1. Limited time only! Sign up for a free 30min personal 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!

Gauss law in divergence form

  1. Apr 26, 2013 #1
    Hey. I want to use integrals-math to get from Gauss law in divergence form to the one in integral form. I know you can do it by simply accepting ∇*E dV = ρ/ε => ∫ ∇*E dV= ∫ρ/εdV = Q/ε = ∫E*dA, but I wanna do it another way. I want to begin with ∫∇*E*dV and end up with Q/ε.

    So: E = Q*v/(4pi*ε*|r|2), where v is the directional vector of r: v= D/|r| = [x,y,z]/√(x2+y2+z2).


    ∫∇*EdV = ∫∇*v*[Q/(4pi*ε*|r|2)] dV = Q/(4pi*ε)*∫ ∇*D*|r|-3 dV =

    Q/(4pi*ε)*∫(1 + 1 +1)*|r|-3 dV = 3*Q/(4pi*ε)*∫|r|-3 dV.

    However, I can't seem to solve the integral ∫|r|-3dV using spherical coordinates, as I get that it is infinitely large.. So can you guys assist me? Does perhaps ∫|r|-3dV = |r|-3 ∫ dV = 4pi/3?
    Last edited: Apr 26, 2013
  2. jcsd
  3. Apr 26, 2013 #2


    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Education Advisor

    You didn't take the divergence correctly. You can't ignore the 1/r3 part when differentiating.
  4. Apr 27, 2013 #3
    ahh, yeah. Thanks allot! :)
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted