1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    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!

Homework Help: Zero Electric Flux Through a Closed Surface

  1. Sep 24, 2014 #1
    • Warning! Posting template must be used for homework questions.
    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?
  2. jcsd
  3. Sep 24, 2014 #2
    I guess in that case you would get a non zero net electric flux, therefore the net electric charge enclosed in your closed surface wouldn't be zero. That is, if the strength of E is different in both sides, then maybe the presumption of no enclosed charge breaks down.
  4. Sep 24, 2014 #3
    A field can only be terminated ON a charge, so ANY field line that enters the cube must exit the cube if there is no charge inside for the field line to terminate on.
  5. Sep 24, 2014 #4
    Hello and thank you for the responses.

    I've seen this as the standard explanation and I am not quite satisfied with it, due to the fact that it boils the problem down to single field lines. By definition of flux, the contribution of a single field line is zero because it enters the surface at a point, which has area zero.

    This is what Guass's law tells us. I'm under the assumption that my field cannot be a real physical electric field, but don't really have a way to verify it at the moment. My main problem is that in every case of the cube example that I have seen, the instructor assumes that the field strength is equal at both sides, which is a very specific case. My "counterexample" is a very simple situation, and I'm wondering why exactly it is breaking down.

    I wrote up a better explanation of the situation: We have a cube and an electric field that runs perpendicular to four of its faces and parallel to the other two faces, which I'll call face 1 and face 2. The electric field strength at face 1 is the constant value of [itex] E_1 [/itex] and the strength at face 2 is the constant value of [itex] E_2 [/itex]. Let [itex] A [/itex] denote the area of each face.

    Thus, the flux through face one is [itex] \Phi_1 = -E_1 A [/itex] (assuming outward facing orientation for the faces) and the flux through face two is [itex] \Phi_2 = E_2 A [/itex]. The total electric flux is

    [itex] \Phi = \Phi_1 + \Phi_2 = -E_1 A + E_2 A = ( E_2 - E_1 )A [/itex].

    If [itex] E_1 \not= E_2 [/itex], then we have nonzero flux.
  6. Sep 24, 2014 #5
    If the field is not uniform then the field lines are not parallel and you will have nonzero flux through more than two faces.
    Parallel field lines means uniform field.
  7. Sep 24, 2014 #6


    User Avatar
    Homework Helper

    Gauss' law should not be an obvious thing. It is an experimentally observed law. So don't worry if an intuitive explanation is not fully convincing to you. I prefer to think of it like the charges are a source (or sink) for the electric field. So, if there are no enclosed charges, we get no net flux. I fully realise that this is not a convincing logical argument for why Gauss' law should be true. It is not meant to be. It is just meant to serve as a way to intuitively think of the problem.
  8. Sep 24, 2014 #7
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted