The Electric field is conservative. This means that integrals are path independant. So any path will lead to the same result.

You don't integrate the field directly. You perform the integral for the work done in moving a (unit) charge from point A to point B.

The electric field (I prefer to call it a potential field) has an origin. this is normally used as our point of reference, common or ground point or what you will.

Somewhere back along there was another thread about all this, I will try to dig it out.

The thing to take away is that 'voltages' are voltage differences between two points or across a component. There is no such thing as absolute voltage.

 

If so what would happen if I disconnect the croc clip from the power rail ( but not from the battery) and touched it to the negative power rail? would I see sparks? (I would certainly see sparks if I touched a lead connected to the positive rail to the negative one.)

My thought is that there is so little current flowing for that split second, you won't notice a spark. If you changed your +60V rail to a +60 Billion V rail, and repeated the experiment, I think you would see a spark. You might even see a spark when you first hooked up the battery to the +60 Billion V rail.

These are the lines along which I am thinking: http://www.eskimo.com/~billb/emotor/stmiscon.html#nine

According to that article, an imbalance of charge between two regions means a voltage between them, right? I don't think an origin for the electric field matters since we're making all our measurements as differences (difference in charge, difference in electric field, and potential difference)

 

Have you answered all the questions in sequence?

 

Have you answered all the questions in sequence?

Sorry. Let me see if I can do this incorporating my understanding of static electricity.

Take a 9volt battery.
Connect the negative terminal with croc clips and a flying lead to the +60 rail of my power amplifier.

What is now the 'voltage' of the battery negative terminal?
What is now the voltage of its positive terminal?

The voltage of the battery's negative terminal is now +60V, the voltage of the positive terminal is +61.5V

Have I 'charged up the negative terminal' to this new voltage?
In other words has any current passed from the amplifier power rail to the battery?

I know you want me to answer "no" here, but let skip this one for now. We can come back to it later if you'd like.

If so what would happen if I disconnect the croc clip from the power rail ( but not from the battery)

Now the negative side of the battery is still at +60V.

touch it to the negative power rail?

At this point current flows. Granted, it's a super small amount of current, because the capacitance between the negative power rail and the negative side of the battery is so small. So, you don't notice any spark. But current did flow, for a moment. After this current flows, the negative side of the battery is at the same voltage as the negative power rail.

I realize the magnitudes of current we're talking about is worthless at these low voltages, but it seems to me that if you used a super high voltage, in theory you could get a spark with the experiment you've described.

Anyway, I hope I'm making more sense now. Don't get me wrong - I agree with all your points _completely_ from a practical level. I'm just trying to reconcile my understanding of static electricity with electric circuits (and this, I think, was at least a part of CanElec's original confusion)

Also, I forgot to mention:

The Electric field is conservative. This means that integrals are path independant. So any path will lead to the same result.

Completely correct. I suspected I was forgetting something as I was writing "shortest path", but it's been a while since my last physics class. oops!

 

I just realized something I've got incorrect that might be (at least part of) what is bugging you, studiot.

I've been thinking that a just the difference in charge between two objects dictates the voltage between them. This is plain wrong. The voltage between two objects is related to the electric field between them, which in turn is related to not just the charges, but their distance apart, medium that they're in, and the shape/size of the objects.

So, back to the two balls in space, one more charged than the other. These two balls have a voltage between them related to their charge, distance, etc. In fact, you could just think of the two balls acting as a capacitor - and you could measure their potential difference without a third reference point the same way the voltage between two leads of a charged capacitor can be measured without a third reference point.

Thanks again for bearing with me here, studiot - I feel like this conversation is really improving my understanding.

 

The voltage of the battery's negative terminal is now +60V, the voltage of the positive terminal is +61.5V

OK.

60 volts and 61.5 volts. (I presume you missed where I specified a 9v battery and thought it was a 1.5v?)

Suppose I asked you to measure this.

You would connect the red lead of your voltmeter to the terminal/rail junction.

Where would you connect the black lead?

 

Oops, you're right - silly me, I'm stuck in 1.5V battery land.

Lets say that our voltmeter is ideal, so it doesn't draw any current. Lets also say our wire is ideal with zero impedance.

So, using the nine volt battery -

I have the black lead of the voltmeter connected to the negative rail of the supply. It will be connected here for the duration of the experiment. I connect the red lead to the positive rail of the supply to measure +60V. I connect the battery's - lead to the supply's +60V. I touch the red lead to the battery's - lead and measure +60V. I touch the red lead to the battery's + lead and measure +69V.

