Gauss's law for electrodynamics

AI Thread Summary
Gauss's law relates electric charges to electric fields and can be qualitatively proven by demonstrating that the electric field inside a charged closed surface is zero. It is also integrated into Maxwell's equations, which assert its validity in electrodynamics. Experimental verification of Gauss's law, particularly for time-varying electric fields, remains challenging, with no direct tests identified. The law is derived from Maxwell's equations and is supported by the continuity equation for charge density. Overall, Gauss's law's inclusion in Maxwell's framework provides indirect proof of its general validity in various scenarios.
hercules68
Messages
11
Reaction score
0
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 !
 
Physics news on Phys.org
Gauss's law, is a specific case of Stokes's theorem.
http://en.wikipedia.org/wiki/Stoke's_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.
 
Last edited:
Khashishi said:
Gauss's law is a specific case of Stoke's theorem.
http://en.wikipedia.org/wiki/Stoke's_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.
 
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%27s_law_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.
 
Last edited:
hercules68 said:
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.
 
  • Like
Likes davidbenari
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)
 
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 said:
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
 

Similar threads

Replies
1
Views
1K
Replies
5
Views
4K
Replies
83
Views
4K
Replies
9
Views
4K
Replies
30
Views
2K
Replies
5
Views
3K
Replies
13
Views
2K
Replies
2
Views
1K
Back
Top