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Flux through a pyramid

  • Thread starter origamipro
  • Start date
  • #1
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Homework Statement



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


Homework Equations



EA cos theta

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?
 

Answers and Replies

  • #2
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Area of face of pyramid is 1/2 base x height = (.5)(6)(4) = 12
The height of the faces is not the height of the pyramid.
 
  • #3
14
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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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