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Gauss's Law for a Cube

  1. May 14, 2013 #1
    1. The problem statement, all variables and given/known data

    When is the net flux for a cube 0 and when is it not?


    2. Relevant equations

    ∫EdA= Q/8.85E-12

    3. The attempt at a solution

    if you have no test charge then the flux of a cube is zero. but if you have a test charge then the net flux is the charge divided by 8.85E-12. I just want to make sure I am thinking about this correctly.
     
  2. jcsd
  3. May 14, 2013 #2

    SammyS

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    It has nothing to do with a test charge.

    What does the Q represent in Gauss's Law? ##\displaystyle\ \int \vec{E}\cdot\vec{dA}=\frac{Q}{\varepsilon_0} \ ? ##
     
  4. May 15, 2013 #3
    Q is the total charge...
     
  5. May 15, 2013 #4

    SammyS

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    The total charge of what? ... Where is this charge located?
     
  6. May 15, 2013 #5
    So far you are on the right track... Just think about where the charge is in relation to your Gaussian Surface. It's very important that you understand this!!!
     
  7. May 15, 2013 #6
    Look closely at the form of the Equation:


    [itex]\oint[/itex][itex]\vec{E}[/itex][itex]\cdot[/itex][itex]\vec{dA}[/itex]


    What are the implications of the closed integral?
     
  8. Jun 10, 2013 #7
    Q is the total charge enclosed by the surface. its located in the center of the surface...Im not sure I understand.
     
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