How Does Motional EMF Affect Electron Distribution in a Conductor?

[sophiecentaur]

Okay, let's try this one:

View attachment 86841
I charge up a capacitor (value discussed below) to 1 volt and then insert it as shown in the circuit above. The resistor is 1 ohm, so the current is initially 1 amp. A larger capacitor will maintain the flow longer, but will not increase the voltage. The voltmeter shows the voltage starting at 1 volt and dropping from there. It seems to me:

1. That the voltage between A and B is directly related to the ratio of excess/deficit electrons on the top/bottom of this circuit. Yes?

2. That the number of excess/deficit electrons may be substantial. For example, if the capacitor is large enough to sustain a 1-amp flow for 1 second, then the number of excess electrons must have been at least 6.2e18 at the start. Yes?

3. That a larger capacitor can maintain the 1-volt reading longer, not because the ratio of excess/deficit electrons is different, but simply because it takes longer to move enough electrons to significantly change that ratio. Yes?

So far so good? I hope so. Now for the $64,000 question:

4. What's the relationship between the various ratios of excess/deficit electrons that exist in this circuit at various points in time, and the voltage readings we see on the meter?

That's a really good example of how an alternative, personal model of Electrical Circuit Operation really doesn't help. If you want to talk about electrons in this context then you have to talk Quantum Mechanics and Statistics. This is why EE is taught in terms of Charge and Current. Only when you are fully conversant with the standard approach to EE theory can you afford to get into the questions you seem to be wanting to ask for starters.

 

[Gerry Rzeppa]

I am truly sorry but I find that statement ["The ultimate goal is to explain the workings of a basic vacuum tube guitar amp circuit to a ten-year-old in electron-flow terms."] very disturbing. You are here on PF trying to get together a very personal view about Electricity, which is, from what I have read, full of misconceptions and you are actually contemplating passing these on to a tender young brain...

On the contrary, my questions here are an attempt to clarify my understanding so I don't pass misconceptions on to the little guy. The way you and others can help is to simply answer the questions I ask, in the order asked, stopping (with an explanation) when an error is found.

This is the kid we're talking about (he's a bit older now):

This is the "No-Solder Banana Jack Amplifier Kit" I've developed for the course:

This is our text ("The clearest and most entertaining introduction to basic electronics ever written," https://www.amazon.com/dp/0962781592/?tag=pfamazon01-20):

And the four questions I'm working on at the moment appear below:

I charge up a capacitor to 1 volt and then insert it as shown in the circuit above. The resistor is 1 ohm, so the current is initially 1 amp (assuming a sufficiently large capacitor). The voltmeter shows the voltage starting at 1 volt and dropping from there. It seems to me:

1. That the voltage between A and B is related to the ratio of excess/deficit electrons on the bottom/top of this circuit. Yes?

2. That the number of excess/deficit electrons may be substantial. For example, if the capacitor is large enough to sustain a 1-amp flow for 1 second, then the number of excess electrons must be at least 6.2e18 at the start. Yes?

3. That a larger capacitor can maintain the 1-volt reading longer, not because the ratio of excess/deficit electrons is different, but simply because it takes longer to move enough electrons to significantly change that ratio. Yes?

4. What's the relationship between the various ratios of excess/deficit electrons that exist in this circuit at various points in time, and the voltage readings we see on the meter?

[Please refrain from replying unless you have specific answers to the specific questions asked; extraneous posts only make it difficult for others to find the most recent unanswered questions. Thanks.]

 

[sophiecentaur]

1. That the voltage between A and B is related to the ratio of excess/deficit electrons on the bottom/top of this circuit. Yes?

Q=CV

That the number of excess/deficit electrons may be substantial. For example, if the capacitor is large enough to sustain a 1-amp flow for 1 second, then the number of excess electrons must be at least 6.2e18 at the start. Yes?

Q=∫Idt over the limits t →0

That a larger capacitor can maintain the 1-volt reading longer, not because the ratio of excess/deficit electrons is different, but simply because it takes longer to move enough electrons to significantly change that ratio. Yes?

4. What's the relationship between the various ratios of excess/deficit electrons that exist in this circuit at various points in time, and the voltage readings we see on the meter?

