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Wed 23 Jan, 2008 09:34 pm
The fluid within a living cell is rich in potassium chloride, while the fluid outside it predominantly contains sodium chloride. The membrane of a resting cell is far more permeable to ions of potassium than sodium, and so there is a transport out of positive ions, leaving the cell interior negative. The result is a voltage of about -63.0mV across the membrane, called the resting potential. If the membrane is 7.85nm thick, and assuming the electric field E across it is constant, determine the magnitude of E.
Assume that the membrane's cytoplasmic (interior of cell) charge can be attributed to the presence of a certain fraction of negatively charged phospholipid molecules, each with a cross-sectional area of 0.550e-9m2. If each negatively charged lipid carries -1.60e-19C, how many such molecules are found in 1.0e-6m^2 of the inner surface of the membrane? What percentage of the membrane's inner surface do these molecules cover?
ok, I got E which is 8.03e6 N/C that's easy... but how do i start from here? I am so lost...
Find the total charge necesaary to get the voltage (or E), then divide that by the charge per lipid. That will give you the number of lipids. You can check this by doing the units: charge/(charge/lipid)=no. of lipids.
Then you have the number of lipids, and the cross-sectional size of one lipid, and you can find the density required to get the voltage, etc.
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