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I'm trying to understand from basic principles why a resistor causes a drop in potential. Going beyond Ohm's law (which doesn't really explain it microscopically), it seems that each side of the resistor gets a different charge density due to accumulation of charges as electrons have more difficulty going through the resistor.

This looks like a promising explanation Why is there an "electric" potential drop across the resistor? but I feel like this buildup of charges would in the end "disassemble" due to the accumulation of repulsive forces. Would it be better to see it as having these repulsive forces constantly push some of the electrons to the resistor so these effects cancel out and there's a steady flux of electrons?

And also, while the "left/negative" side explanation feels somewhat reasonable, I don't see why the "right/positive" side would have a positive charge density. As the electrons go faster because of having a repulsion force along their direction of motion, it could make sense that, relative to the other side, they go faster and thus pile up less. But why isn't it negative with a lower absolute value, that is for example -1C/m^3 vs -0.5C/m^3?

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  • $\begingroup$ If you accept the fact the Ohm's laws is locally obeyed in the low conductivity resistor and the high conductivity wires, you reach the conclusion that since current density j is the same in both (we must obey the continuity equation), the electric field E = j/sigma must be higher in the resistor and smaller in the wires. Then all you need is Gauss law applied to a closed surface enclosing the interface between the materials to show that you need charge at the interface. It has to be there in steady state to justify the nonzero electric flux. $\endgroup$
    – Peltio
    Commented Jan 29, 2023 at 14:53

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One way to think about this is to compare with the case of a device such that there is oppertunity for charge buildup, but (up to a voltage where failure occurs) there is a barrier to charge making it across.

A capacitor has those properties.

A capacitor is like a resistor with in effect infinite resistance. (Of course, every capacitor has its failure point, after that the resistance is no longer infinite.)

A parallel plate capacitor is an instance of the category of capacitor where the feature that gives the capacitor its properties is in plain sight. So for a vivid picture: think of a plate capacitor.

In the absence of a voltage the plates remain uncharged.
When there is a voltage across the capacitor there is buildup of charge in the plates; one plate will have a surplus of electrons, the other plate will have some degree of depletion of electrons, such that the capacitor as a whole is electrostatically neutral.

So: how much charge will the plates accumulate?
As charge accumulates in the plates a voltage across the gap builds up. As long as that voltage is still less than the applied volgate accumulation of charge continues.

The accumulation of charge levels off at the point where the voltage across the plates reaches the same value as the applied voltage (opposite in direction).


A resistor has negligable capacitance, of course, but the point of the comparison is that at the instance that voltage is applied there will be some degree of charge buildup. Due to the capacitance being negligably small the amount of charge buildup is minimal.

The factor that matters is that there is charge buildup. Voltage across the resistor increases until it reaches the same value as the applied voltage (opposite in direction).

On one side of the current resisting interface there will be a surplus of electrons, and on the other side a corresponding degree of electron depletion.




Later edit:
Responding to request for additional discussion: the case where the resisting barrier is long. One example of that would be a heating element, of say, an electric toaster.

So that is an instance of a resistor with a lot of length to it. We know that from one end of the resistor to the other the voltage drops off. If the resistance if uniform along the length then the voltage drops off linear with length along the resistor. I infer there must be a corresponding linear drop-off in electron density.

I don't have experimental corroboration for that. I'm inferring that electron density gradient; it seems to me that is the only self-consistent interpretation.

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  • $\begingroup$ Okay, it more or less makes sense, but how is this visualised if we have a non ideal wire with a constant resistance? There would be no "interface" at all. I think it's modelled as having a single resistor with an equivalent resistance, but how would it work really? Or what justifies this model? $\endgroup$ Commented Jan 29, 2023 at 17:30
  • $\begingroup$ @JaimeYepesdePaz yes, there would be two interfaces between the battery electrodes and the resistive wire. You can remove the battery by using a uniform resistive ring placed in a changing magnetic field. Look up "Lewin ring", there is a paper/essay by Lewin and Belcher where this uniform ring is treated. $\endgroup$
    – Peltio
    Commented Jan 29, 2023 at 20:39
  • $\begingroup$ @JaimeYepesdePaz About the resistive wire case: I added discussion to the answer. $\endgroup$
    – Cleonis
    Commented Jan 29, 2023 at 23:24
  • $\begingroup$ @JaimeYepesdePaz The interface charges are not the only charges that need to appear to create the electric field complyimg with Ohm's law in the circuit. There are also surface charges on the lateral surface if the wires and the resistor that create a field that superpose to the original battery field to produce E directed along the axis of the wires and is constant in magnitude (where conductivity and cross section is constant). Look up "Chabay Sherwood surface charge" $\endgroup$
    – Peltio
    Commented Jan 29, 2023 at 23:33

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