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Flux of a Cube with a Corner Charge

  • Thread starter breez
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1. The problem statement, all variables and given/known data

A point charge is placed at a corner of a cube. What is the net flux through all faces of the cube?

2. Relevant equations



3. The attempt at a solution

[tex]\phi = \frac{q}{\epsilon_0}[/tex]

or perhaps

[tex]\phi = \frac{q}{8\epsilon_0}[/tex]

The latter solution is based on the fact that for 3 faces on the cube, no flux will be present as the e field is parallel to the faces. Place 7 other cubes about this cube to force a cube with the charge in the middle, and each cube individually will only have flux through 3 of the cubes. A charge enclosed in a gaussian surface must result in a flux of [tex]\phi = \frac{q}{\epsilon_0}[/tex] for the surface. There are 8 cubes, and our original situation is 1/8 of the entire cube, and hence: [tex]\phi = \frac{q}{8\epsilon_0}[/tex]
 

Doc Al

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3. The attempt at a solution

[tex]\phi = \frac{q}{\epsilon_0}[/tex]
That's the total flux from the point charge.

or perhaps

[tex]\phi = \frac{q}{8\epsilon_0}[/tex]
That's the one you want.

The latter solution is based on the fact that for 3 faces on the cube, no flux will be present as the e field is parallel to the faces. Place 7 other cubes about this cube to force a cube with the charge in the middle, and each cube individually will only have flux through 3 of the cubes. A charge enclosed in a gaussian surface must result in a flux of [tex]\phi = \frac{q}{\epsilon_0}[/tex] for the surface. There are 8 cubes, and our original situation is 1/8 of the entire cube, and hence: [tex]\phi = \frac{q}{8\epsilon_0}[/tex]
Sounds good. I'd view it as a 2x2x2 stack of 8 cubes with the charge at the center corner. By symmetry, the flux is equally distributed among the 8 cubes. Done!
 

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