Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Gauss's Law

  1. Aug 19, 2011 #1
    Hello all... I have been working on this problem that I just am not being able to solve.

    I've been spending my spare time learning some vector calculus and non-euclidean geometry (my aim is to be able to finally tackle relativity). After learning some basic things about the del function, I found that I had sufficient mathematical knowledge to be able to derive Maxwell's equations (well I thought I did).

    I had a go at the first of the four. The way I am trying to do it is by taking Coulomb's law, writing it as a vector in three dimensions and then taking the divergence. I am hoping to get this from it:

    [tex]\nabla{\mathbf{.E}} = \frac{\rho}{\epsilon}[/tex]

    But I keep on getting 0. And I have no idea why...
    Here is a scanned copy of my working, I would be very grateful if you could point out my errors (other than the minus sign maybe) to me.

    Cheers.

    http://postimage.org/image/q44ym4ro/ [Broken]
     
    Last edited by a moderator: May 5, 2017
  2. jcsd
  3. Aug 19, 2011 #2
    For the hypothetical situation you described, with a single point charge at (0,0,0), the charge density should be zero everywhere except at (0,0,0).

    The charge density would be q times a 3 dimensional (generalized) Dirac delta. This way the volume integral of [itex]\rho[/itex] for a volume containing (0,0,0) would be equal to q.

    Perhaps it would be simpler to just use the integral forms when dealing with point charges.
     
    Last edited: Aug 19, 2011
  4. Aug 19, 2011 #3

    K^2

    User Avatar
    Science Advisor

    If you mean taking divergence of E from point charge, you have to use Cauchy Formula, in which case you will get exactly q δ(x)/ε = ρ/ε.
     
  5. Aug 19, 2011 #4
    28eb9f07e333ea24556d4914c0991799.png
    Definition of divergence.

    If you assume the charge has a continuous distribution over space, the divergence won't be 0. So if you put q = ρV in Coulomb's law, where ρ is the charge density and V is the volume of some arbitrary region, you'll be able to derive the equation. The volume V must be a function of x, y, and z of course.
     
    Last edited: Aug 19, 2011
  6. Aug 20, 2011 #5
    So, the problem lies in not the mathematics but my interpretation of it?

    Why does a point charge yield a zero divergence but a charge enclosed in finite volume (say a sphere of radius R) yield finite divergence when the electric fields generated by them at point satisfying [tex]x^2 + y^2 + z^2 \geq R^2 [/tex] is the same. An imaginary sphere around the point charge has field lines leaving the volume enclosed through the surface, and the field lines generated by those two configurations of charge are the same at a large distance.
     
    Last edited: Aug 20, 2011
  7. Aug 20, 2011 #6
    Also, is there no other way to deal with it (other than using dirac delta function)? Does this mean that there is a singularity there? How does classical electromagnetism deal with this singularity? Or are the equations inadequate (meaning that point charges are the wrong way to look at what is really going on)?
     
  8. Aug 21, 2011 #7

    vanhees71

    User Avatar
    Science Advisor
    2016 Award

    If you misinterpret the [itex]\delta[/itex] distribution as a function, there's a singularity, but it's not a function but a functional on the space of sufficiently smooth and sufficiently quickly falling test functions, and that's the adequate description of a point particle in a field theory.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Gauss's Law
  1. Gauss's Law (Replies: 9)

  2. Gauss's Law (Replies: 3)

  3. Gauss's Law (Replies: 2)

Loading...