# How do Electrons perform work in circuits?

1. Mar 25, 2008

### nuby

Do electrons flowing in a circuit perform work with their mass (or kinetic energy)? For example like a water flowing in a creek through a turbine. If so does E = 1/2 mv^2 apply.

2. Mar 25, 2008

### pmb_phy

The kinetic energy of a non-relativistic electro, such as those in an electric circuit, is K = mv^2/2. A deeper analysis requires quantum mechanics where the concept of force, and thus work, is meaningless.

Pete

3. Mar 25, 2008

### nuby

What does the "/2" represent in that equasion (in reality)? K = mv^2/2

Last edited: Mar 25, 2008
4. Mar 25, 2008

### pmb_phy

Division. E.g. the expression x/2 means that x is divided by two.

And you're very welcome. Glad to be of service.

Best wishes

Pete

5. Mar 25, 2008

### nuby

i knew it was divided by 2 :P .. but I was wondering more why it is divided by two. :)

6. Mar 26, 2008

### Snazzy

$$KE=\int_0^Vmv dv = m\int_0^V v dv = \frac{1}{2}mv^2 \ |_0^V =\frac{1}{2}mV^2$$

7. Mar 26, 2008

### nuby

I'm guessing the /2 means it is only 1/2 the force used to put an object in motion. Is that correct?

8. Mar 26, 2008

### chroot

Staff Emeritus
As Snazzy showed, the 1/2 is the result of taking an integral of a first-order quantity, velocity.

- Warren

9. Mar 26, 2008

### Staff: Mentor

Whenever you see constants like 1/2 in an equation you can be pretty sure that the equation was derived from some more basic calculus expression. Once you get a few weeks of calculus those terms will make sense.

10. Mar 26, 2008

### nuby

Any other opinions on this topic? The basic question is: Are moving electrons' mass the only source of energy in an electronic circuit.

11. Mar 26, 2008

### Staff: Mentor

I would think their electric field is much more relevant than their mass.

Your reference earlier to water in a turbine is close. I would actually relate it to water in an enclosed pipe. Hydraulic energy is pressure times volume. The analogy would be of "electricity" being like an incompressible fluid where the energy is voltage times charge. The mass doesn't really factor in explicitly for either case.

Last edited: Mar 26, 2008
12. Mar 27, 2008

### nuby

An electric field in my opinion is just a force that makes electrons move (measured in volts), and not necessarily the main 'work' force in a circuit.

If you have two magets sitting on a table (N/S), with say a unipolar magnet between them (o), and a non-magnetic barrier (I).

like this:

..N<.....I.....o......<S..

From a simple point of view, the mass of the 'o' will be the main factor of the work actually done to push over (I). I believe this is the same concept occuring in most electronic circuits, and the electron push/pull force then would be the energy 'transmitters' and the mass of the electron, which changes due to electrical resistance (giving off heat/light) is doing the actual 'work'.

If this is true, then voltage is actually a measurement of the potential speed of the electrons in a circuit.

Last edited: Mar 27, 2008
13. Mar 27, 2008

### Staff: Mentor

First, there is no such thing as a unipolar magnet.

Second, the mass of 'o' is irrelevant to the work done in this example. Remember, work is the (scalar) product of force and distance, mass does not enter in. If 'o' experiences 1N of force and travels a distance of 1m then the work done on 'o' will be 1J regardless of the mass of 'o'.

14. Mar 27, 2008

### nuby

Ok, I understand there is no such thing as unipolar magets, and this is probably a bad example to show my thought... which is, electron mass contributes to most of the work done in electronic circuits

DaleSpam,

So if you throw an object in space, and it travels 1M at 1N force, is the work still 1J regardless of the mass?

15. Mar 27, 2008

### Staff: Mentor

Correct. An object with 1kg mass will end with a velocity of 1.41 m/s and an object with 2kg mass will end with a velocity of 1 m/s, but the work is 1J in each case and the KE is also 1J in each case.

16. Mar 27, 2008

### nuby

DaleSpam,

Can you give me the simple equation for that, as well as the equation for energy required to slow down that same object?

Thanks

17. Mar 27, 2008

### Ulysees

Sorry to interrupt you here, but reading your initial question I wonder something has been misunderstood. You said:
They'd have to slow down to perform work out of their kinetic energy. But they don't slow down to a halt, as they move from atom to atom. Instead they gain some kinetic energy under the influence of the field, they lose some (turned into heat), they gain some, they lose some, and so on. Gaining and losing overlap. But they don't stop, so any expression like 1/2 mv^2 is irrelevant to the work actually done to turn their potential energy into heat and warm up the conductor.

Last edited: Mar 27, 2008
18. Mar 28, 2008

### Staff: Mentor

W = f.d
KE = 1/2 m v^2
dW = dKE

19. Mar 29, 2008

### nuby

Can KE = 1/2mv^2 and F=ma or be applied to electrons.

20. Mar 29, 2008

### Snazzy

In some circumstances, it can. (cathode ray tube questions and sorts).

21. Mar 29, 2008

### Staff: Mentor

Yes. However, remember that in most circuits (other than cathode rays etc. as mentioned), the electrons never move with any significant velocity, never attain any appreciable amount of KE, and give up any energy imparted to them almost as soon as they get it.

22. Mar 29, 2008

### nuby

I've heard electrons travel very slowly in standard circuits before (cms per hour), however, I've never seen any proof of this. How was this measurement taken?

Is it possible that conduction band ("free") electrons move rapidly through a conductor in a spiral path, but only 'outputs' as much current is being drawn... This could give the appearance of slow moving electrons.

Last edited: Mar 29, 2008
23. Mar 29, 2008

### awvvu

Yes, individual electrons (or rather, I mean the drift velocity) actually move very slowly. See this:

http://hyperphysics.phy-astr.gsu.edu/hbase/electric/ohmmic.html

24. Mar 29, 2008

### Snazzy

The current in a circuit is defined by the amount of charge moving past a certain point each second:

$$I=\frac{\Delta Q}{\Delta t}$$

Where:

$$\Delta Q=nA\Delta x q$$

Where n is the density of charge carriers (a very large number) and is a property of the material you are running a current through, ie. a constant. Therefore:

$$I=\frac{nA\Delta x q}{\Delta t}={nAv_dq$$

Where v_d is the drift velocity.

25. Mar 30, 2008

### nuby

Is it possible that conduction band electrons, and valance, etc. move a difference speeds?