Not quite. On forum after forum I see beginners ask, "Do electrons accumulate in front of resistors?" and invariably the answer is, "NO!" My answer is: "Yes! The electrons that account for the current flow don't accumulate and are uniform everywhere in a stabilized circuit; but the electrons that account for voltage differences exist in greater and lesser concentrations all over any given circuit (and not just in capacitors).

Correct

And these accumulations are significant. Consider, for example, my earlier guitar pickup example (alternating current):

View attachment 88833

Here are my "electron maps" for time t1 and t2, respectively.

View attachment 88834
As you can see, there is a trickle of current that makes it through the resistor. But the music is conveyed to the amp not by this current, but by the much larger quantities of electrons that never make it through the resistor. It is these accumulations that account for the voltage swings that we see on the scope.

And until somebody can draw me a picture that makes more sense, "That's my story, and I'm stickin' to it."

I thank you all for your time, effort, and help.

Incorrect. The charges are insignificant. I still say you are unable to calculate the charge. If you can't calculate the charge, and if the charge matters, how could you design a circuit? The voltage matters, but the charges don't matter. One person's circuit might have 10 times more charge accumulation than another person's, but if the voltages, resistances etc are the same, then basically the same circuit operation is seen. The charges are incidental and are a response, not a cause.

 

The charges are insignificant.

I don't see how you can say that. Consider this circuit:


When the grid is positive, electrons cross from the cathode to the plate, exit the valve, and -- because they can't get through the resistor as fast as they are coming out of the tube -- there is a build up of negative charge at the point marked with the blue arrow. This buildup of electrons can be seen on a scope (the first half-cycle of the wave shown) and is typically on the order of 50 to 100 volts. Is that insignificant?

Keep in mind that I'm not making this stuff up all on my own. Others have said essentially the same thing:

I still say you are unable to calculate the charge.

If the NEETS manual is correct where it equates a deficiency (or surplus) of 6.28x10^18 electrons with 1 volt, then the number of electrons necessary to produce the voltages mentioned above are easily calculated. Here's the quote from the manual:

"When a charge of one coulomb exists between two bodies, one unit of electrical potential energy exists, which is called the difference of potential between the two bodies. This is referred to as ELECTROMOTIVE FORCE, or VOLTAGE, and the unit of measure is the VOLT."

"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."

So the lowest point on the trough of the wave shown, if we assume it is 100 volts below ground, indicates a surplus of 100 x 6.28x10^18 free electrons (relative to ground) at the spot indicated by the arrow at that point in time.

 

There is no point in me trying to explain it to you. You are wrong and you can't see it because you don't have the background. You are misinterpreting those references. The references are correct but your interpretation of them is wrong. This is exactly why we don't teach this stuff to amatuers and beginners.

I've actually already answered all of these questions if you really are interested in my answers. Just go back and fail to understand them again.

 

Gerry,

The plate surface area, the internal lead to the exterior tube terminal, the conductor that goes to the bottom of the resistor................all of this surface area has the same density.

All of my comments so far have been for DC steady current.

When a current changes magnitude, the current generates an electric voltage field and a magnetic field.................that opposes the power supply field.

When you sum those two voltages................you will get the output(signal) voltage at the bottom of the resistor.

The current's E and M field, working against the power supply E and M field is the basis for circuit operation.

I think at this point, changing current would be hard to grasp.

Neets is meant to explain electronics, not physics(everyone is looking for that book).

If you want to think that electrons pile up at the entrance to a resistor.............that's fine.

Many people work with electronics all their lives and not know. They do not need to know.

It works and they can work.

If your need for understanding is greater, you must study the very basics.

Your time is never wasted there.

By the way......the number of electrons that fly from the cathode to the plate, they flow thru the resistor. They do not get in line and wait.

Here's a tip. A power supply, conductors and electronic components............does not make a circuit.

Only current, can make a circuit. Current has entity.

 

If you want to think that electrons pile up at the entrance to a resistor.............that's fine.

It's fine as long as he does not engage in fantasy. When he claims that there are many Coulombs of charge accumulation, he is misleading himself and others. So, that is not fine at all. He is going backwards with this approach. The more he studies, the less he knows and the further he gets from the truth.

