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Derivation of drift speed question

  1. Mar 3, 2014 #1
    1. The problem statement, all variables and given/known data
    First off, this is NOT a homework problem. This is a conceptual question I have regarding the derivation of the drift velocity

    [tex]v_d =[(qE)/m] \tau[/tex]

    Typically, when this formula is derived, you first calculate the acceleration of a particle in the electric field (qE/m) and then it is noted that using v = v_o + a*t you can rearrange to get your drift velocity. However, and here is my question, it is usually noted that in the absence of an electric field the velocity of a charged particle is random and thus v_o = 0. However, this derivation assumes that the particle in a conductor experiences an electric force F = qE and thus you have to have an electric field, which I would imagine means you can NOT let v_o = 0. Am I missing something? Hopefully my question is clear, if not, I'll try to clarify. Thanks.

    2. Relevant equations
    See above.


    3. The attempt at a solution
    N/A
     
  2. jcsd
  3. Mar 3, 2014 #2

    BvU

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    is not correct. It should say <v_0> = 0, meaning: the average (x-) component of the velocity of the electrons is zero. They move like crazy all over the place, though.
     
  4. Mar 3, 2014 #3

    BvU

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    is somewhat unclear to me. If you take a piece of wire and apply an emf by connecing both ends to the two ends of a battery, that emf is present, irrespective of the v0.

    If it helps, you could think of a conducting wire as if it were a long pipe filled with short springs of a little less than the inner diameter: push in a spring at one end, out comes one at the other. But it takes quite a while before the spring you press in re-appears. (warning: this analogy doesn't go very far...)
     
  5. Mar 3, 2014 #4
    Ah yes, of course, <v_o> = 0
    Thanks for pointing that out. However, I think my question still holds: in the absence of an electric field the velocity of a charged particle is random and thus <v_o> = 0. However, to derive the acceleration you have to assume an electric field since F = qE. How do you reconcile the two statements? Thanks.
     
  6. Mar 3, 2014 #5

    BvU

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    I have difficulty understanding where exactly do you think there is a contradiction ? The electric field comes from outside, not from the electrons themselves.
     
  7. Mar 4, 2014 #6
    I know the electric field is external. So perhaps another way of wording the question is: <v_o> = 0 in the presence of an E-field? I don't believe that is true. That is why we say in the absence of an electric field the velocity of a charged particle is random and thus on average = 0. However, here we are assuming an external E-field and thus <v_o> should not be equal to 0.
     
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