# Electrons inside batteries

1. May 28, 2013

### Q_Goest

Let's say you have 2 batteries, both 1.5 V. If you put your red voltmeter lead on the + of one battery and your black voltmeter lead on the - of the other battery, and assuming the two batteries are not touching, the voltage you read is zero. I know this is a simple question, but why? Why don't you read some voltage, say 1.5 V?

I've always thought of voltage as a potential field that is more negative on one end and more positive on another. I'd envision the battery having excess electrons on the negative end of the battery and they would be attracted to any more positive electrical potential, kinda like this picture:

http://www.qrg.northwestern.edu/projects/vss/docs/power/2-how-do-batteries-work.html

It seems like the electrons on the negative end of the battery should be able to flow to the positive end of another battery when given a conductive path, even when the batteries aren't connected - kinda like lightning. But clearly that isn't the case. So why don't the electrons go from one battery to another unless they're connected?

Last edited by a moderator: Apr 30, 2017
2. May 28, 2013

### Forensics

Well, the little that I know about batteries is that they do not really generate current until there is a need for them to do so. They maintain a state of balance until there is an actual demand placed on them, forcing the chemical reaction to begin and current to flow again.

For example, putting multiple batteries in series will not cause current flow because each battery, assuming they are all new or equally used, because they will be in a state of balance. It is only when you place a demand on these batteries that their balance is upset and the chemical reaction in the dry cell batteries starts.

The chemical reaction will continue until the device is turned off (balance) or the chemicals within the battery become depleted and are no longer able to react (and create a flow of current.)

Also, I believe there is no flow because the voltmeter is not placing a demand on the batteries, it is not trying to draw current. The voltmeter does not upset the balance within the batteries as far as I know and if it does it is not a measurable amount. This keeps the voltmeter accurate when measuring voltage.

Anyway, I may be completely off and I would love to be corrected if that is the case. I learn best when people correct me. :)

Last edited: May 28, 2013
3. May 28, 2013

### Averagesupernova

I always like to answer questions like this with another question. Why would you pick 1.5 volts? Why not 1000 volts? Or if your reason is that because you have a pair of 1.5 volt batteries then my question would be what if you have on battery that is 1.5 volts and the other is a 6 volt battery? So what does this tell you? I guess it is my way of pointing out the fact that you are missing some criteria to get an answer to your question.
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The short and direct answer is that electrons CAN flow for an instant because the two batteries with an insulator in between form a capacitor very similar to how layers in the atmosphere and the earth form a capacitor. Lightning is always for an instant right? Same thing. But to get continuous current, there has to be a conductive path which your diagram shows.

4. May 28, 2013

### davenn

Hi Q_Goest

both the other guys missed an important part of your statement

it should be pretty obvious why you are not measuring any voltage
... There's a hole in the circuit between the 2 batteries so no current can flow ;)

if that was the case, then we would never need wires to connect anything up.
you havent given a conductive path

Lightning is a whole different ball game, you need to liken that to EHT voltages 220kV etc transmission lines, where the electric field is high enough to cause ionisation and a discharge across a significant gap

Dave

Last edited: May 28, 2013
5. May 28, 2013

### Q_Goest

Thanks. Voltmeters actually draw current right? But isn't a voltmeter a device with a very high resistance and the higher the resistance the better? Ultimately, an infinite resistance would be best, right? So in that case, there is no current draw like there is with a conventional voltmeter, yet you should still measure voltage, right?

6. May 28, 2013

### davenn

only when its across opposite ends of the same battery ( or batteries that connected)

Dave

7. May 28, 2013

### Q_Goest

Sorry, I don't see the point regarding different voltages. I picked 1.5 because it's common.

If electrons flow for an instant, why not continuously until the battery discharges? Having the voltmeter on the two different batteries doesn't change the fact they both maintain the same voltage over time. So if the electrons flow for an instant, why not continuously until they've completely discharged and come to equilibrium like lightning?

8. May 28, 2013

### Q_Goest

Sorry, I'm still not getting it. We have a bunch of electrons here and a bunch of protons over there. We put a conductor (wire) between them. Shouldn't the electrons want to go over to the protons? I would have thought so. But because the electrons are in one battery and the protons in another, the electrons don't flow. If they were in the same battery they would flow. What's the difference?

9. May 28, 2013

### Forensics

I tidied up my original post, 3 hours of sleep is not working out for me.

10. May 28, 2013

### davenn

they need a path, you havent provided one. Touch those terminals of the battery together or put a wire between them. Then you have a path for the electrons (charge) to flow

Ohh also you dont have all the electrons in one battery and protons in the other battery

Dave

11. May 28, 2013

### Forensics

Without a load on the batteries, wouldn't there be no flow of electrons through the wire? This is what keeps batteries from draining exceptionally fast while turned off, I assume. Batteries will slowly lose their charge I admit, but I am referring to losing their charge at a rate that would be equal to if the device was on.

Also wow, I missed that point haha. I'm going to bed now. :D

12. May 28, 2013

### davenn

he's going to get a current flow through the meter which IS the load BUT only once the 2 batteries are connected ( in this case we are referring to which is in series)
Then and only then will he measure the 3V total of the 2 x 1.5 volt batteries in series

Dave

13. May 28, 2013

### Q_Goest

Put a wire between the positive of one battery and the negative of another - there's no current flow. Aren't there electrons at the negative end of the battery and protons on the positive end?

