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Symmetry Arguments-a finite cylindrical can

  1. Jan 13, 2009 #1
    Symmetry Arguments--a finite cylindrical can

    1. The problem statement, all variables and given/known data
    Consider a finite cylindrical can shape that has charge uniformly distributed on its surface. Symmetry does allow us to say some things about the electric field of this distribution
    A) at points along the can's central axis
    B) At points lying on the plane that cuts the can in half perpendicular to its central axis
    C) at the can's exact center

    What does symmetry tell us in these three cases?

    Hint: be sure to consider both the direction of the electric field and on what variables it might or might not depend


    2. Relevant equations



    3. The attempt at a solution

    Ok, first off, I'm not even sure what I'm supposed to be doing exactly, and is the shape just like a soda can, and the surface the outside not including the top/bottom, ie the body?

    a and b to me are confusing. I could see them being zero, but at the same time, they depend on the point you're using on the plane, so I believe you'd have to use an integral, but that'd be a double integral and that seems to complex for this problem.
    For A it seems as if it's completely symmetrical across the plane, and it seems to be true for B as well. Meaning there'd be no net charge on either. hmm

    c) I'd imagine this has to be zero. Perfect symmetry in all directions and uniform distribution throughout.

    So I guess I'm just looking for a start on where I should be heading for and what I should be trying to solve. Thanks, as always, in advance for the great help. Truly appreciated.
     
    Last edited: Jan 13, 2009
  2. jcsd
  3. Jan 13, 2009 #2

    Redbelly98

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    Re: Symmetry Arguments--a finite cylindrical can

    Hmm, I interpreted a "can shape" as including the top and bottom. However, I don't think that will affect these questions.

    (a) and (b) have more to do with what direction the field points toward. They don't want the actual value of the field, just its direction, so no integration is necessary.

    You're correct about (c).
     
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