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Divergence of electric field

  1. Jun 3, 2012 #1
    The divergence of electric field at a point is proportional to the charge density at the point. Divergence is the rate of change with distance, the rate of change of electric field due to a distant charge is not zero, so how can it be said that the divergence at a point depends only on the charge density there?
     
  2. jcsd
  3. Jun 3, 2012 #2
    No, the divergence is a particular combination of partial derivatives. You should compute the divergence of the electric field of a point charge yourself. You'll find that it is zero except at the position of the charge, where it is singular.
     
  4. Jun 3, 2012 #3
    Divergence is not about the rate of change with distance.
     
  5. Jun 3, 2012 #4

    rbj

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    i dunno if i agree with you guys (Duck and Leland). divergence is about rate of change with distance. it is about how rapidly the field strength changes with distance outward from the point where the charge density is.

    divergence is essentially a microscopic version of Gauss's Law. and Gauss's Law works only for inverse-square fields and it gives you the amount of charge contained in the volume surrounded by a closed surface. then squeeze that volume and closed surface down to a teeny-little differential volume. then the field diverging out of that differential volume is equal to the teeny charge contained inside which is the charge density times the teeny differential volume.
     
  6. Jun 4, 2012 #5
    rbj, I agree with your definition, but the way I interpret it your definition is quite different from 'rate of change with distance'. The word 'outward' is the key.
     
  7. Jun 4, 2012 #6
    Isn't divergence defined as the difference between total flux (or field strength) coming out of a volume versus flux (or field strength) going into it? (a source-sink relationship)
     
    Last edited: Jun 4, 2012
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