Nikitin
- 734
- 27
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?
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: