Why is there an "electric" potential drop across the resistor?

Question

An electric potential difference is created when two charges are separated. In a capacitor, there is a clear accumulation of opposite charges on the two separated plates, therefore work has to be done in moving an unit charge from one plate to another, against the electric field.

But, according to this answer, a potential difference is created in a resistor due to the heat produced by the collision and increased lattice vibrations. So there is an energy loss across the resistor, yes, but how does this energy loss due to heat create an electric potential difference. In other words, do electrons accumulate on one or both sides of the resistor? If not, why should we do electric work in moving from one end of the resistor to another. If the difference in energy is not due to electric work, then that's not electric potential, is it?

Answer 1

Usually I don’t post to questions that have an accepted answer. However, this specific aspect of your question has been consistently addressed incorrectly in the existing comments and answers.

In other words, do electrons accumulate on one or both sides of the resistor?

Charges do, in fact, accumulate on both sides of the resistor. The role of surface charges in circuits is rather under appreciated. Part of that is due to the fact that the surface charges depend on the geometry of the circuit and the geometry is abstracted away in circuit theory. Then, in electromagnetics ordinary circuits become too difficult to model well and simplified toy models are considered instead.

There are some papers that attempt to bridge this gap. See:

A semiquantitative treatment of surface charges in DC circuits:

https://pdfs.semanticscholar.org/9818/9465eefebcbc5ac1af967c7ed50894228f0d.pdf?_ga=2.70346073.856866084.1575031981-2051020381.1575031981

The whole paper is worthwhile but figure 2 specifically addresses your question. At the interface between two materials with different conductivities you do get a surface charge. It is this surface charge which establishes the E-field inside the resistor. The dissipation (collisions) does not produce the E-field inside a resistor, but rather provides an outlet for the energy such that the charge carriers do not gain much KE.

The E-field itself (inside the resistor) is established through the configuration of surface charges at the interface. This must be the case. A resistor does not violate Maxwell’s equations, and from Maxwell’s equations in a DC circuit it is clear that the sudden change in the E-field from wire to resistor must be associated with a charge distribution at the interface. That is the only way to satisfy Maxwell’s equations, specifically Gauss’ law.

Another relevant paper is:

Energy flow from a battery to other circuit elements: The role of surface charges

http://depa.fquim.unam.mx/amyd/archivero/El_flujo_de_energia_de_una_bateria_a_otros_elementos_de_un_circuito_20867.pdf

There is, in fact, accumulation of charges at the interface between materials of different conductivities. Of course, a resistor is a passive element so it does not create the energy, but given a source of power it does take that power and use it to produce a surface charge distribution leading to a given voltage and current.

Answer 2

but how does this energy loss due to heat create an electric potential difference.

The energy loss due to heat doesn't create an electrical potential. It represents a loss, or drop in electric potential (electrical potential energy per unit charge) in moving the charge between the end points of the resistor.

A rough gravity analogy is the drop in gravitational potential (gravitational potential energy per unit mass) when a mass $m$ slides down from a height $h$ on an incline with friction (the resistor) at constant velocity (constant current). The loss in gravitational potential ($mg$) (electrical potential, $V$) equals the friction work done on the mass (charge) moving down the incline. The mechanical friction work is dissipated as heat. The electrical analog is electrical work dissipated in the resistor as heat.

Hope this helps.

Answer 3

What does it mean for a resistor to cause a "potential drop"? This means that the potential of an electron after going through a resistor is lower than the potential of the electron before it went through the resistor. Why is this the case?

First, let us go over what electric potential means. The electric potential of an electron is defined as the energy that would be needed in order to bring this electron to its present location from a distance of infinity. This means an electron has a higher electric potential when it is close to a large amount of other electrons (it would take more energy to overcome the repulsive forces of all these electrons in order to bring our electron to them).

Now, remember, we're trying to explain why electrons that have not yet entered the resistor have a higher potential than electrons that have exited it. It turns out that the reason is that electrons end up being more concentrated on the side of the resistor that they are entering, and less concentrated on the side of the resister that they are exiting. Let's look at why this happens.

Let's take a simple circuit consisting of a battery, a wire, and a resistor. Now, we know the resistor is made of material that is less conductive than the wire, so electrons aren't able to move as quickly in the resistor as they are in the wire.

In the diagram above, the white circles represent the electrons in the wire at their initial state (before the battery is connected). The red circles represent the electrons a moment after the battery is connected to the wire.

When the electron closest to the negative end of the battery moves, it causes the next electron to move as well (because of the repulsive forces between electrons). This creates a cascading effect.

However, electron A in the above diagram cannot move as much (in the same timeframe) as all of the electrons behind it. As a result, electron A will be closer to all the electrons behind it than it was before. Hence, it repels these electrons more strongly. This slightly cancels the repulsive force on these electrons due the negative end of the battery, causing them to slow down.

Now, since the repulsive force an electron exerts on another electron decreases with distance, it is clear that those electrons that are closest to electron A will be slowed down the most and those that are furthest will be slowed down the least. Since the electrons that are in the front (meaning electron A and those that are slightly behind it) are the ones that are moving the most slowly, the electrons will start bunching up a bit near the entrance of the resistor.

Let's look at what happens on the other side of the resistor.

On the other side of the resistor, electron B cannot move as much as the electrons in front of it. Hence, electron B will be farther from these electrons than it was before, so its repulsive force on these electrons decreases. Therefore, these electrons also slow down, since the force with which B is pushing them away is now smaller.

However, the increase in distance from B to the electrons ahead of it will cause a greater decrease in repulsive force for electrons that are further from B. (This can be shown by basic math.) Hence, the electrons that are closest to electron B will slow down the LEAST. Hence, electrons will not bunch up nearly as much on the exiting side of the resistor as they will on the entering side.

Since electrons are more concentrated on the "entering side" of a resistor, an electron will have higher potential when entering the resistor and being among all these close-together electrons than upon exiting the resistor.

I hope that made sense!

Answer 4

There is a potential drop across the resistor because the resistor creates an electric field that resists the motion of the charges inside the circuit.

A battery creates an electric field in the resistor pointing from the positive end of the battery to the negative end. If the electric field from the battery were the only thing there would be a constant force on the charges in the resistor and they would accelerate constantly and the current would increase forever.

This doesn't happen because the material inside the resistor resists, like friction or air resistance. An Ohmic material is a material that creates a resistive electric field that is proportional to the amount of current that goes through it. $E_{resist.} = I \rho $. This is kind of like air resistance: the faster the charges inside the material are moving, the stronger the resisting force.

When you first switch a circuit on the electric field from the battery accelerates the charges inside the circuit. As they accelerate, the resistive electric field increases until it balances the electric field from the battery. When that happens the current stops accelerating and continues at a constant level.

I have described all of this in terms of electric fields and forces, but we can also describe it in terms of electric potentials. Electric potential is related to electric field by $\Delta V = -E\Delta x$. A battery provides a constant voltage $V_battery$ which causes the current to accelerate until it is balanced by the resistive potential difference across the resistor of $V_R = IR$ where $R = \rho\Delta x$.

Energy conservation requires that a circuit in a steady state has to have all the electric potential differences across all the elements balance out. If the voltages don't balance out there will be a leftover net force somewhere that will cause the current to speed up or slow down until the voltages do balance.