It is just that some battery can be discharged to a very low level and recover. some battery cannot recover and they becomes hi inpedance devices. the voltage is there OK but without a load. as soon as you apply a load the voltage drops.

 

Italo, please go read a standard textbook and come back when you can contribute to the thread.

Please don't post erroneous information where beginners might actually believe it.

One wire is not a circuit.

A potential field can easily exist without charge flow (current). Many do.

Yes you can discharge a battery by shorting across its terminals. So what?

 

A battery does not have a current until it has a load. The chemicals are resting and are doing almost nothing in a battery if there is no load. When there is a load then the chemicals create a voltage and a current.

A capacitor stores a charge. It also does not have a current if it does not have a load.

 

Why do you expect to read any voltage since there is no current? we need both to measure anything.

 

Oh my, I didn't expect this to get so out of hand.

I'm not really sure where to begin with all the incorrect things that have been said.

The point that I was trying to make is outside of the assumptions and approximations that are made in basic circuit analysis. It is involving a non-ideal behavior of the system. So when you try to mix basic circuit analysis techniques with my point, you will get funny results. The points I'm making here are really more applicable in a physics forum. Let me echo studiot here: if you are just beginning in basic circuit analysis, this is way out of your league. I'm sorry; I didn't mean it to get that way, but I didn't realize the full implications of what I was saying when I made my first reply (and was evidenced by the fact that my post had some flaws because my understanding of capacitance wasn't fully developed).

So let me make some things very clear:

If you connect one end of a battery to a lump of metal or ground, the battery will not eventually be drained from current continually flowing into the metal or ground. The point that I have been very unsuccessfully trying to make is simply that if you connect two objects with different amounts of static charge, a small amount of current will flow for a split second and _stop_. So if the end of the battery is statically charged and you connect it to ground, current will flow for a split second and then stop. Like when you shock yourself with static electricity on a doorknob, a small amount of current flows for a fraction of a second. Current does not continue to flow - otherwise, you could get seriously electrocuted after petting your cat.

I'm sticking strictly with ohms law, that in effect says a voltage can only exist when a current is flowing through a resistance. V=I*R

In basic circuit analysis, you assume that things that are not connected have infinite resistance between each other. If they are not connected, no current flows. V = 0 * inf = indeterminate. Ohms law gives you no information here. As long as you have infinite resistance between two points, you can have whatever voltage between them you want and no current will flow. Take a battery for example: V = 1.5V, leads are not connected so R = inf. I = V/R = 1.5/inf = 0. You do not need current to have a voltage.

If you do not really understand the physics that studiot and I are using, please save your comments until studiot have come to an agreement. After studiot and I get on the same page, we'll be happy to help clear up your confusion.

---------------------------------------------------------

First forget capacitance.

studiot, my entire point here is based on capacitance. You have to include capacitance to understand what I am talking about. All your arguments are neglecting capacitance - in other words, you are saying that the voltage of an object can change instantly. If you keep this assumption and try to analyze the two plates of a capacitor, you will be forced to say that they are a nothing more than an open circuit: in other words, when you connect the plates of a cap to a supply, you have to say that their potential difference is instantly the potential difference of the supply. As I'm sure you know, this is not the case - it takes a finite amount of time for the potential difference between the plates of a cap to reach that of the supply. And all the while, current flows. Once the potential differences are equal, current stops flowing.

All I'm saying, is that this same behavior occurs when you move the connection of a metal object from the positive rail of a power supply to the negative rail of a power supply because of the capacitance between the object and the negative rail of the power supply. But since this capacitance is so small, for all intents and purposes the voltage changes instantly and no current flows.

 

Hobbyist,

Whilst Ohm's Law is widely applicable it is not appropriate in every situation. Devices or situations where it is applicable ar referred to as 'ohmic' .

One area where it is definitely not applicable is Electrochemistry. If you wish to know more I suggest you look up the following.

Standard elctrode potentials
Redox potentials
the Nernst Equation
Polarisation

 

Khusmann,

I was just wondering why it isn't possible for current to flow from the -ve of one battery to the +ve of another? And why when you measure the voltage between the two, you get zero? The -ve of one battery should still be attracted to the +ve of another I would think.

My remarks have been tempered with the fact that I have been trying to address the original question throughout this thread and provide an understandable explanation for Canelec's clear appreciation of the truth and quest for an explanation.

It is normal to simplify when seeking understanding of the underlying physical processes.

However if you really wish to introduce capacitance:

In my example where I introduce a single neutral object into the field surrounding an isolated charge the capacitance is zero since it involves ∫dA, which is zero around a point charge.

You have not yet explained how you think charges could appear on a neutral object, isolated from the rest of the universe?

Now introduce a second isolated object at a different position.

There will be a difference in potential between the two objects.
Charge will not flow or be separated.

Connect the bodies together and charge will indeed be separated (flow) until the porential due to the separation equals the potential difference in the original field.

Further, let the object have physical dimensions.
Then charge may well be separated within it, creating a dipole.
Much theory of molecular behaviour depends upon this.

Returning to the other examples.

If I connected one plate only of a capacitor to either one terminal of a battery or to one rail of the power supply and left the other plate to 'float'

What is the voltage across the capacitor?
What is the charge separated?
What is the work therfore done?

Answer in all cases zero

If I now disconnect the capacitor.

What is the voltage across the capacitor?
What is the charge separated?
What is the work therfore done?

Answer in all cases zero.

No charge has flowed into or out of the capacitor.


Refining my experiment with the croc clips, battery terminal and psu rail,
Connect a perfect voltmeter between the negative rail and the flying croc clip.