When the charge might be 1 nanoCoulomb, for example, it is better to get it wrong and think the answer is zero than to get it wrong and think the answer is 10 Coulomb's. Why do I say this? Because if the charge is zero, forces are zero. If the charge is 1 nanoCoulomb, the forces are insignificant and we would never feel them, and we need a super-sensitive measurement system to even know about. If the charge is 10 Coulombs, the electrons are ejected from the surface of the metal in a corona discharge, or an arc discharge. If the 10 Coulomb could be held on the metal, which is impossible of course, the circuit would literally explode to pieces.

How is this level of misconception fine?

 

Because the only flow he knows is matter flow, i.e......water, sugar or sand. And he is new to the study. And it works for him. As he learns more and the need arises, he will re-think it.

How do you explain the flow of something that is self repulsive and invisible?

You can explain it with measurements and equations to justify the measurements, or with law and words,................but how do we all picture it in our mind?

I think the only time we will all have the same picture..............is when we can watch that flow.

Many people now study, analyze, and design working circuits without considering real electron flow.

Many only consider positive flow. And that's fine, cause it works for them.

We have been using electricity for over one hundred years. In that time, on one has explained what an electron is. In my opinion, Ampere and Parson did.

Until we know what an electron is, how can anyone know?

 

OK, I guess I approach science in a different way. I like to get the right answer AND have a good physical understanding of what is happening. lol

 

How do you explain the flow of something that is self repulsive and invisible? ... We have been using electricity for over one hundred years. In that time, on one has explained what an electron is... Until we know what an electron is, how can anyone know?

Well said.

All language (except that tiny portion that deals with direct sensory experience) is metaphorical. And metaphors are always, by their very nature, incomplete and inexact. A good metaphor, therefore, is not something that is identical with a particular reality; rather, a good metaphor conjures up in one's mind an image that is useful for understanding, analyzing, and predicting the behavior of a particular reality.

I think the metaphor I've describe above (with the blue balls in the wires) meets that criteria. With that picture in mind, the kid can understand why, for example, when his guitar string goes up and down, a corresponding sine wave appears on his scope; with that picture in mind, he can analyze the circuit not only piece by piece, but can see how all the pieces work together to make the music play; but most importantly, with that picture in mind, he can predict what will happen if, say, he uses a larger plate resistor or a smaller coupling capacitor. In other words, it's a single, simple, consistent metaphor that can be used throughout the whole discussion; it's an anchor point -- something that is visual and sensible that doesn't change as we move from topic to topic.

If anyone has a better metaphor for this kid and this circuit, please describe it. I'm more than willing to throw out good for better. But I need a picture first, then words, and then numbers. In that order. Because that's what the kid needs.

 

If anyone has a better metaphor for this kid and this circuit, please describe it. I'm more than willing to throw out good for better.

OK, my advice is to keep your picture if you like it, but you need to scale down the level of the charge buildup and the importance of charge buildup in your picture. The charge is a response and not the cause. The amount of charge is not one number (6x10^18) for every volt. The charge is much smaller than that and depends on the geometry of the structure.

The best metaphor for a kid that has ever been invented is the water analogy. Yes, the water analogy is not perfect, but neither is yours, and the water analogy does much better to get the important things right. EMF is like water pressure and current is like water flow rate. Power for water is pressure times flow rate and power in electrical circuits is EMF times current. The metaphor works as a picture and it also works as a physical model that gets quite a bit of the physics correct in terms of energy, power, force and movement.

I can't imagine that any kid would even worry about charge buildup or would understand the significance of it. But, when I was 10 years old hooking up batteries to light bulbs and motors, I know I understood the water analogy and was very happy with it until I was 13 or 14 and started to study electronics formally.

 

The best metaphor for a kid that has ever been invented is the water analogy.

I prefer the Blue Ball analogy over the water analogy because it makes the concept of pressure (which is where beginners typically get stuck) more familiar, more quantitative, and especially more visible -- and thus more accessible to the student. All we have to do is tell the kid that electrons don't like to stand next to each other, and he immediately gets the right picture: widely-spaced electrons will tend to stay where they are, while closely-crowded electrons will be agitated and anxious to move someplace else. Then, when we show him a picture like this:


he immediately (and rightly) concludes that there's more pressure (voltage) on the bottom than the top -- simply because there are more electrons on the bottom trying to get away from each other.

It's actually the same picture of voltage that an oscilloscope gives us, only "distance above or below the x axis" has been replaced with "quantity of blue balls above or below the center of the circuit." But it has this advantage: when we get around to explaining how a vacuum tube works, we can use the very same imagery both inside and outside of the device -- which we can't do with the scope since the image on the scope says nothing about current. In other words, the Blue Ball analogy, by making both voltage and current visible with a single graphical entity (blue balls), gives us a clear educational advantage over both the scope and the water analogy.

