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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.

Sean
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TheQuantumMan
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1 Answers1

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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.

BowlOfRed
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  • 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. – TheQuantumMan Mar 17 '15 at 23:46
  • 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. –  Mar 17 '15 at 23:51
  • http://physics.stackexchange.com/questions/102930/why-doesnt-the-electric-field-inside-a-wire-in-a-circuit-fall-off-with-distance – TheQuantumMan Mar 17 '15 at 23:54
  • 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. – TheQuantumMan Mar 17 '15 at 23:59
  • have you seen the link? – TheQuantumMan Mar 17 '15 at 23:59
  • Yes. The link shows an ideal battery attached to an ideal conductor. The system is not at equilibrium. Rather than constant current, current is increasing. In a real circuit, charges would rearrange to the point that $E=0$ inside the wire. – BowlOfRed Mar 18 '15 at 00:07
  • @BowlOfRed So,the link shows the system before reaching its steady-state.But when it finally reaches steady-state,i still do not understand why E=0.Why would they rearrange in a certain way as to give the system E=0?also,this is not electrostatics.Because the link that you posted above point out that E=0 because the charges do not move on the surface.In this situation,the charges DO move. – TheQuantumMan Mar 18 '15 at 13:00
  • In steady state the charges move but the flow of charges is constant and uniform so that it gives the same results as in static case. I gave you the same answer about electric field being 0 inside a conductor in equilibrium on your other question on the same topic yesterday. – Ronan Tarik Drevon Mar 19 '15 at 09:57