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Flux - general not hard question

  1. Sep 10, 2006 #1
    Question: A pyramid with horizontal square base, 6.00m on each side and a height of 4.00m is placed in a vertical electic field of 52.0 N/C. Calculate the total electric fluc through the pyramids four slanted surfaces.

    I know you can figure out the flux through the base = 36 m * 52.0 N/C because it's a unform electric field so flux in base = flux out pyramid sides, but if you were to do it without using that fact, actually using the geometry of the sides and such, what angle should you be using and cosine or sine? I can't seem to work it out. Thanks
  2. jcsd
  3. Sep 11, 2006 #2


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    The electric field and the angle it makes with the normal to the surface is still constant over the entire pyramid side, so the flux is [itex]\int\vec E\cdot d\vec A=EA\cos(\theta)[/itex]. Where theta is the angle between the normal and the vertical electric field.
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