- #1
loom91
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Hi,
I'm just learning about Maxwell's equations in high-school and was playing around with them. Supposedly they are 4 independent and self-sufficient equations that when connected with the Lorentz force law will predict classical electrodynamics in its entirety. But then, it appears to me that Gauss's law can be derived from Ampere's law! How can this be possible?
Ampere's law with Maxwell's correction:-
[tex]
\int_{\partial S} \vec{B} \cdot \vec{dl} = \mu i_S + \mu \epsilon \frac{d\Phi_{E,S}}{dt}[/tex]
Now let us take the limit as S becomes a closed surface. In this limit, the line integral of B vanishes (a closed surface has no boundary), while on the right side we have the net current coming out of the closed surface plus the electric flux over the closed surface. Note the the net current coming out = - rate of change of charge enclosed by the surface. This gives us
[tex]\epsilon d\Phi_{E,S} = dQ_S[/tex]
Integrating, we get
[tex]\Phi_{E,S} = \frac {Q_S}{\epsilon}[/tex]
Which is Gauss's law for electricity.
How can this happen? Maxwell's equations are supposed to be independent! The only way out I can personally see is to take Gauss's law as simply the boundary condition that the net flux through a closed surface enclosing zero net charge is zero, a condition that is used to do the integral in the above derivation. But this boundary condition is nowhere near as strong as Gauss's law itself! What is the matter?
Thanks for your help. I will also appreciate it if you kept this discussion within the boundary conditions of my knowledge (for example, I don't understand the relativistic formulation of classical electrodynamics)
Molu
I'm just learning about Maxwell's equations in high-school and was playing around with them. Supposedly they are 4 independent and self-sufficient equations that when connected with the Lorentz force law will predict classical electrodynamics in its entirety. But then, it appears to me that Gauss's law can be derived from Ampere's law! How can this be possible?
Ampere's law with Maxwell's correction:-
[tex]
\int_{\partial S} \vec{B} \cdot \vec{dl} = \mu i_S + \mu \epsilon \frac{d\Phi_{E,S}}{dt}[/tex]
Now let us take the limit as S becomes a closed surface. In this limit, the line integral of B vanishes (a closed surface has no boundary), while on the right side we have the net current coming out of the closed surface plus the electric flux over the closed surface. Note the the net current coming out = - rate of change of charge enclosed by the surface. This gives us
[tex]\epsilon d\Phi_{E,S} = dQ_S[/tex]
Integrating, we get
[tex]\Phi_{E,S} = \frac {Q_S}{\epsilon}[/tex]
Which is Gauss's law for electricity.
How can this happen? Maxwell's equations are supposed to be independent! The only way out I can personally see is to take Gauss's law as simply the boundary condition that the net flux through a closed surface enclosing zero net charge is zero, a condition that is used to do the integral in the above derivation. But this boundary condition is nowhere near as strong as Gauss's law itself! What is the matter?
Thanks for your help. I will also appreciate it if you kept this discussion within the boundary conditions of my knowledge (for example, I don't understand the relativistic formulation of classical electrodynamics)
Molu