# thought experiment on battery terminals and capacitance

Discussion in 'Physics' started by meemoe_uk, Aug 1, 2012.

1. ### meemoe_uk Thread Starter New Member

Jul 31, 2012
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If you connect some copper wire to a 9V battery terminal ( but don't make a circuit ), we say the wire now has a PD of 9V compared to the other battery terminal. But does it look any different to it's 0V self under nanoscopic inspection?

To answer this I use the water pressure analogy.
- 2 identical volumes of water.
- different pressures.
-> The one with higher pressure than the other has more water, mass , density.
-> Similarly, the wire at 9V has more charge in it

Everything is a capacitor
The wire is a capacitor
When attached to the 9V terminal, it'll charge up
Also the other terminal is a capacitor, and it's charge will adjust according to the new capacitance of the 9V terminal and connected wire.

Use VC = q

So I reckon yes, the wire is physically and detectably different from its 0V self.

Also, in theory its possible to discharge\ use up a battery just by altering the capacitance of the terminals by connecting random items, e,g, a piece of wire, your finger, a coin, to one terminal but leaving the circuit open.
It would take a long time because the capacitance of such objects with the other terminal is so low hence a low charge hence a low energy usage to charge.

e.g.
I'm guessing by touching the 9V terminal with your finger you'd create a roughly 1 pico farad capacitor with your body and the 0v terminal.
By VC = q
9 * 1E-12 = 9E^-12 coulombs.
Energy required to charge capacitor ( human body )= C*V^2 = 1E-12*9*9 = 81E-12 J

This is very rough calculation. But the argument is a few pico joules are expended from the battery cell every time the capacitance between the 2 unconnected battery terminals changes.

Is this correct?

2. ### MrChips Moderator

Oct 2, 2009
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I don't think this is correct.

Let us say we had a very large piece of metal which in theory has a very large capacitance. If you connected this to one of the terminals of the battery would you discharge the battery?

I don't think so.

The earth has very large capacitance. Connect one terminal to earth. Will you deplete the battery?

3. ### MrChips Moderator

Oct 2, 2009
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5,366
Here is another experiment you can try.

Take a capacitor, say 1000μF electrolytic and charge it to say 5V.

Alternate between touching the +ve and -ve leads to earth ground, never shorting the capacitor.
Measure the voltage across the terminals with a DVM and see if you have managed to discharged the capacitor.

4. ### davebee Well-Known Member

Oct 22, 2008
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I think you're correct that connecting discharged capacitors could conceivably discharge the battery.

Mr. Chip's is also correct as far as his example goes in that charging a single capacitor once from a battery won't completely discharge the batttery.

But if he were to substitute connecting many discharged capacitors to the battery, letting them charge, then removing them, then yes, that would eventually discharge the battery.

5. ### MrChips Moderator

Oct 2, 2009
17,379
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I think you misread the question.

The capacitor is not being repeatedly recharged by the 9V battery.

The op is connecting a capacitor to only one terminal of the battery.

6. ### Wendy Moderator

Mar 24, 2008
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DC flow depends on circuits, a loop. Indeed, the definition of a circuit is a loop, when using its non-electronic definition.

Even a capacitor depends on both leads being connected to a battery to charge.

I see what the OP is trying to say, but it doesn't work.

Take a 10pf and a 1000µF cap and put them in series, then connect the capacitor to the battery. The 10pf gets the bulk of the charge, but it is tiny. If we were talking a wire in open air, it (the capacitance) is immeasurable. It would be hard to measure the amount of charge the 1000µF cap got through the 10pf cap, with an immeasurable capacitance it would be no charge.

Last edited: Aug 1, 2012
7. ### davebee Well-Known Member

Oct 22, 2008
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This is like physics vs. conventional electrical engineering.

You're right in that for all practical purposes, if you don't complete a conventional circuit, no practical amount of DC current will flow.

