How is Potential Difference Created across a Resistor?

• I
• Dario56
Dario56
In a simple circuit consisted of a battery and a resistor, how is potential difference actually established on the resistor?

My understanding is that battery creates the electric field which propagates through space at the speed of light. Resistor is put inside this field and therefore potential difference exists on that resistor (as potential difference is just the line integral of the electric field), causing current according to Ohm's law.

However, I also know that potential difference on resistors are established through charges on the conductor surface.

I'm not sure how to connect these two views as in the former case it seems that potential difference is simply created by the electric field of the battery (in the same way how voltage exist between two points in the electric field of the single charge, supposing that two points have a different distance from that charge) while in the latter case there is a charge redistribution on the conductor surface creating that potential difference the resistor.

In another words, in the former case, potential difference has nothing to do with surface charges and their redistribution while in the latter case that holds.

Therefore, according to my understanding, these two views aren't supporting each other. What am I missing?

PhDeezNutz
What a great paper! If now he'd also have added the full calculation with the magnetic field and drawing also the Poynting vector describing the energy flow! Then I'd force all my colleagues in the physics-didactics institute to read it carefully! May be they think over their nonsensical "water analogy" for circuit theory, which they hammer into the heads of the teacher students I try to teach Maxwell electrodynamics in their theoretical-physics lecture course :-(.

Dale
Dale said:
You are right, the connection between these two is not obvious at all. This is the best paper I know for addressing this concept

Rainer Muller. A semiquantitative treatment of surface charges in DC circuits.

https://www.tu-braunschweig.de/inde...oken=2cc8a71e4fdbf159121c6b8ef8348952a2e0c197
Thank you, Dale. I remember when I first found this paper and how much I liked it. It seems that the gap is bridged by the non-stationary state where the initial electric field created by the battery is changed by the local charge redistribution on different interfaces. This redistribution changes the electric field as long as the charge accumulation at all points inside the conductor becomes zero, ##\frac {\partial c}{\partial t} = - \nabla \cdot j =0##. Stationary state is reached, current inside the different conductors in serial connection (branch) is the same.

vanhees71 and Dale
Yes. Note that during that non-stationary state you cannot use circuit theory as it violates the assumptions of circuit theory.

PhDeezNutz, DaveE, cianfa72 and 1 other person
Reading the aforementioned paper, I have a basic doubt.

In the circuit and in space surrounding it, the sources of the electric field ##E## are the charges localized on the battery terminals and on the surface of the wires (surface charge density ##\sigma##) while the current density ##J## inside the wires is the source of magnetic field ##B## (according Maxwell's equations).

Assuming stationary conditions, the continuity equation says that the current density ##J## inside the wires is constant (the charge density ##\rho## inside the wires is null).

Therefore the flux of electrons inside the wires is not a source for the electric field, right ?

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cianfa72 said:
Assuming stationary conditions, the continuity equation says that the current density J inside the wires is constant (the charge density ρ inside the wires is null).
The charge density inside the wires can be non-zero. In particular wherever the conductivity of the wires changes. The continuity condition (in a magnetostatic situation) only says that the conductivity is constant with respect to time.

cianfa72 said:
Therefore the flux of electrons inside the wires is not a source for the electric field, right ?
Correct. The current density is not a source for the electric field.

Dale said:
The charge density inside the wires can be non-zero. In particular wherever the conductivity of the wires changes. The continuity condition (in a magnetostatic situation) only says that the conductivity is constant with respect to time.
In stationary conditions, however, the non-zero charge density ##\rho## inside the wires (wherever the conductivity change) is constant w.r.t. time. Hence from continuity equation one gets ##\nabla \cdot J = 0##, i.e. the density current is the same along the wires.

Dale said:
Correct. The current density is not a source for the electric field.
Here the point is, even though there is a flow of electrons inside the wires, at any given instant in time the "average" of moving electrons's charge and atom's nuclei bound in the metal "lattice" is actually zero.

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Dale
cianfa72 said:
In stationary conditions, however, the non-zero charge density ##\rho## inside the wires (wherever the conductivity change) is constant w.r.t. time. Hence from continuity equation one gets ##\nabla \cdot J = 0##, i.e. the density current is the same along the wires.

