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Zero divergence in an enclosed point charge

  1. Apr 11, 2007 #1
    Why does an enclosed point charge have zero divergence/flux? Mathematically I can see that when the divergence operator is applied to E=q/r^2 (pointing in the r direction) I get zero, but what is the physical explanation for what is going on? I am confused because other enclosed electric fields yield non-zero values for divergence.
  2. jcsd
  3. Apr 12, 2007 #2
    I understand that your problem is not with mathematics but with the physical meaning of "divergence". Then let's forget the math and formulas.
    You could ask why people christened the divergence with such name, instead calling it "convergence", "curl" or "bla-bla". Let's see why.
    To "know" if a zone of space has a positive, zero or negative divergence, wrap the volume in a closed surface. Draw all (well, just a few) lines of field ("lines of force") and count the number of lines that cross the surface to enter the volume and the number of lines that cross the surface to exit the volume.
    If there are more lines that exit than lines that enter, on average, the arrows of the field diverge from the volume. And vice-versa, if there are more lines that enter than lines that exit, en average the arrows converge to the volume; or said otherwise, the arrows have a negative divergence from the volume.
    Does it help?

    A word about math. What I asked to do was to "calculate" the flux exiting the surface [tex]\int E\cdot ds[/tex] and then to use the Gauss theorem which can be read as: "all things created within a volume must exit traversing the surface".
    Last edited: Apr 12, 2007
  4. Apr 12, 2007 #3

    Claude Bile

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    Divergence is only non-zero at points where there is a non-zero charge density. This is what Gauss' Law in differential form tells us. For a point charge, the divergence will be zero everywhere, except for the single point where the point charge is located.

    The flux through any surface enclosing a non-zero net charge must be non-zero - this is what Gauss' law in integral form tells us.

    There is a complication with using a point charge with Gauss Law as you can get inconsistent results if you do not properly account for the singularity. To account for the singularity, you need to represent the charge density as a delta function.

  5. Apr 16, 2007 #4
    Actually, you don't get zero. Since the field depend only on radial distance, divergence is
    [tex]\mathbf \nabla \cdot \mathbf E = \frac{1}{r^2} \frac{\partial (r^2 \mathbf E)}{\partial r}[/tex]
    which should be handled carefully at [itex]r=0[/itex], because of the denominator. Using divergence theorem, it can be shown that it's [itex]4 \pi \delta(r)[/itex] there.

    http://en.wikipedia.org/wiki/Dirac_delta" [Broken] has an inifinely sharp peak at [itex]r=0[/itex], and zero everywhere else, such that it's integral all around is equal to 1.

    Writing down the the Gauss' law,

    [tex]\mathbf \nabla \cdot \mathbf E = \frac{\rho}{\epsilon_0} = 4 \pi \delta(r)[/tex]

    which's not too much surprising since you've squeeze a finite charge to zero volume to yield infinite charge density at that point, and zero everywhere else.
    Last edited by a moderator: May 2, 2017
  6. Oct 12, 2007 #5
    thats a good explanation to resolve the contradiction of the maxwell equation D.E = 0 and P/E, where D=del, .=dot product, P=density, E=epsilon.

    However, according to that derivation, D.E is 0 for any r>0. But, if a point charge is at the origin, then the flux at, say r=5 meters, is not 0. But using, the divergence theorem, the flux is 0 since D.E is 0 at r=5 meters. This is a contradiction. I'm uncertain as to how to resolve this
  7. Oct 12, 2007 #6

    Doc Al

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    What do you mean by "the flux at r"?
  8. Oct 12, 2007 #7
    like I said, at say r=5 meters, where r is the distance of the charge from the guassian sphere, and the value of the flux at that distance
  9. Oct 13, 2007 #8

    Doc Al

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    If you mean the flux through a gaussian sphere of radius r surrounding a point charge, why do you think that the divergence being zero at r implies that the flux through that surface is zero? You have to integrate the divergence over the complete volume enclosed by the surface, which includes the point charge.
  10. Oct 13, 2007 #9
    I think I get what you're saying. The total flux at r is nonzero simply because the flux at r=0 is, which is because the divergence is nonzero at r=0.
  11. Oct 13, 2007 #10

    Doc Al

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    Flux only has meaning through a surface, so I'd phrase it a bit differently: The total flux through any closed surface containing a point charge is nonzero because the integral of the divergence of the field over the entire enclosed volume is nonzero (which is due, as you point out, to the non-zero divergence at the point charge).
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