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I've attached a fairly benign example from the fourth edition of Griffiths's Introduction to Electrodynamics. My only problem with the example, which was something that bothered me in other parts throughout the section, was the very first step. The problem sets us up with two electrodes at constant potential, and asks for the current flowing between them in a given length. To calculate this from first principles requires that we use $\mathbf{J}=\sigma\mathbf{E}$ and so Griffiths proceeds to obtain $\mathbf{E}$ between the electrodes. My question though, is how he justifies saying (although he doesn't say it, he just writes it down) the field between two electrodes at a fixed potential difference (and with the given geometry) is necessarily given by the field due to some electrostatic charge distribution. Up to this point in the book, we've strictly studied electrostatic situations where the electrostatic field is sourced by charge distributions that have been nailed down in place. Indeed, every single $\mathbf{E}$ field that we have computed thus far has come directly from Coulomb's Law which has been laid down as an axiom and which applies to stationary charges only (I imagine it's more common to, equivalently, lay down Gauss's Law as the axiom, but that doesn't change the point). Here, I am told only the voltage, and not necessarily that it is indeed a static charge distribution causing this voltage difference. So why can I calculate the field as though it is?

An idea I had was that, assuming that it is reasonable to say that $\mathbf{E}$ is irrotational (is this reasonable to assume a priori? If so, why?) so that $V$ is well-defined, we can argue that Laplace's equation will hold in the region $a<s<b$ (because the material is ohmic and we may assume a divergenceless $\mathbf{J}$ flow by symmetry, so $\mathbf{E}$ is divergenceless). Then the field Griffiths writes down does indeed correspond to a potential that solves Laplace's equation in that region and which meets the boundary conditions given an appropriate choice of $\lambda$, and so we can make an appeal to some appropriate uniqueness theorem for $V$ given symmetry considerations. However, what I have described is quite convoluted and so I would be surprised if my explanation is right because its omission from the problem seems strange.

Essentially, my question boils down to why can we calculate $\mathbf{E}$ fields in certain cases as though they were sourced by a given static distribution, even though we don't know that there is physically actually any static charge there doing the "sourcing". For example, in a circuit with a single (DC) battery and resistor, is there really a surface of charge at each end of the resistor that is setting up the large $\mathbf{E}$ in the resistor? enter image description here

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The problem sets us up with two electrodes at constant potential, and asks for the current flowing between them in a given length. ... My question though, is how he justifies saying (although he doesn't say it, he just writes it down) the field between two electrodes at a fixed potential difference (and with the given geometry) is necessarily given by the field due to some electrostatic charge distribution.

In Maxwell's equations, only two things produce electric fields: charge and changing magnetic fields.

Since this is a (quasi-)static situation (the voltages aren't changing) then we know there are no significant changing magnetic fields present. Therefore it must actually be charges that are producing the fields. And since this is a steady-state problem, we know these charges are not moving (or are moving slowly enough to be treated as static for the purposes of this problem).

Put another way, it is the presence of charges that makes the electrodes have their potentials.

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  • $\begingroup$ Thank you very much for the response. I can accept the fact that there are some stationary charges, but I might then ask how we can know a priori that the field must be precisely as given? Presumably a fully "rigorous" treatment would have to argue as I did in the last paragraph? Also, when you say that we must be at steady-state, does that come from symmetry implying that $\mathbf{J}$ must have zero divergence, and that $\mathbf{J}$ is not a function of $t$? As far as I can tell, those would be the necessary conditions for the charge density in the region between cylinders to be constant. $\endgroup$
    – EE18
    Commented Nov 20, 2020 at 19:25
  • $\begingroup$ @1729_SR, we're at steady state because you said, "the problem sets us up with two electrodes at constant potential." I don't see what argument you are referring to in the final paragraph. Do you mean about the "surface charge at each end of the resistor"? In that case, yes there is a charge at each end of the resistor to set up the field. But I don't think you have to assume this a priori --- you can derive it from the Laplace equation and the boundary condition that the electrodes contacting the two ends are kept at constant potentials. $\endgroup$
    – The Photon
    Commented Nov 20, 2020 at 19:41

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