What is the flux through (a) each cube face

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    Cube Flux
In summary, the conversation discusses finding the flux through different faces of a cube when a point particle with charge q is placed at a corner of the cube. Gauss's law and symmetry arguments are used to determine the flux, with the final answer being \Phi=q/(24\epsilon_0).
  • #1
danago
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A point particle with charge q is placed at a corner of a cube of edge a . What is the flux through (a) each cube face forming that corner and (b) each of the other cube faces?

I tried modelling the problem by positioning the charge such that it is at point (0,0,0) in the x-y-z space. The cube then extends out in the positive x, y and z directions.

I then defined the unit vectors i, j and k to extend in the x, y and z directions respectively. Now I am not sure if my next step is valid, but it seemed to be a decent thing to do. I said that the electric field due to the charge is given by:

[tex]
\overrightarrow E = \frac{{kq}}{{x^2 }}\widehat{\underline i } + \frac{{kq}}{{y^2 }}\widehat{\underline j } + \frac{{kq}}{{z^2 }}\widehat{\underline k }
[/tex]

I then evaluated the integral of the dot product of vectors E and dA over one of the cubes faces, and came up with the expression for flux as kq, however this wasnt the correct result.

Anyone able to help with the question?

Thanks,
Dan.
 
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  • #2
That expression for the field strength is not correct. The field points radially outward from the point charge; its magnitude is kq/r^2.

In any case, you don't need that at all. Instead, put on your thinking cap. :wink:

Picture the field emanating uniformly in all directions. Using symmetry arguments and Gauss's law (and the meaning of flux) will tell you the flux without the need for any calculation.
 
  • #3
Sorry forgot to mention, i got part (a) correct as 0.

It was part (b) i was having trouble with. Is (b) still solvable using symmetry arguments?
 
  • #4
danago said:
Sorry forgot to mention, i got part (a) correct as 0.
Good.
It was part (b) i was having trouble with. Is (b) still solvable using symmetry arguments?
Sure!

Hint: What's the total flux through all sides of the cube? Which side has the most flux? :wink:
 
  • #5
Doc Al said:
Good.

Sure!

Hint: What's the total flux through all sides of the cube? Which side has the most flux? :wink:

Will the total flux be [tex]\Phi=q / \epsilon_0[/tex]? And will the flux through each of the other faces be the same, therefore making the flux through each face just be [tex]\Phi=q / (3\epsilon_0)[/tex]?
 
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  • #6
You got it!
 
  • #7
Thanks very much :smile: That made it much easier than whatever i was trying to do originally haha.
 
  • #8
Hmm i tried inputting that answer, but it isn't showing up as correct. Apparently the answer is supposed to be [tex]\Phi=q / (24\epsilon_0)[/tex]. Any idea what's going on there?
 
  • #9
Yes. Big whoops! (:redface:) [itex]\Phi=q / \epsilon_0[/itex] is the total flux in all directions. How much goes through that cube?
 
  • #10
Doc Al said:
Yes. Big whoops! (:redface:) [itex]\Phi=q / \epsilon_0[/itex] is the total flux in all directions. How much goes through that cube?

Oh i think i see what's going on here. I can picture the charge as being in the center of 8 smaller cubes, so the flux through just the cube will be 1/8th of the total flux; and then dividing by 3 gives the correct answer :smile:

Thanks again Doc Al!
 
  • #11
Exactly. (Sorry about not catching that the first time around. Asleep at the wheel.)
 

What is the flux through each cube face?

The flux through each cube face refers to the amount of a vector field that passes through each face of a cube. It is a measure of the flow of that vector field through the surface of the cube.

How do you calculate the flux through each cube face?

The flux through each cube face can be calculated using the formula Flux = ∫∫S F · dS, where F is the vector field and dS is the differential area element of the surface of the cube.

Is the flux through each cube face always the same?

No, the flux through each cube face can vary depending on the vector field and the orientation of the cube. It can also vary if the vector field is not constant throughout the surface of the cube.

What is the unit of measurement for flux through each cube face?

The unit of measurement for flux through each cube face is determined by the unit of the vector field, multiplied by the unit of the surface area of the cube. For example, if the vector field is measured in Newtons and the surface area of the cube is measured in square meters, the unit of flux would be Newton-meters.

Why is it important to calculate the flux through each cube face?

Calculating the flux through each cube face is important in many scientific and engineering applications, such as fluid dynamics, electromagnetism, and heat transfer. It allows us to understand and analyze the flow of a vector field through a given surface, which can provide valuable insights and help make predictions about the behavior of the system.

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