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Electric field & Parallel plate Capacitor

  1. Jan 8, 2006 #1
    Question When a nerve impulse propagates along a nerve cell, the electric field within the cell membrane changes from 7.0 x 10^5 N/C (pointing in one direction) to 3.0 x 10^5 N/C (pointing in the other direction). By approximating the cell membrane as a parallel-plate capicator, find the magnitude of the change in charge density on the walls of the membrane
     
  2. jcsd
  3. Jan 9, 2006 #2
    Has anyone ANY ideas?
     
  4. Jan 9, 2006 #3

    berkeman

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    Staff: Mentor

    What are the typical dimensions of such a cell? What is the typical thickness of the membrane? Do you know the formula for the capacitance of a parallel plate capacitor in terms of C, epsilon, Area and separation Distance? (ignoring fringe capacitance effects)
     
  5. Jan 10, 2006 #4
    Oh sorry, the thickness is 0.12 micrometres
    and the parallel plate capacitor has a surface charge density of 5.9×10^-6 C/m2
     
  6. Jan 10, 2006 #5

    berkeman

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    Staff: Mentor

    You still need to address this part....
     
  7. Jan 10, 2006 #6
    No I don't. What is it?
     
  8. Jan 10, 2006 #7

    berkeman

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    Staff: Mentor

    C = epsilon * A / d

    epsilon = permittivity
    A = area of one parallel plate
    d = separation of the parallel plates.

    So you can take the typical dimensions for the cell membrane configuration, and calculate the capacitance (you'll need to approximate the epsilon of the cell membrane -- I have no idea what it is), and that will give you the equivalent capacitance. Given the change in electric field, you can calculate what charge change it takes to make that happen. That and the area will give you the charge densities....
     
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