Does the first maxwell equation apply only....

AI Thread Summary
The first Maxwell equation applies to both static and dynamic electric fields, including those created by moving or accelerating charges within a closed surface. The electric flux through a closed surface is determined solely by the total charge enclosed, regardless of the motion of the charges. This characteristic makes Maxwell's equations more universal than Coulomb's law, which does not account for the asymmetrical electric fields of moving charges. The discussion confirms that the law remains valid even when considering changing magnetic fields and electric fields. Understanding this principle clarifies the broader applicability of Maxwell's equations in classical electromagnetism.
physics user1
For static electric fields? does it apply also for a generic electric field? Let's say, moving charges inside a closed surface?
If there are charges accelerating in a closed surface a part of the electric field is due to the charges, and a a part is due to the fact that the charges acceleration produce a changing magnetic Flux the a changing electric field, but these electric field lines are closed then these lines should give 0 contribute to the net Flux, then the law is still valid

Can you tell me if what I said is Correct?

Sorry for bad grammar
 
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All of Maxwells equations hold for any classical EM phenomena.
 
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That the integral of electric flux through a closed surface surrounding charges, including charges moving in our frame of reference, depends only on the sum of the charges is what this makes this Maxwell equation more universal than Coulomb's law (as the E field of a moving charge is not spherically symmetric and so doesn't follow Coulomb's law).
 
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Philip Wood said:
That the integral of electric flux through a closed surface surrounding charges, including charges moving in our frame of reference, depends only on the sum of the charges is what this makes this Maxwell equation more universal than Coulomb's law (as the E field of a moving charge is not spherically symmetric and so doesn't follow Coulomb's law).
Thank you, now i got it
 
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