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Consider a simple circuit with a battery and a resistor, connected by wires.

We usually treat the wires (excluding the resistor) as ideal conductors (no resistance). Therefore, we conclude that the voltage drop happens across the resistor only, meaning that the potential across any two points of the same wire (either from one end of the battery up to the start of the resistor, or from the end of the resistor to the other end of the battery) is the same.

If the potential across a wire is constant, and the electric field is the derivative of the potential, it means the electric field on any point on the wire is zero.

If the electric field is zero, it means no electric force is acting on charges in the wires and accelerate them.

The charges in the wires do move, though, so it appears electric field does act on them. What am I missing?

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    $\begingroup$ If the wires really had zero resistance, then the elections wouldn't need a force in order to keep moving. Remember $F=ma$. $\endgroup$
    – DanielSank
    Commented Sep 30, 2017 at 18:19
  • $\begingroup$ I had that in mind, but we expect an object at rest to stay at rest. Why would a single electron even start flowing? (I understand why that would happen at the moment of connection, but not why it happens in the middle of the process. Is the kinetic energy given by the chemical process of the battery?) $\endgroup$
    – Yiftach
    Commented Sep 30, 2017 at 18:24
  • $\begingroup$ @Yiftach There is a potential difference between the two ends of the battery which gives the push $\endgroup$
    – jim
    Commented Sep 30, 2017 at 19:09
  • $\begingroup$ @jim But a potential difference between two "far" points does not cause charge to start flowing from one to the other- the field at the charge's exact spot does. And even though there is a potential difference between these two points, there is apparently no field. $\endgroup$
    – Yiftach
    Commented Sep 30, 2017 at 19:17
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    $\begingroup$ @Yiftach If there is a potential difference, then there is a field. $\endgroup$
    – J. Murray
    Commented Sep 30, 2017 at 19:33

3 Answers 3

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We usually treat the wires (excluding the resistor) as ideal conductors (no resistance)

This is true in ideal circuit theory where, to model a physical wire with resistance, we actually add an ideal resistor to the circuit to account for the voltage drop across the physical wire (we can also add an ideal inductor for the self-inductance of the wire etc.)

For a physical circuit, the zero resistance wire is a good approximation in the case that the resistance of the connecting wires is insignificant compared to the total resistance.

If the electric field is zero, it means no electric force is acting on charges in the wires and accelerate them.

Correct and, for the DC case that you've stipulated, this is exactly what we want otherwise the current would be changing in contradiction to the stationary assumption.

The charges in the wires do move, though, so it appears electric field does act on them.

Why? In the previous quote, you correctly state that an electric field accelerates the (mobile) charge. But move is not a synonym for accelerated.

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You can think of $R$ as a parameter that models the resistance of the entire wire. In real circuits there is a nonzero electric field in every point of the wire, and each tiny section of the wire has some resitance.

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  • $\begingroup$ I understand what happens in reality, my question regards theory. Only in theory do we treat the potential across the wire as constant, and I'm asking how that model can be correct even if we truly had wires with zero resistance. $\endgroup$
    – Yiftach
    Commented Sep 30, 2017 at 19:25
  • $\begingroup$ Not if a wire is a superconductor. $\endgroup$
    – safesphere
    Commented Sep 30, 2017 at 23:38
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The problem here is that you give physical relevance and interpretation to the non-physical approximations of your model. A real wire has finite resistance and a finite voltage across it. However, if the resistance (and thus voltage) is very small in comparison to the voltages across other components then this voltage can be ignored to simplify calculation. The approximation of 'treat small voltages as zero' has no physical meaning for the components effected by these voltages, and as such trying to figure out how an object behaves with zero voltage across it is not the correct interpretation.

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  • $\begingroup$ Superconductors are very much physical and have a zero resistance. $\endgroup$
    – safesphere
    Commented Sep 30, 2017 at 23:43
  • $\begingroup$ Ok, but we're not assuming copper is a superconductor, so my assertion that this is not a physical property is correct. $\endgroup$
    – Eddy
    Commented Sep 30, 2017 at 23:45
  • $\begingroup$ I believe the OP does assume superconductivity in his question by saying, "We usually treat the wires (excluding the resistor) as ideal conductors (no resistance)." $\endgroup$
    – safesphere
    Commented Sep 30, 2017 at 23:47
  • $\begingroup$ Well, maybe, but my guess would be that the OP is looking at standard circuit diagrams where the resistivity of copper cables is ignored, and this is what's confused them. If they are looking at superconducting wires I would have thought that would be more explicit. $\endgroup$
    – Eddy
    Commented Sep 30, 2017 at 23:53
  • $\begingroup$ See en.m.wikipedia.org/wiki/Perfect_conductor?wprov=sfla1 $\endgroup$
    – Eddy
    Commented Sep 30, 2017 at 23:53

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