Electric field of a cell membrane

[throllen]

Homework Statement

he 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?

Q=CV
E=sigma/e0
sigma=Q/a

ok, I got E which is 8.03e6 N/C that's easy... but how do i start from here? I am so lost... I tried using Q=CV, and i got 1410 as the answer for how many of those molecules are in 1e-6 m^2 of membrane, but isn't the % 77.6%? I keep getting it wrong...

 

[cwoo123456]

How did you get the answer for the molecules??

 

[thomate1]

I would approach this problem by first understanding the concept of electric fields and their role in cell membranes. The electric field is a physical quantity that describes the intensity of an electric force at a given point in space. In the case of a cell membrane, the electric field is created by the separation of charged ions on either side of the membrane.

Based on the given information, we know that the resting potential across the membrane is -63.0mV. This potential is created by the movement of positively charged potassium ions out of the cell, leaving the inside of the cell negatively charged. This results in an electric field that is constant across the membrane.

To determine the magnitude of this electric field, we can use the equation E=sigma/e0, where sigma is the charge density and e0 is the permittivity of free space. From the given information, we can calculate the charge density sigma as Q/A, where Q is the total charge and A is the area of the membrane. Using the given values, we can calculate the charge density as 8.03e6 N/C.

Moving on to the second part of the problem, we are asked to determine the number of negatively charged phospholipid molecules on 1.0e-6m^2 of the inner surface of the membrane and the percentage of the inner surface that these molecules cover. To solve this, we can use the formula Q=CV, where Q is the total charge, C is the capacitance, and V is the potential. We know that the potential is -63.0mV and the capacitance can be calculated as C=e0A/d, where e0 is the permittivity of free space, A is the area of the membrane, and d is the thickness of the membrane.

Using the given values, we can calculate the capacitance as 1.92e-19 F. Substituting this value and the potential into the equation Q=CV, we can solve for the total charge Q. From this, we can calculate the number of negatively charged phospholipid molecules using the charge of each molecule (-1.60e-19C). The result is 1410 molecules on 1.0e-6m^2 of the inner surface of the membrane.

To determine the percentage of the inner surface that these molecules cover, we can use the formula Acovered/Atotal * 100%. The total area of the