I don't think there's much to object to in that. And I think I could also say, without fear of contradiction, that if each ball represents 6.28x10^18 electrons, then one ball flowing past a certain point per second would represent one amp of current.

The trouble begins when we ask, "What is the relationship between the number of extra balls in one spot versus another, and voltage?" What do we mean by "extra", and exactly what is that relationship?

 

I prefer the Blue Ball analogy over the water analogy because it makes the concept of pressure (which is where beginners typically get stuck) more familiar, more quantitative, and especially more visible -- and thus more accessible to the student.

Ok, it's your choice. Apparently you don't think much of a kid's experience sucking liquid through a straw, or using a squirt gun, or filling a balloon with water, or turning on a water hose to water the garden. I did all of these things as a kid and had a very good idea of what water pressure was and how flow rate and pressure were related to each other. Maybe kids are too busy playing video games nowadays.

 

Ok, it's your choice. Apparently you don't think much of a kid's experience sucking liquid through a straw, or using a squirt gun, or filling a balloon with water, or turning on a water hose to water the garden.

It's not that he lacks those experiences; it's that they don't relate directly to the movement of electrons in (and outside) a vacuum tube like the Blue Ball analogy does. And they don't relate both current and voltage back to the mutual repulsion of electrons like the Blue Ball analogy does. So I choose, at least for now, to further investigate the Blue Ball analogy.

Let's get back to some questions. If each blue ball represents 6.28x10^18 electrons, the relationship between blue balls and current is easy: one amp of current is the equivalent of one blue ball moving past a given point per second. The question is, What is one volt in "blue ball" terms? Given what the NEETS manual says, I think it's a surplus or deficiency of 1 blue ball per volt. Yet you say I'm misinterpreting the manual. Okay, then, here's what the manual says:

"When a charge of one coulomb exists between two bodies, one unit of electrical potential energy exists, which is called the difference of potential between the two bodies. This is referred to as ELECTROMOTIVE FORCE, or VOLTAGE, and the unit of measure is the VOLT."

"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."

1. What's the correct interpretation of those paragraphs?

2. What's the relationship between extra/missing Blue Balls and voltage?

 

It's not that he lacks those experiences; it's that they don't relate directly to the movement of electrons in (and outside) a vacuum tube like the Blue Ball analogy does. And they don't relate both current and voltage back to the mutual repulsion of electrons like the Blue Ball analogy does. So I choose, at least for now, to further investigate the Blue Ball analogy.

Let's get back to some questions. If each blue ball represents 6.28x10^18 electrons, the relationship between blue balls and current is easy: one amp of current is the equivalent of one blue ball moving past a given point per second. The question is, What is one volt in "blue ball" terms? Given what the NEETS manual says, I think it's a surplus or deficiency of 1 blue ball per volt. Yet you say I'm misinterpreting the manual. Okay, then, here's what the manual says:

"When a charge of one coulomb exists between two bodies, one unit of electrical potential energy exists, which is called the difference of potential between the two bodies. This is referred to as ELECTROMOTIVE FORCE, or VOLTAGE, and the unit of measure is the VOLT."

"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."

1. What's the correct interpretation of those paragraphs?

2. What's the relationship between extra/missing Blue Balls and voltage?

OK, i guess it won't kill me if I try to explain and you still don't understand. I really feel you don't have the background to understand this and I already answered this, but you missed it or didn't understand it.

As I mentioned before, the wording of the manual is not good. It is confusing, to say the least. Actually it is horrible and inexcusable to say the most.

In order to understand what they are saying ( or a least should be saying) you need to understand capacitance. Capacitance is defined as the charge per unit voltage. So, that is your answer to your question 2. Since 1 coulomb of charge is equal to 1 blue ball, then capacitance is measured in blue balls per volt. A blueball per unit volt, which is also a Coulomb per unit volt is also called 1 Farad, denoted as 1 F. So, the voltage you will get for a given charge is determine by the capacitance. So, there is no "one number" that tells you the voltage for a given charge. The answer depends on the geometry and this is the theory of capacitance. If you have not studied the theory of capacitance, then I doubt this can make sense to you.