Here the point is, even though there is a flow of electrons inside the wires, at any given instant in time the "average" of moving electrons's charge and atom's nuclei bound in the metal "lattice" is actually zero.
Yes and yes

Therefore in a sense the charge density inside in the wires (wherever the conductivity changes) is actually a bound charge density ##\rho_b## - is it a surface or volume charge density ?

cianfa72 said:
Therefore in a sense the charge density inside in the wires (wherever the conductivity changes) is actually a bound charge density ##\rho_b## - is it a surface or volume charge density ?
No. Bound charge is not simply a charge density that is not moving. Bound charge is the charge associated with polarization by ##\rho_b =-\nabla \cdot \vec P##. Since ##\vec P=0## we have ##\rho_b=0## also.

Note, the lattice of protons in a typical wire are free charges, not bound.

Dale said:
No. Bound charge is not simply a charge density that is not moving. Bound charge is the charge associated with polarization by ##\rho_b =-\nabla \cdot \vec P##. Since ##\vec P=0## we have ##\rho_b=0## also.
Ah ok so, wherever the conductivity changes through the wire, where the non-zero free charge density ##\rho## comes from (as you said the polarization ##\vec P = 0##) ?

Dale
cianfa72 said:
Ah ok so, wherever the conductivity changes through the wire, where the non-zero free charge density ##\rho## comes from (as you said the polarization ##\vec P = 0##) ?
It is established during the initial dynamic phase when ##\partial \rho /\partial t \ne 0##

cianfa72
Dale said:
It is established during the initial dynamic phase when ##\partial \rho /\partial t \ne 0##
Ok, the charge density ##\rho## inside the wire you were talking about is the Type-I surface density as described in the aforementioned paper.

We said they are not bound charges (zero polarization inside the wires) therefore I believe they actually correspond to a deficit (positive) or excess (negative) of free electrons at the interface between conductors with different conductivity (i.e. at the interface between a wire and a resistor).

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Yes, and even when it is a deficit of electrons in a wire, the resulting positive charge is still a free charge.

cianfa72
When it comes to energy balance, assuming wires with infinite conducibility, one can consider a closed surface around the resistor (load). In stationary conditions EM fields do not change in time, therefore the EM energy stored in the volume enclosed from that closed surface doesn't change in time as well. However ##J \cdot E \gt 0## inside it, hence energy is transferred from EM field to the matter (Joule effect). From Poynting theorem then the flux of Poynting vector field over that closed surface is exactly the amount of the volume integral of ##J \cdot E##.

Same kind of reasoning applies to a closed surface around the battery, however in this case ##J \cdot E \lt 0## in the same amount of the above EM energy transferred to the resistor.

The above shows that the battery is the source of energy transferred to the load through the EM fields.

Dale
The above questions arise from a basic doubt about how to use Maxwell's equations in any specific scenario.

From my textbooks, Maxwell's equations are solved by taking the sources ##\rho## and ##J## as assigned/fixed functions of spacetime coordinates. On the other hand, EM fields act themselves on the charges constituing the sources through the Lorentz force.

Therefore, to get a complete and self-consistent solution, one should also take in account that behavior in the calculation.

Yes, that is correct.

cianfa72 said:
Therefore, to get a complete and self-consistent solution, one should also take in account that behavior in the calculation.
Therefore the complete set of equations should include: 4 Maxwell's equations, Lorentz equation, Newton 2nd law and other equations to describe for example the laws of the forces that transform EM energy into heat (plus of course initial conditions for charge and current densities as well as boundary conditions for EM fields).

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Dale
cianfa72 said:
and other equations to describe for example the laws of the forces that transform EM energy into heat
In this regard, I believe Ohm’s law in the form ##J=\sigma E## in some sense "models", from a macroscopic perspective, the forces and the processes involved in the transformation of EM energy into heat. Indeed the electric field ##E## inside the wires/resistor would accelerate the free electrons. However, in stationary conditions, the electrons "drift" velocity is constant (the current density ##J## is constant) hence the Lorentz force from ##E## is balanced out from a non-conservative force responsible for the EM energy transformation in heat (the magnetic field ##B## doesn’t do work). At the equilibrium the Ohm’s law establishes a macroscopic linear relationship between ##J## and ##E## through the conductivity ##\sigma##.

cianfa72 said:
In this regard, I believe Ohm’s law in the form J=σE in some sense "models", from a macroscopic perspective
Yes. There is Ohm’s law and other equations that describe the response. Broadly these are called “constitutive equations”. We have them for polarization and magnetization too.

cianfa72

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