Flux of a Cube with a Corner Charge

In summary, the net flux through all faces of the cube with a point charge at one corner is \phi = \frac{q}{8\epsilon_0}. This is based on the fact that the total flux from the point charge is \phi = \frac{q}{\epsilon_0}, but due to symmetry and the placement of 7 other cubes around the original cube, each individual cube will only have flux through 3 of its faces. Therefore, the total flux is divided among the 8 cubes, resulting in a net flux of \phi = \frac{q}{8\epsilon_0}.
  • #1
breez
65
0

Homework Statement



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

Homework Equations





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]
 
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  • #2
breez said:

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!
 
  • #3




I would like to clarify that there is a difference between electric flux and net electric flux. Electric flux is a measure of the electric field passing through a given area, while net electric flux takes into account the direction of the electric field and the orientation of the surface. In this case, the net electric flux through all faces of the cube would depend on the direction of the electric field at each face.

If the electric field is perpendicular to a face, the net flux through that face would be zero. However, if the electric field is not perpendicular to a face, there would be a non-zero net flux through that face. Therefore, the net electric flux through all faces of the cube would depend on the orientation of the electric field at each face and cannot be determined solely based on the placement of the point charge at a corner of the cube.

Additionally, the equation \phi = \frac{q}{\epsilon_0} is the electric flux through a closed surface enclosing the point charge, not through individual faces of the cube. The equation \phi = \frac{q}{8\epsilon_0} assumes that the electric field is perpendicular to all faces of the cube, which may not be the case.

In summary, the net electric flux through all faces of the cube cannot be determined without knowing the orientation of the electric field at each face. It is important to consider the direction of the electric field and the orientation of the surface when calculating net electric flux.
 

1. What is the definition of flux?

Flux is a measure of the flow of a quantity through a surface. In physics, it is often used to describe the flow of an electric or magnetic field through a surface.

2. How is the flux of a cube with a corner charge calculated?

The flux of a cube with a corner charge can be calculated by using Gauss's Law, which states that the electric flux through a closed surface is equal to the enclosed charge divided by the permittivity of free space. In this case, the enclosed charge would be the charge at the corner of the cube.

3. What factors affect the flux of a cube with a corner charge?

The flux of a cube with a corner charge is affected by the magnitude of the charge at the corner, the size of the cube, and the distance between the cube and the charge.

4. How does the orientation of the cube affect the flux?

The orientation of the cube can affect the flux because the surface area of the cube that is perpendicular to the electric field lines will have a higher flux compared to the surface area that is parallel to the field lines.

5. What is the significance of the flux of a cube with a corner charge?

The flux of a cube with a corner charge can help us understand the behavior and strength of electric fields. It also has practical applications in fields such as electrical engineering and physics.

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