But the OP is correct in that any conductor has capacitance. If you connect a piece of metal to a battery terminal, a tiny burst of charge will flow onto that conductor until it is at the same potential as the battery terminal.

He's really not talking about DC flow; his discussion is more like the mechanism used by the electrophorus for transferring charge in that a conductor is able to pick up a collection of charge from another charged conductor.

8. ### Wendy Moderator

Mar 24, 2008
21,421
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I understood that, but the point is a really immeasurable capacitance might as well not exist. With fancy electronics it can be used, but only as a sensor.

9. ### meemoe_uk Thread Starter New Member

Jul 31, 2012
12
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thanks for the quick replies. Interesting.

The point I'm trying to get at, is that I would find it unsatisfactory that a open circuit wire at 9V is physically identical to itself at 0V. There must be some physical difference, without needing to reference it to another object, however slight.

Now I know the electro-magnetic force is extremely strong and the capacitance of a typical wire used to carry 9V is very small, so it would be sufficient to my mind by VC = q if the physical difference between 9V and 0V was only a dozen electrons missing.

>
The earth has very large capacitance. Connect one terminal to earth. Will you deplete the battery?

This is a fair point Mr Chips, and one I've mused over.
The answer is yes, but only by an amount on the same magnitude as would using a small piece of wire.
The crucial question I came to is :
Would any discharge into the Earth ( or any whatever object we're using as a capacitor ) effect a equal and opposite intake of charge at the other terminal? ( i.e. act like varying a typical variable electronic capacitor? )

Bear in mind the Earth is significant as a capacitor independently of the capacitor formed by unconnected battery terminals. So by adding the Earth we've now got 2 capacitors in the experiment.
The capacitor I'm talking about is the one that occurs between any 2 open circuit terminals with different voltages.

The above crucial question annoyed me. As I said at the start, I want to consider the voltage at a terminal independently from the other terminal.

So what if the 2 terminals were ideally EM sealed off so there could be no electro static force between the 2 terminals ( capacitor plates )?
Again, I'd like think they'd be a physical difference between a terminal of 9V and itself at 0V of a displaced few electrons.

But lets rephrase the orignal question...
Consider a normal circuit with just battery and capacitor. If the capacitance of the capacitor is zero, will there be any charge in the plates? the wires? If there is no charge in the wires, then by VC = q , there is no voltage, or at least voltage becomes meaningless.

Before I was modeling the terminals as capacitor plates and working out my logic from that, but now that the capacitor is null I need another starting point...
The other reference point I can start at is to consider the double layer at the electrode and electrolyte in one of the half cells in the battery. This division of charge creates an electrostatic field, and a voltage. This voltage ( energy per charge ) is in the form of electro-static potential energy ( the electrons are pulled away from their zero voltage positions and are held at a 9V position ).
It then follows that any object connected to the terminal will also have electrons pulled up from zero potential up to 9V.

So in conclusion, yes, an object is 'charged up' on contact with an isolated single battery terminal, even if it doesn't alter the capacitance between the 2 terminals.
So in theory, if you were to grasp a very high voltage wire you would notice an electric shock, even if you weren't connected to a circuit.
Note it would have to be very high. Even to 10kV, the energy required to charge a human capacitor is only 50 micro joules.

The crucial difference I see between this zero capacitance experiment and one where the 2 terminals form a capacitor, is that all the energy required to charge an incident object on one of the terminals, must be returned if the object is taken away from the terminal.
To visualize it, consider before the object is connected. The voltage in the battery pulls off electrons to 9V distance from their 0V position. When the object is connected, it effectively increases the electron reservoir over which the 9V pulls. The 9V pull is redistributed amongst the increased number of electrons, so that the ones already in the terminal have less pull than before. The only alternative I see to this is that increasing the free electron reservoir allows more electrons to pass over the double layer, increasing the voltage. But I don't think this is right, is a battery's voltage dependent on the size of the electrode? It would be equivalent to saying that connecting a battery terminal to a large lump of conductor will increase the voltage.

