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Gauss's Law

  1. Aug 18, 2008 #1
    y is this statement wrong?

    The electric field on a gaussian surface is generally not influenced by charge that is not enclosed by the surface.

    ps: isn't the electric field proportional only to the ENCLOSED CHARGE, but not the outside charge?? :confused:

    This is getting me so fRustrated
     
  2. jcsd
  3. Aug 18, 2008 #2

    Defennder

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    Gauss law is a law of electric flux, not E-field. If the situation is symmetric enough, you may extract the value of the E-field from it. Consider a Gaussian surface enclosing an electric dipole. Are you able to determine the E-field at any point on the surface? Now imagine the distance between the positive and negative charges increasing along with the size of the Gaussian surface. What is the E-field at any point on the surface?
     
  4. Aug 18, 2008 #3
    hmm, first, thx Defender! I no that u need a constant E throughout the Gaussian surface in order to extract E from the integral. but how do you explain why the E- field on Gaussian surface is affected by a charge outside of the surface?? cuz the ENCLOSED Q is the key right? or do u mean the size of the Gaussian surface is changing when a charge is placed outside of it?

    THXXX :)
     
  5. Aug 18, 2008 #4
    ohh, I think I got it now! So the distribution of E field is changing on the Gaussian surface, but the total E dA is the same. yAy
     
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