Now here's the key. I disconnect the battery's - lead from the supply. I say that because of the capacitance between the battery's - terminal and the - lead of the supply, the battery's - lead will remain +60V. You can measure this with the voltmeter's black lead on the supply's - rail and the red lead on the battery's - terminal.

 

So you have connected the black lead to a reference point from which you are measuring!

And you move the red lead about measuring 60 volts, 69 volts etc etc, all referenced to the black lead conection.

Now let me tell you that I have secretly connected the point where the black lead joins the amplifier power supply to the positive of your mythical 60 billion volt supply.

Leave the black lead where it is and tell me what readings you get on your voltmeter with the red lead?

Now tell me why you think voltage is not always measured between two points?

 

Now tell me why you think voltage is not always measured between two points?

Woh there, I have never mean to convey that point. I agree - voltage is always measured between two points. So, of course, nothing changes our measurements when you reveal that our reference position is at 60 billion volts. Our problem here is looking more like a miscommunication than a misunderstanding:

One of the points I'm making is that you don't need a third reference point to measure the potential difference between two arbitrary bodies. In the context of our two differently charged balls in space example it means that you can measure the potential difference between them without connecting them.

Looking back at the conversation, I think the miscommunication started when I stated that these two charged balls are at "different voltages". Your response was that you need a third reference point in order to measure the voltage. And here's where our brains got on different tracks: I thought you meant that you need a third reference point to measure the potential difference or voltage between the balls - not that you need a third reference point to say what the voltage of each body is (relative to that reference point).

So when I was saying that these two bodies were at "different voltages", I was only concerned about the difference in voltages, which will be the same no matter which coordinate system you choose, as I'm sure you'd agree. So again, I did not mean to convey that I was thinking that the bodies themselves had some absolute voltage.

In the context of your battery/power supply example, here is the point at which I was thinking we disagree: When we disconnect the neg terminal of the battery from the +60V supply, I say that it is still at +60V (relative to the negative rail of the supply). I'm not sure what you would say the voltage of the battery was relative to the negative rail of the supply at this point. When we connect this neg terminal of the battery (at +60V), to the neg rail of the supply, I say that a small amount of current flows for a split second, until there is no potential difference remaining between the neg terminal of the battery and the neg terminal of the supply. From what I am understanding from your points, you're arguing that absolutely no current flows. Am I understanding you correctly?

The last thing I wanted to convey is that I think there is such thing as "absolute voltage". Let me quote myself from my first reply to CamElec:

Your problem is that you are seeing the positive and negative terminals of batteries as absolute voltages. But that's the problem - there's no such thing as an absolute voltage. Voltage is always measured relative to something else.

Anyway, I hope this reply clears up some of the confusion.

 

Looking back at the conversation, I think the miscommunication started when I stated that these two charged balls are at "different voltages". Your response was that you need a third reference point in order to measure the voltage. And here's where our brains got on different tracks: I thought you meant that you need a third reference point to measure the potential difference or voltage between the balls - not that you need a third reference point to say what the voltage of each body is (relative to that reference point).

So when I was saying that these two bodies were at "different voltages", I was only concerned about the difference in voltages, which will be the same no matter which coordinate system you choose, as I'm sure you'd agree. So again, I did not mean to convey that I was thinking that the bodies themselves had some absolute voltage.

Well I'm glad we finally got there.

I am attaching a potted summary of the nearest we get to defining 'absolute voltage'.
Do ask if that raises any questions.

 

Seeing the responses makes me feel a little better that I didn't ask a stupid question There's a lot of info to sort through here which will probably take me a while. Thanks for the help from all who contributed.

 

It's an understandable question, not a stupid one.

Ask if you are still not sure after digesting the stuff on offer.

 

Well I'm glad we finally got there.

Whew! Me too. Thanks for the attachment - I didn't think about defining "absolute voltage" in that way, but it all makes sense.

I'm still wondering about this part though:

In the context of your battery/power supply example, here is the point at which I was thinking we disagree: When we disconnect the neg terminal of the battery from the +60V supply, I say that it is still at +60V (relative to the negative rail of the supply). I'm not sure what you would say the voltage of the battery was relative to the negative rail of the supply at this point. When we connect this neg terminal of the battery (at +60V), to the neg rail of the supply, I say that a small amount of current flows for a split second, until there is no potential difference remaining between the neg terminal of the battery and the neg terminal of the supply. From what I am understanding from your points, you're arguing that absolutely no current flows. Am I understanding you correctly?

What's the proper way of thinking about that situation? Does current flow?

