current returning to a different source

[winnie]

[crossman's questions in post 191]

1-7: Yes
8: No
9: Unclear
10: Yes
11: Yes

I have to recognize that I have been conflating a physicist's definition of capacitance with an EE definition of capacitor. But while a single electrode can have capacitance, it takes two electrodes to make a capacitor. The two definitions are not inconsistent; if I have two spherical electrodes separated by a large distance, I can treat that as a capacitor. Furthermore, I can take the formula for capacitance of a sphere, apply it to the two electrodes, and then apply the formula for series capacitances, to calculate the capacitance of this strange capacitor. (And this is a strange capacitor; as the electrodes separate, the capacitance goes down to a finite, non-zero value; once the distance is large enough, the capacitance does not really depend upon distance.)

I'd like to propose another circuit, in the vein of your examples and Smart$'.

In this circuit, current will clearly flow each time the switches get cycled. Current flows out of the positive and negative terminals of the battery during each switch cycle, in an essentially balanced fashion. But with one switch open, when you close the other switch, you don't see current flow on the wires leading to the other switch. The current flow in different parts of the circuit is different. As I said in a previous post, over the portion of the circuit where the electric current is not flowing, the electric field is changing, in a way that accounts for the lack of current flow.

Back to your example, I believe that you are suggesting that the open switch itself will act as a capacitor, and that the entire charge flow to the large conductive object would be balanced by charge flow right up to the switch contacts. I believe that in fact the battery itself will act as a large electrode; and that the entire battery assembly will become slightly negative as the conductive body becomes slightly positive. I believe that the electric field over the entire circuit will redistribute once you close one of the switches.

-Jon

 

[jghrist]

3) Do any electrons move from point F move toward the Conductive body?

Electrons move from point F to displace those that the battery chemical action pushes to the negative side of the battery.

Note that to satisfy Kirchoff's point rule, "The algebraic sum of the current toward any branch point is zero", requires the definition of a current through the capacitance between F and G. I=C?dV/dt. Is this current real or does it exist only to satisfy Kirchoff? Who is this guy Kirchoff anyway and why do we have to satisfy him?

 

[Rick Christopherson]

Because of the ultra-high voltage, there will be a massive electric field around the power line radiating outward. A charged particle (aka electron) immersed in an electric field will have a force acting on the particle: F = Eq. Because these charged particles are within a conductor, if there is a force on them, they are free to move from that force. This will result in a bunching up of charge on one side of the helicopter. When the electric field reverses for the second half cycle, the charge will bunch up on the other side of the helicopter. Because we are close enough to the power line for arcing to occur, a lot of this charge movement includes the charge transfer to and from the line.

When I wrote this several days ago, you dismissed it and said the phenomenon was capacitive. You are trying to prove that a circuit exists instead of disproving my original premise that no circuit exists.

What is the E-field outward from a long conductor, and what impact would it have on the electrons in another conductive body? If I am wrong, it should be very easy to disprove my premise.

 

[crossman]

Jghrist: Thank you for your concise answers to my questions.

Looking again at question 10:

Concerning this experiment and initial conditions, is there any mechanism which could pull significantly more electrons off the conductive body than the number of electrons which left the negative side of the battery and traveled into wire G-H?

You answered: I don't think it is useful to talk about how many electrons are pulled off of the conductive body. The same number are pulled off as are pushed on the other side. At the end, the conductive body is electrically neutral.

So actually if you think about it, aren't you answering this one with a "no"? If the conductive body ends up neutral, then apparently there is not a mechanism in the experiment which could leave the conductive body with a positive charge.

BTW, I am in complete agreement that the conductive body ends up neutral. Of course this is based on my initial parameters and exclusions.

I agree with the rest of your assessment. We are pretty much on the same page.

 

[Rick Christopherson]

Who is this guy Kirchoff anyway and why do we have to satisfy him?

Kirchoff's Current Law:
This law is also called Kirchhoff's first law, Kirchhoff's point rule, Kirchhoff's junction rule, and Kirchhoff's first rule.

The principle of conservation of electric charge implies that:

At any point in an electrical circuit where charge density is not changing in time, the sum of currents flowing towards that point is equal to the sum of currents flowing away from that point.

It is this changing charge density that permits this current to flow without a return path without violating Kirchoff's first rule.

 

[crossman]

What is the E-field outward from a long conductor, and what impact would it have on the electrons in another conductive body? If I am wrong, it should be very easy to disprove my premise.

