Why is Potential Across a Conductor in a Circuit the Same if we Neglect its Resistance?

I know that when we put a metal conductor inside an external electric field, free electrons will move through its influence towards the one side of the conductor until internal electric field balances the external one. So, net field is zero and thus potential is constant.

Consider a circuit where a battery is connected to the conductor, there is also some resistor in a circuit to impede the current. My question was how is potential across conductor the same if we neglect its resistance given the fact that charges can't build up on the edge of a conductor which is connected in a circuit since they can move until they get to the other terminal of the battery, so internal electric field can't develop.

But, in circuits charges aren't constrained to stay on any side of a conductor because they can move until they get towards the other terminal of the battery and thus charge can't really build up to create internal electric field. Any constructive answer is appreciated.

• What is the question? Commented Apr 18, 2021 at 20:43
• What don't you understand about it? Commented Apr 18, 2021 at 20:58
• It's not that I don't understand what you are saying, it's just that I don't see a question being asked. Commented Apr 18, 2021 at 21:06
• I also don't see a clear question. But I don't understand the part about the battery. I am not sure about if the battery is connected to the metal conductor or if the metal conductor is free floating near the terminals of the battery or what. Please add some clarification and a direct question.
– Dale
Commented Apr 18, 2021 at 21:46
• @Dale Battery is connected to the conductor, there is also some resistor in a circuit to impede the current. My question was how is potential across conductor the same if we neglect its resistance given the fact that charges can't build up on the edge of a conductor which is connected in a circuit since they can move until they get to the other terminal of the battery, so internal electric field can't develop. Commented Apr 19, 2021 at 8:25

So if a conductor is in an electrostatic situation then $$J=0$$ so the E field must be entirely perpendicular.
In contrast, if a conductor is in a circuit then $$J\ne0$$ so there will be some E field parallel to the surface. The better the conductor the smaller that E field for a given current. But even so you can still get surface charges forming. Those surface charges will lead to the component of the E field perpendicular to the current. Because it is perpendicular to the surface the charges cannot move that direction even though they are free to move in a conductor.