Q=Q₀ Exp(-t/RC)

If you approach it this way then the verbal description / arm waving can be checked against the formulae. Science without some Maths is, imo, a waste of time. The ancients found this during the Enlightenment period.

 

[Gerry Rzeppa]

Q=CV... Q=∫Idt over the limits t →0... Q=Q₀ Exp(-t/RC)... Science without some Maths is, imo, a waste of time...

Thanks for the specific answers. You are obviously an intelligent, educated, and well-spoken person. Unfortunately, your answers are of little use to me. I'm not looking for formulas that are beyond the kid's reach. I'm looking for plain English descriptions of how the electrons behave in the circuit, and how that behavior is related to what the kid sees on the scope and his multi-meter.

I understand the critical role of mathematics in science -- I have a degree in mathematics myself. But having spent most of my adult life teaching children (I'm now 62), I have developed a firm conviction that the human brain learns best when it is fed pictures before words, and words before formulae.

If you approach it this way then the verbal description / arm waving can be checked against the formulae

Actually, the "checking" should go both ways. Errors can easily slip through the cracks between pictures and words, to be sure. But formulae can also be easily misunderstood and misapplied (as when a kid thinks there cannot be a potential difference across an open switch, because V = IR, and if there is no I there cannot be any V; or when a kid agrees, on theoretical grounds, that the current leaving a light bulb has to be equal to the current entering the bulb, but argues that the current in the bulb filament itself has to be much less, since V = IR, and the bulb filament has more R than the neighboring wires so it must have less I). It thus takes all three to really get an idea into a kid's head: pictures, words, and formulae -- in that order. I'm working on the pictures and words here.

Want to try again?

I charge up a capacitor to 1 volt and then insert it as shown in the circuit above. The resistor is 1 ohm, so the current is initially 1 amp (assuming a sufficiently large capacitor). The voltmeter shows the voltage starting at 1 volt and dropping from there. It seems to me:

1. That the voltage between A and B is related to the ratio of excess/deficit electrons on the bottom/top of this circuit. Yes?

2. That the number of excess/deficit electrons may be substantial. For example, if the capacitor is large enough to sustain a 1-amp flow for 1 second, then the number of excess electrons must be at least 6.2e18 at the start. Yes?

3. That a larger capacitor can maintain the 1-volt reading longer, not because the ratio of excess/deficit electrons is different, but simply because it takes longer to move enough electrons to significantly change that ratio. Yes?

4. What's the relationship between the various ratios of excess/deficit electrons that exist in this circuit at various points in time, and the voltage readings we see on the meter?

 

[sophiecentaur]

Unfortunately, your answers are of little use to me. I'm not looking for formulas that are beyond the kid's reach.

I'm looking for plain English descriptions of how the electrons behave in the circuit, and how that behavior is related to what the kid sees on the scope and his multi-meter.

Well, this is the problem. You are trying to tell people about things that are bound to be beyond their understanding (which is why such things are usually taught to more mature students). You seem to think that talking in terms of 'electrons' makes the subject more accessible. Bearing in mind that they (and even you) have a limited appreciation of the behaviour of electrons - particularly when they are in a metal, I suggest that an alternative description is less likely to confuse their take on Physics. I could suggest the term "Charge" would be good, for a start. It has satisfied Engineers and Scientists at all levels for generations. If you insist on presenting your own model then you are 'on your own' (also true when people try other personal approaches to Science). The very fact that you are asking for step by step help in justifying your model, shows that it is not in a state to present to school kids.
You may feel that my reaction is over the top. But I have spent many hours of 'remedial' work to put right the misconceptions that A level students have gained from their previous Science teaching in lower school. Kids need consistency. Please wait until your ideas have been truly tested against accepted theories before you unleash them on young minds. It is essential that you should learn the basics of what you present thoroughly. Until then, I recommend that you stick to the limits of the curriculum (flawed as it may be).

What they see on a scope is a nice concrete experience for them, which they may remember for the rest of their lives. Formal thought does not develop for several years and the concepts involved in Electricity theory are very very Formal. If you think that your method will work then perhaps you should devise a test, to determine just what they have gained from your intended lessons. The results you get back could be revealing.

 

[Jeff Rosenbury]

Q=CV means the charge (in this case electrons) is equal to the voltage times some value we call the capacitance. The capacitance is defined by this equation. Now you might question just what capacitance is, but this equation is correct -- by definition.

