Gauss's law for electrodynamics

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SUMMARY

Gauss's law relates electric charges to electric fields and is a specific case of Stokes's theorem. It holds true even for electrodynamics as stated in Maxwell's equations. The discussion highlights the relationship between Gauss's law and the divergence theorem, emphasizing that the electric field inside a charged closed surface is zero. Experimental verification of Gauss's law, particularly for time-varying electric fields, remains a challenge, although its inclusion in Maxwell's equations provides indirect proof of its validity.

PREREQUISITES
  • Understanding of Maxwell's equations
  • Familiarity with Stokes's theorem
  • Knowledge of divergence theorem
  • Basic concepts of electric fields and charge density
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  • Research experimental methods for verifying Gauss's law in electrodynamics
  • Study the implications of the divergence theorem in electromagnetism
  • Explore the relationship between electric and magnetic fields in varying conditions
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hercules68
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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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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 Gauss. Even in Gauss 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.

Gauss 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.
 
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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 Gauss 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
 

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