 

When we connect this neg terminal of the battery (at +60V), to the neg rail of the supply

I didn't ever say that you connect the negative terminal of the battery to the negative rail.

I did say you connect the negative terminal of the battery to the positive rail of the power supply. Since the terminal is a lump of metal (a conductor) it instantly attains the potential of the rail, at 60 volts.

No charge flows.

Since I didn't connect the positive terminal to anything it is left to 'float'. Since it is fixed at 9 volts more positive than the negative terminal it is forced to attain 69 volts, relative to the negative terminal of the supply.

To see why no charge flows look at the EFI calculation I posted.

Now ask yourself

If I place a lump of metal (a conductor) at distance r(1) from a charge, what is its potential?

Answer = Er(1) as in my calculation.

Now move the lump of metal to a new distance r(2) .

Have I expended any energy in electrical terms?

No, but the potential is now Er(2).

Since the potential has changed, no charge was involved or the electrical work done would not be zero.

ie work done = energy change = qΔE = 0

Since ΔE ≠0, q = 0

The situation is essentially the same for the battery connection to the power supply.

If you could continuously feed energy into one terminal of a battery by connecting to one terminal you would break the law of conservation of energy and have a perpetual motion machine.

If I moved a circuit in an electric field, eg a loop the answer would be different, as would the answer if I connected a circuit to the power supply rail.

And before you ask I should mention that I am ignoring the energy of the conductor acting as a generator by moving in a field - whereby some of the Kinetic energy of the conductor is turned into electrical energy. I am concentrating upon energy changes due to change in the field intensity due to position, which equal zero.

 

I didn't ever say that you connect the negative terminal of the battery to the negative rail.

I'm refering to this part of your experiment:

If so what would happen if I disconnect the croc clip from the power rail ( but not from the battery) and touched it to the negative power rail? would I see sparks? (I would certainly see sparks if I touched a lead connected to the positive rail to the negative one.)

No charge flows.

My problem is that I think you are ignoring the capacitance between the negative terminal of the battery and the negative rail. Can you come to the same conclusion while including this?

If I place a lump of metal (a conductor) at distance r(1) from a charge, what is its potential?

Answer = Er(1) as in my calculation.

Isn't Er(1) an instantaneous change in potential? Since you didn't supply a reference, I'll assume you're using your definition of "absolute potential". Then the potential at the lump of metal would be Vr(1). But in order to say that, you'd also have to assume that the lump of metal had no excess charge.

Since the potential has changed, no charge was involved or the electrical work done would not be zero.

ie work done = energy change = qΔE = 0

Since ΔE ≠0, q = 0

Shouldn't this be W = qΔV?

Also, when you say "no charge was involved", what you mean is that no charge moved through an electric field, right?

If I may, let me propose a similar thought experiment that I think will be easier to analyze, because we can talk about relative voltages, and we can use the simple parallel plate capacitor model.

If I have two plates that are of area A, charges ±Q, separated by a distance d and in a medium with elect. permitivity e, the voltage between them will be Qd/(eA). (http://en.wikipedia.org/wiki/Capacitor#Parallel_plate_model)

If I increase the distance between the two plates, the voltage between them increases. This checks out with the W = qΔV, because even though the voltage changed, no electrical work was done because no charges moved in an electric field (because the electric field on the outside of the two plates is zero in our idealized approximation because they cancel. See diagrams here: http://www.ac.wwu.edu/~vawter/PhysicsNet/Topics/Capacitors/ParallCap.html)

So now lets hook up the plates to our power rail. First, lets hook them both to the negative rail, to get them both at the same potential. Now, while leaving one plate on the negative rail, lets hook the other plate to the +60V rail. Current flows to put charge on this plate. After a moment, the potential difference between this plate and the negative rail is +60V. Now we disconnect the positive plate from the supply. The charge remains on the plate, so the potential difference between the positive plate and the negative rail is still +60V. Now we connect the positive plate to the negative rail. Current flows again, bringing the potential difference between the plates back down to zero.

This is why I'm saying that after you disconnect the negative terminal of the battery from the positive rail of the supply, it is still at +60V relative to the negative rail of the supply. The negative terminal of the battery is like one plate of a capacitor, and the negative rail is like the other plate of the capacitor. When you connect the negative terminal to ground, current flows.

I realize that for +60V this is all insignificant because the capacitance between the negative rail and the negative battery terminal is so small. But for different configurations, perhaps ones using high voltages, ones that are sensitive to really small currents, or ones where the negative terminal of the battery (or whatever) is really really close to the negative terminal of the rail, I think it might be necessary to take this sort of thing into account. I'm not sure how practical it is though, since my experience with electronics (especially with high voltages, small currents, and microscopic distances) is pretty limited at this point. I just want to make sure I'm thinking about this whole capacitance thing correctly, in case I run into a situation where I do have to take it into account (and even if I never end up using this knowledge for a real project, this understanding helps satisfy my intellectual curiosity).

