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How do Electrons perform work in circuits?

  1. Mar 25, 2008 #1
    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. jcsd
  3. Mar 25, 2008 #2
    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
     
  4. Mar 25, 2008 #3
    What does the "/2" represent in that equasion (in reality)? K = mv^2/2

    Thanks for your response.
     
    Last edited: Mar 25, 2008
  5. Mar 25, 2008 #4
    Division. E.g. the expression x/2 means that x is divided by two.

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

    Best wishes

    Pete
     
  6. Mar 25, 2008 #5
    i knew it was divided by 2 :P .. but I was wondering more why it is divided by two. :)
     
  7. Mar 26, 2008 #6
    [tex]KE=\int_0^Vmv dv = m\int_0^V v dv = \frac{1}{2}mv^2 \ |_0^V =\frac{1}{2}mV^2[/tex]
     
  8. Mar 26, 2008 #7
    I'm guessing the /2 means it is only 1/2 the force used to put an object in motion. Is that correct?
     
  9. Mar 26, 2008 #8

    chroot

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

    - Warren
     
  10. Mar 26, 2008 #9

    Dale

    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.
     
  11. Mar 26, 2008 #10
    Any other opinions on this topic? The basic question is: Are moving electrons' mass the only source of energy in an electronic circuit.
     
  12. Mar 26, 2008 #11

    Dale

    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
  13. Mar 27, 2008 #12
    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
  14. Mar 27, 2008 #13

    Dale

    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'.
     
  15. Mar 27, 2008 #14
    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?
     
  16. Mar 27, 2008 #15

    Dale

    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.
     
  17. Mar 27, 2008 #16
    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
     
  18. Mar 27, 2008 #17
    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
  19. Mar 28, 2008 #18

    Dale

    Staff: Mentor

    W = f.d
    KE = 1/2 m v^2
    dW = dKE
     
  20. Mar 29, 2008 #19
    Can KE = 1/2mv^2 and F=ma or be applied to electrons.
     
  21. Mar 29, 2008 #20
    In some circumstances, it can. (cathode ray tube questions and sorts).
     
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