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Why do Gaussian Surfaces work?

  1. Sep 12, 2013 #1
    I was doing a problem:

    An infinite conducting plane has a uniform surface charge density of 30 μC m‾².
    Find the electric field strength 7.0 mm from the plane.

    so we can use a gaussian surface (e.g a cylinder), and come to the conclusion that E = 30 μ / ε

    but that got me thinking, doesn't the electric field strength depend on distance?

    I understand that the electric field within the gaussian surface is 0, and that only the top of the gaussian surface we used (cylinder) would have flux passing through it.

    But from my working, wouldn't this mean if we were finding the electric field strength at any distance from the conducting plane e.g 100 metres away , E would still be the same?

    How does that even work?

    Thanks for your time!
     
  2. jcsd
  3. Sep 12, 2013 #2
    Yes, you're right, the electric field due to an INFINITE plane with constant charge distribution is a constant field that does not depend on the distance to the plane.
     
    Last edited: Sep 12, 2013
  4. Sep 12, 2013 #3

    jtbell

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    Staff: Mentor

    The key word here is "infinite", of course. There's no such thing as a truly infinite plane of charge. Even for a very very large plane, you can eventually get far enough away from it that the field it produces does fall off with distance. In fact, eventually you get far enough away that the plane "looks like" a point for all practical purposes, and from that distance outward the field falls off like 1/r2 just like for a point charge.

    Nevertheless, the concept of an infinite plane is a useful approximation when you are "close enough" to a "large enough" plane of charge. You just have to be careful not to use the approximation outside the region where it is "good enough" for your purposes.
     
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