Microscopic explanation of potential drop in resistors 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?
 A: 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.
