# I Questions about electric fields, voltage and electric energy

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1. May 15, 2016

### Scooby2016

1. If e.m.f is NOT a force, but actually energy – and e.m.f is measured in Voltage – then why is p.d. (V) NOT used up? Since we know energy *does* get used up by the component.

I know the electrons are just the medium – energy carriers – but Voltage doesn’t mean number of electrons; it means the “energy” they carry right? So energy *should* be used up.

I know this is THE classic misunderstanding – but I’ve read tons online – but still not clear on this.

2. If a circuit diagram goes from:
Battery (one terminal) = 6V –> before component = 6V – > after component = 0V – > return to battery (other terminal) = 0V THEN it looks like the “V” got used up right there at the component!

3. How DO electrons end up holding on to some of their energy when there are 3 components in a series - as if they’re “experiencing” that there’s a total of 3 components?

A lot of people have asked this exact question – but answers usually end up saying that there’s a bunch of electrons choosing x or y path – rather than ONE path. In one path (series) they are having to hang on to some of that energy for a bit longer. It’s as if the component can only “ask” for a photon of a certain lower frequency if there’s 3 components present. And if there was 1 component it could now take the higher frequency photon in the final energy-transfer. Extreme lack of clarity here.

4. Why don’t electrons take a shortcut through the electrolyte? From the side they’re sitting in to the side they “want” to be in – I mean; why go all the way round the wire if it’s an actual closed circuit?

5. If p.d. is just an electric field created – then what’s the “range” of this field? Electrons sitting at any given part of the circuit will experience the force and move, so do we assume the range is the entire wire? And nothing else. Maybe some EMR escapes outside the wire too?

6. Big question: WHAT is the INITIAL energy? Is it somehow energy stored in chemical bonds? It all MUST have started with some chemical energy – middle-schoolers know that much. I know it’s a redox reaction whereby something like Cu gets reduced, and after becoming ions the electrons flow towards Cu+ ions. I don’t mean to be stupid – but how is redox releasing ENERGY – it’s not like some chemical bond has been made. Is there?

7. I’m having a hard time visualizing this “electric field”. It’s nice to visualize a tiny formula unit of say Na+ and Cl- each ion then exerting a small field of attraction towards each other – but this electric field in a wire is not nice. Doesn’t want to be visualized. Does the field pull the electrons for example 1 nanometer at a time (like a knock-on effect like Newton’s Cradle) – or the entire distance to the other electrode? By “pull” I mean the “location they’re trying to get to”. I guess I’m saying is: how exactly is it produced – and what does the field look like and how do charges experience this field precisely, going from anode to cathode?

2. May 15, 2016

### Staff: Mentor

The electrons don't carry the energy, the fields carry the energy. If the electrons carried the energy then AC lights could never work because the electrons don't move from the power supply to the lights.

In general, it is not terribly useful to think about electrons in classical EM, and even less useful in circuits. Really, for circuits all you should worry about is KVL, KCL, and the voltage vs current relationship for the various circuit components. If you are asking about electrons then you are asking a question which belongs to a more complicated theory.

Last edited: May 15, 2016
3. May 15, 2016

### Scooby2016

OK, so the electrons aren't like "excited" - that's a relief! I did find that quite troubling to try to picture a lone electron traveling alone being "excited" since it ought then to be in one of the "allowed orbitals" around an actual ion.

So the electric field IS carrying the energy, much like any old EMwave would... I guess the moving electrons are just the guys creating the electric field throughout the wire? Thanks!

I would really love to understand HOW the electric field is generated though - thinking about the starting point of the redox reaction of say some metal with an acid - and finishing with electrons returning to the other end of the battery.

4. May 15, 2016

### Staff: Mentor

Yes, the energy flux is given by the Poynting vector, just like in an EM wave.

There are three things to think of here. There is the bulk metal, the bulk electrolyte, and the electrode surface. At the surface the redox reaction occurs which results in some charge being built up in the electrode and the opposite charge being built up in the electrolyte. The charge makes a field and then by Ohms law there is a current.

Last edited: May 15, 2016
5. May 15, 2016

### Scooby2016

Yay! Finally I have picture - sketchy as h**k probably - but still, a picture from energy source to lighting the bulb! :) (Been on this for a while.)

I had been thinking along the lines of stored chemical energy released from bonding in some way - but I guess the ionization of the metal is where the energy more or less "starts" - and then an electric field is produced, which is converted into say, light/heat. The electrons are only doing their thing, going from X to Y, but once they stop, the electric field "dies" and the battery is flat.

I found an image of the electric field: http://www.furryelephant.com/lib/img/intro_pages/poynting.png

I assume it's accurate - there were more than one and they look more or less the same.

Thanks a lot for your time and help,
Scoob

6. May 15, 2016

### jartsa

I have a question: Why can one power supply force electrons to move trough a 1 mm air gap, but another power supply just can't do that?

Let me try to answer: Let's say there is an electric field and a charged object, and my task is to push the object though the field. I try to push with one hand, object does not budge, then I use both hands and the object moves. The charge pushing ability of my two hands is about two times larger than the charge pushing ability of my one hand. Right? And the charge pushing ability of two 1.5 V batteries connected in series is two times the charge pushing ability of one 1.5V battery. Right?

