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Derivation of gauss's law for electrical fields please help

  1. Feb 3, 2007 #1
    This is a very stupid question. extremely stupid. In fact I'm extremely embarassed.

    I was reading a text on electromagnetism, and it said that since the flux due to a charge does not depend on the radius of the sphere then the formula, q/permitivitty applies to all closed surfaces. This is where i got confused. why does that make it apply to all closed surfaces?

    please don't laugh at me.

    Please I need an answer!
    Last edited: Feb 3, 2007
  2. jcsd
  3. Feb 4, 2007 #2


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    Try calculating the electric flux of spheres with various radii using [itex]\phi = \int\vec{E}\cdot d\vec{A}[/itex] and see where that gets you. Intuitively, one can consider the definition of electric flux (number of electric field lines passing through a unit area) and the definition of the electric field and how it relates to the density of electric field lines...
  4. Feb 4, 2007 #3


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    After calculating for spheres, try (say) a "northern" hemisphere at radius r1 and a "southern" hemisphere at radius r2 with an http://mathworld.wolfram.com/Annulus.html" [Broken] joining their "equators". Note the flux through this annulus.

    Then, approximate ANY closed surface by portions of spheres and portions of annuli. ...take limits.
    Last edited by a moderator: May 2, 2017
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