- #1
Quicksilvr
- 7
- 0
I'm trying to understand how the integral form is derived from the differential form of Gauss' law.
I have several issues:
1) The law states that ## \nabla\cdot E=\frac{1}{\epsilon 0}\rho##, but when I calculate it directly I get that ## \nabla\cdot E=0## (at least for ## r\neq0##).
2) Now ## \iiint\limits_\nu \nabla\cdot E d\tau ## should be zero no matter what the value of the divergence is at 0, since the divergence is zero everywhere but 0 (in contrast to the law which states it is non-zero).
3) a. The proof itself goes on to use the divergence theorem to state that for any volume ##\nu##, ## \iiint\limits_\nu \nabla\cdot E d\tau = \iint\limits_{\partial\nu} E d a ##, however the divergence theorem requires E to be continuously differentiable everywhere in ##\nu## (it is not differentiable at 0, let alone continuously differentiable there).
b. The function cannot be corrected at any way at 0 since the derivative goes to infinity around 0.
c. The point 0 cannot be removed from the integrated volume because the divergence theorem requires that the volume of integration be compact.
d. In light of the former I don't see how the divergence theorem can be used here.
I have several issues:
1) The law states that ## \nabla\cdot E=\frac{1}{\epsilon 0}\rho##, but when I calculate it directly I get that ## \nabla\cdot E=0## (at least for ## r\neq0##).
2) Now ## \iiint\limits_\nu \nabla\cdot E d\tau ## should be zero no matter what the value of the divergence is at 0, since the divergence is zero everywhere but 0 (in contrast to the law which states it is non-zero).
3) a. The proof itself goes on to use the divergence theorem to state that for any volume ##\nu##, ## \iiint\limits_\nu \nabla\cdot E d\tau = \iint\limits_{\partial\nu} E d a ##, however the divergence theorem requires E to be continuously differentiable everywhere in ##\nu## (it is not differentiable at 0, let alone continuously differentiable there).
b. The function cannot be corrected at any way at 0 since the derivative goes to infinity around 0.
c. The point 0 cannot be removed from the integrated volume because the divergence theorem requires that the volume of integration be compact.
d. In light of the former I don't see how the divergence theorem can be used here.