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Homework Help: Electric potential inside a living cell

  1. Jan 22, 2006 #1
    Electric potential inside a living cell is 0.075V lower than the electric potential ouside the cell. Thickness of membrane is 120nm.

    a) What is the magnitude and direction of the electric field within the cell membrane?

    Possible Ans: Is this just use the forumula

    E = -dV/ds = 6.25 x 10^5 V/m and its directed out

    b) For a cell of thickness 150nm and the same electric potential as above, calculate the work done (in joules) to move one sodium ion from inside of the cell to outside.

    Possible Ans: Is this using the forumla W = -qEs
    If it is what charge will I use for q? or is it just +1?
  2. jcsd
  3. Jan 22, 2006 #2


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    Homework Helper

    a)You've got the magnitude of the electric field correct, but the direction is inwards - which can be explained as follows. The electric field points from a region of high potential to a region of low potential - think of its gravitational equivalent - you have to push a mass (a positive test charge) up to a region of higher potential energy, that is up against the gravitational field. The natural motion of the mass is to move from a region of high potential to a region of low potential (for positive electrical charges) - that is in the direction of the field.

    b)The electric potential difference between two points in the electric field (inside cell to outside the cell) tells us how much energy or work is involved in moving a unit of electric charge (that is one coulomb) between the two points. Since a sodium ion, [itex]Na^+[/itex], has a positive charge of [itex]1.60 \times 10^{-19}\ C[/itex] it follows that the work that has to be done in order to push it out of the cell is given by
    [tex]W=\Delta Vq_e[/tex]
    this energy will have to come from the chemical processes inside of the cell in order to drive the ion out of it.
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