0
$\begingroup$

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.

  1. On one hand, PD depends on amount/density of charge separated.

As example: distribution of voltage during transients, like on this image: enter image description here

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.

  1. 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: enter image description here

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.

$\endgroup$
0
$\begingroup$

But potential of conductor (and voltage between them) is considered as equal everywhere along its length (neglecting some drop due resistance).

The potential of a conductor is equal everywhere in the static case, i.e., in the absence of currents.

When a steady state current flows in a conductor, the potential is continuously changing from one end to the other, dropping faster over high resistance areas and slower over low resistance areas.

$\endgroup$
  • $\begingroup$ Sorry, but it seems we misunderstood each other. I completely rewrote the post, can you please check it again. $\endgroup$ – Steve T. Sep 12 '18 at 19:03
  • $\begingroup$ @SteveT. "But potential of conductor (and voltage between them) is considered as equal everywhere along its length (again neglecting drop due resistance)." It is considered and, in fact, is the same everywhere in the static case, i.e., when no current is flowing. When a current does flow, it creates voltage drops on resistors (V=IR). If resistance was zero everywhere, there would not be any voltage drops and the potential everywhere would be the same like in a superconductor, but that is not the case on your diagram, which has non-zero resistors. $\endgroup$ – V.F. Sep 12 '18 at 19:36
  • $\begingroup$ Thanks for help, but my question is not about resistance actually. Lets neglect it at all. Amount of charges separated on conductors itself is small, but across resistor it's much higher. More charge separated means higher voltage. In this way voltage across resistor must be higher than between wires itselfs. So why in this case voltage across resistor is same as between wires itselfs ? $\endgroup$ – Steve T. Sep 12 '18 at 20:12
  • $\begingroup$ @SteveT. "So why in this case voltage across resistor is same as between wires itselfs ?" Could you be more specific. Which resistor are you talking about and what makes you think that the voltage drop across a resistor is the same as the voltage drop across a conductor carrying the same current. Also, if you are referring to your diagram, it is a bit confusing (not sure if it is correct) and I would not use it to study the basics of electricity and charge distribution. $\endgroup$ – V.F. Sep 12 '18 at 20:55
  • $\begingroup$ @SteveT. If we have a resistor, say, 100 ohm, connected to a battery with two copper wires and the current flows, most of the voltage of the battery will drop on the resistor. If the resistor is 0.01 ohm, the some voltage will drop on the wires (depending on their gauge and length), some on the resistor, some on the internal resistance of battery. Bending wires would not change that. That's all there is to it. All you need to know here is Ohm's law. Perhaps, it is simpler than you think it is. $\endgroup$ – V.F. Sep 12 '18 at 21:57
0
$\begingroup$

If by steady state you mean steady current flowing through a wire then in that case the potential is of the battery which causes the charge to flow. What kind of voltage is applied across the conducting wire would dictate the flow of the charge(subject to the resistances applied).

But just taking a static case(no currents flowing) of a conductor of which as you pointed out, the potential is considered to be constant irrespective of the geometry of the conductor. It is important to note that as we consider smaller and smaller size scales, the bends or the physical macroscopic deformations hardly affect the local charge densities at specific points essentially ensuring PD remains static to a significant extent throughout.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.