GIVEN: The fluid within a living cell is rich inpotassium chloride, while the fl
ID: 1740117 • Letter: G
Question
GIVEN: The fluid within a living cell is rich inpotassium chloride, while the fluid outside it predominantlycontains sodium chloride. The membrane of a resting cell is farmore permeable to ions of potassium than sodium, and so there is atransport out of positive ions, leaving the cell interior negative.The result is a voltage of about -75.5mV across the membrane,called the resting potential. If the membrane is6.95nm thick, and assuming the electricfield E across it is constant, (1.)determine themagnitude of E.(2.)The membrane may be thought of as acapacitor: 2 charged surfaces filled with oil (dielectricconstant=2.94). Find its capacitance per unit area.
(3.)What is the restingmembrane's surface charge density?
(4.)Assume that themembrane's cytoplasmic (interior of cell) charge can be attributedto the presence of a certain fraction of negatively chargedphospholipid molecules, each with a cross-sectional area of0.555nm2. If each negatively charged lipid carries-1.60×1019C, how many such molecules arefound in 1.0m2 of the inner surface of themembrane?
(5.)What percentage of themembrane's inner surface do these molecules cover?
GIVEN: The fluid within a living cell is rich inpotassium chloride, while the fluid outside it predominantlycontains sodium chloride. The membrane of a resting cell is farmore permeable to ions of potassium than sodium, and so there is atransport out of positive ions, leaving the cell interior negative.The result is a voltage of about -75.5mV across the membrane,called the resting potential. If the membrane is6.95nm thick, and assuming the electricfield E across it is constant, (1.)determine themagnitude of E.
(2.)The membrane may be thought of as acapacitor: 2 charged surfaces filled with oil (dielectricconstant=2.94). Find its capacitance per unit area.
(3.)What is the restingmembrane's surface charge density?
(4.)Assume that themembrane's cytoplasmic (interior of cell) charge can be attributedto the presence of a certain fraction of negatively chargedphospholipid molecules, each with a cross-sectional area of0.555nm2. If each negatively charged lipid carries-1.60×1019C, how many such molecules arefound in 1.0m2 of the inner surface of themembrane?
(5.)What percentage of themembrane's inner surface do these molecules cover?
Explanation / Answer
a) Formula for calculating the value of the magnitude of theelectric field is E = V / d = (75.5mV) / (6.95nm) = (75.5*10-3V) / (6.95*10-9m) = 10.86*106 V/m b) We know the formula for the capacitance with dielectric isplaced in the between the plates of the capacitor is C = kAo / d C/A = ko / d ......................(1) Here k = 2.94 o =8.85*10-12C2/N.m2 d =6.95*10-9m c) Formula for calculating the value of the surface chargedensity is = Eo = (10.86*106V/m)(8.85*10-12C2/N.m2) = (10.86*106V/m)(8.85*10-12C2/N.m2)Related Questions
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