

Homework Statement

A charge q sits at the back corner of a cube. What is the
flux of E through the opposite (front)side?

Homework Equations

Flux=q/ε_0

The Attempt at a Solution

As the flux through the whole cube must be q/ε_0, I thought that for one side it would just be q/6ε_0, but the answer is q/24ε_0. I don't know how the position of the enclosed charge affects the flux through the side of the cube.

 

[HallsofIvy]

Why would you think that? The charge is NOT symmetrically situated with respect to the sides of the cube. You should expect more flux through those sides closer to the charge. The problem I have is that if the charge is situated at one corner then there are 3 different faces you could reasonably call "opposite". Strictly speaking what you need to do is integrate the field over the face of the cube:
\int\int \vec{e}\cdot d\vec{S}
where \vec{e} is the field strength and d\vec{S} is the (vector) differential of surface area.

 

[Dick]

If the charge is located AT the corner then only 1/8 of the total flux of the charge enters the cube. There is no flux through the three faces which touch the charge since the field is tangent to those faces. So the 1/8 of the total flux exits through the three opposite faces. Their orientation is symmetric with respect to the charge, so the same amount of flux passes through each.

 

[JamesMay]

ok, so would I put the whole thing into cartesian coordinates? I don't know how to integrate that when dS is not spherical.

 

[JamesMay]

Great thanks, that makes sense.

 

[FedEx]

If the charge is located AT the corner then only 1/8 of the total flux of the charge enters the cube. There is no flux through the three faces which touch the charge since the field is tangent to those faces. So the 1/8 of the total flux exits through the three opposite faces. Their orientation is symmetric with respect to the charge, so the same amount of flux passes through each.

The reason that 1/8th of the flux is considered is cause while considering such problems we enclose the charge symmetrically. In this case if the side length is A. We add seven more cubes and make a cube of side 2A. Now flux passing through this is q/epsilon. So now flux passing through one cube is q/8epsilon.