The inner and outer
surface of a cell membrane carry a negative and a positive charge, respectively. Because of these
charges, a potential difference of about 0.070 V exists across the membrane. The thickness of
the cell membrane is 8.0 × 10−9m. What is the magnitude of the electric field in the membrane?
If the cell membrane were a vacuum or air, you could write:
E field = (voltage change)/(thickness)
The units are V/m, which is the same as Newtons/Coulomb
However, you need to know the dielctric constant of the cell membrane to solve this problem. It is not 1. It is probably in a 2-80 range, and will depend upon cell type. The E field gets reduced in inverse proportion to the dielectric constant.
so can I not solve this problem? My instructor assigned this and didn't give any other information. Alse the same problem is in the textbook with the same amount of information. Is there possibly another way to do it?
Ok assuming the dialectric is air the constant = 1.00054. What do I need to do after I find the Efield?
To find the magnitude of the electric field in the membrane, we can use the formula for electric field:
Electric Field (E) = Potential Difference (V) / Distance (d)
Given:
Potential Difference (V) = 0.070 V
Distance (d) = 8.0 × 10^(-9) m
Substituting these values into the formula, we get:
E = 0.070 V / (8.0 × 10^(-9) m)
Now, we can solve this equation to find the value of the electric field.