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Gauss's law for electrodynamics

  1. Dec 2, 2012 #1
    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 !
     
  2. jcsd
  3. Dec 4, 2012 #2
    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: Dec 4, 2012
  4. Dec 4, 2012 #3
    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.
     
  5. Dec 4, 2012 #4
    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 [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.
     
    Last edited: Dec 4, 2012
  6. Dec 5, 2012 #5

    Meir Achuz

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    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.
     
  7. Dec 5, 2012 #6
    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)
     
  8. Dec 5, 2012 #7
    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]
     
  9. Dec 6, 2012 #8
    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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