If you start with a capacitor at 1 V through a resistance of 1Ω, an instant later, the voltage will no longer be one volt and the current will no longer be 1 Amp. That should be obvious to someone with a degree in mathematics. Since you have been teaching children for many years, I can understand your calculus being a bit rusty, but even so you should be able to grasp that if Q=CV and Q drops while C remains constant, V must drop as well. And since V=IR, and V is dropping (with R remaining constant) I must drop as well. So there is no capacitor value large enough to do as you claim.

In Electrical engineering we use a concept called the RC time constant. Basically the value τ=RC gives the time for a capacitor to discharge by ≈63%. This acts as a half-life function (actually a 63%-life function) with the capacitor discharging an additional 63% every τ. Tau can be made larger by increasing either R or C or both. But under nearly all circumstances the voltage and current start to fall instantly. (The major exception is when the resistance is variable such as in a transistor, tube, or other gain providing device.)

Finally, isn't there some unwritten rule (or perhaps a written one) about posting the pictures of minors on public forums? This kid could grow up to join a biker gang and be terribly embarrassed about his good grades as a child. The internet is forever.

 

[sophiecentaur]

Q=CV means the charge (in this case electrons) is equal to the voltage times some value we call the capacitance

I deliberately left out the 'explanations' to my equations because, without knowing what they satnd for and all the associated background, trying to compose a model of Electricity is a hopeless task. Scientists use equations because that is basically the only way they can communicate and make progress. A waving arm can cover less than one metre squared.

 

[sophiecentaur]

Finally, isn't there some unwritten rule (or perhaps a written one) about posting the pictures of minors on public forums? This kid could grow up to join a biker gang and be terribly embarrassed about his good grades as a child. The internet is forever.

I think it's a matter of honi soit qui mal y pense, quite frankly. We spend too much time second guessing things these days. That picture was a breath of fresh air afaiac.
If he joins a biker gang he can always say "My name is Sue, how do you do."

 

[Gerry Rzeppa]

You are trying to tell people about things that are bound to be beyond their understanding...

I think that describes pretty much everything. We don't fully understand how sunflowers grow, but that didn't stop the kid from growing one this summer and from grasping the "great circle of life" -- seed to plant to flower to seed, nor did it stop him from grasping the fact that the sunflower must be made mostly of water and air since the seed was tiny and the flower big and we didn't see that much decrease in the amount of soil "consumed." I'm pretty sure a guitar amp can be explained at that sort of level.

You seem to think that talking in terms of 'electrons' makes the subject more accessible.

Yes. No kid I have ever known has had trouble with the electron-flow description of a vacuum tube. But lots of them have trouble with the proper application of Ohm's Law (as illustrated in my previous post).

Bearing in mind that they (and even you) have a limited appreciation of the behaviour of electrons - particularly when they are in a metal, I suggest that an alternative description is less likely to confuse their take on Physics. I could suggest the term "Charge" would be good, for a start.

If I was interested in making things up, I'd simply stick with Amdahl's "greenies."

The very fact that you are asking for step by step help in justifying your model, shows that it is not in a state to present to school kids.

Granted. But that doesn't mean I should abandon it. It means I should keep pursuing it until either (a) it is ready for the kid, or (b) I become convinced that it's making things harder rather than easier. So if you're really interested in the kid's welfare, and you think abandoning the electron-flow perspective is the thing to do, help me get to (b) as quickly as possible -- by simply answering the questions I'm asking. I may be more unconventional and persistent than the average person, but I'm neither stupid nor stubborn.

Please wait until your ideas have been truly tested against accepted theories before you unleash them on young minds.

As I said before, that's exactly what I'm doing right here, right now.

If you think that your method will work then perhaps you should devise a test, to determine just what they have gained from your intended lessons. The results you get back could be revealing.

I have performed a preliminary test. I gave my wife (who knows nothing about electricity) a standard basic text on electrons, and Amdahl's book. She quickly rejected the standard text, but couldn't stop reading Amdahl's work. And, thanks to that, she is now able to understand these posts when I print them off and discuss them with her. So it seems to me that Amdahl was on the right track when he wrote his book. I'm simply trying to fill in a couple of the blanks in his presentation.