Now what is the manual really saying? It is giving an example in which the capacitance is 1 Farad. So, naturally you think it is 1 blue ball for every volt because of the poor way they worded the text. The meaning of their statement "one body is deficient of 6 coulombs (representing 6 volts)" is that this particular body has a capacitance of 1 Farad, so a deficiency of 6 coulombs corresponds to 6 V in this example only. The more I read their text, the more I realize how absolutely horrible it is. It is pitiful and inexcusable. First, the wording is confusing, and second they use a ridiculously large value of capacitance for the example. I suppose it is even possible that I may be giving the author more credit than he deserves. Maybe he doesn't really understand it.

So, you are free to think I'm an idiot just making stuff up, and your interpretation is correct, or you can realize that maybe I actually know what I'm talking about. I could give you my credentials, but appealing to authority is not a good way to do science. Rather, I would ask all others in the forum to correct anything I said which is incorrect. You should have confidence in the collective view of the many great professionals in this forum, more than just one individual, such as myself. So far no one else objected to my statement and I can assure you that when I make a mistake (which I certainly do from time to time being a human) people jump all over it fast, and correct it.

Does anyone in this forum think that 1 Coulomb of charge represents 1 Volt always? I sure hope no one else does.

Let's give just one example where this is clearly not true. How about the Van de Graaff generator? See the following reference.

https://en.wikipedia.org/wiki/Van_de_Graaff_generator

One of these can generate tens of thousands or hundreds of thousands of volts (depending on the size) and the charge would only be a few nanoCoulombs. According to you, the charge would be tens or hundreds of thousands of Coulombs.

 

...the wording of the manual is not good. It is confusing, to say the least. Actually it is horrible and inexcusable to say the most... The more I read their text, the more I realize how absolutely horrible it is. It is pitiful and inexcusable. First, the wording is confusing, and second they use a ridiculously large value of capacitance for the example. I suppose it is even possible that I may be giving the author more credit than he deserves. Maybe he doesn't really understand it.

I've been staring at the units of measure on the standard definition of voltage (ie, joules/coulomb), and at another definition, V=kQ/r^2, and I see now that the NEETS manual must be just plain wrong on this point. I won't mention in again. I'm sorry for frustrating you on this point, and I thank you for your perseverance.

That still leaves me with the question of how to picture and describe potential differences (voltage), and I have to admit that I am still and attracted to the general idea of "electrical osmosis": the diffusion of molecules [electrons] through a semipermeable membrane [wires, resistors, etc] from a place of higher concentration to a place of lower concentration until the concentration on both sides is equal. In other words, I still like the fact that the kid can immediately see from the relative concentration of blue dots in this diagram that there's a potential difference favoring current flow from bottom to top:


My mistake has been in trying to equate voltage with the number of blue dots, or even with the ratio of blue dots top versus bottom, rather than with the concentration of blue dots (whether the number is large or small). Which, of course, is what "joules/coulomb" suggests. In kid talk, I might put it this way: "It takes more energy to get three people into a phone booth than it does to get thirty people into a football stadium. And the people in the phone booth are going to be much more agitated and anxious to get out."

But the above drawing still works (and works well, I think) -- as long as the kid realizes it's not the number of dots that represents a greater potential, but rather the lack of space around the dots; it's not how many dots there are, it's how tightly packed they are. Which would lead the kid to think that the pressure below the resistor is greatest, the pressure in the resistor is less, and the pressure on the top is least; which is a correct conclusion for him to draw.

Better?

 

I think that is better now. I'm glad the issue of the charge/voltage relationship is resolved, as that was a very big flaw in the conceptualization.

I can't say I like the approach personally, but maybe it works well enough for a kid. I guess you have to try it on the kid and see if it makes sense to him. This is far from perfect, but the water analogy is not perfect either. For that matter, Maxwell's equations are not perfect either because they do not take into account quantum effects.

 

Okay, great. Progress. Now let me see if I can get a little further in quantifying voltage without getting myself into trouble again.

The educational problem is still the fact that "energy" cannot be seen. But the effects of energy can be seen. For example, if I see one phone booth with three people crammed in, and another with just two, I know that more energy was expended creating the first situation than the second. And that there's more "stored energy" in the first booth than the second. In other words, I can kind of "see" the potential difference in the booths simply by comparing the number of people versus the space available.

