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I know Kirchhoff's law and that current must be constant. But what exactly is the phenomenon, from first principles, that causes the electric field to be stronger in a region with lower carrier mobility/concentration.

I thought it might be repulsion from electrons that pushes the electrons in front ahead (as charge instantaneously accumulates) but I can't really wrap my head around how exactly more work is done to move a charge through a higher resistance.

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  • $\begingroup$ I'm not an expert at this, so if someone could tell me if this is right, I'd appreciate it: From what I've understood, what will happen when connecting a resistor to a battery with 0-resistance cables is that the charge from the battery immediately distributes itself in the wire, cancelling out the electric field that started to form (basically, the wire becomes a "lengthened arm" of the battery). I suppose that for the general case, it's this redistribution of charge that makes the electric field stronger when resistance is higher. $\endgroup$
    – Max
    Commented Apr 5, 2017 at 13:09

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Are you speaking specifically about currents in wires? If you look at (the simplest version of) Ohm's law you will see that $\mathbf{J} = \sigma \mathbf{E}$, where $\sigma$ is the conductivity (technically it's a tensor, but we'll assume a constant scalar for now). In this case, $\eta$ = $\sigma^{-1}$, which is the resistivity. Thus, we can show that $\mathbf{E} = \eta \mathbf{J}$.

So if the current, $\mathbf{J}$, is constant, then when the resistance (= resistivity per unit length) is high, the electric field $\mathbf{E}$ should be high, if this simple approximation of Ohm's law holds.

Side Note
Generally, it is true that the electric field and current density are coupled to each other, but Ohm's law can be much more complicated. So you should not blindly assume that when the resistance is high, the electric field will be high because of this relationship. I make this point because in a plasma, the resistance is often supplied by the electric fields from electromagnetic and electrostatic waves acting as an effective "drag force" on the particles that are attempting to stream relative to each other.

Thus, in this case a larger resistance to flow is caused by the electric fields having a larger impact on the particles, not a larger electric field caused by a larger resistivity.

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    $\begingroup$ I understand that it must be so in order for the current to be the same throughout the conductor. But what is the physical phenomenon that results in this higher electric field. Is it some kind of charge build up? What automatically changes the electric field based on the total resistance on the circuit? $\endgroup$
    – Dr Coconut
    Commented Jun 11, 2015 at 13:09
  • $\begingroup$ I apologize, I am not entirely sure what you are asking. The requirement that current be constant and the assumption that Ohm's law holds set a requirement on the value of the electric field in order for this to be true. You could also think of this in terms of electric potentials, if one envisions a simple circuit. Then the external source (e.g., a battery) does work on the system by "pushing" charges through the wire. In limit that time variations go to zero (i.e., DC), then one has a constant current and potential… $\endgroup$
    – honeste_vivere
    Commented Jun 11, 2015 at 16:25
  • $\begingroup$ The potential is set by the external source, the resistance is set by the properties of the circuit, and then the current adjusts to force the potential drop across all resistive components to match the potential drop across the external source. Does that make more sense? $\endgroup$
    – honeste_vivere
    Commented Jun 11, 2015 at 16:27
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    $\begingroup$ I guess what my question really is then is how do the results of Ohm's law correspond with Coulomb's law. What is the mechanism for the increase or decrease in electric field. $\endgroup$
    – Dr Coconut
    Commented Jun 11, 2015 at 16:28
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    $\begingroup$ Not quite, my question is more about the exact physical process that results in the variations in electric field strength in circuits $\endgroup$
    – Dr Coconut
    Commented Jul 3, 2015 at 12:01
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$E = - \frac{\mathrm dV}{\mathrm dx}$ means that when voltage changes rapidly with distance inside the resistor, the electric field will be high. You could have two, cylindrical, 1000 ohm resistors, one of which is 5 mm long and the other is 50 mm long. If they are in the same circuit (carrying the same current), the 5 mm resistor will contain the higher electric field because the voltage changes more rapidly across it.

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I believe what occurs is that surface charges accumulate on the surface of the resistor, preferentially toward the side with the negative potential (assuming the charge carrier is negative), inducing an electric field in that direction. And the gradient of surface charge density would be higher in regions of higher resistance.

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