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Ben Niehoff
#54
Feb26-12, 01:57 PM
Sci Advisor
P: 1,588
Right. So, with sources, we have

[tex]d * F = * J[/tex]
So now, merely integrate this over some 3-volume V:

[tex]\begin{align*} \int_V d * F &= \int_V * J \\ \int_{\partial V} * F &= \int_V * J \end{align*}[/tex]
which is Gauss' Law.

The reason I focused on the source-free equations, is because it is only when [itex]J = 0[/itex] that the result of integration doesn't care about the choice of surface. Obviously, if [itex]J \neq 0[/itex], then different surfaces might contain different amounts of charge, hence giving different results. Gauss' Law still holds, though.