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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).

    Thus,

    ∫∇*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

    vela

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    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! :)
     
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