7. May 15, 2016

### Staff: Mentor

That is essentially correct. To do it quantitatively requires some rather hairy computations solving Maxwell's equations. Usually it is not necessary to do that for practical purposes.

That certainly is correct, but to actually show it requires some quantum mechanics which is beyond me and usually not necessary. It is not usually even necessary to know about the electrodes and electrolyte and the redox reaction, unless you are specifically designing the battery itself.

Last edited: May 15, 2016
8. May 16, 2016

### Scooby2016

Thanks for the feedback. I realize some of what I asked is not "necessary" - and probably out of my league, especially the math. But it seems important to have a picture in your head that at least isn't wrong. It's about that mental image - then later there won't be a ton of re-learning.

BTW - I'm a newbie on here - I couldn't believe how amazing PhysicsForums.com is - people on here are so genuinely smart and helpful. Awesome site. :)

9. May 16, 2016

### jartsa

I have never seen a calculation of forces in an electric circuit. So let's calculate:

Let's say the drift velocity of electrons inside a battery is 10-5 m/s when the battery is doing work at rate 1 W. Power is force times speed, so the force the battery exerts on the drifting electrons must be 1 W / 10-5 m/s = 105 N.

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

Let's say all the energy delivered by the aforementioned battery is used in a light bulb at rate 1 W, and let's say the drifting speed of the electrons in the thin wolfram wire of the bulb is 10-3 m/s. Power is force times speed, so the drag force the electrons exert on the rest of the wire must be 1 W / 10-3 m/s = 103 N. So something is pushing the electrons with that same force.

What happens if we connect two of these batteries in series? Forces double, drifting speeds double, powers become four times larger. Power is force times speed, you see. (When I say that forces double, I mean forces inside one battery do not change, but the two batteries together exert a force on the drifting electrons inside the two batteries, and that force is double the force that one battery alone exerts on the drifting electrons inside of it.

So my point is that we can do electricity calculations using Newton's mechanics. So maybe we can even understand electricity using Newton's mechanics.

10. May 18, 2016

### Scooby2016

It sounds very interesting. More than one path to the same answer always gives a much deeper understanding, but the math is beyond me. Will read again in a year from now!!

11. May 19, 2016

### fizzle

No, you generally think of the wires as 100% conductors and the load, e.g. a resistor, as the place where the electromagnetic energy is dissipated (converted to heat). As Dale noted, it's the electromagnetic field that carries the energy from place to place. The most obvious example is the Sun. When you go outside and feel the warmth of the Sun on your face, there are aren't any wires connecting the Sun to your face! It was the electromagnetic field from the Sun that carried the energy that warmed your face.

Newtonian mechanics are used in Maxwell's Equations (forces, conservation of energy, etc.). The key addition were electromagnetic fields.

12. May 19, 2016

### jartsa

When I go outside and feel the warmth of the Sun on my face, there aren't any wires connecting the Sun to my face, it is the high speed electromagnetic fields emitted by the Sun that punch my face, and the punches cause some warming.

When I grab two live wires ... can you explain what happens? If you say that there is an electric field that is standing still and carrying energy, then I must very skeptical about such claim. Because things that stand still don't carry stuff.

13. May 19, 2016

### Staff: Mentor

I agree completely. In my opinion getting wrong pictures in your head is such a problem that it is better to deliberately leave blanks than to give wrong pictures.

I also do not believe that a correct picture is possible without actually learning the theory that gives the correct picture. This is what is attempted by pop sci authors, with universally poor results.

For the purpose of a circuits class (which is what I assume you are studying) I would treat a battery simply as a voltage source. That is a correct picture that won't have to be unlearned later, it just leaves the details of the internal working blank. I would explicitly tell the students that I would not cover that, but it would be covered in future courses.

14. May 19, 2016

### Staff: Mentor

The phrase "an electric field that is standing still" doesn't make any sense. The fields have a magnitude and a direction at each point in space, they do not also have a velocity. You cannot say either that they are moving or that they are standing still, they just have a magnitude and direction at each point.

The EM field carries energy according to the Poynting vector $S=E\times H$ and does work on matter according to Poynting's theorem
https://en.m.wikipedia.org/wiki/Poynting_vector
https://en.m.wikipedia.org/wiki/Poynting's_theorem

Before you touch the wires there is a nonzero E, but H is zero. After you touch the wire there is both an E and H field transporting energy, and more importantly inside the body there is work done according to $E\cdot J$

15. May 19, 2016

### Staff: Mentor

The fields at each point can change their magnitude and/or direction as time passes, and can do it differently at different points. This can cause the pattern formed by the fields to move as in e.g. an electromagnetic wave. However, we don't associate that motion with the fields themselves. A common analogy is a "wave" formed by football fans in a stadium. The fans don't move (horizontally), but their "wave" does.

16. May 19, 2016

### jartsa

I see. But a neutral push rod and a charged push rod are still both very much mechanical things, although according to Poynting's theorem in the case of a charged push rod some of the mechanical energy is transferred through the surrounding air. So I see no reason to not talk about forces when talking about push rods, or when talking about electric circuits.

17. May 20, 2016

### Staff: Mentor

I have no problem talking about forces in EM, but the point is to learn how to quantify this additional "some" energy that cannot be accounted for strictly mechanically. It can be small in certain scenarios, but it can be large in others.