1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Point-Source derivative form Gauss's law valid?

  1. May 9, 2012 #1
    I reviewed Maxwell's equations today and I was upset to find that the volume integral of the derivative form of Gauss's equation seems to be 0 for point sources when it should come out to be Q/(electro constant). This comes from the fact that the gradient is equal to 0 and anything else you do to 0 appears to make it stay zero.

    By the way I checked both versions of Gauss's law on valid field equations that did not have singularities and everything seemed to work out fine. (For instance, I used the integral form of gauss's law to create a field equation for gravity inside a sphere of uniform density to check the derivative form of Gauss's law)

    What I did (and sorry I don't know how to make the math look pretty like you other guys do):

    It looks like the equation for the field of a point source is -kq(x/r^3, y/r^3, z/r^3) where r^2 = x^2 + y^2 + z^2 and k = 1/4/pi/(electro constant). You can use Pythagoras's theorem to check that the magnitude of this vector is kq/r^2. When I took the divergence for x I got -1/(x^2 + y^2 + z^2)^1.5 + 3x^2/(x^2 + y^2 + z^2)^2.5 .... (pattern continues for y and z). Which equals -3/r^3 + 3(x^2 + y^2 + z^2)/r^5 = 3(1/r^3 - 1/r^3) = 0. Maybe you should check to make sure I did this right. This is what you would expect from Gauss's theorem I think because the charge density is zero everywhere except at the point source. However, taking the volume integral of this should give you kq4pi = q/(electro constant).

    1. Have I made a mistake?

    2. If not, is there a way to resolve this issue with the singularity to make Gauss's law valid in this case?
    Last edited: May 9, 2012
  2. jcsd
  3. May 9, 2012 #2

    rude man

    User Avatar
    Homework Helper
    Gold Member

    The problem is there is a singularity at the origin. The diivergence is not definable there.

    Using spherical coordinates to make life easier, with vectors in bold,

    D = Dr r where r = unit vector in r direction,
    div D = (1/r2)∂/∂r(r2Dr)
    which at r = 0 gives ∞ for the 1/r2 part and zero for the partial derivative, and of course Er= Dr0.

    So what is zero times infinity? Trouble, that's what ...

    Possibly one can fool around with the δ function to get around this. Any takers?

    Somehow you did not take div D correctly. I suspect you "used" the ∞ times zero thing and got beached. As you probably know, you can "prove" 1 = 2 by falling in that trap.

    P.S. the divergence theorem is OK because it multiplies div D by volume element dv = (4/3)πdr3 surrounding the point charge. dv is infinitesmally small but not zero, whereas a point charge has identically zero volume.
    By the same token, the surface element dA surrounding dv is non-zero.
    Last edited: May 9, 2012
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Similar Discussions: Point-Source derivative form Gauss's law valid?