Deriving Gauss law from divergence form using ∫|r|⁻³ dV

[Nikitin]

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 want to 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?

 

[vela]

You didn't take the divergence correctly. You can't ignore the 1/r³ part when differentiating.

 

[Nikitin]

ahh, yeah. Thanks allot! :)