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.
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.
