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Gauss's law for electrodynamics

by hercules68
Tags: electrodynamics, gauss
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hercules68
#1
Dec2-12, 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 !
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Khashishi
#2
Dec4-12, 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.
the_emi_guy
#3
Dec4-12, 07:06 PM
P: 585
Quote Quote by Khashishi View Post
Gauss's law is a specific case of Stoke's theorem.
http://en.wikipedia.org/wiki/Stoke%27s_theorem
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
#4
Dec4-12, 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.
Meir Achuz
#5
Dec5-12, 06:50 AM
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P: 1,992
Quote Quote by hercules68 View Post
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 !
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
#6
Dec5-12, 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)
yungman
#7
Dec5-12, 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]
andrien
#8
Dec6-12, 04:51 AM
P: 1,020
Quote Quote by andrien View Post
I just want to say that gauss law follow immediately from maxwell's fourth eqn when combined with continuity eqn for charge density.
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


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