You say that an E-field from one conductive body can have an effect on electrons in another conductive body. You are absolutely correct.

Now, keeping your postulate in mind, go back and review the diagrams in Post #139 on page 14 of this thread. But don't just stop at the helicopter and the electrons being pushed around in it. You must also consider what the charge is in the earth. And you must consider that these charges combine to form the overall electrostatic field that the helicopter flies into.

Both the earth and the powerline affect the movement of the electrons in the helicopter. This is what you have been ignoring all along. You are not looking at the entire electrostatic field. You are looking at it only in the immediate vicinity of the helicopter.

My Physics texts explain that electrostatic fields point away from positive charges and point into negative charges. There are drawings which show how this is. The electrostatic field from a positive charge will extend to a negative charge. On this earth, and most likely in this universe, there are no totally isolated charged bodies. Electrostatic fields extend from positive bodies to negative bodies. So again, you cannot isolate the powerline-helicopter and base your opinion on that. You must look at the entire situation including the earth and any other bodies. They all affect each other.

The electrons in the helicopter which are repelled away from the negative powerline will have an electrostatic connection to the positive charges in the earth. You cannot ignore this.

Edit: typo

 

[crossman]

I'd like to propose another circuit, in the vein of your examples and Smart$'.

In this circuit, current will clearly flow each time the switches get cycled. Current flows out of the positive and negative terminals of the battery during each switch cycle, in an essentially balanced fashion. But with one switch open, when you close the other switch, you don't see current flow on the wires leading to the other switch. The current flow in different parts of the circuit is different.

Winnie, believe it or not, the circuit in your diagram is no longer a series circuit. It is a combination circuit and it is perfectly acceptable to have different current flows in different parts of the circuit.

Redraw the circuit and consider the capacitor voltages, and all of the ordinary laws for circuits still hold.

 

[Rick Christopherson]

Your pictures in that posting were actually on the right track. Take a look at picture #4 and think about a couple of things.

1. What impact is the helicopter having on the charge at the ground in this picture?

2. How does this picture change when the voltage swings to the next half cycle of the power line?

Remember, the probe has not yet touched the power line, so the probe is arcing.

 

[crossman]

1. What impact is the helicopter having on the charge at the ground in this picture?

The right hand side of the helicopter is relatively negative. The earth is relatively positive. There is an electrostatic field between the helicopter and the earth. There is also a potential difference between the helicopter and the earth.

2. How does this picture change when the voltage swings to the next half cycle of the power line?

All polarities are reversed. This includes the wire, both sides of the helicopter, and the earth. The earth-helicopter electrostatic field and the helicopter-wire electrostatic field are reversed. The current flow in the arc changes direction.

 

[Rick Christopherson]

The right hand side of the helicopter is relatively negative. The earth is relatively positive.

What about the left hand side of the helicopter, isn't that impacting the earth too?

The answer is, no. The net charge from the helicopter as seen from the ground is zero. As far as the ground is concerned, the helicopter isn't even there.

So given the capacitor theory, where would the capacitor connect to at the helicopter? Does it selectively pick just the left or right side? If it doesn't selectively pick one side or the other, then how can capacitance be the primary mechanism between the helicopter and ground?

I am not trying to be antagonistic. I am just trying to give you enough thoughts about this to steer you in the right direction.

 

[winnie]

Winnie, believe it or not, the circuit in your diagram is no longer a series circuit. It is a combination circuit and it is perfectly acceptable to have different current flows in different parts of the circuit.

I entirely agree (presuming that 'combination circuit' means the same thing as 'parallel circuit').

However I believe that the circuit that you drew is _also_ a parallel circuit. There is not only capacitance across the switch terminals, there is capacitance all over the system.

In other words, when you close the switch, electrons move from the large conductive body to the battery terminal. Chemical reactions happen, ions move about, etc. The net result is that the battery as a whole will take a slight negative charge.

I think that you want to ignore the capacitance represented by the battery (which is a conductive object of finite size), but I believe that for the example that you gave, which explicitly looks at distributed capacitances (the 'large conducting body') you can't look at only some of the distributed capacitances.

-Jon

 

[crossman]

I need to be clear about something:

Is the diagram below a "completed" circuit?

If the answer is yes, then I can continue with the discussion. If the answer is no, then we have been arguing semantics all along.

 

[zog]

Without going back to look does anyone remmember the OP question? I am guessing not. Everyone just admit Rick is right and move on.

