current returning to a different source

[rattus]

Correct, except that the above formula is an approximation that ignores 'fringe effects'. Fringe effects are significant extending outward from the edges of the plates by about the plate separation. When the plates are infinitely far apart, then all you have are fringe effects.

-Jon

Sure Winnie, but the fringe lines are even longer than the straight lines. "d" is still infinity, therefore "C" is zero.

I see what you are trying to do with Crossman, but I won't spoil his fun--not yet anyway.

 

[winnie]

Rattus, with you I was just trying to make the point that a single electrode in free space, infinitely far away from anything else, has a defined capacitance. This defined capacitance means that if you move a unit charge to this electrode, its potential difference relative to your reference zero will change by a definite amount.

See:
http://hyperphysics.phy-astr.gsu.edu/hbase/electric/capsph.html#c2

-Jon

 

[jghrist]

In Diagram 1, the positive terminal of the source has pulled some electrons out of the earth and has sent some electrons into the powerline. This is current flow due to the capacitance between the earth and the wire and this current could not exist without the electrostatic field that is set up between the earth and the wire.

Well, I guess you could look at it this way, but it seems bass ackwards to me. To me, the source is causing the current flow, not the capacitance. The source, let's say a generator, converts mechanical energy into an electromotive force that pushes the electrons away from the negative terminal. The electrons set up an electrostatic field between the wire and earth. This electrostatic field exists across the capacitance, which I would look at a passive element.

As the distance between the chopper and wire diminish, there comes a point when the resistant of the air can no longer prevent the excess electrons in the wire from jumping into the holes in the chopper. Current flows from wire to chopper, and the chopper becomes the same potential as the wire.

And now, when the source changes to make the wire positive, the electrons are pulled out of the helicopter. As the voltage in the wire alternates between positive and negative, the electrons go in and out of the helicopter as it equalizes potential with the line. Actually, most of the electrons never actually have a chance to go in and out of the helicopter because they don't move that far before the voltage reverses. The electrons do move back and forth, an alternating current.

I like Winnie's explanation better. The current flows in the open circuit, creating a changing electric potential.

 

[crossman]

Well, I guess you could look at it this way, but it seems bass ackwards to me. To me, the source is causing the current flow, not the capacitance. The source, let's say a generator, converts mechanical energy into an electromotive force that pushes the electrons away from the negative terminal. The electrons set up an electrostatic field between the wire and earth. This electrostatic field exists across the capacitance, which I would look at a passive element.

Agreed. I should have been more precise with my wording. What I mean is that the source voltage causes the current flow but the capacitance allows the current flow, in essence, the capacitance completes the circuit.

I like Winnie's explanation better. The current flows in the open circuit, creating a changing electric potential.

I like mine.

 

[crossman]

The source, let's say a generator, converts mechanical energy into an electromotive force that pushes the electrons away from the negative terminal.

Also:

Something we often forget. The positive side of a voltage source PULLS electrons through the circuit just as hard as the negative side of the source PUSHES the electrons.

So in my example, I was trying to show that both the negative part of the circuit and the positive part of the circuit is what makes current flow to and from the helicopter. And whenever we have positive charges and negative charges seperated by an insulator, we have capacitance.

 

[jghrist]

Agreed. I should have been more precise with my wording. What I mean is that the source voltage causes the current flow but the capacitance allows the current flow, in essence, the capacitance completes the circuit.

That's better. I can wrap my mind around that. The capacitance represents the physical place to store the electrons that get pushed around.

If you had a very small capacitance between the helicopter and earth, then it wouldn't take very many electrons to be pushed into the helicopter before the voltage between the helicopter and earth matched the line voltage because V=Q/C. Without the "capacity" of the helicopter to accept electrons, there can be no current. I can buy that.

 

[crossman]

Now in retrospect, this post by TwinCitySparky was humorous and right on target! I didn't catch the significance on the first go-round.

This is all semantics

Very interesting thread though. To me, this is like a climatologist crashing a plumbers website to argue that there is "water currect" flowing around us continuously without pipes via evaporation. Clouds as storage units etc.

For some reason though, I think the plumbers would just accept this proven theory and go back to work. :roll:

We are a fiesty bunch arent we! :grin:

 

[ELA]

I think that most of us agree that a capacitor can be part of a closed circuit. A capacitor does not have actual flow of charge between its electrodes; instead it has a changing electrostatic field. So rather than saying 'current can only flow in a closed circuit', it is probably more accurate to say 'when current flows in an open circuit, a changing electric potential develops'. This changing electric potential acts to block the flow of current; the capacitor charges up and the current flow drops.

