# Voltage as electron pressure

1. Sep 4, 2015

### jaydnul

The water analogy can be used as a rudimentary tool to describe an electronic circuit, but in the case of voltage on a wire, it is exactly analogous. When a voltage on a wire is induced, the power supply or battery is shoving excess charge (which moves to the surface of the conductor) into the copper. Yes there is an electric field running through the wire, but that field is a product of the extra charge on the surface of the conductor. The more charge you shove onto the surface of the wire, the higher the voltage (with a constant reference).

So why isn't voltage explained exactly like this to beginners? Instead of starting with the electric field running through the wire, why not just explain it as electron pressure? Is this just a subjective case where it's more intuitive to only myself?

2. Sep 4, 2015

### Staff: Mentor

The excess charge is not proportional to anything useful or relevant in the circuit, and it is extremely tiny in most setups.

3. Sep 4, 2015

### axmls

They're not exactly analogous. Pressure is an absolute quantity: the force per unit area. Voltage is relative and only makes physical sense when speaking of a voltage difference.

4. Sep 4, 2015

### sophiecentaur

That isn't the way it's taught. Voltage is usually presented as the Energy transferred for a unit charge passing. (1V = 1Joule per Coulomb); hence it corresponds a change of potential energy (a scalar) and no force (a vector) is involved. A field doesn't "run through" anything.
The water analogy always seems totally pointless to me. It merely puts off the problem of acknowledging that is is not complete enough and gives a false sense of 'understanding'. This is a common topic on PF and there is a split between people who want to do it 'properly' and people who think you can get away with it. IMO, those people have just not followed it deeply enough to find the serious flaw.
I think that's exactly the problem. The personal model is not a valid one because you have not used the terms in their conventional sense.

5. Sep 6, 2015

### jaydnul

Ok, but where does the electron get the extra potential energy? What does the power supply actually do to the charge in the copper wire?

6. Sep 6, 2015

### davenn

It comes from the potential difference between the terminals of the power supply, measured in volts

7. Sep 6, 2015

### jaydnul

Yes, but how does the power supply bestow this difference in potential, physically? A generator moves the electrons through a magnetic field, which gives them a higher potential energy and pushes them along the wire. But what about the electrons that aren't submerged in the magnetic field? Where do they get the higher potential energy? The only logical conclusion I can come to is that they are being pushed along by their neighboring electrons.

8. Sep 7, 2015

### sophiecentaur

In a conceptual sense, that's ok but does it help in any way with 'understanding' and using circuits?
It is true that charges will flow when there is an unbalanced charge. I think it's valid to talk of an 'individual' electron being attracted more in one direction than another so you have to be able to make an explanation in terms of the local forces on all the electrons (and Protons) around the circuit. The problem with approaching electric circuits in this way is that the calculations would become very involved and to what end? Fields (Volts per metre) must exist in a wire and must cause net electron drift. If you want to include this in the understanding of circuit behaviour then the routing and length of the conductors would need to be involved for every circuit you considered. Conventional circuit theory can be hard enough and I suggest that all the extra ideas that your model introduces do not simplify anything - just make it harder. This approach is just not a good way to work out what circuits are doing. There are many papers published that study the details of transfer of power in circuits in this way - but the guys who are doing the theory measurements will still base the design their circuits in the conventional way and will 'understand' what they're doing from the standpoint of the Elelctrical Theory they learned originally.

9. Sep 8, 2015

### jaydnul

The energy per electron (voltage) is inversely proportional to the distance between electrons $V=k\frac{e}{r}$ (for a 1 dimensional wire with an idealized setup). And the distance between electrons is inversely proportional to the density of electrons (excess charge) on the wire $D_e=\frac{C}{m}$. So the voltage IS directly proportional to the excess charge.

I realize that pressure is the incorrect terminology, but the general idea is that the electrons in a wire are separated by an average distance. As you shorten that average distance, the voltage becomes higher. If two wires at different voltages come into contact, the respective average distances will reach equilibrium until the combined wires are the same voltage (which is now the average of the two separate voltages).

