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Gauss' law cubes, how to prove

  1. Feb 9, 2012 #1
    So I was doing some practice problems and one of them asked for the flux through one side of a cube that has a point charge(Q) at its center, it seems intuitive that it would be 1/6 of the charge but how would i show this?

    if I do

    ∫E dA → E∫dA
    the surface area of a cube is 6L2 so i would get

    E*6L2=Q/ε0

    then?

    thanks.
     
  2. jcsd
  3. Feb 9, 2012 #2

    jtbell

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    Staff: Mentor

    You can't do that because (a) different points on the surface of the cube are at different distances from the center, so the magnitude of E isn't constant; (b) the direction of E is different at different points on the surface so you have to take that into account:

    [tex]\vec E \cdot d \vec A = E dA \cos \theta[/tex]
     
  4. Feb 9, 2012 #3
    Ok so one of the problems says its Q/6e0

    how did they arrive at this?
     
  5. Feb 9, 2012 #4

    Born2bwire

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    By symmetry. The electric field from a point charge is spherically symmetric and being placed at the center of the box we deduce that each side of the cube must have the same amount of flux through it. Now we know from Gauss' Law that the total flux over the surface of the cube will be proportional to the total charge enclosed, so the flux through one side of the cube is 1/6th of this.
     
  6. Sep 14, 2013 #5
    What would the cosθ be in terms of x,y and/or z? I am trying to do a proof of this that shows the result of Q/ε for the whole cube. On my first run through I did not account for the spherical distribution of the e-field and I was on the way to get a value of 0. I figured cosθ need to be included but it seems that this will only account for one dimension on the face of the cube. For example if I said that

    cosθ=[itex]\frac{x}{\sqrt{x^{2}+z^{2}}}[/itex]

    How would this correspond to the whole face of the cube or is that taken care of by dA. If so dose it even matter if I define cosθ with x,z or x,y?

    Edit: Just to be clear I'm using the example of the face of the cube on the positive x-axis and that a charge q is located at the origin at the center of the cube. The cube is the Gaussian surface.
     
    Last edited: Sep 14, 2013
  7. Sep 14, 2013 #6

    Born2bwire

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    The \theta he is using is the angle between the electric field and the plane of the box's surface that it is intersecting. Its dependence on the Cartesian coordinates will differ for each of the 6 sides of the box.

    But this is immaterial. This is a basic undergraduate problem that even I came across when I did my undergrad. You are supposed to solve it via the symmetry argument that I put forth earlier. Doing it explicitly mathematically would be very complicated. But perhaps it may be easier to figure out how to parameterize the integration so that you can do it in spherical coordinates.
     
  8. Sep 14, 2013 #7
    Here is the question statement:
    Construct a Gaussian surface S that consists of a cube of side 2, centered at the origin.
    Calculate [itex]\oint E\bullet d\vec{a}[/itex] for the point charge, using this surface, thus verifying (or not?) Gauss’ Law.

    So I don't think that the symmetry argument will be sufficient. Do you agree?
     
    Last edited: Sep 14, 2013
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