But if you had an _infinite_ capacitance, then current could flow forever through this 'open' circuit.

-Jon

Don't we often just refer this as the "capacitive reactance" to make our understanding easier? (With it understood that current doesn't actually flow through the capacitor.)
When you say "infinite capacitance" is that the same as saying a very large capacitor with 0 ohms capacitive reactance?

 

[Smart $]

I think you guys are going way too far up the theoretical ladder. The concept under discussion is much, much more elemental than what y'all are proposing.

Examine the circuit depicted below...

View attachment 1239

The hatched rectangle represents a substantially large "chunk" of nominally high conductive material. If switch 1 is closed, will the ammeter to its left register current flow? The answer is yes. However, the current is near instantaneous and will not be sustained. Now open switch 1 and close switch 2. Will the ammeter to the left register current flow? Again, yes. And again only instantaneously. The current in both instances is a result of equalizing electrical potential difference. Is this a capacitor... NO!

Now the difference between this circuit and the power line helicopter scenario is that the electrical potential difference is high enough to break down the dielectric property of air prior to physical contact.

 

[Smart $]

Exactly.

You are using a metal-insulator-metal chain, rather than an insulating belt.

This lets you get to higher voltage gradients, and is cleaner for operation in a vacuum chamber.

But it is the exact same principal of operation.

-Jon

A somewhat similar concept (which might be quite a stretch in some circles :grin: ) is used in some car audio power amplifiers, known as PWM (Pulse Wave Modulation) power supplies. TTBOMK, these utilize an electronically switched power signal sent through a torroidal transformer. Coupled with the high inrush current of the transformer, higher voltages than the supply or typical step-up transformer supplies are possible.

If I'm not mistaken, this might even be the method used in automotive inverters.

 

[crossman]

Smart $:

Concerning your diagram and statements:

I disagree. It isn't quite that simple.

1) Batteries operate through chemical reactions to produce an electromotive force. For this chemical reaction to take place, an electron can enter the positive terminal only if another electron leaves the negative termminal.

If my statement above is incorrect, then it would be possible to drain the voltage out of the battery by simply connecting one terminal to some huge metallic object. This is not what happens in reality.

2) Considering that #1 is true, if the ammeter near switch 1 reads a current, then the ammeter near switch 2 also reads a current. The reason why is explained below in #3 and #4.

3) With Switch 1 closed, switch 2 becomes a capacitor. Everything connected to the positive terminal of the battery becomes the positive plate. Everything connected to the negative terminal becomes the negative plate. The air gap in the switch is the dielectric. Of course, there is also capacitance across any of the air gap between the positve components and the negative components.

4) The metal body and circuitry all the way to the right hand contact of switch 2 will give up electrons to the positive terminal of the battery causing the ammeter at the top to register. The wiring and contact on the left side of switch 2 will have electrons pushed/pulled onto it, causing the bottom ammeter to read.

5) The 2 ammeters will read the same amount of current.

6) The conductive metal object does not act like some sort of "electron donor" unless the metal body is acted on by both the negative and positive parts of the electrostatic field.

Edit: Added Smart's diagram

 

[Smart $]

Smart $:

I disagree. It isn't quite that simple.

Ahh, but it is that simple.

You are too wrapped up in believing electrostatic fields plays a major role in this matter. Yes they exist, there is some capacitance in the open switch when the other is closed, but as Rick pointed out, it is essentially insignificant to the effect.

No, the ammeters will not read the same amount of current when one switch is open, the other is closed. Think in terms of free electron density. If this were a closed or loop circuit, that would be the case. But as long as one switch is always open, it is an open circuit and functions accordingly.

The biggest problem you are having is that this principle is not taught conventionally because it is trivial in most circumstances.

 

[crossman]

So then you disagree with my supposition #1?

 

[crossman]

And have you noticed since the engineers in this thread have basically described what is happening in the past few days, that rick has been MIA?

 

[winnie]

Hmm. Disagreement to go around It would be quite interesting to try Smart$ circuit as an experiment, but I can't imagine measurable results with reasonably sized components, voltages, and frequencies.

Smart$: of course your large hunk'o'metal is a capacitor. Current can flow into it, changing its potential, and current will continue to flow as long as there is a driving electric field to push the electrons. The larger your hunk'o'metal, the more charge it can accept for a given change in potential.

Crossman: a battery works by having two separate reactions going on at the positive and negative terminals. One of these reactions will only complete if it accepts electrons; the other reaction will only complete if it rejects electrons. The common thread is that ions need to move between the two electrodes, taking part in both reactions. For example, one reaction might reject a pair of electrons into the external circuit, leaving a positive ion floating about, and the other reaction might consume a positive ion and require a pair of electrons from the external circuit. The positive ions diffuse through the electrolyte from the source electrode to the consuming electrode.

