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Charge Density and Neuronal Cells-Help!

  1. Jan 10, 2007 #1
    Charge Density and Neuronal Cells--Help!

    The cell membrane if a typical nerve cell consists of an inner and outer wall separated by a distance of 0.10 µm. The electric field within the cell membrane is 7.0 x 105 N/C. Approximating the cell membrane as a parallel plate capacitor, determine the magnitude of the charge density on the inner and outer cell walls.




    E=charge density/Eo
    Q=charge density*Area



    I tried doing this problem by using E=Charge density/Eo and just solving for charge density. I'm not really quite sure how to find the area to use in the equation (I know its cylindrical, but I'm not sure how I can find the area with just the distance)

    Thank you!
     
  2. jcsd
  3. Jan 11, 2007 #2
    bump for answer :smile:
     
  4. Jan 11, 2007 #3

    Kurdt

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    Well using the first equation rearrange for charge density because thats all that is asked for. You know the E-field and epsilon nought is a constant. It is however a very crude approximation as the cell is more cylindrical than anything else and the gap is not a vacuum.
     
  5. Jan 11, 2007 #4
    That is what i did. I solved for charge density by the equation derived in my text book for cylindrical approximations and got 1.239x10^-5 C/m^2 when using the equation, Charge Density = E*2Eo. However, when I type this answer in (it's online homework), it keeps saying it is incorrect, which is making me wonder if I am missing something??
     
  6. Jan 11, 2007 #5

    Kurdt

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    The question states treating it as a parallel plate capacitor not a cylindrical one. For a parallel plate the charge density will be half the value you have obtained.
     
  7. Jan 11, 2007 #6
    Nevermind, I got it, FINALLY!

    I had to solve for Q using E=k*(Q/r^2) then find Area and solve for charge density by Q/A.
     
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