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Gauss's Law

  1. Sep 28, 2007 #1
    consider a region where charge is distributed then we have by gauss' law
    [tex] \nabla \cdot \vec E =\frac{\rho}{\epsilon_{o}}[/tex]
    what is [tex]\vec E[/tex] here.if it is the net electric field then how is that the field is defined at a point where charges are itself present
    Last edited by a moderator: Sep 29, 2007
  2. jcsd
  3. Sep 28, 2007 #2


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    The electric field is defined as the force/charge ratio on a small "test charge", see for instance

  4. Sep 28, 2007 #3
    so how come u introduce a test charge at a place where ther i s already presence of charge
  5. Sep 29, 2007 #4
    Electric and Magnetic fields in matter are the macroscopic fields ... which means the average over regions large enough to contain many atoms .... The actual microscopic fields will fluctuate strongly inside matter ...

    This is what I understand from Griffiths.
  6. Sep 29, 2007 #5
    because by definition a test charge's charge is vanishingly small. you need this to argue that the field of the charge itself doesn't interfere with the ambient field at that point.
  7. Sep 30, 2007 #6
    I think what the issue is trying to get at is that if we calculate the electric field of a charge distribution in matter then what happens to E at the exact position where we have the charge (just based on classical electrodynamics).
    Last edited: Sep 30, 2007
  8. Sep 30, 2007 #7


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    A charge does not exert a net electric force on itself. Otherwise a charge could set itself in motion by way of its own electric field. Therefore, when calculating the electric force on a particular charge, we include only the fields produced by the other surrounding charges.

    An infinitesmal piece of the charge distribution does not exert an electric force on itself. However, the remainder of the charge distribution does. A distribution of (say) positive charge flies apart by mutual repulsion unless there are other forces holding it together.
    Last edited: Sep 30, 2007
  9. Sep 30, 2007 #8
    yes this is my question and not whether the charge exerts force on itself or not.
    if we have a continious charge distribution then hoe is that w e find the net field at a point(what is the charge we have to leave out)
  10. Sep 30, 2007 #9

    I tried to be clear that we are not talking about test charges here at all. The issue is we all know that charge is quantized. So what does one exactly mean by volume density of charge. This is made clear by the question here where we are discussing the field at the very point we have an electron sitting. So as I said without getting into Quantum effects what does classical electrodynamics have to say about this.

    So a charge distribution having a volume density [tex]\rho[/tex] is really a macroscopic average density and the electric field that we calculate can only be a macroscopic average. The field calculated using a volume distribution cannot be expected to be uniform in the tiniest scale. This is an idealization to enable ease of calculation.

    Griffiths talks about this exact problem in his book "Intro. to electrodynamics". Just look for "macroscopic" in the index if you have the book.
  11. Oct 1, 2007 #10
    yes exactly that's what my point is also another question .any book has a derivation of a field inside a solid sphere by using gauss's law on a concentric sphere .But what's the meaning of field inside the sphere .yes one could argue that if we remove some charge and then calculate the field the'r but will that give u the actual net field it does but how do we know that .
    for ex had that distribution be surface charge clearly we would have got only half of our desired value
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