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Electric flux through a cube

  1. Feb 3, 2009 #1
    1. The problem statement, all variables and given/known data
    [​IMG]
    [​IMG]


    2. Relevant equations

    E * A

    3. The attempt at a solution

    I need to find the electric flux through each face. I am a bit confused. I believe the flux through sides 3 and 1 are zero because those sides are parallel to the electric field. The whole flux through the cube should be 0, I am not sure how to find the electric flux through the other sides because the electric field has 2 components.
     
    Last edited: Feb 3, 2009
  2. jcsd
  3. Feb 3, 2009 #2

    LowlyPion

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    Homework Helper

  4. Feb 3, 2009 #3
    Would the angle be Tan(3.27/5.20) for the electric field?
     
  5. Feb 3, 2009 #4
    Ok, if I start to write out the area vectors I get.

    A1 = -L[tex]^{2}[/tex]j
    A2 = L[tex]^{2}[/tex]k
    A3 = L[tex]^{2}[/tex]j
    A4 = -L[tex]^{2}[/tex]k
    A5 = L[tex]^{2}[/tex]i
    A6 = -L[tex]^{2}[/tex]i

    Now I need to multiple these by the electric field, what is the best way to do it?
     
  6. Feb 4, 2009 #5
    anyone?
     
  7. Feb 4, 2009 #6
    Use the definition of flux. It is the dot product of the electric field and the outwardly directed area vector. Since the electric field is non-uniform, you will need to do a little calculus. You should notice that the x-coordinate is constant on two faces, and the z-coordinate is constant on two faces.
     
  8. Feb 4, 2009 #7
    I am still confused, I haven't really done a problem before with a non-uniform electric field. My textbook doesn't really have any good examples of this.
     
  9. Feb 4, 2009 #8
    Ahhh, I think I understand it now.
     
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