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Homework Help: Top view of a Gaussian surface in a uniform electric field

  1. Mar 1, 2013 #1
    The field makes an angle θ with side 1 and the area of each face is A. In symbolic form, find the electric flux through (a) face 1, (b) face 2, (c) face 3, (d) face 4 and (e) top and bottom.

    My professor got:
    e= 0
    I understant why e=0 but for the other sides, for some reason I am not understanding why she got for example a=EAcosθ and not sinθ. Which part of the triangle am I trying to find out? Would it be the adjacent for each or the opposite side for each?

    I hope my question makes sense.

    1. The problem statement, all variables and given/known data

    2. Relevant equations

    3. The attempt at a solution
  2. jcsd
  3. Mar 1, 2013 #2


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    For a uniform electric field, and a flat surface, the electric flux [itex] \Phi_E [/itex] through a surface with area A is

    [tex] \Phi_E = \vec E \cdot \vec A [/tex]

    Notice I'm using vector notation. The direction of [itex] \vec A [/itex] is normal to the surface. And there's a loose convention that for a closed surface, the surface vector points out of the surface. (That's just a convention though, although that conversion is used here.)

    Another thing to notice is that we're dealing with the "dot" product. Another way to represent the dot product is

    [tex] \vec E \cdot \vec A = EA \cos \theta [/tex]

    The dot product measures (in part) how parallel two vectors are. If they are perfectly parallel, the dot product is simply EA. If they are perpendicular the dot product is 0. If they are in perfectly opposite directions, the dot product is -EA.

    If it helps, imagine the situation where [itex] \theta [/itex] is 0, and other situation where it is 90o. If [itex] \theta [/itex] is 0, a maximum amount of flux would pass through side 1. How many flux lines would pass though side 1 when [itex] \theta [/itex] is 90o? So would cosine or sine be used to represent that?
    Last edited: Mar 1, 2013
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