The electron producing reaction could occur without the electron accepting reaction...but this would cause the negative electrode to take a stronger negative charge, holding the positive ions from diffusing to the other electrode. If you somehow had a method for preventing the change in potential (say by connection to a large hunk'o'metal) then the reaction could continue until an excess of positive ions builds up, again stopping the reaction.

You can think of the battery as a capacitor with inverse leakage. There is some potential for a bit of charge movement from one side only, but very quickly the movement of charges will cause the reaction to stop.

So I can see how current would flow over time in Smart$' circuit, though I wonder if it would be more fair to describe the system as a multi-electrode capacitor where the battery electrodes are also part of the capacitor electrodes.

-Jon

 

[Smart $]

I almost responded to Crossman's post but noticed you had already.

Hmm. Disagreement to go around It would be quite interesting to try Smart$ circuit as an experiment, but I can't imagine measurable results with reasonably sized components, voltages, and frequencies.

Yes, that does present a challenge in itself.

Smart$: of course your large hunk'o'metal is a capacitor. Current can flow into it, changing its potential, and current will continue to flow as long as there is a driving electric field to push the electrons. The larger your hunk'o'metal, the more charge it can accept for a given change in potential.

If this hunk'o'metal is considered a capacitor simply by the fact current can flow into and out of it, I would most certainly have to revise my understanding and definition of a capacitor. I think of a capacitor as two conductive bodies intentionally separated by a dielectric specifically for the purpose of capacitive effect, and not where conductive elements simply exhibit a capacitive property.

 

[crossman]

Smart$: of course your large hunk'o'metal is a capacitor. Current can flow into it, changing its potential, and current will continue to flow as long as there is a driving electric field to push the electrons. The larger your hunk'o'metal, the more charge it can accept for a given change in potential.

Winnie, some "yes or no" questions:

1) Considering the term "driving electric field" that you used, does this "driving electric field" have both a positive side and a negative side?

2) In Smart's diagram Don't both the positive side of the battery and the negative side of the battery play a role in strippng electrons off the metal body?

The electron producing reaction could occur without the electron accepting reaction...but this would cause the negative electrode to take a stronger negative charge

I'm still going to say that on a scle greater than one or two electrons, if we have electrons leaving the negative terminal, there will be electrons entering the positive terminal, even if these electrons are pulled out of the very terminal itself.

So I can see how current would flow over time in Smart$' circuit, though I wonder if it would be more fair to describe the system as a multi-electrode capacitor where the battery electrodes are also part of the capacitor electrodes.

Exactly!!!

The electrostatic field is not some insignificant theoretical intity that we can just ignore. It is an EXTREMELY important aspect of electrical phenomena. Capacitance is everywhere.

If a voltage source has a positive terminal and a negative terminal, BOTH terminals play a role in any electron movement outside of the source, because there is an electrostatic field set up between the terminals even without anything connected to them.

 

[winnie]

During freshman physics we derived the capacity of an isolated sphere. I posted a link to a similar derivation earlier in this thread. Not a practical capacitor; a sphere the size of the Earth has a capacity of 700uF.

-Jon

 

[crossman]

But that capacity is in relation to the universe as a whole?

I don't know guys. I thought I was on target earlier, but now I am wondering what each of us is thinking.

Are we simply arguing semantics?

I am going on record that the electrostatic field which exists between the positive terminal and the negative terminal* of a voltage source will be the very driving force that will push/pull electrons onto or away from remote bodies, thereby giving them an electrical charge.

*or any extension thereof such as a wire connected to it

 

[rattus]

I think you guys are going way too far up the theoretical ladder. The concept under discussion is much, much more elemental than what y'all are proposing.

Examine the circuit depicted below...

View attachment 1239

The hatched rectangle represents a substantially large "chunk" of nominally high conductive material. If switch 1 is closed, will the ammeter to its left register current flow? The answer is yes. However, the current is near instantaneous and will not be sustained. Now open switch 1 and close switch 2. Will the ammeter to the left register current flow? Again, yes. And again only instantaneously. The current in both instances is a result of equalizing electrical potential difference. Is this a capacitor... NO!

Now the difference between this circuit and the power line helicopter scenario is that the electrical potential difference is high enough to break down the dielectric property of air prior to physical contact.

Smart,

As the schematic is drawn, there would be no current. Now if you add the parasitics to make it complete, then there is the possibility of current. You would also have to specifiy initial conditions.