Gauss's Law

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Hi, can you derive Gauss's Law without using Coulomb's Law? If so, how?

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  • #2
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Hi, can you derive Gauss's Law without using Coulomb's Law? If so, how?

Thanks

That depends on what you mean by Gauss Law: There is the Gauss law in matematics, also known as the Divergence Theorem, and then there's is Gauss law in electrodymanics, which is a special case of the Divergence theorem applied to the vectorfunction [tex]\mathbf{\vec F}=kr^{-2}\mathbf{\hat r}[/tex] (which essentally is Coulomb's Law)

The former can be derived mathematically, but that you should be able to find in virtually any undergraduate math textbook.
 
  • #3
Dale
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You can derive Gauss's law from Coulomb's law and vice versa. In other words, they are simply different ways of stating the same thing. However, I don't think that you can "derive" either without the other. They were determined experimentally and not from more fundamental considerations.
 
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Well Gauss law it si just Laplace-Poisson eqn revisited in another sauce. It is obvious that coulomb <-> gauss. just because on the mathematical nature of the electrostatic potential.
Well the first principal is that we are trying to solve laplace eqn assuming isotrpy of the universe. so you finf that in a 3-dimensional world, the correct potential is V=-k/r.
Try everybody to solve (with distribution theory it is easier) D(f)=delta. Where D is the laplacian and delta is Dirac one. Physically its like to imagine a point charge/mass genarating the field...

you can figure out also the dipendence of the potential versus the space dimension, for example if i remeber righ, if we were leaving in flat land the electric potential would have been v=Cost*log(r)... somethinh like that.

I think you may apply the same result to the gravitation potantial sinc the mathematical nature is the same.... y
 
  • #5
Meir Achuz
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Hi, can you derive Gauss's Law without using Coulomb's Law? If so, how?
Thanks
Maxwell's equation[tex]\nabla\cdot\{\vec D}=4\pi\rho[/tex] can be taken as a starting point for electrostatics, rather than Coulomb's law. Then Gauss's law follows from putting this into the divergence theorem.
 
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Thanks very much.
 

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