I'm must be missing something again but isn't there some contradiction in 2 statements below ?
Edit: I've rewrote the post and added images for clarification.
- On one hand, PD depends on amount/density of charge separated.
Voltage here is directly correlate with amount/density of charge in certain place of conductor.
Or when incident pulse of voltage and current reach an open end there will be short-term increase of voltage. This is explained due pile up of charges at the end since current have to stop. So here we again have relation: more charge - higher potential of conductor in this place.
In this way, if source voltage doesn't change at all, charge distribution on conductor must be uniform to maintain same potential along it length.
Note: actually there will be some potential drop due resistance and hence amount of surface charges is slowly decreasing but it doesn't matter here.
- But, on the other hand, if we see at DC steady state (with steady current), distribution of charges on surface of conductors, actually, is non-uniform. I.e. in some place amount of charge is pretty small, on bends it larger, and really huge accumulation of charge in the area of wire - resistor interface.
As example, image from A semiquantitative treatment of surface charges in DC circuits:
But potential of conductor (and voltage between them) is considered as equal everywhere along its length (again neglecting drop due resistance).
I mean there is area at the end of conductor (at resistor boundary) with significantly more amount of charge than other parts of the wire. If remember transmission line example then potential of this area must be higher.
It sound strange, but in this way voltage across resistor must be higher than across wires itselfs. But this is nonsense and here it obviously same as between other parts of the conductors.
Why voltage across resistor in this case same as across wires despite the fact that there is much more charge accumulation than on wires itselfs ?
Maybe it somehow related to capacitance, but I don't know.