# How do resistors form electric fields and thus potential drops?

$$V_{ab}=-\int_b^a\vec{E}*\vec{dl}=\int_a^b\vec{E}*\vec{dl}$$

By Kirchhoff's loop rule, the sum of the potential drops across a closed loop is zero. Going in the direction of conventional current across a battery (negative to minus, thus in the opposite direction of $$\vec{E}$$) we have a positive voltage term. Subsequently, resistors must have potential drops to make the loop sum to zero. Therefore, for resistors our conventional current will cross resistors going in the direction of the electric field. Instead of gaining potential going across a resistor, our conventional current is having work done on it by the electric field across the resistor and thus having a potential drop in the loop.

My main question

So I've read a lot on stackexchange and quora about beliefs that resistors slow down electrons are overly-simplified, and even a post detailing the fact that resistors are materials with low free-charge carrier density. But I still haven't figured out how a resistor can create an electric field. I do understand it basically in the case of a battery being the result of chemical processes generating an EMF, but not in the case for resistors.

Previous beliefs I would say that the resistor causes a clogging of electrons on one side thus making the other side +ve. However, not only is this overly simplified but intuitively it seems that the side of electron clogging will be the side before entering the resistor thus by even that logic the resistor would form an electric field going opposite the flow of conventional current. In return, this resistor would not cause a potential drop but instead a raise.

• The electric field is from the voltage supply, the resistors just distribute it. – PhysicsDave Apr 20 at 4:12
• @PhysicsDave that does help my knowledge. Thank you. Do you mind being more specific on how the resistors seem to redistribute the field to be a vector opposite of the conventional current unlike the battery? – ebehr Apr 20 at 4:53
• @PhysicsDave Would this be a valid explanation? Based on physics.stackexchange.com/a/127775/227444 I would say that a resistor, relative to the normal wire, has less free electrons. Electrons moving in here are resisted more from bumping into the resistor's lattice and from E&M repulsion. The original EMF is from the battery's field, and the energy the electrons had gained from there is dissipated along each resistor in the form of heat. As energy is dissipated from the electrons, there is a potential drop as those electrons do work (or have work done on them to reaccelerate them?) – ebehr Apr 20 at 5:34
• @PhysicsDave Sorry this is long, but to add onto my previous assertion, the electric field is redistributed to do work on the electrons moving because their energy is dissipated in the resistor's lattice. And because the current must be constant throughout the wire (in total flow) an electric field is established inside the resistor that maintains constant current density throughout the circuit (non-parallel circuit assumption for simplicity) – ebehr Apr 20 at 5:38
• Yes all your explanations are good. The E field does exist even without the free electrons as explained in the answer below. – PhysicsDave Apr 20 at 15:43

$$\let\s=\sigma$$ I'd like to give an upside-down answer. Resistors don't produce or distribute electric field. It would exist independently of resistor. On the contrary, given a field a resistor determines the current it allows to traverse itself. The main equation is $$j=\sigma E \tag1$$ where $$\sigma=1/\rho$$ is conductivity.
Let me explain it with a simple example. You're given a plane capacitor, but it instead of a dielectric contains a conductor, of significant resistivity. If that peculiar "capacitor" is connected to a battery of negligible internal resistance, then in the space between plates you'll observe an electric field $$E=V/d$$ (obvious symbols) exactly like if it were a true capacitor, with vacuum, air or dielectric in between.
That field would accelerate free electrons if they were really free, but it's never so in a solid matter (superconductors apart). Electrons experience a friction due to collisions with fixed positive charges and the average effect is that they acquire a drift velocity proportional to $$E$$: $$v=-\mu E$$ where $$\mu$$ is called mobility (electrons are negative and move in opposite direction to the field). If there are $$n$$ electrons per unit volume it easily follows that electrons' drift causes a current density $$j = -nev = ne\mu E.$$ Thus we get to eq. (1) with $$\s=ne\mu$$.
If capacitor has plates of area $$A$$ we have $$I = A\,j = \s\,A\,E = {\s\,A \over d}\,V = {V \over R}$$ which is Ohm's law, with $$R = {d \over \s\,A} = {\rho\,d \over A}.$$