Are electric circuits considered as electrostatics or electrodynamics?

  • #1
Anti Hydrogen
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Thanks in advance!
 

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  • #2
Gordianus
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I'd say quasistatic models
 
  • #3
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Summary:: hello there, in the electromagnetic theory, in what subbranch are both DC and AC circuits ?

Thanks in advance!
Electrostatics or electrodynamics is kind of a false dichotomy. There are at least 6 different branches: electrostatics, quasielectrostatics, quasimagnetostatics, circuit theory, ray optics, electrodynamics.
 
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  • #4
vanhees71
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Well, let's systemize it!

As far as we know, the complete theory explaining electromagnetic phenomena is (quantum) electrodynamics. In everyday life we almost always only need the classical Maxwell theory.

Since the full theory is a relativistic field theory and thus for full consistency one needs to describe everything relativistically, usually one teaches the theory in another order, i.e., starting from the most simple special cases and describes the charged matter non-relativistically, which is also almost always well justified for everyday-life phenomena.

The most simple special case is electrostatics. This are the full Maxwell equations for the case of charged matter strictly at rest and the electromagetic field time independent. Then the magnetic field is strictly 0 and you only have an electric field.

The next case is magnetostatics, which is a bit a misnomer, because it rather means to describe time-independent fields, charge distributions and current distributions, i.e., steady flowing matter is taken into account. Then you have both (and that's why I think "magnetostatics" is a misnomer, but I've no better idea to name it either; maybe simply static fields?). The advantage of this is that in the non-relativistic limit the electric and the magnetic field completely decouple.

Both electrostatics and magnetostatics are still in principle exact special cases of Maxwell's equations (except the non-relativistic approximation of the charged medium, which however usually is a very good approximation in everyday-life situations).

Then there is the socalled quasistationary approximation, which is however a bit more subtle than originally thought. In principle you have two limits of "Galilean electrodynamics" as this is phrased more modernly: it's the electric and the magnetic limit, depending on the situation you want to describe. Roughly speaking it boils down to neglect almost always the displacement current and thus retardation. This makes the theory applicable in regions around the sources small compared to the wavelength of the typical em. field under consideration (speaking in terms of Fourier transformed fields). A special application is AC circuit theory. Here in fact you use both limits of "Galilean electrodynamics". The point where you cannot neglect the displacement current is when it comes to capacitors. For details, see

https://itp.uni-frankfurt.de/~hees/pf-faq/quasi-stationary-edyn.pdf
 
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  • #5
Motocross9
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DC circuits in "equilibrium" so to speak (meaning a steady current) can be described using electrostatics. See, for example, J. A. Hernandes and A. K. T. Assis, Electric potential for a resistive toroidal conductor carrying a steady azimuthal current. Griffiths problem 7.42 is also a DC circuit problem that can be solved analytically. For a more general description, see Rainer Muller, A semiquantitative treatment of surface charges in DC circuits.
 
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  • #6
vanhees71
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DC circuit in "equilibrium" are magnetostatics, i.e., the full set of Maxwell equations for the special case of time-independent fields. If you want electrostatics you must also have vanishing current densities: ##\vec{j}=0##.

Of course in macroscopic magnetostatics the electric and magnetic fields decouple in the "non-relativistic approximation" of Ohm's law.

Then the complete set of equations is
$$\vec{\nabla} \times \vec{E}=0, \quad \vec{\nabla} \cdot \vec{E}=\frac{1}{\epsilon_0} \rho, \quad \vec{\nabla} \times \vec{B}=\mu_0 \vec{j}, \quad \vec{\nabla} \cdot \vec{B}=0, \quad \vec{j}=\sigma \vec{E}.$$
I.e., you can solve the electrostatic problem first, given the charge distribution and then you can solve the magnetostatic problem with the then given ##\vec{j}##.
 
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  • #7
Motocross9
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I agree with what you've said. I said what I did because I don't think many people are aware that the reason DC circuits work/behave as they do is surface charge on wires and interfaces. That is purely an electrostatics problem, but then of course there is also a magnetic field since ##\vec j## is not zero or vanishingly small.
 

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