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Ion drift speed

  1. Oct 23, 2011 #1
    Help much needed! Thanks!


    So this problem has been dealt with here on the physics forums. But the fact is that answers either appear to be false and/or they differ from what other people have done.

    First, here's the problem:

    In a certain region, the atmospheric electric field strength is 120 V/m and the field is directed vertically downward. This field causes singly charged positive ions, at a density of 620cm^-3, to a drift downward and singly charged negative ions, at a density of 550cm^-3, to drift upward. The measured conductivity of air in that region is 2.70x10^-14 (ohm*m)^-1. Calculate the magnitude of the current density and ion drift speed. Assume it is the same for positive and negative ions.



    3. The attempt at a solution

    [1] Here's the first attempt:

    v = J / (e(n1-n2))


    J = conductivity * E = 2.7*10^-14 * 120 = 3.24*10^-12 <<------- correct
    v = J / ne
    Here we have two values for n, one of positive ions and one of negative ions. So I figured I could write:
    v = J / (e(n1-n2))
    where n1 represents the density of the positive ions and n2 of the negative.

    v = 0.393258427 m/s

    [2] Second attempt:
    J=nevd = (n+ n-)*e*vd // this time we're adding n.

    3.24*10^-12 = (620-550)*10^-6*1.602*10^-19*vd

    v=0.289m/s.

    [3] Third attempt:
    (n_total)*e*vd = J // adding n again.

    v= 3.24x10^-12/ [(6.2+5.5)*10^8 (1.6x10^-19)]
    v =1.73cm/s


    So what is v (the velocity of the drift)? Some people add the n's. Some subtract. Which one to use? Other times, I can't tell if my algebra is off (other theirs). Cause I can't get their answer.

    Also, how do we know when to add n- and n+) and when to subtract. (n is the number of carriers per unit volume).
     
    Last edited: Oct 23, 2011
  2. jcsd
  3. Oct 23, 2011 #2

    SammyS

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    Why isn't it (e)n1-(-e)n2 = (e)(n1 + n2), because they go in opposite directions.

    Looking at it another way, positive current flowing downward is equivalent to negative current flowing upward.
     
  4. Oct 23, 2011 #3
    Thanks.

    But that still doesn't get to the heart of the matter. It turns out that we are adding (via the proffered equation) because the of the double negative, ie., we are subtracting and e is negative. Well, e is always going to be negative when talking about electrons. I assume that is why it is (-e)n2, rather than en2. Given that, the question remains:
    why are subtracting in the first place? Why isn't it en1+(e)n2?

    And what physical circumstance does that apply to?

    Lastly, is positive current flowing upward equivalent to negative current flowing downward?
    (Btw, the last part of your response made no sense to me because it would seem that the positive and negative current would start to cancel each other out, so you would subtract the two.)

    In short, I have two options: just accept the formula or try to see if there is something deeper, more intuitive that explains the formula. So far, the result you have given me (which I know to be correct) seems unintuitive and "out of the blue" so-to-speak. I would like to correct that.
     
  5. Oct 24, 2011 #4

    gneill

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    You know the net current density, J, as specified by the electric field and conductivity. In this case it represents a net drift of conventional-current charge carriers (e+) downwards.

    It is a true statement that a downward flow of positive charge carriers is equivalent to an upward flow of negative charge carriers. This is at the heart of the equivalence between conventional current flow and electron current flow in wires -- everyone knows it's really electrons flowing in the opposite direction, but the math just doesn't care! This carries over into situations where the charge carriers really are positive charges (like ions in solutions in electrochemistry, or in plasmas), or when there's a mix of charges both positive and negative moving in opposite directions (such as electrons and "holes" in semiconductors).

    What matters is the net movement of charge from one point to another, and we usually do our sums in terms of conventional current. In this case we have positive charges moving downwards (just like conventional current assumes), and negative charges moving upwards --- which is, for all intents and purposes to the accountants, the same as MORE positive charges moving downwards; If the flow of positive charges from top to bottom leaves the top less positive and the bottom more positive, that's the same as saying it leaves the top more negative and the bottom less negative, a result that could be achieved equally well by moving negative charges from the bottom to the top.

    So, on to the accounting. The total current density, specified as a conventional current per square meter of cross section, will be given by

    [itex] J = \sigma_{e^+}(e)V + \sigma_{e^-}(-e)V [/itex]

    where the σ's are the volumetric charge densities, e the elementary charge, and V the drift velocity.
     
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