Gaussian surface - cube


by quantum13
Tags: cube, gaussian, surface
quantum13
quantum13 is offline
#1
Oct6-09, 07:18 PM
P: 66
1. The problem statement, all variables and given/known data
A particle charge of q is placed at one corner of a Gaussian cube. What multiple of [latex] q/\epsilon_0 [/latex] gives the flux though each cube face not making up that corner?

The solution is amazing - stack up eight cubes around the corner and find the flux through each individual cube and individual face of the cube
[tex]
(1/8 * 1/3) q/\epsilon_0 = 1/24 * q/\epsilon_0
[/tex]

However, I don't see how this makes sense. There is no charge included in the three opposite faces of the cubes meaning there should be no flux there, even though there definitely is flux from the stacked cubes method. How can this contradiction be explained?
2. Relevant equations
Gauss's Law


3. The attempt at a solution

Uh.. ??
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Delphi51
Delphi51 is offline
#2
Oct6-09, 07:54 PM
HW Helper
P: 3,394
Whew, confusing to say "q is placed at one corner of a Gaussian cube".
Leave out the word Gaussian for this cube!
The Gaussian surface you must consider is the outside surface of the set of 8 cubes surrounding q. From symmetry, you get the same flux through all the outside faces of those 8 cubes as through the three remote faces of that first cube. Looks like the Gaussian surface has 24 faces so each one gets 1/24 of the flux. And you are asked for the flux through 3 of those.


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