Physics Forums (http://www.physicsforums.com/index.php)
-   Classical Physics (http://www.physicsforums.com/forumdisplay.php?f=61)
-   -   Gauss's law for electrodynamics (http://www.physicsforums.com/showthread.php?t=656255)

 hercules68 Dec2-12 07:44 AM

Gauss's law for electrodynamics

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 Dec4-12 06:44 PM

Re: Gauss's law for electrodynamics

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 Dec4-12 07:06 PM

Re: Gauss's law for electrodynamics

Quote:
 Quote by Khashishi (Post 4184558) 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 Dec4-12 07:58 PM

Re: 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 $\nabla \cdot \vec B = 0\;$ where it states there is no mono magnetic pole.

Guass law is mainly used in Divergence theorem where $\nabla \cdot \vec E=\frac {\rho_v}{\epsilon}$ Where:

$$\int_v \nabla\cdot \vec E dv'=\int_s \vec E\cdot d\vec s'=\frac Q {\epsilon}$$

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:

$$\int_s \nabla X\vec B\cdot d\vec s'=\int_c \vec B \cdot d \vec l'= \mu I$$

that relate current loop with field through the loop.

 Meir Achuz Dec5-12 06:50 AM

Re: Gauss's law for electrodynamics

Quote:
 Quote by hercules68 (Post 4181439) 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 Dec5-12 08:28 AM

Re: Gauss's law for electrodynamics

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 Dec5-12 11:09 AM

Re: Gauss's law for electrodynamics

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 $\int_s \vec E\cdot d\vec s'$.

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:

$$\nabla X \vec E=-\frac{\partial \vec B}{\partial t}$$

 andrien Dec6-12 04:51 AM

Re: Gauss's law for electrodynamics

Quote:
 Quote by andrien (Post 4185194) 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

 All times are GMT -5. The time now is 06:04 PM.