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Gauss's law for electrodynamics 
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#1
Dec212, 07:44 AM

P: 11

Gauss's law can be proved qualitatively by proving that the field inside a charged closed surface is zero. However Maxwells' equations says that gauss's law holds true even for electrodynamics. how can this be verified experimentally? Thanks in advance !



#2
Dec412, 06:44 PM

P: 886

Gauss's law, is a specific case of Stokes's theorem.
http://en.wikipedia.org/wiki/Stoke%27s_theorem edit: I interpreted Gauss's law to mean the divergence theorem, which is a mathematical statement. My mistake; that would probably be called Gauss's theorem. 


#3
Dec412, 07:06 PM

P: 585

Stoke's theorem is a purely mathematical statement, like the commutative property of addition. 


#4
Dec412, 07:58 PM

P: 3,883

Gauss's law for electrodynamics
I am not good in definitions but I did look into Gauss Law. I really don't see the relation of Stokes and Guass. Even in Guass law for magnetism:
http://en.wikipedia.org/wiki/Gauss%2..._for_magnetism It only said [itex]\nabla \cdot \vec B = 0\; [/itex] where it states there is no mono magnetic pole. Guass law is mainly used in Divergence theorem where [itex]\nabla \cdot \vec E=\frac {\rho_v}{\epsilon}[/itex] Where: [tex]\int_v \nabla\cdot \vec E dv'=\int_s \vec E\cdot d\vec s'=\frac Q {\epsilon}[/tex] http://phy214uhart.wikispaces.com/Gauss%27+Law http://phy214uhart.wikispaces.com/Gauss%27+Law The only one that remotely relate magnetic field through a surface is: [tex] \int_s \nabla X\vec B\cdot d\vec s'=\int_c \vec B \cdot d \vec l'= \mu I [/tex] that relate current loop with field through the loop. 


#5
Dec512, 06:50 AM

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PF Gold
P: 1,992

2. I don't know of any direct experimental test for a time varying E field. The fact that its inclusion in Maxwell's equations leads to many verifiable results is an indirect proof of its general validity. 


#6
Dec512, 08:28 AM

P: 1,020

I just want to say that gauss law follow immediately from maxwell's fourth eqn when combined with continuity eqn for charge density.(just take the divergence)



#7
Dec512, 11:09 AM

P: 3,883

I can't think of any direct prove on Guass surface with varying charge inside. But I cannot see anything wrong that the total electric field radiate out of a closed surface varying due to vary charge enclosed by the closed surface still obey [itex]\int_s \vec E\cdot d\vec s'[/itex].
The difference is with varying charges generating the varying electric field, a magnetic field MUST be generated to accompany the varying electric field according to: [tex]\nabla X \vec E=\frac{\partial \vec B}{\partial t}[/tex] 


#8
Dec612, 04:51 AM

P: 1,020

c^{2}(∇×B)=j/ε_{0}+∂E/∂t now, c^{2}{∇.(∇×B)}=∇.j/ε_{0}+∂(∇.E)/∂t USING ∇.j=∂ρ/∂t and the fact that gradient of curl vanishes. one gets, ∇.E=ρ/ε_{0} 


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