Initially the croc clip is free and the meter measures zero.
When I clip onto the psu rail it jumps to +60.
When I remove it from the rail it immediately returns to zero, before I clip onto the negative rail.

No charge flows

 

I'm not going to answer the other parts of your post yet because I want to zero in on one critical misunderstanding that I think will clear up the other parts.

If I connected one plate only of a capacitor to either one terminal of a battery or to one rail of the power supply and left the other plate to 'float'

This is not what I'm saying. The other plate is connected to the other rail of the power supply. Now, before you tell me that I am contradicting myself, consider this:

First imagine our setup with two the two plates - one plate is connected to the positive rail, the other plate is connected to the negative rail. The plate connected to the positive rail is +60V relative to the other plate. When I disconnect the positive plate from the supply, it is still at +60V. You agree up to this point right?

Ok, now reset the experiment. Two plates, one connected to the positive rail, the other connected to the negative rail. Start shortening the wire connecting the negative plate to the negative rail, moving the negative plate closer in distance to the negative rail, while leaving the positive plate where it is. As you do this, also imagine this negative plate shrinking in size until it becomes the negative terminal of the supply. Now imagine the positive plate changing shape until it becomes a lump of metal. There is still a capacitance between the lump of metal and negative rail of the supply just as there was between the parallel plates; it is just really really small.

So, when you disconnect the lump of metal from the positve rail of the supply it remains at that positive rail voltage, because of the capacitance between the metal and the negative rail of the supply - no different than the two parallel plates.

 

K,
Of course if you connect both terminals of a capacitor to a supply it will charge up.

You have a circuit !

The original poster only made one connection and I have been addressing this situation all along.

 

You have a circuit !

Well, I suppose you do, even though electrons aren't flowing between the plates. I suppose, since we agree on that, we are as much on the same page as we're going to get.

Anyone have any other comments?

 

A circuit is a path, starting at any point and passing through components, including batteries, earths and power supplies, switches and off-transistors, which eventually returns to the original point. some of these components may block or restrict DC or AC or some frequency.

If you are interested there is a 'current' which flows in capacitors. It is called displacement current as distinct from conductive or ohmic current, and can be shown to be necessary to complete a vector analysis of Coulomb's and Maxwells laws.

I do not have time to delve further tonight.

 

I'm on wikipedia now reading about displacement current. Interesting stuff! Thanks for pointing me in that direction, studiot!

 

This was a good, thought-provoking thread.

I agree with Khusman's points - all conductors have capacitance, defined as C=Q/V, so even an isolated conducting object such as a piece of wire or a human will experience charge transfer if it is connected to another conductor of a different voltage. However, as studiot points out, without a circuit connected to both ends of a battery, its 1.5 or whatever volt difference between its terminals from its internal chemical reaction will not maintain a voltage difference between the overall battery object and the external object, and doesn't contribute to the circuit when only one end of the battery is considered.

But actually, if you touch two conductors, allow their voltages to equalize, then separate them, the separated object will not necessarily maintain that voltage. What will be constant is the charge on the isolated object, not necessarily its voltage to some reference point. Think of a variable capacitor that has been charged, then separated from any external circuitry. Because V=Q/C, if you decrease the capacitance, you increase the voltage, since charge is constant.

In the real world there will always be some charge imbalance between any two insulated conductors, so touching them will always produce some charge flow until the voltages equalize. I don't think its all that small of a charge flow either - it only takes a few thousand volts from a Van De Graaff generator to cause a large spark to jump to a person, even if he's standing on an insulated platform.

 

wow, excellent summary davebee - you've really added a lot of clarity to the discussion.

I completely agree with everything you said there, and I feel like studiot and my points are now fully reconciled. I also appreciate the added info about the Van De Graaf generator, since it helps me get a better intuition for the relative magnitude for this phenomena.

thanks for your input!

 

Davebee,

all conductors have capacitance, defined as C=Q/V,

Perhaps you would like to elaborate. What charge and What voltage?

If I take a completely discharged 1 microfarad capacitor out of my components box, is it not still a capacitor despite the fact the Q = 0?

And what if I apply zero volts to it, does it now possess infinite capacitance?


K,

I'm on wikipedia now reading about displacement current. Interesting stuff!

The theory of displacement current can be pretty heavy. If you like I will post what I consider to be an accessible introduction.

 

OK, suppose I said that the capacitance of an object "may be measured" as C = Q/V
such that if a charge imbalance Q is applied to it, V is the voltage across it. So a capacitor is still "a capacitor" even if it contains no charge imbalance. Is that better?

 

V is the voltage across it.

Across what? This implies it has at least two terminals.

Q is applied to it,

Applied to what? One of the terminals, both of the terminals somewhere in between?

 

but it does have two "terminals". One is the conducting object, and the other may be
another nearby object, like the second plate of a conventional capacitor, or it may
be the rest of the universe, as for the case of an isolated object like a conducting
sphere in space.

And if you start with a neutral object, the only way to get an unbalanced charge is to separate some positive and negative charges, putting some on one terminal, so the other "terminal" is wherever the other charges went.

 

Exactly so Dave.

I'm not trying to get at you, just get you to think about more complete/accurate definitions please.

What you are saying sounds interesting, but needs amplification.

 

No problem; it's good to get at the heart of this question.

Going back to CanElec's original question, I think maybe it could be answered by not so much thinking in terms of voltage but of charge. The battery separates charges, after which there are electric field lines connecting those charges (and only those charges). Since the field lines due to the charge separation of one battery don't extend to the other battery, there is not a voltage between the two batteries due to each battery's internal battery chemical activity.

If each battery as a whole was given a charge, then there could be electric field lines (and a voltage) between the batteries, but independently from the battery chemical action.

(But studiot, please don't ask me to define electric field lines.)