If you have a thunderstorm with lots of electrons in the air and another area is positively charged, then as soon as you put a conductor between them, don't the electrons flow? Why isn't it the same for two batteries? Remember, I'm suggesting we connect a wire from the positive of one battery to the negative of another. That's all. No other wires. But the electrons don't want to flow. Why?

14. May 28, 2013

### 256bits

You are forgeting about the chemical reaction that goes on in the battery to continously supply electrons to the negative terminal.

15. May 29, 2013

### SteamKing

Staff Emeritus
The voltmeter is a bit of a red herring. What happens if you put the positive terminal of one battery against the negative terminal of the second battery? Nothing, because you haven't provided a complete circuit in which the current can flow.

16. May 29, 2013

### davenn

yup

which is what we have been trying to convince him of ;)

Dave

17. May 29, 2013

### willem2

What happens with a battery is that you remove electrons from one side, but you don't add electrons to the other side. The entire battery will get a positive charge, and this will produce a field that will prevent any more electrons from leaving the battery.
The potential from this field will be the charge that has flown out of the battery, divided by the self capacitance from the battery, a few picofarad (10^-12) at most.
Once this is equal to 1.5 V, the current will stop. So we have Q = C V ~ 10^-11 coulomb,
so you could have 1 Ampere current for 10^-11 seconds, or 1mA for 10^-8 etc.

18. May 29, 2013

### Q_Goest

Guys, I understand that you need a complete circuit. That's the answer to the wrong question. The question is why does the circuit have to return to the original battery? Or why do electrons have to go from one side of a battery to the other side of the same battery? Why can't they go to another battery? Isn't there a potential difference between the ends of the same battery? If that's the case, then isn't there a potential difference between the end of one battery and the end of another one? Why/why not?

19. May 29, 2013

### Q_Goest

Hi willem. I think you might have something here but I'm not understanding it. Yes, if we moved electrons from one battery to another then one of the batteries would become negative and the other would become positive. Perhaps that's stopping the flow? Could you elaborate?

20. May 29, 2013

### Staff: Mentor

Each time you have separated charges there exist a potential difference, doesn't it?

When you connect these batteries there is some initial electron flow, but it stops as soon as the charges build up. When the charges build up, you have a potential difference that counteracts battery ability to build the potential difference. Net effect is zero.

-[]+ : single battery, with charges build up on both ends, these charges are pretty small, but large enough so that we can measure the potential difference.

-[]+[]- or +[]-[]+ : two batteries, with charges build up on the ends and in the place where the batteries touch each other. There is a potential difference between every + & -, but both ends (+ & + or - & -) are charged in such a way potential difference between them is zero.

I feel like in this case as long as the circuit is open, for a simple phenomenological analysis batteries can be modeled as capacitors.

21. May 29, 2013

### Q_Goest

Ok, thanks Borek. That kinda makes sense.

Another question then, is there something you could do that would drain a battery without having the electrons leaving the battery, go back to the battery at the opposite pole? In other words, can you isolate a battery and discharge it by removing electrons from one pole and adding other electrons from a different source to the opposite pole? Or is the only way to get these electrons out of the battery to provide a closed circuit?

22. May 29, 2013

### Staff: Mentor

Closed circuit is the simplest way. In a long term most (if not all) batteries will discharge on their own, but it is often process quite different from the intended one. You can also break the battery and mix the chemicals so that they can react directly (electrons move from one molecule to the other directly, not taking a detour through the external circuit).

23. May 29, 2013

### Q_Goest

Thanks, but what I'm really interested in is confirmation of the separability principal of classical mechanics which would say (assuming the phenomena exhibited by a battery are classical which I have to believe they are) that the battery can be isolated from its surroundings and made to act as if it were not isolated from its surroundings. That means it should be possible to remove electrons from the negative end and put others back in at the positive end without having a closed circuit.

24. May 29, 2013

### Staff: Mentor

Doable, but technically challenging. We are talking about supplying and removing charges measured in kilocoulombs for a 1 Ah battery (AAA type). Separating such charges (which would be necessary if we don't close a circuit) means dealing with insane forces (think levitating the Great Pyramid of Giza).

25. May 29, 2013

### Averagesupernova

1.5 volts is common only in your mind when using 1.5 volt batteries. It has nothing to do with the voltage that MAY exist in the setup you describe. You could take 2 metal conductors that are in no way shape or form batteries and do the exact same thing. Voltage is a relative thing. Voltage is often described as a difference in potential. It is in no way implied for how long. I could take your 2 batteries and space them several inches apart by insulators and hook a 5000 volt supply as you describe hooking the voltmeter and then remove it. Hooking a voltmeter to them would give you a reading of around 5000 volts for an instant until this homemade capacitor discharges. It would be such a brief amount of time it would be difficult to measure but that is irrelevant. However, if you measured the voltage of each individual battery you would find they are 1.5 volts before or after you hooked the 5000 volt supply to them.
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The reason I try to answer your question with several other questions is to try to get you to think in a certain manner.

Last edited: May 29, 2013