Gauss' Law -- How did he come up with it?

[amjad-sh]

How did Gauss came out with his law εΦ=Qenc,where Φ is the flux and Qenc is the net charge inside the gaussian surface?Was it an experimental work or just a theoretical one? thanks.

 

[ZapperZ]

How did Gauss came out with his law εΦ=Qenc,where Φ is the flux and Qenc is the net charge inside the gaussian surface?Was it an experimental work or just a theoretical one? thanks.

It is unclear how far back do you want to start here. First of all, do you know the differential form of Gauss's Law, which is one of the 4 equations that are collectively known at "Maxwell Equations"?

If you do, then do you know the Divergence Theorem that allows you to go from the differential form into the integral form?

Zz.

 

[EM_Guy]

It was actually Faraday's experiment (in 1837). Gauss stated the generalization of the law with mathematical sophistication.

http://www.globalspec.com/reference/62200/203279/chapter-3-density-of-displacement-flux

http://www.docstoc.com/docs/74579527/Unit-3-Electric-Flux-Density_-Gauss's-Law-and-Divergence

 

[amjad-sh]

It is unclear how far back do you want to start here. First of all, do you know the differential form of Gauss's Law, which is one of the 4 equations that are collectively known at "Maxwell Equations"?

If you do, then do you know the Divergence Theorem that allows you to go from the differential form into the integral form?

OK.Yes I know Maxwell's equations and the divergence theorem but i didn't go deep to them.I just know them.
So do you mean that to understand deeply how Guass makes his law I need to grasp Maxwell's equations and the divergence theorem? and it is mostly a theoretical work?

 

[Noctisdark]

One of the most important formulas you get out of electromagnetism is the coulomb force F = kQq/r^2 where k = 1/(4πε₀), from that fact we define the electric field to be E = kQ/r^2, where you can notice the inverse square, 1/r^2 if you sketck the electric field R/r^3, it seems to be just divergent in any point but, when applying the formula of divergence, you'll be shocked that it's exactly zero, by the time the delta function was born it became widely known that this isn't quite true and ∇.(R/r^3) = δ³(r), r is the position, so ∇.E = 4πδ³(r)*kQ, by the definition of k ∇.E = Qδ³(r)/ε₀, so ∫∫∫∇.EdV = ⊂∫∫⊃E.dS, this is the gauss's famous divergence theorem, ∫∫∫∇.E dV = Q/ε₀*∫∫∫δ³(r)dV = Q/ε₀*1, so the flux Φ = ⊂∫∫⊃E.dS = Q/ε₀,Cheers