Charge distribution in a resistor with a current

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Discussion Overview

The discussion revolves around the charge distribution in a resistor within an idealized circuit containing a constant voltage emf and perfectly conducting wires. Participants explore the nature of potential drops across resistors, the role of electric fields, and the behavior of charges in relation to resistance.

Discussion Character

  • Exploratory
  • Technical explanation
  • Debate/contested

Main Points Raised

  • One participant questions the idea that charge redistribution occurs in a resistor, arguing that a perfectly resistive element should not allow some electrons to drift while others remain static due to resistance.
  • Another participant asserts that surface charges are necessary to account for the discontinuity of the electric field at the surface of the resistor and the wires.
  • A later reply supports the claim about surface charge distribution being critical for circuit function, referencing a paper that discusses this concept.
  • Further, a participant shares a different paper that provides a more quantitative analysis of the energy flow in circuits, suggesting that it complements the previous discussion.

Areas of Agreement / Disagreement

Participants express differing views on the role of charge distribution in resistors, with some supporting the idea of surface charges while others remain skeptical about the necessity of static charges within the resistor itself. The discussion does not reach a consensus.

Contextual Notes

Participants reference various papers to support their claims, indicating a reliance on external sources for further exploration of the topic. There is an acknowledgment of the complexity of the electric field behavior at the surfaces of resistors and wires.

fluidistic
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Consider a very simple idealized circuit, with a constant voltage emf, perfectly conducting wires and a resistor all in series. There is a potential drop across the resistor, given by Ohm's law: ##V = -IR##. I have read on the Internet that many people say that the potential drop is caused by a redistribution of charges either at the entry and exit of the resistor (a bit like a capacitor), or linearly distributed throughout the resistor. These charges would be held responsible for the electric field that counters the one produced by the emf.

However this doesn't make sense to me. I see no way that a simple perfectly resistive element would cause some electrons to pass through it with a drift velocity and others to just sit unaffected by the E field, solely due to the resistance.

Mathematically, the electric field inside the resistor should be worth ##\vec E = - \nabla V##, where ##V## is linear with respect to ##x##. So it means the electric field is constant through the resistor. I see no problem in thinking that it's the emf that is causing this constant electric field across the resistor, why on Earth would it need to be created by static electrons?

From a solid state point of view, the electrons are not well localized, and the very few ones that do feel the electric field of the emf (the ones near the Fermi surface), move at speeds near the Fermi velocity, i.e. about 2 orders of magnitude slower than light speed in vacuum. Mathematically, to describe the current, it's the same as if all the electrons had acquired a very small net drift velocity, while in reality what happens is quite different. But I see no way for resistance to redistribute the electrons. I see resistance directly linked to resistivity, i.e. scattering of the very few (quasi)electrons that can be affected by the ##\vec E## field with phonons, impurities and the boundaries of the resistor.

Is there a way to prove or discard that some charges are indeed static across the resistor?
 
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The charges are surface charges. They are necessary because the E field is discontinuous at the surface. Note, the surface charges exist on the perfectly conducting wire also, it has little to do with the resistance.
 
Last edited:
Dale said:
The charges are surface charges. They are necessary because the E field is discontinuous at the surface. Note, the surface charges exist on the perfectly conducting wire also, it has little to do with the resistance.
So the claim is essentially correct? Because the surface of the resistor at the junction between the wires and the resistor is actually inside the material. See fig.2 of that freely available paper: https://aapt.scitation.org/doi/10.1119/1.4731722
 

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