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Electric field of infinite plane

  1. Jun 26, 2009 #1
    Hallo,

    Why does the field of a infinite plane does not depend on r? i know it's equal to

    two [tex]\pi\sigma[/tex] but why does his "infinite" makes it independent on r?

    thanks

    Omri
     
  2. jcsd
  3. Jun 26, 2009 #2

    tiny-tim

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    Hallo Omri! :smile:

    (have a pi: π and a sigma: σ :wink:)
    The field lines from a point charge spread out, so the field decreases with increasing distance.

    But the field lines from an infinite plane are straight, and don't spread out, so the field is the same at any distance. :smile:
     
  4. Jun 26, 2009 #3

    rcgldr

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    The math for field strength from a solid disc is explained here:

    http://hyperphysics.phy-astr.gsu.edu/Hbase/electric/elelin.html#c3

    as [tex]R^2 \ \rightarrow \infty [/tex]

    then [tex] \frac{z} {\sqrt{z^2 + R^2}} \ \rightarrow \ 0 [/tex]

    and you end up with [tex]E_z = k \ \sigma \ 2 \ \pi [/tex]

    An alterative approach is to consider the field from an infintely long line (= 1/z), then integrate an infinitely large plane composed of infintely long rectangles that approach infinitely long lines as their width approaches zero.
     
    Last edited: Jun 26, 2009
  5. Jun 26, 2009 #4
    thanks everyone,

    it was very helpful.
     
  6. Jun 26, 2009 #5

    rcgldr

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    You can do a web search for "guass law infinite plane" and see a few various approaches for this.
     
  7. Jun 26, 2009 #6

    Born2bwire

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    One thing to note is that the problem is invariant in two-dimensions. If the plane lies in the x-y plane, then you know right away that the field cannot rely on x or y since the source is infinite and invariant along those directions. So all you need to do is convince yourself that the resulting fields will also be invariant with regards to z, the remaining dimension. Jeff's post is good for showing this part.
     
  8. Jun 26, 2009 #7
    The 'math' answer: The derivation of the field strength, using Gauss's Law, shows that the Electric Field is independent of z. Not a very satisfying answer, but a truthful one.

    The Intuitive Answer: Imagine you are a point charge hovering in a balloon over an infinite plane of charge. The plane extends infinitely, as far as the eye can see, in every direction below you. The view will be exactly the same, no matter how close or far you are from the plane. In fact, for a blank, empty, infinite plane with no points of reference on it, you will have no way of telling how far from the plane you are from where you sit in your balloon.
     
  9. Jun 28, 2009 #8
    about the ituitive answer, i can claim the same argument over aninfinite wire but is field does depend on location.
     
  10. Jun 28, 2009 #9

    tiny-tim

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    he he :biggrin:

    (the field lines from a line charge also spread out)

    omri3012 1

    JazzFusion 0 :wink:
     
  11. Jun 28, 2009 #10
    ....so if you move laterally away from the line charge, the 'view' changes.
     
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