We know that when the system reaches steady-state(current does not change with time),the electric filed inside the circuit is constant. In many textbooks and lectures,professors make a graph like this one:

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So,in the "wire" sections of the graph,shouldn't the potential drop due to the existance of the electric field?Because we know from electrostatics that potential drops when we move away from the point of reference.So i think the potential should drop even when there is no resistor!


1 Answer 1


Technically the resistance of a wire is never 0. Typical wires of copper have electrical conductivity of $10^{7} S/m$ at room temperature. So there is indeed a very slight potential drop but highly negligible compared to the one that would be experience at a proper resistor.

As for the problem of electrostatic, indeed from a single charged particle the field vanishes far away from this charge. Though keep in mind that in a conductor a lot of charges (around $10^{24}$ or more) are actually moving under the influence of the electric field provided by the accumulator. When steady state is reached the conductor(wire), if considered as a ideal, has organised its charge distribution so that the potential is constant on its surface which explains why there is no potential drop along the wire.

  • $\begingroup$ I have read from several books that the electric field is the one that is constant once the system reaches steady-state.So,if the field is constant(and non-zero) then how s the potential also constant?I can't understand.The only way to have constant potential is by having E=0 which is certainly not the case. $\endgroup$ Commented Mar 18, 2015 at 16:24
  • $\begingroup$ The field is constant, but it is very close to 0. If it was non-zero, then a large current would flow in the wire, and charge would build up at the terminal of the resistor (charging the parasitic capacitance of the resistor). This would not be the steady state. In the steady state, the charge has built up sufficiently on the resistor terminal to oppose the field produced by the battery, resulting in near-0 field within the wire. $\endgroup$
    – The Photon
    Commented Mar 18, 2015 at 16:45
  • $\begingroup$ Exactly! A conductor is such that it always organizes its charge distribution to reach equilibrium i.e. the charges force the field to be zero inside a perfect conductor hence the potential is constant. The field is not zero between the 2 poles of the battery and the resistor. $\endgroup$ Commented Mar 18, 2015 at 16:48

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