Is this correct?

perhaps it would be easier for me to study all this properly in a electrical engineering degree, rather than try and puzzle it out over a forum!

( yes I know most half cells are about 1.5 to 2 V, my hypothetical 9V half cell is washing over detail )

10. ### studiot AAC Fanatic!

Nov 9, 2007
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Why so?

I think you are mixing up the ideas of charge and potential, which are different things.

There is an electric potential which exists in the space between the battery terminals, in the space between two charged plates or other bodies or in the space around any charged body.

Introduce a conducting body into that space without making physical contact and the conducing body will acquire the potential existing in that space.

This is fundamental and measurable.

There has been no exchange of charge, and the body is the same.

This is, of course the same as saying that a football is the same ball at the top of a hill as at the bottom, although its gravitational potential is different in each position.

11. ### meemoe_uk Thread Starter New Member

Jul 31, 2012
12
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>
I think you are mixing up the ideas of charge and potential, which are different things.

Yes, with VC = q. Where's there's a voltage, there's a charge. Everything is a capacitor. But for typical 'non capacitor' components we don't usually bother considering the charge because most objects have such low capacitance, and the quantities involved are negligible wrt the rest of the setup. However it is there, it is very slight.

>There is an electric potential which exists in the space between the battery terminals, in the space between two charged plates or other bodies or in the space around any charged body.
Yes, i agree.

Introduce a conducting body into that space without making physical contact and the conducing body will acquire the potential existing in that space.
Not without it becoming charged. Only a charged body has potential in an electric field. Again the issue is about the charge being so tiny that it is usually negligible. i.e. in practice you can assume a conducting body will have potential in a electric field because it usually always has some finite charge ( i.e. it has a different number of electrons to protons ).

pressurized water is a better analogy to voltage than gravity. When closely inspecting 2 identical full water vessels, one with high pressure water, the other with low pressure, we see that the high pressure vessel has a few more water molecules in. It's the same with voltage. A high voltage wire has a few more charged particles in , by VC = q, and so has a charge and a electro static field.

Another way of expressing this point I'm trying to make :
For most practical purposes, voltage is relative, but in theory it is absolute.
i.e. in most practice the only significant aspect to voltage is the potential difference between objects. But in theory you could count the number of electrons and protons in an object and calculate an absolute voltage, i.e. weigh up the charge of an object, then use VC = q to work out its absolute voltage.

12. ### meemoe_uk Thread Starter New Member

Jul 31, 2012
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For the sake of correctness, I've got to correct a mistake I've made. There's an effect that may cause a neutral body to be attracted to one of the plates. In an electric field, opposite component charges in a body are pulled in opposite directions. For example the average electron in a body will be closer to a electrically positive plate than the average proton in the body. Therefore by coulomb force, the electrons will pull the body towards the plate with more force than the protons are pushed away, therefore a neutral body will be pulled towards a charged plate.

But I don't think this was important to what I was talking about before.

13. ### studiot AAC Fanatic!

Nov 9, 2007
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meemoe_uk,
I assumed you were seeking some sort of explanation when you made your original post.

However it seems you wish to teach Mr Faraday how to suck eggs.
He thanks you for your insight and wishes me to remind you that capacitance applies to dielectrics , not conductors.

In point of fact what will happen to the isolated conductor is that there will be an electric voltage of opposite polarity but magnitude equal to the local field induced in the conductor. So the net effect will be that the conductor maintains zero voltage relative to the outside world.

14. ### meemoe_uk Thread Starter New Member

Jul 31, 2012
12
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>capacitance applies to dielectrics , not conductors.
Are you aware of Van_de_Graaff generators? They use conductor material to store electric charge.
http://en.wikipedia.org/wiki/Van_de_Graaff_generator

And in general, everything has the capacity to store charge, not just dielectrics.

