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Considering that an electric field exists outside a battery and inside a circuit, shouldn't the potential drop while we move along the wire even if there is no resistor ($E=\nabla V$)?

I am asking this because when I see diagrams of potential along the wire they all show a constant potential along the wire until it reaches a resistor in which the potential drops.

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At equilibrium, the field inside an ideal conductor is zero. http://hyperphysics.phy-astr.gsu.edu/hbase/electric/gausur.html#c2

A charge moving through such a conductor neither gains nor loses energy.

We can't attach an ideal conductor to an ideal voltage source. Something has to give. There will be a voltage drop along a real wire due to non-zero resistance, and there will be a reduction in voltage from the battery that we can attribute to a non-zero internal resistance. That's assuming neither one bursts into flame from overheating first.

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  • $\begingroup$ Is it zero or non-zero and constant?Because i have read that the current auto-correct it self before reaching steady-state where it has electric field non-uniform and when it reaches steady-state the field is uniform but not zero. $\endgroup$ Commented Mar 17, 2015 at 23:46
  • $\begingroup$ Where have you read this, @Landos Adam? Because BowlofRed is right in the context which he speaks of. Apply Gauss's law of electricity in integral form to obtain the result that the field inside an ideal conductor is zero. $\endgroup$
    – user70720
    Commented Mar 17, 2015 at 23:51
  • $\begingroup$ physics.stackexchange.com/questions/102930/… $\endgroup$ Commented Mar 17, 2015 at 23:54
  • $\begingroup$ i think that when you apply gauss law,then the Q enclosed inside the gaussian surface is non-zero.So the electric field is not zero.If the electric field is uniform then the E comes out of the integral of E.dS and then you only have E multiplied by integral of dS and it is equal to Q enclosed/εο.But the key here is that Q enclosed is not zero. $\endgroup$ Commented Mar 17, 2015 at 23:59
  • $\begingroup$ have you seen the link? $\endgroup$ Commented Mar 17, 2015 at 23:59

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