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What happens when the field when it did the work?

  1. Nov 2, 2009 #1
    I recently brake my brain on the following problem:


    -------------------------------------- (+)

    H~> light

    ^
    e |




    In uniform electric field the free electron (or some ion) gets accelerated to the point that it smashes against the hydrogen atom on the way. Hydrogen emits light and the work done by the electric field get converted into light.

    the question is:

    what happens to the electric field after it did a work? obviously if the charge has not reached the positive electrode the field has the same strength as it had before and it is ready for unlimited cases as above. work has been done and who payed for it? not the field obviously?
     
  2. jcsd
  3. Nov 2, 2009 #2
    Who ends up paying in this case is the fella who put the electron there in the first place.
     
  4. Nov 2, 2009 #3
    The total field gets weakened anyway. So the total system goes to an equilibrium. It is not correct to separate the "extermal filed" from the total: the field equations contain all sources of field, including the dispalced charge.
     
  5. Nov 2, 2009 #4
    I will change the situation a bit :) I understood that the electron came to the system and its negative charge weakens the positive electrode field.Now:


    --------------------------------- (+)

    XY-> X + Y
    ^
    |
    e




    In this case electron hits a molecule XY ,gets X knocked out and stays with X forming a neutral atom. Y is now neutral too because XY shared a single electron as an atomic bond.
    Obviously energy was spent on destruction of the XY bond.

    Electron comes with the wind :) so is it that:

    1) effectively wind energy splits XY?
    2) no change in the field strength occurs?
     
  6. Nov 2, 2009 #5
    As an electron moves (accelerates) in an electric field between two electrodes (e.g., anode and cathode), image charges on the electrodes move from one electrode to the other via the external network and the power supply. Voltage times charge = work.
    Bob S
     
  7. Nov 2, 2009 #6
    @Bob

    There is no negative electrode here and neither is positive plate connected to any external source. Consider it a charged plate capacitor after removing its negative plate from the system.

    PS:Anyhow you have answered my other question i had in my mind :)

    Anyone?
     
  8. Nov 2, 2009 #7
    If there is an electric field, there have to be electric image charges (with rare exceptions) that create the field. Conventionally this is similar to a capacitor. If a capacitor C has charge +/- Q on its plates with separation d, then the voltage and field is V=Q/C, and E = V/d. When an electon moves from the neg plate to the pos plate, it reduces the charge on the capacitor, and reduces both the voltage and field. The energy stored in a capacitor is (1/2)C V2. The discussion of what is happening as the electron is moving in the field is a little more complicated.

    Bob S
     
  9. Nov 4, 2009 #8
    Bob, what happens on a normally connected capacitor was clear to me already. When there is a wire between electrodes i see no problem. Whenever an electron jumps from one electrode to the other one (surge) i also understand that positive and negative plates will loose a charge.

    What i ask is what happens when electron gets captured on its way to the positive electrode, so it never has a chance to reduce its charge. Electrode is not connected to anything, just a charge is resting on it so it attracts electrons from the environment, speeding them up, but they always get caught on the way. So when there is an interaction that disappears at one moment, but the work done by field is dissipated. However apparently, the field is never getting weaker after it did some work unlimited number of times...
     
  10. Nov 4, 2009 #9
    The best way to calculate what happens when a small charge q leaves one plate of a capacitor with charge Q and plate separation X is to consider the following situation:

    One plate has charge Q-q on it . At a very close at distance x is an intermediate plate with charge -(Q-q) on one side and +Q on the other (net charge difference +q). The other plate of the main capacitor is a distance X-x away, and has charge -Q on it.

    Calculate the total stored energy as the intermediate plate moves from on electrode to the other.

    Bob S
     
  11. Nov 4, 2009 #10

    Dale

    Staff: Mentor

    Here is a hint: are there an unlimited number of electrons available to do work? How do you think a large number of those "working electrons" would affect the field and the initial energy required? What is the difference in principle between a large number of working electrons and a single electron?
     
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