If a battery has wires connected to its poles then there is going to be movement of electrons from the wire connected to the positive pole towards the wire connected to the negative pole. If there is a completed circuit then in steady state this will result in a (roughly) constant current flow at all points of the circuit and an equal distribution of electrons everywhere. However if there is a break in the circuit then there will still be the force acting on the electrons but rather than having a current flow at steady state there will instead be a greater number of electrons on the wire connected to the negative pole than on the wire connected to the positive pole. I believe there being a "potential difference" is just the common way of referring to this exact situation.

Another analogy would be a tube full of water with a pump in the middle. If the tube is a complete circle then the pump will constantly push the water around the circle. If the tube is blocked some distance away from the outflow of the pump then that portion of the tube will fill with water while the inflow dries up. If the blockage is removed then the water will start to flow again, even if the pump is also turned off at the same moment.

That’s the part that doesn’t make sense to me. Only differences in voltage matter. It’s defined with an integral equation so it’s indeterminate up to a constant. I also don’t get how the wire connected to the negative end has more electrons. Whichever wire is longer has more electrons.

Basically the water analogy is fine. But you can think of one end of the battery at a height of 5 meters and one at 0 meters. If you don’t have a path from 5 meters down to 0 you don’t have any flow because water just sitting there at 5 meters or 0 meters is an arbitrary choice of units of gauge in mathematical language.

Quora really is the gift that keeps on giving.

The gravity version is fine but water will absolutely flow until it reaches the break in the path. If you think of the battery as a gravity pump on a lazy river, the pump raises the water from the bottom to the top and then the water flows down entirely due to gravity. If there is a blockage halfway down the river then water will flow until it gets to that point and once steady state is reached there will be no flow and all the water will be stuck at the top. Remove that blockage and you immediately get flow due to the gravitational potential being converted to kinetic.

Where the pump adds gravitational potential a battery adds electrical potential and that will result in a flow from a point of high potential to low potential. Having done some more research on it the electron density term I was using might have been a bit inaccurate (I'm still not entirely sure) but I am still confident in saying that connecting two wires with different electrical potential will immediately cause a current to flow between them. The rest of the circuit is only relevant for having created that situation and for determining what will happen in a longer time frame.

So you have an open wire connected to a 24 volt bus. You add another open wire segment to this you are saying you’ll measure current because it’s going from 0 to 24 V?

I’m saying you won’t because absolute energy levels are arbitrary and voltage is an energy level. It’s something like the line integral of the electric field on the path from A to B. It doesn’t make sense unless you are comparing it at 2 different points in the same circuit and your integral becomes a definite integral and any choice of arbitrary constant for the anti derivative cancels

. In the water analogy, if you have a cup that holds water on the ground, it won’t spill out if you slowly put it on a table even though it’s at a higher energy state now. For it to spill you have to create a difference in energy between the 2 ends of the cup and the results are the same whether you pour it out on the ground or on the table. Maybe we agree with all of this, in the electrostatic approximation?

Guys, anonymous Quora user who doesn't know any maths has just solved one of the millennium prize problems and here you are arguing about integrals!

You shouldn’t any help convincing people you’ve proved P=NP if you actually have. Just give the correct answers for various instances of 3-sat etc that are considered intractable with modern technology unless P=NP.

Damn, if only Quora OP had thought of that, he'd be a millionaire by now!

Essentially yes. Voltage is electric potential and in order to increase electric potential there has to have been a charge added to it and that happens in the form of an electrical current. Obviously the current is only transient if the circuit is not complete but in the time that the potential is being balanced there will be a measurable current.

One more attempt at a water analogy - consider two tanks that are identical and placed at the same height. There is a pipe connecting the bottom of the tanks together with a tap that can be open or closed. Before t=0 the tap is closed and tank A has been filled with water while tank B is empty. At t=0 the tap is opened and because the water in tank A has gravitational potential this causes water to begin flowing from tank A into tank B. The flow will continue until the gravitational potential of the water in each tank is equal. If A is constantly being topped up while this is happening (equivalent to a battery) then there will be a flow of water until both tanks are completely full and the water in each tank will have the same potential that the water in tank A had at t=0.

Well I’m pretty sure that is wrong but not sure how I can convince you. When a bird lands on a power line it goes from 0 V to 1000 V. But it doesn’t gain any charge from the wire. It just flies off and doesn’t get electrocuted when it lands on a 0 V piece of metal as it would if it had more electrons than before. Of course if you touch the bird from the ground with a metal poll you’ll both be electrocuted because you created a closed circuit from the 1000v bird to ground which now creates current through the bird . Perhaps someone else can chime in as this seems like a pretty elementary circuits 101 type thing.

Oh and I think your tank example is misleading because there water flows from the “battery” tank to the other one. But that’s not how electrical current works. The charges that power electronic devices are already in the wire before you hook it up to a power source. They are actually already moving also, but they can’t power anything because at any given moment the same number of electrons move in 1 direction as in the opposite. What the power source does is create a slight bias in favor of 1 direction so more charge goes one way than the other. But the actual charge was always in the wire the whole time.

Maybe I'm not explaining something very well but literally everything I've seen about this topic has agreed that there would be some amount of current flow - the argument is always centred around the definition of "on" because the induced current through the bulb would be tiny. For example the person who uploaded the "Veritasium is Wrong!" video that d2 linked responds in the comments that he agrees that there would be a "transient pulse across the bulb" but that it wouldn't light up.

Perhaps a dipole antenna is a more useful reference, since that is wholly reliant on current flowing through it to function at all, despite not being part of a complete circuit. It works by constantly switching polarity so even though is not a complete circuit electrons are moving back and forth and a current would be measurable. The switch/bulb circuit can be thought of a single switching of the polarity of an antenna.