Also capacitors are typically made of 2 conductor plates, with a dielectric in between. The symbol for a capacitor is 2 conducting plates with air in between.

Only if the conductor was a perfect conductor, otherwise, it'll exhibit a negligible but non-zero di-electric effect, resistance, and capacitance.
Free charge in non-perfect conductor will come to rest at the near side to a positive exterior voltage and by the coulomb force this nearer charge will be attracted more strongly then the ions are repulsed, therefore the conductor will move closer to the exterior voltage.

I agree though, the free conductor is at zero volts. At least until it gets close enough to the voltage source to transfer charge.

Where did I say the free conductor was not at zero voltage?

15. ### nsaspook AAC Fanatic!

Aug 27, 2009
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Thought experiments are great because they almost always fixate on the "possible" not the probable. Due to random quantum fluctuations there is a negligible but non-zero chance that a 1968 pink convertible could materialize out of thin air in my driveway. The effects of a few electrons missing compared to the huge number of electrons available for conduction would be completely masked by background noise.

Free electron concentration in copper n = 8.5 × 10e28 per m3

16. ### meemoe_uk Thread Starter New Member

Jul 31, 2012
12
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Well done nsaspook, within your own context you're right.
However, wrt the thread subject, I think this is the pertinent reply.

OP " I want to look at an effect that is negligible for most practice. "
Reply " Well for most practice the effect is negligible, so problem solved "

17. ### studiot AAC Fanatic!

Nov 9, 2007
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Yes of course.

Are you aware that you cannot make a VDG out of a single conductor?

The system comprises a conductor, charge elevator and earth.

The capacitance exists in the dielectric between the earth and the charged ball.

This is inherent in the observation that you cannot have a voltage of one point.

The definition of capacitance is C = q/v and since v requires two points so does C and it also includes the space and material (dielectric between them)

18. ### Austin Clark Active Member

Dec 28, 2011
413
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I had the exact same thought experiment awhile ago. I knew that the wire connected only to the 9V terminal would also have that same potential, so I figured it would gain this potential by "taking on" a few electrons from the positive terminal of the battery. HOWEVER, this is not the case. When you've only connected to one terminal, NO current flows, because there's nowhere for it to flow to. I like to understand this by thinking of the battery as a closed system, with a specific number of charges (electrons), the problem is, how would you force electrons to be expelled from this system (creating an electrical charge)? If electrons DID flow without a complete circuit, and you connected the positive terminal to ground, the battery would suddenly take on a HUGE negative electrical charge, obviously this doesn't happen.

Basically, the complete circuit is required. I like to think that there is actually no potential on either terminal until the circuit is actually plugged in. If there's no flow of current (like when the battery isn't plugged into anything) then, using ohms law, 0A*ohms = 9v, and because nothing multiplied by 0 is going to be 9 volts, we look and find that the only way for the equation to work out is if V = 0.

EDIT: I just had a thought. Maybe a few charges DO flow, but not very many and not for very long because eventually the electrical charge equally opposes it.

19. ### meemoe_uk Thread Starter New Member

Jul 31, 2012
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>This is inherent in the observation that you cannot have a voltage of one point.

But objects aren't one point, and they have capacitance, and charge, so by VC = q , they also have voltage.
Perhaps this 'self voltage' of an object isn't much practical use in circuits but I think its valid.
Maybe it comes down to that objects are made of electrons and protons at distances from each other, and with distance and charge you can derive voltage, as is the case for single atoms with their electron energy ( voltage ) levels.

This is an interesting side topic from my original question.

20. ### nsaspook AAC Fanatic!

Aug 27, 2009
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The "self voltage" like other quantum interactions are useful but not at the circuit level directly but in the devices that use the effects. At the atomic level it's possible to tunnel across a otherwise impossible barrier because interactions occur within the wave probability space of matter. To utilize this in a circuit you must design a device with a region of space that in at least one dimension (thickness) is quantum scale, the other dimensions can be arbitrarily large.