Scientists use equations because that is basically the only way they can communicate and make progress.

That's just silly. Anything that can be expressed in a formula can be equivalently expressed in a sufficiently complete natural language. The one may be more concise and convenient at certain times, but the reverse may be true at other points. Sometimes a picture (or a formula) is worth a thousand words, sometimes it's the other way around. Pictures, words, and formulae are all necessary for effective communication. Consider, for example, this page from one of Einstein's notebooks, where he was trying to get his thoughts down in the most concise and effective manner:

Note that he naturally and intuitively uses all three tools.

 

[sophiecentaur]

Anything that can be expressed in a formula can be equivalently expressed in a sufficiently complete natural language.

Now that really is silly. You clearly don't appreciate the advanced Maths that is necessary to describe even a simple servo loop. The fact is that Maths is just the complete natural language that's called for.

Einstein's name frequently comes up as someone who did things just by intuition and who couldn't do Maths. Both are misconceptions. He used rigorous Maths to arrive at his conclusions. I can't read that attachment because it stays the same size whatever I do but his note books were intended for his personal use and you can have no idea about what messages were there for him to understand.

How would you know how much your wife 'understands' of these posts if she has rejected standard texts in favour of a very entertaining book? She is not in a position to understand anything abut the validity of what Amdahl's book says and it would be unfair to subject her to the sort of questions that even a simple electrical circuit would present. Entertainment is not synonymous with basic learning - even if it is fun.

You seem to think that your alternative approach to things is valid and sufficient, which suggests to me that you probably have not had a vast experience of the conventional approach. This boring and lacklustre regime has been responsible for all the advances in the Physical Sciences and in Engineering since Gallileo. That sort of convinces me that it has a certain worth. What evidence do you have that your approach could achieve the same?

 

[Gerry Rzeppa]

If you start with a capacitor at 1 V through a resistance of 1Ω, an instant later, the voltage will no longer be one volt and the current will no longer be 1 Amp. That should be obvious to someone with a degree in mathematics. Since you have been teaching children for many years, I can understand your calculus being a bit rusty, but even so you should be able to grasp that if Q=CV and Q drops while C remains constant, V must drop as well. And since V=IR, and V is dropping (with R remaining constant) I must drop as well. So there is no capacitor value large enough to do as you claim.

Excellent point. I stand corrected. I was only trying to establish that the number of electrons we're talking about may be in the 10¹⁸ range and not mere millions, even in ordinary circuits. I'll reword that point after I get answers to some similar questions regarding a simpler configuration, below.

The pictures below show two capacitors (little and big) in various states of "charge". (The blue dots are not individual electrons, but large yet equal masses of free electrons; the arrangement of the dots is not literal, but figurative for easy counting):

The caps on the left (little and big) are uncharged -- the number of free electrons is equal top and bottom. The caps in the middle are charged, but not to capacity. The caps on the right are more fully charged, but still not quite to capacity. The charging process, it can be easily seen, is a matter of pulling electrons from the top of each capacitor and and pushing an equivalent number of electrons into the bottom of each.

Based on those pictures, I think the kid will be able to see:

1. That the voltage between the top and the bottom of these capacitors is directly related to the ratio of free electrons on the top and bottom of a cap. (The ratio, not the absolute number of electrons: the bottom left drawing has more electrons on the bottom, and twice as many altogether, than the top right drawing; yet the voltage is greater in the top-right capacitor because the ratio, bottom-to-top, is greater.) Yes?

2. That it takes more electrons to charge up a large capacitor to the same voltage level as a small capacitor. A large capacitor will therefore discharge more slowly than a small capacitor, all other things being equal. Yes?

3. That for any given ratio the voltage will be the same, no matter how big or small the capacitor. And vice-versa: for any given voltage, the ratio will be the same, no matter how big or small the capacitor. Yes?

So we've got some nice pictures and some meaningful words to go with them. What's needed is a formula to complete the set. We know that q=C/V (correction V=q/C), but based on point (3) above, it seems we ought to be able to relate q to V without mentioning C. The NEETS (Navy Electricity and Electronics Training Manual) says the relationship is this (http://www.tpub.com/neets/book1/chapter1/1k.htm):

""A difference of potential can exist between two points, or bodies, only if they have different charges. In other words, there is no difference in potential between two bodies if both have a deficiency of electrons to the same degree. If, however, one body is deficient of 6 coulombs (representing 6 volts), and the other is deficient by 12 coulombs (representing 12 volts), there is a difference of potential of 6 volts. The body with the greater deficiency is positive with respect to the other."

