Charge distribution in a resistor with a current

  • #1
fluidistic
Gold Member
3,961
266
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?
 
Physics news on Phys.org
  • #2
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:
  • #3
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
 
Back
Top