So let's go back to the very first experiment in this thread -- a copper rod moving through a uniform magnetic field at a constant velocity. And let's say the rod is just three atoms long, and we're moving the rod downward through the field. Before we move the rod, all three of the atoms in the rod will look the same (one electron in each atom's outer shell):

But once we start moving the rod, the leftmost valence electron will move to the center atom, which will kick the center valence electron into the right atom and we'll end up with no valence electron on the left, and two on the right, like so:

Now there's a potential difference (a voltage) between the ends of our rod. And while we can't see the energy it took to force the electrons into this configuration, we can "see" the effect of that energy expenditure -- and the stored potential energy -- in the number of missing or extra electrons in the atoms at the ends of the rods. So it seems to me we ought to be able to describe voltage in such terms: we ought to be able to say something like, "Voltage is the ratio of the average number of missing or extra electrons per atom between two spots." Or at least something like, "Voltage is proportional to the ratio of the average number of missing or extra electrons per atom between two spots." Questions:

1. Yes?

2. If so, what's the formula for that?

 

Question 1: I don't think that works in general. There may be certain circumstances that this might work out approximately. I can give it more thought if no one else provides an opinion on this.

The key thing is that there is no one formula in general. It really is a case by case, or geometry situation that determines the answers. So question 2 is hard to answer in general. However, a formula can be found on a case by case basis and we generally figure this out when we derive formulas for capacitance.

However, let's look a the example you provided and try to approximate what we might derive for potential. You showed three atoms and one electron moving from the first atom to the last atom. Well, if we allow all proton/electron pairs to cancel out, and consider one left over proton on the first atom and one extra electron on the last atom, then we basically have what is called an electric dipole. The following reference shows the formula for the potential for a dipole.

http://www.phys.uri.edu/gerhard/PHY204/tsl100.pdf

I expect that this is not the answer you are looking for because you probably want one formula to work in all cases. The closest we can get to this is to realize that the potential for any one point charge is V=kq/r where q is the charge and r is the distance of separation from the charge. Then the total potential is found by adding up the contribution from all individualy charges. Notice in the reference that the derivation for the potential of a dipole is found by adding the potential of the + charge and the - charge.

 

Two things. First, Let's say we take a rod that has just two atoms and we run it through the field -- but cut it in half once we get it moving. We would end up with two pieces, one missing an electron and one with an extra one. Like so:



(I've made the valence electrons blue so they're easier to see.)

It seems to me that there's a potential difference between these two pieces regardless of the distance between them and without regard to their orientation. Specifically, one has "one electron missing" and one has "one electron too many" -- and given the chance, the one will suck up an electron just as fast and with the same force that the other one will spew one out. Yes?

Secondly, I'm unclear about what happens with more atoms in the rod. Say they're copper, we have five, like this:



Assuming a strong enough field and a fast enough velocity, what is the most extreme electron distribution we can get out of this arrangement? Will atoms A, B, C, and D all end up missing an electron while E gets all four extras?

 

You are missing the fact that many structures in matter are crystalline, or related. Amorphous forms of matter exist, but are the minority. Silicon as glass (amorphous) is an insulator, as a crystal it is a semiconductor. Trying to analyze atoms by themselves is out of context, and is a mistake.

 

You are missing the fact that many structures in matter are crystalline, or related... Trying to analyze atoms by themselves is out of context, and is a mistake.

I'm only talking about copper here, and the context is the very first post in this thread where I was given this picture:



And this text, which is common to scores of textbooks and websites about electronics:

The magnetic force acting on a free electron in the rod will be directed upwards. As a result, electrons (the blue dots) will start to accumulate at the top of the rod. The charge distribution of the rod will therefore change, and the top of the rod will have an excess of electrons (negative charge) while the bottom of the rod will have a deficit of electrons (positive charge). This will result in a potential difference between the ends of the rod.

I'm simply trying to fill in the blanks in the mental image I've been given. And so I ask again:

1. Let's say we take a rod that has just two atoms and we run it through the field -- but cut it in half once we get it moving. We would end up with two pieces, one missing an electron and one with an extra one. Like so:

(I've made the valence electrons blue so they're easier to see.) It seems to me that there's a potential difference between these two pieces regardless of the distance between them and without regard to their orientation. Specifically, one has "one electron missing" and the other has "one electron too many" -- and given the chance, the one will suck up an electron just as fast and with the same force that the other one will spew one out. Yes?

2. I'm unclear about what happens with more atoms in the rod. Say they're still copper, we have five, like this:

Assuming a strong enough field and a fast enough velocity, what is the most extreme electron distribution we can get out of this arrangement? Will atoms A, B, C, and D all end up missing an electron while E gets all four extras? What's the maximum number of electrons that a copper atom can have in it's outermost shell?