Gauss's law for electrodynamics

hercules68

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 !

Khashishi

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.

the_emi_guy

Gauss' law is a law of physics that relates electric charges to electric fields.

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

yungman

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.

Meir Achuz

1. The charged closed surface must be a conductor.
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.

andrien

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)

yungman

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]

andrien

Let us see,
c²(∇×B)=j/ε₀+∂E/∂t
now,
c²{∇.(∇×B)}=∇.j/ε₀+∂(∇.E)/∂t
USING ∇.j=-∂ρ/∂t and the fact that gradient of curl vanishes.
one gets,
∇.E=ρ/ε₀