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Current, I'm not sure about my solution

  1. Dec 13, 2008 #1
    Earth's lower atmosphere contains negative and positive ions that are produced by radioactive elements in the soil and cosmic rays from space. In a certain region, the atmospheric electric field strength is 120V/m and the field is directed vertically down. This field causes singly charged positive ions, at a density of 620cm-3, to drift downward and singly charged negative ions, at a density of 550 cm-3, to drift upward. The measured conductivity of the air in the region is 2.70*10-14. Calculate a) the magnitude of the current density and b) the ion drift speed, assumed to be the same for positive and negative ions.

    E=120V/m; n+ = 620cm-3=620*106m-3; n- = 550cm-3 = 550*106m-3

    σ=2.7*10-14 (Ω*m)-1

    σ = 1/ρ; ρ=E/J then 1/ρ = J/E

    2.7*10-14 = J/120

    J=2.7*120*10=3.24*10-12 A/m2 ===== a)


    J=nevd = (n+ - n-)*e*vd

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

    vd=0.289m/s

    Am I correct? Is there anything wrong in my solution. Please tell me.

    Thank you.
     
  2. jcsd
  3. Dec 13, 2008 #2

    Redbelly98

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    Looks good. :smile:

    Just a minor point, it's generally preferred to manipulate symbols algebraically instead of the numbers. For example, make it:

    J=Eσ

    and then plug in the numbers.

    You won't get any points off in an introductory physics class for doing it your way, but generally you're less likely to make an error if the number-plugging is done as the final step.

    Good job!
     
  4. Oct 23, 2011 #3
    How does this solution look good? I have plugged in these same numbers and you do not get the same number.
     
  5. Oct 24, 2011 #4

    Redbelly98

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    Can you show what you did? There was a typo in the OP, but the final answer was correct as far as I can tell. The correct value of (n+-n-) is really

    (n+ - n-) = (620-550) cm-3 = (620-550)*10+6 m-3
     
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