So this extremely tiny excess charge is all you need to get usable voltages in electronics, and all a power supply does is keep the voltages from reaching equillibrium by continually sourcing/sinking charges to keep the electron densities on each lead constant.

Please share your thoughts and also thanks for all the input so far!

10. Sep 8, 2015

### Staff: Mentor

There are no ideal 1D wires. Your wire would still have a logarithmic divergence, and an actual 3D object has an even worse divergence. That approach does not work. You can use capacitance to find the excess charge.

11. Sep 9, 2015

### jaydnul

Ok, this is one of those things that I thought I had down for a while now and I liked the model because of its simplicity (to me). It's great when something I think I know is proven completely wrong (even something as elementary as this) because it reminds me that nature doesn't care what I like and just is the way it is.

Anyways, thanks for all the help!

12. Sep 10, 2015

### jaydnul

Sorry to post on this thread again but I didn't wan't to start a new one just for supplemental questions.

I think I have it worked out, I just want to make sure it's correct. So the charge density of the electron sea (the charge that moves and makes up the current) in the copper is more or less constant and has no significant correlation to anything in the circuit. However, the density of extra charge on the surface of the wire does vary, and this is what causes the electric field through the wire that gives the electron sea a reason to move (voltage). A power source essentially just provides extra charge to go onto the surface of the wire to create an electric field throughout. Is this correct?

13. Sep 10, 2015

### ZapperZ

Staff Emeritus
What exactly are you basing this on? This is not what the Drude model is, nor is it what is used in the Boltzmann transport model, both of which are extensively used in Solid State physics.

Are you making up your own theory?

Zz.

14. Sep 10, 2015

### Staff: Mentor

As discussed in the last posts: not in a way that would be useful for circuit analysis in any way.

15. Sep 10, 2015

### jaydnul

So if the extra charge doesn't cause the potential, what does?

It's what I took from this text. Not saying it's right, hence the question "is this correct" at the end of the post.

I understand you have to be abrasive with people to deter crackpots, but sometimes it can be unwarranted.

http://www.matterandinteractions.org/Content/Articles/circuit.pdf [Broken]

On pg.10, last paragraph: "The surface charge density is proportional to the circuit voltage..."

http://www.matterandinteractions.org/Content/Articles/circuit.pdf [Broken]

Pg. 15, part A:

"Inside a resistive wire 1 meter long connected to a 3 volt battery, the electric field is only 3 volts/meter, which is tiny compared with typical fields encountered in electrostatic phenomena (for example, the breakdown strength of air is about 3 million volts/meter). Therefore, the amount of surface charge on the wires of a typical circuit is extremely small compared to typical electrostatic charges, which is why it requires high-voltage circuits to observe electrostatic effects. However, these small amounts of charge are responsible for driving the current inside the wire"

Last edited by a moderator: May 7, 2017
16. Sep 11, 2015

### sophiecentaur

Consider a different context for potential energy. Would you say it is the strain energy in a spring that 'causes' the potential energy of the mass that you're lifting with the spring or is it your arm muscles? The spring is an intermediate step in the energy transfer and the stretched spring will actually store some of the energy input. If the spring has a very high k, you can ignore this energy. Likewise, the equivalent 'k' for the wire full of electrons is equally high and you may as well ignore it. That 'k' is actually the dielectric constant of the metal, which is virtually infinite.

17. Sep 11, 2015

### jaydnul

So in your analogy, the battery is the arm muscle, and it does work to increase the electric field within the wire. But my problem is still this; the definition of dielectric constant is "a quantity measuring the ability of a substance to store electrical energy in an electric field".

My question is what is the wire doing to "store electrical energy in an electric field"? The links I referred to in previous posts claim that the small amount of excess charge on the surface of the wire IS the cause of this electric field. But mfb says that the excess charge has nothing to do with anything useful.

Hopefully you can see my confusion. Thanks for the help so far!

18. Sep 11, 2015

### sophiecentaur

That is a qualitative description. Used in a formula, it is a constant relating charge displacement and applied voltage.
Have you thought of approaching this Mathematically instead of arm waving? I am always wary of the "what's really happening?" approach instead of the "Model to fit the facts" approach. I have no argument at all with mfb's approach. I don't think it clashes with anything I wrote (?). The "excess charge" in a wire, due to the PDs in the circuit is not relevant because the wire is the equivalent of "a very strong spring" which will not stretch.