My problem with that statement is that it doesn't appear to take the ratio idea into account: it would make the large cap in the center of my drawing have a voltage equal to the rightmost small cap (since both have a difference of 4 globs of electrons).

Amdahl, I think, explains it better:

"Voltage is the reason electrons move... It's a percentage thing. It's the ratio between two charges. If there are 10 times as many electrons one place as another, it will have certain voltage, whether you're talking about 10 electrons or a million electrons."

Thoughts?

 

[analogdesign]

Amdahl, I think, explains it better:

"Voltage is the reason electrons move... It's a percentage thing. It's the ratio between two charges. If there are 10 times as many electrons one place as another, it will have certain voltage, whether you're talking about 10 electrons or a million electrons."

Thoughts?

Are you sure this is right? Lots of practical circuits work on the basis of charge balance (or charge sharing). Imagine if you had an isolated capacitor of 1F with 1V across it. By definition the charge Q stored on the capacitor is 1C. Now if you magically moved the plates closer by a factor of two, now the capacitor would have a capacitance of 2F and the voltage would decrease to 0.5V since Q couldn't have possibly changed (remember this system is isolated). Where is the
ratio between two charges?" There is one charge. The charge on the capacitor.

Since you brought up Einstein earlier remember it was he who said "Everything should be made as simple as possible, but not simpler". The concepts of charge, current, and voltage are quite simple and they work. Very well.

 

[davenn]

We know that q=C/V,

No, that is incorrect, you missed what was told to you several times
Q=CV ( Q = C x V)

I have been reading through the thread several times ...
You came asking for help, and people have given you the better way to do this teaching to the kids,
but you still want to ignore that and go with your own way which, to be honest, I find more confusing than the correct way

Take a step back and a deep breath
and consider a better way, which won't lead to further confusion for them later in life if they decide to pursue electronics

Dave

 

[Gerry Rzeppa]

No, that is incorrect, you missed what was told to you several times
Q=CV ( Q = C x V)

You're right, I typed that in wrong. I've corrected it above.

You came asking for help, and people have given you the better way to do this teaching to the kids,
but you still want to ignore that and go with your own way which, to be honest, I find more confusing than the correct way.

I think, by "correct," you mean usual. Many people agree that the usual ways don't work as well as they should. Here are two interesting papers on the subject:

http://files.eric.ed.gov/fulltext/ED287730.pdf
http://www.matterandinteractions.org/Content/Articles/circuit.pdf

Clearly, some change of approach is called for.

 

[Gerry Rzeppa]

Imagine if you had an isolated capacitor of 1F with 1V across it. By definition the charge Q stored on the capacitor is 1V. Now if you magically moved the plates closer by a factor of two, now the capacitor would have a capacitance of 2F and the voltage would decrease to 0.5V since Q couldn't have possibly changed (remember this system is isolated). Where is the ratio between two charges?" There is one charge. The charge on the capacitor.

Excellent! What you've done (by magically moving the plates closer by a factor of two) is to change one of my little capacitors into one of my big capacitors -- but without changing the number of electrons. So the ratio stays the same. Like so:

So it's clearly not: (a) the absolute number of electrons that determines the voltage; we eliminated that earlier.

And now -- thanks to your little thought experiment -- we can say that it's not just (b) the ratio of electrons in one spot versus another.

So what is it? What's changed? Seems to me it's the concentration of electrons that has changed -- the number of electrons per cubic inch, or the number of free electrons per proton, or something like that. (There's obviously more room in the bottom of the larger capacitor and it thus would take less energy to cram five extra globs of electrons in there. Or, after the fact, we could say that the electrons on the right bottom are less compacted than the ones on the left bottom, and thus the mutual repulsion between them is less. Either way, the voltage will be lower.)

So it seems, then, that voltage must be (c) a measure of the relative concentration of electrons in two spots. Better?