 

[LarryFine]

I find it a bit hard to believe that the electrostatic fields as shown in Rick's Diag. 4 are 60Hz AC.

 

[winnie]

When I wrote this several days ago, you dismissed it and said the phenomenon was capacitive. You are trying to prove that a circuit exists instead of disproving my original premise that no circuit exists.

What is the E-field outward from a long conductor, and what impact would it have on the electrons in another conductive body? If I am wrong, it should be very easy to disprove my premise.

I have come to the conclusion that you are not wrong in the way that you are considering the E-field and the way that changes in the E-field permit current to flow in 'non-closed' circuits.

At the same time, you _are_ wrong in the way that you dismiss capacitance.

This is a semantic argument being made on both sides, I reconcile the above two statements by noting that _capacitance_ is all about energy being stored in the changing electric field created by moving charges.

For days I've been describing the _capacitance_ of an isolated sphere. Very early on in this thread, you suggested treating the helicopter as a 'reservoir of electrons', and I incorrectly dismissed that, focusing on capacitance. But when you consider a _single electrode capacitance_, that exactly describes a reservoir of charge; finite charge can be introduced or subtracted with a finite change in potential.

As I realized yesterday, two conducting spheres separated by a distance much larger then their diameter could be considered a capacitor (you have two electrodes with a space in between them). However the value of this capacitor would not change significantly with distance, as long as you maintained sufficient separation. This 'capacitor' could be reasonably approximated as two _isolated_ capacitances, using the equation for the capacitance of a sphere to figure each.

Consider a circuit consisting of two 100 meter spheres, 10000 meters apart, with a 1V RMS, 1 Hz AC source sitting between the two connected by wires to the spheres, in free space (no magnetic fields, no gasses or plasmas, etc.) How much current flows in the conductor right next to the AC source? Treat the system as a capacitor formed by the two spherical electrodes, or treat the system as two charge reservoirs with electrical potentials that depend upon the geometry and charges moved, and the answer will be the same.

Back to the helicopter: I claim that it makes no difference if you consider the helicopter to be a 'reservoir of electrons' or if you consider the capacitance between the line and the helicopter, in series with the capacitance between the helicopter and the rest of the universe. I absolutely agree that "_capacitive_ current flow" is an accounting trick to make Kirchoff's current law 'work'; where the change in the electric field is used to balance the 'missing' motion of charge. It is thus equivalent to say 'current is flowing in an open circuit with a change in electric field' or 'current is flowing in a circuit closed by capacitive current flow'.

-Jon

 

[winnie]

Oh, and for what its worth, the latest XKCD is right on target.
http://xkcd.com/386/

I've appreciated the discussion so far!

-Jon

 

[crossman]

Rick, here is your answer.

All of the following has come from my research in the physics texts.

This is a crude representation of the electrostatic field which exists around the powerline when the powerline is negative in respect to earth.

The electrostatic field points from the positive charge in the earth to the negative charge in the wire.

Any body which moves into this electrostatic field will have its positive charges moved along the electrostatic lines toward the negative wire, and the body's negative charges will move along the electrostatic lines toward the earth. This is why the left side of the chopper becomes positive and the right side and bottom become negative.

In the diagram below, electrons are repelled from point A leaving point A positive. Electrons are attracted to point B, making point B negative.

It all works together. You have to look at the entire thing. You can't single out the helicopter and wire and ignore the rest of the electrostatic field.

 

[crossman]

I find it a bit hard to believe that the electrostatic fields as shown in Rick's Diag. 4 are 60Hz AC.

Rick's diagram 4 was taken from a previoous post. In that post, it was stated that this diagram was from a precise instant when the 60 Hz sine wave voltage had negative polarity on the wire and positive polarity on the earth.

 

[rattus]

Crossman, your field line drawing is correct with the exception that the lines do not pass through the helicopter. They terminate on its surface and pick up again on the opposite surface. It all amounts to two caps in series. The relative field strengths and capacitance values do not matter. The structure is there, therefore it is!

There is stray capacitance everywhere and your diagram is, as you know, a gross simplification of actuality, but it demonstrates the principle.

 

[crossman]

Rattus, I agree. The electrostatic lines from the earth would terminate at the electrons on the right side of the chopper. New lines would extend from these electrons to the positive charge on the left hand side of the chopper, and then other lines would go from those positive charges to the wire.

I just didn't have the time or desire to draw in the tedious stuff.

But as you mention, it is representative although somewhat crude and elementary.