19. Sep 11, 2015

### sophiecentaur

The wire is not storing any energy - that's my point (a very strong spring that will not stretch and cannot store any energy under the conditions of the experiment).

20. Sep 11, 2015

### jaydnul

http://www.matterandinteractions.org/Content/Articles/circuit.pdf [Broken]

Bruce A. Sherwood and Ruth W. Chabay from Carnegie Mellon University seem to think that the excess charge on the surface of the wire is responsible for the electric field running through the wire. My question is, are they wrong? Or am I not understanding what they are saying.

BTW, I'm not trying to appeal to some other scientific authorities, I'm just genuinely curious about the split in opinion between you guys and these guys.

Maybe there isn't a split and I'm just not getting it...

Last edited by a moderator: May 7, 2017
21. Sep 11, 2015

### Staff: Mentor

Please don't make multiple posts in a row, if you would like to add something you can edit your post. I merged some of them.

It is a possible view, but I don't think it is a useful one.

22. Sep 11, 2015

### jaydnul

Oh ok, sorry. Probably says that somewhere in the forum rules, it's just been a while.

So it's not supported by evidence?

Edit:

You said:

And Bruce A. Sherwood and Ruth W. Chabay in this http://www.matterandinteractions.org/Content/Articles/circuit.pdf [Broken] on pg.10, last paragraph said:

"The surface charge density is proportional to the circuit voltage..."

This specific claim can be tested in the lab, and you directly contradict them. So, I'm sorry, but saying...

...doesn't really fly in this situation. Now I haven't done, nor do I have the means to do the experiment, which is why I'm asking on PF.

Last edited by a moderator: May 7, 2017
23. Sep 11, 2015

### sophiecentaur

What do you mean by the word "responsible"? Surely it is the emf at the source end that is "responsible". Logically, that isn't the end of it, of course and you could take the responsibility right the way back to the Big Bang. I can only repeat my previous argument in a slightly different form. The field and the charge situation are both 'just there' because of the existence of an emf somewhere else - which produces a potential difference. Unless you can claim that there is significant difference in the charge density in a wire, when you switch on, then (as in the strong spring), there is no deformation and no storage of energy.
I read this article before and I noted that it was written by 'Educationists", which makes me wonder. I could find just one serious equation in the whole paper and no mathematical proof or evidence of measurements. That's a good reason not to take its message too seriously. It's hardly at a higher level than a chat on PF.
The order of the field / charge thing seems upside down. If you put a conductor in an electric field, charges will flow, to reduce the Potential Energy situation. If you put charges on each end of a wire (briefly) they will flow and eliminate any field that you started off with. To maintain that field, you need to be doing work on moving the charges.
Bottom line is that I don't see where this 'alternative thing' is going.
The reason for the "split in opinion" is, I think, that those guys have not actually tried to apply their ideas to a real situation, needing results. In an arm waving sort of a way, their case is arguable - not complete nonsense ; they want a model that they can present to students (without any health warning which worries me). But they have no quantitative (apparently) evidence or an idea to what degree it applies and no equations. That, to me, is pretty relevant.

24. Sep 11, 2015

### jaydnul

Cool, that's all I was looking for. I'd still like mfb to address my previous post. If the excess charge doesn't proportionally increase/decrease with an increase/decrease in voltage, what is causing the excess charge in the first place?

25. Sep 11, 2015

### sophiecentaur

There will be Capacitance between the wires and components. It isn't surprising that there are - charges detected at the - terminal because Q = CV applies, and so on round the circuit. But that doesn't mean that the charges are the 'cause' of the current. If the circuit has finite resistance then those charges will move due to the Potential Difference.
You may say 'why use an energy rather than a field description of what's happening?' It's because the field based description would involve a total re-write of the circuit equations if the wires were re-routed or shortened just to result in the same answer! That is hardly a good system for circuit analysis and it would be guaranteed to confuse students..