1. PF Contest - Win "Conquering the Physics GRE" book! Click Here to Enter
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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


    User Avatar
    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
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted