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Electric Fields: Sodium Ions pumped

  1. Feb 22, 2009 #1
    1. The problem statement, all variables and given/known data
    Living cells actively “pump” positive sodium ions (Na+) from inside the cell to outside
    the cell. It’s called pumping because it requires work to move the ions from the
    negatively charged inner surface of the cell membrane to the positively charged outer
    surface. It is estimated that as much as 20% of the energy we consume in a resting state
    is used in this sodium pumping. The potential difference across the membrane is 0.070V,
    and the membrane is 0.10μm thick.
    a.) Which is at higher potential, the inside or outside of the membrane?
    b.) How much work has to be done to move one sodium ion from inside the
    membrane to outside?
    c.) If the thickness of the membrane were doubled, how much work would it require
    to pump out one sodium ion?
    d.) Assuming the E field is constant in the membrane, what is the magnitude and
    direction of the E field between the inside and outside of the membrane?



    2. Relevant equations
    None Given


    3. The attempt at a solution
    I have attempted this but I am having trouble starting out correctly. Like which is the proper equation to use...my general understanding is not jump starting this problem.
     
  2. jcsd
  3. Feb 22, 2009 #2

    LowlyPion

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    First of all they give you the answer to a) don't they?
    So what is that?

    Second of all if you know the voltage change and you know the charge don't you know the work required to affect the ion transfer?
     
  4. Feb 22, 2009 #3
    Yea I now know A, which is outside. And the equation I am using is the work function W=qEd...assuming q is the charge of the ion,1. then E=V/d...plugging all that in gets me back to the original voltage of .070V.
     
  5. Feb 22, 2009 #4

    LowlyPion

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    More to the point, don't you think, work is q*ΔV, since work is over the distance?
     
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