 

Shouldn't this be W = qΔV?

Yes indeed it should. Sorry about that I was rushing to knock off for the night.

But it doesn't change the basic argument presented.


Let's start from something you know to be true.

You know that no charge or current flows when you touch a single terminal terminal of a battery to a metal object.

Why do I say this?

Well if you touch the battery terminal to a suficiently large current sink or lump of metal, you should be able to totally discharge the battery.
An earth is an infinite curent sink
When did you last instantaneously discharge a battery by touching one terminal only to earth?

Now to try to understand what is going on.

First forget capacitance. We are talking ideal components here. Using real components only complicates thigs until you have the basic mechanisms sorted. They do not change the basic mechanisms.

Second, my thought experiment involved only one lump of metal. This lump is neutral to start with.

So consider again a single isolated charge with a surrounding electric field in vacuuo.
Introduce a single lumple of metal at a distance r(1) from the charge.

Are you suggesting that the metal somehow acquires a charge by virue of its potential vis a vis the charge and its electric field?

If so are you further suggesting that by repeated withdrawing, discharging and reintroducing the same lump of metal you could either go on for ever dipping at the charge well or would somehow eventually deplete the isolated charge?

Back to my original experiment, if we simply move the metal to some other point does the charge on it change?

What would the equation be to calculate the change in charge as we moved the lump about?

It's very easy to talk about a flow of charge 'so small we don't notice it'. Put some figures to this statement.

Final thing to think about, since I see you are doing circuit theory in another thread.
If I connected one end of a resistor to the battery terminal, instead of my lump of metal, what would be the potential (voltage) at the other end?

I ask because this concept is very very important when appraising circuit diagrams.

 

Most decidedly not

Current sources have (theoretically) infinite internal resistance and maintain the fixed current in an external circuit, regardless of that extenal circuit's resistance, adjusting their own voltage to cause this to happen.

Voltages sources have (theoretically) zero internal resistance and maintain a fixed voltage across an external circuit, regardless of that extenal circuit's resistance, adjusting the current to cause this to happen.

Batteries are voltage sources.

how wrong can one be 100%?. You read but do not understand. CONSIDER THIS PUT A SHORT ACROSS A BATTERY you will measure zero volts and gabs of current. On the other hand ESD cannot biuld up if the internal resistance is zero can it in your book maybe not in my.

 

It's not to do with attraction.

It's that you haven't made a circuit.

A circuit is a complete path starting at one terminal of a battery and passing through a collection of components, external to the battery, to the other terminal.

If one of these other components is a second battery current (electrons) will indeed flow from the -ve terminal of one battery to the +ve terminal of the other.

But you must have a circuit for this to happen.

Remember also that conventionally we think of current as passing from +ve to negative. this is called 'conventional current flow'.

I IMAGINE A CIRCUIT TO BE A WIRE RIGHT?

 

I have attached an image. So if I have a circuit, then current can flow from the neg of one battery to the pos of another?

In the case of (A), current would flow through the load?

In the case of (B), current could flow from the negative terminal of (1) to the positive terminal of (2), going only through 2's load?

Believe it or not the higher battery will not only supply the load current to both load but also supply current to the low battery in your b circuit the a is just doing nothing

 

I am still confused as to why I cannot measure a voltage across the negative terminal of one battery and the positive terminal of another battery.

As far as I know, you don't need any current to flow to measure a voltage. So when you measure 9 volts across a (single) battery, no current is flowing, thus no chemical reaction is happening inside the battery.

If there is no change inside the battery, then why can't I read a voltage when I connect one probe to the negative terminal of one battery and the other probe to the positive terminal of another battery.

When I touch one probe of my meter to the negative terminal of a battery, all of the electrons start pushing against the probe. Now when I connect the other probe to the positive terminal (of the same battery), it has an electron deficiency, so it is attracting electrons. Yet no current is flowing, so it is simply this push or pull that is being measured by the meter. So why can't it detect this push or pull (now between two different batteries) between the negative terminal of one battery and the positive terminal of another?

You cannot measure voltage because there is just floating potential. The sky has millions of volts and millions of amps right but it is there . You cannot measure ESD not until there is cuurnt flow. ONE CANNOT ESISTS WITHOUT THE OTHER,