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Question on zero electric flux through a closed surface.

  1. Sep 24, 2014 #1
    I'm relearning basic electricity concepts and I can't find an answer to a situation I've thought up.

    Imagine a cube with no enclosed charge and an electric field through it parallel to two of its faces. Guass's law says that the flux should be zero because there is no enclosed charge.

    Every example of this that I have seen goes something like this: The four perpendicular faces have zero flux due to the sides being perpendicular to the field, and the flux from the other two sides cancel each other out. I.e if [itex] E [/itex] is the strength of the electric field, [itex] A [/itex] is the area of the sides of the cube, and [itex] \Phi [/itex] is the total flux, then

    [itex] \Phi = EA + (-EA) = 0 [/itex].

    But what if the strength of the electric field is not the same at both sides? Then there won't be any cancelling out and there will be nonzero flux, which is contrary to Guass's law?
     
    Last edited: Sep 24, 2014
  2. jcsd
  3. Sep 24, 2014 #2
    Sorry, I see that this thread should have been posted in the coursework section. I shall post this there and this thread can be deleted.
     
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