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Flux through a cube

  1. Sep 13, 2008 #1
    1. The problem statement, all variables and given/known data
    A cube of side L = 2m is centered at the origin, with the coordinate axes perpendicular to its faces. Find the flux of the electric field E = (15N/C)i + (27N/C)j + (39N/C)k through each face of the cube


    2. Relevant equations

    phi total = (E * n) delta A


    3. The attempt at a solution

    Ok, so I know that
    phi total = phi 1 + phi 2 + ... + phi 6

    what I am confused about is how do I know where the flux is parallel to the faces of cube, where is it zero???

    Thanks
     
  2. jcsd
  3. Sep 13, 2008 #2

    gabbagabbahey

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    (E*n) is the dot product between [tex]\vec{E}[/tex] and [tex]\hat{n}[/tex]. You are given [tex]\vec{E}[/tex], so what is the defintion of [tex]\hat{n}[/tex] for any surface? What is [tex]\hat{n}[/tex] for each surface of the cube? Once you find these, you just compute the dot product and then integrate over the area of each surface to get your phis.
     
  4. Sep 13, 2008 #3
    Yeah I get that
    [((15N/C)i + (27N/C)j + (39N/C)k] . i = (15N/C) a^2
    "" "" "" "" . -i = -(15N/C) a^2
    ...
    ...

    I am just wondering how do I know where one of this calculations will be zero, where will the flux be parallel to the sides of the cube, how do I recognize that???
     
  5. Sep 13, 2008 #4

    gabbagabbahey

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    Flux is a scalar quantity...it will be neither parallel nor perpendicular to any given side. Why do you have an a^2 there? And for which face is [tex]\hat{n}=+\hat{i}[/tex]?Is it the same for all faces? Can you show your whole solution?
     
  6. Sep 13, 2008 #5
    1, 0, 0

    I see, so for this problem

    [(15N/C)a^2 + (-15N/C)a^2 + (27N/C)a^2 + (-27N/C)a^2 + (39N/C)a^2 + (-39N/C)a^2] = 0

    Thanks
     
  7. Sep 13, 2008 #6

    gabbagabbahey

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    If by a you mean L, then yes but you have calculated the TOTAL flux. The question asks for the flux through each side. You should draw a diagram, where you label each side and give the flux through each side.
     
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