Do the ends of a standard constant voltage supply have a positive and negative electric potential relative to infinity?

Question

This is something that's been on my mind since I was in High School; and today, as a first year physics student I'm still quite unsure. What I'm trying to ask, more specifically, is:

I understand that the reference with respect to which we announce the electric potential of either electrodes of a voltage supply could potentially be arbitrary; and that no matter what the true potential of either electrode is (w.r.t. a point where there is 0 electric field) a reference potential can always be found such that electrode A is $V_A$ and electrode B is $V_B$ (where $\Delta V=V_A-V_B$ and $V_A=-V_B$). This way of interpreting the labeling of the ends of a voltage supply, however, only makes sense when it is connected to a complete circuit. For example, what happens if we short out the voltage supply?

Well, an E field will be set up by the electrode A and one will be set up by B; here, our choice of a reference electric potential should be irrelevant as an E-field simply dictates a force on a particle at a point. Luckily this doesn't interfere with our understanding of the circuit based on our initial labeling of the voltage supply. The more negative electrode (let's say B) will overpower the less negative electrode (A); and thus, a current will flow from A to B (this makes sense as the less negative electrode will always be labeled as the positive end of the voltage supply).

Now, what if I snap the conductor that's shorted out my voltage supply?

The E-field set up by A and B will no longer oppose one another. This means that the current in the piece of the wire connected to A (I'll call it conductor A) will be solely determined by the electric potential of electrode A w.r.t. infinity; the same will be true for conductor B (the current I'm referring here to is the current that is present for the short amount of time until the entirety of conductors A and B gain the same electric potential as the electrodes they're connected to; I'm unsure whether these currents will be present if we jump from case 1 to case 2 by simply snapping the conductor in case 1. If they aren't, assume that case 2 was constructed separately). Not knowing the true electric potentials of the electrodes has now presented a problem to us; we can't work out the direction of the current in either of conductors A or B (for all we know A and B may have both been negative or positive with A being more positive than B; or in the ideal case, they may have been positive and negative).

Does any of my rambling make sense? Or, are the electric potentials of the electrodes of a voltage supply always given w.r.t. infinity and are such that $V_A=-V_B$ (which is the ideal case)? If they're not, is my argument invalid; or do we just not care if we don't know the direction of these currents?

This is a part of a larger question I have which is an RC circuit (where it's my understanding that the capacitor in some sense creates an open circuit; in other words snaps the shorting conductor). I intend to ask that in reference to this question; I'd be immensely grateful if someone could clear this whole mess of an understanding for me.

Answer 1

no matter what the true potential of either electrode is (w.r.t. a point where there is 0 electric field)

You should think again about this. A point where there is 0 electric field is not necessarily at a "true" potential of zero (i.e. w.r.t. infinity).

what happens if we short out the voltage supply?

If the voltage supply is ideal and the wire you use to short it is not ideal (has non-zero resistance), then you get an electric field in the conductor and a very high current (assuming a reasonably good wire) through the conductor.

If the voltage supply is non-ideal and the wire is ideal (zero resistance), then the potential across the wire goes to zero and the current through the wire is limited by the internal resistance of the supply.

In the real world what happens is usually either the wire or the supply fails due to over-heating, a fuse blows in the supply, the supply's crowbar circuit operates to lower the supply output voltage, etc.

The more negative electrode (let's say B) will overpower the less negative electrode (A);

This doesn't make any sense. The positive terminal of the source is acting to attract electrons. The negative terminal of the source is acting to repel electrons. The fields they produce act in the same direction, so there's no need to think of one "overpowering" the other. They reinforce each other, not oppose each other.

what if I snap the conductor that's shorted out my voltage supply?

I'll assume when you say "snap the conductor" you mean cut or break it.

The E-field set up by A and B will no longer oppose one another.

They never did oppose each other, so there's nothing changed when you break the short from the situation before you connected the short circuit to begin with.

This means that the current in the piece of the wire connected to A (I'll call it conductor A) will be solely determined by the electric potential of electrode A w.r.t. infinity; ... (the current I'm referring here to is the current that is present for the short amount of time until the entirety of conductors A and B gain the same electric potential as the electrodes they're connected to...)

This isn't true. The current will largely depend on the parasitic capacitance between the two cut ends of the wire. Possibly on the parasitic capacitance between the wires and any other physical structures around them.

This means that most of the remainder of your question is based on a false premise.