 

[sophiecentaur]

I think, by "correct," you mean usual. Many people agree that the usual ways don't work as well as they should. Here are two interesting papers on the subject:

http://files.eric.ed.gov/fulltext/ED287730.pdf
http://www.matterandinteractions.org/Content/Articles/circuit.pdf

I had a quick read of those two papers - written by Educationists rather than Scientists, I think.
I was amused that a suggested experiment with a single filament bulb and with two in series, ignored the fact that the resistance of bulbs changes significantly with brightness (temperature). That's just the sort of howler you get when a rigorous treatment is avoided. Also, there was a really ham-fisted discussion of electrons flowing through a metal with a serious innacuracy in reasoning.
So I think that neither of those papers can be relied on to get it right in other matters.

Clearly, some change of approach is called for.

Did you mean 'your' approach or the approach of the teaching establishment?
I would always agree that improvements could be made and the pendulum seems to have swung a long way away from the rigorous approach to study. I believe there is some recent movement to a more academic system. This, I think is due to the fact that employers want to employ graduates who actually know their stuff in the hard disciplines and who can produce useful results rather than waving their arms about.
But, until they pay for a better informed and qualified class of teacher, the average standards will remain mediocre.

 

[sophiecentaur]

So it seems, then, that voltage must be (c) a measure of the relative concentration of electrons in two spots. Better?

Or Q = CV ?

 

[Jeff Rosenbury]

Your diagram lacks protons. The idea that negative charges attract positive charges is critical and lacking.

I do like your idea of redefining physics to make it easier to learn rather than be correct. Hollywood writers make far more than engineers. Clearly the value to society by physics is far less than simplicity and a good story.

[For the humor impaired: Sarcasm.]

 

[analogdesign]

So what is it? What's changed?

What's changed is the voltage, obviously.

Remember the electric field E = V/d. We didn't add or remove charge, so E is constant. We did reduce d by 1/2. Therefore V reduces by one half.

 

[sophiecentaur]

What's changed is the voltage, obviously.

Remember the electric field E = V/d. We didn't add or remove charge, so E is constant. We did reduce d by 1/2. Therefore V reduces by one half.

Please don't cloud the issue by introducing correct Science.
PS Those poor children.

 

[Gerry Rzeppa]

I do like your idea of redefining physics...

I'm not trying to redefine physics; I'm simply trying to illustrate some aspects of electronics in a manner similar to the way the operation of a vacuum tube is invariably described (ie, from an electron concentration/flow perspective).

What's changed is the voltage, obviously.

From the electron concentration and flow point of view, the change in voltage is an effect, not a cause. Bear with me...

Remember the electric field E = V/d. We didn't add or remove charge, so E is constant. We did reduce d by 1/2. Therefore V reduces by one half.

Your diagram lacks protons. The idea that negative charges attract positive charges is critical and lacking.

Yes. We need a simple way to include those aspects in our picture so that the student will be naturally led to correct intuitive inferences. I've got some things I need to attend to right now. Let me think about this a bit. I'll post a (hopefully) improved diagram later.

 

[Jeff Rosenbury]

Yes. We need a simple way to include those aspects in our picture so that the student will be naturally led to correct intuitive inferences. I've got some things I need to attend to right now. Let me think about this a bit. I'll post a (hopefully) improved diagram later.

And don't forget some moving positive charges are virtual/imaginary. For example, in P-type semiconductor material the lack of electrons moves (rather than the electrons simply moving away) even though the protons are locked in a lattice. How do we know this? The Hall effect.

I'm curious how you explain nothing as causing voltage? Me, I study a (very) little quantum theory. But explain it to a ten year old? Not without lots of loaded dice and a stacked deck of cards on hand. (Adult probability with calculus is better, but what kid knows that?)

The basic problem is that the universe is not simple. This shouldn't come as a shock to people, but apparently it does.

Instead of teaching all about electricity, how about teaching more math? Then when the child understands ratios and geometry advanced science will make more sense. But that would be hard and boring I suppose.

 

[Gerry Rzeppa]

Thank you all for your time and interest. I'm going to seek help elsewhere.

 

[davenn]

Thank you all for your time and interest. I'm going to seek help elsewhere

translated as

" I will go and find people that will agree with my misguided ideas, rather than listen to real science and learn from that"

very sad

Dave

 

[Drakkith]

I think this thread has had it. Thread locked before it degenerates into name calling and such.