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Homework Help: Flux through a pyramid

  1. Oct 24, 2013 #1
    1. The problem statement, all variables and given/known data

    base of pyramid is a 6m x 6m square height of pyramid is 4meters. a 52 N/C E field orthogonal to the base of the pyramid

    2. Relevant equations

    EA cos theta

    3. The attempt at a solution

    if flux through base = EAcos theta = 52 (36m^2) cos(0) = 1872 N m^2 / C

    Area of face of pyramid is 1/2 base x height = (.5)(6)(4) = 12

    angle of normal to surface and E field is cos-1(3/5)

    Why is flux through each surface not equal to the following:
    so flux through one surface is = 52 N/C x 12m x cos(53.1) = 374.6
    374.6 x 4 = 1498 N/ C

    how do i derive flux through each face?
  2. jcsd
  3. Oct 24, 2013 #2


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    2017 Award

    Staff: Mentor

    The height of the faces is not the height of the pyramid.
  4. Oct 24, 2013 #3
    right on thanks.

    height is sqrt(3^2+4^2)

    Area of face of pyramid is 1/2 base x height = (.5)(6)(5) = 15
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