As for the bird on a wire there are a few things going on. Mostly though the answer is that there is a current, it's just tiny because the bird has a very low capacitance. Additionally the constant charge/discharge due to it being AC means the bird doesn't ever actually store much electric charge so when it subsequently lands on the ground there's just the same tiny current again.

An inverse example is touching a car and getting a slight shock. There is no complete circuit because of the insulation from the rubber tyres but the car has built up electrical potential and when you touch it that potential causes a current to flow through you to the ground. I wish I'd thought of this one earlier actually because it seems to be pretty much exactly analogous - current flowing from high to low potential without a closed circuit.

Dipoles and stuff are electrodynamics and more complicated than what we are talking about which is electrostatics. Do you disagree with this explanation of why birds don’t get electrocuted? It doesn’t say anything about capacitance or AC vs DC.

https://sciencedemonstrations.fas.ha...nsmission-line

It says the same thing I said. The bird can be at 20000V with no current as long as there is no path to ground. If it was at 20000V because it gained charge how does it ever get back to 0 V after flying away without discharging and dying?

A 10kV discharge isn’t necessarily lethal.

I think this might be the whole issue to be honest. As I understand it electrostatics are about systems that are at steady-state. That's simply not the situation in the moments after the switch is closed.

Similarly that explanation is fine from an electrostatics perspective because at steady-state with a DC voltage there is no current flowing in the bird. However it is a simplification that (probably deliberately) doesn't go into the more complicated effects that arise from things like any object having a capacitance/ability to store electricity and it being an AC transmission (so the bird is constantly charging/discharging). Also while it's quite rare birds absolutely can be electrocuted by the capacitive charging effect alone when landing on a wire if the bird is large enough and/or the voltage is high enough.

Getting a shock from a car seems like the perfect analogy to me. In fact since the charge being positive or negative is irrelevant you can even consider the ground/person as the battery/connected wire and the care as the extra piece of wire attached when the switch is closed. When you touch the car/close the switch there is a transient flow of current because of the difference in potential.

A bird on an AC wire is not constantly charging and discharging. When the link I posted talks about attaching both ends of a bulb to an AC hot wire it simply doesn’t come on. It it was constantly charging and discharging it would.

Charging is not a synonym for having a current running through it. In this context the bulb charging does not mean the same thing as the bulb being powered/on. The former is talking about electrical energy flowing to and being stored in the object, the latter is talking about it drawing power from a current flowing through it. In that example the bulb connected to the same wire will absolutely be charging/discharging (if it's AC) but it won't be drawing more than a trivially low current to do so (as it has a tiny capacitance) and won't turn on.

The relevant equations for the capacitance and current drawn by the bulb charging/discharging in this example would be (modelling the bulb as a sphere with radius 0.05m):
C = 4pi * (vacuum permittivity in pF/m ~ 8.854) * r ~= 5.6pF
I = 2pi * frequency of AC * C * V ~= 253uA
so we're talking fractions of a milliamp required for it to charge/discharge but it does draw a non-zero current.

So instead of just a bulb put a capacitor and a bulb and connect it back to just the hot end. You are saying it will light up now because you can make the capacitance as large as you want? But if that works it should also work on the ground wire also because the voltage drop from hot to hot is the same as from ground to ground.

I'm sort of out of my depth here, but I think what Willd is saying is that a naked length of wire hooked up to a terminal of a battery is in some sense a different "state", or, I guess "energy level" or, "potential" or whatever than the same naked length of wire that is not hooked up to a terminal of a battery - and e d'a, you are saying that as long as there is no closed circuit, it matters not whether one end of it is hooked up to a battery or not. Is that correct?

If my understanding above is correct, I am really tempted to agree with e d'a here just intuitively, but I have no knowledge or understanding to back it up.

Actually, I mean, thinking about it, what about a Van De Graaff generator? What about a Faraday cage? Don't both of those exhibit properties of non-closed circuits? Or am I just getting confused again?

Not exactly my wheelhouse either, but this is a decent restating. Mathematically it’s definitely true, just by how potential is defined as an integral. But it’s possible that doesn’t tell the whole story in real situations so when you talk about real wires that aren’t just perfect conductors etc. But on something like a bird sitting on a high voltage DC wire, which is sort of the simplest static case, I’m not really seeing any answer other than everything is at 1 voltage so you don’t have a circuit to generate current.

It's not really my area either and I think I got ahead of myself with some of the bird on a wire type stuff (trusting one possibly dodgy source and applying some formulas that work for capacitors in series rather than parallel).

My hang-up on the current in AC systems is that there must be some physical manifestation of what it means for an object to go from being at +120V to being at -120V. I've messed about with a couple of online circuit simulators and they seem to indicate there is always some amount of current flowing through anything with capacitance connected to an AC power source (even in an open circuit) but it's generally in the nA or pA range unless at least one of the voltage/capacitance/frequency are huge.

I still think that the car example is the closest analogy to the thought experiment that started this whole thing though (at least in terms of whether it's possible for there to be a measurable current at the switch immediately). It seems to me like that is a fairly definitive example of a current resulting from a difference in potential without having a complete circuit.

Actually, I think you might be right but the explanation you have isn’t the dominant answer. If we consider just a bird on a DC wire, there is a closed circuit created by current going up one leg, through the body and down the other. Cables don’t have 0 resistance so the 2 feet of the bird aren’t at exactly the same voltage. The current will be very small since the resistance of a bird is so much higher than a few inches of cable, but non zero. I think this will be larger than any capacitive charging effect. So the bird isn’t “charged” it just has the same voltage drop across it as a few inches of wire, which is close to 0.