Assume a nerve cell is shaped like a long cylinder enclosed by a thin lipid bila
ID: 1486993 • Letter: A
Question
Assume a nerve cell is shaped like a long cylinder enclosed by a thin lipid bilayermembrane of thickness d. Assume the nerve cell is 0.01mm in diameter and 10 mm inlength. You can treat the cylindrically shaped lipid bilayer as a capacitor with capacitance.
Where A is the surface area of the cylindrical nerve cell, D is the dielectric constantwithin the lipid bilayer and d is the thickness of the lipid bilayer. Assume D=2, andd=5nm. Calculate the capacitance of the nerve cell. How many unipositive charge carriers must cross the cell membrane to give the interior of the cell a potential of 0.1V?
For the first part of the question, I don't know what epsilon is so I'm not sure how to get capacitance. As for the second part about unipositive charge carriers, not sure how to approach that one.
Explanation / Answer
surface area = 2*pi*r*L
= 2*pi*(0.01*10^-3/2)*10*10^-3
= 3.14*10^-7 m^2
D = 2
d = 5 nm = 5*10^-9 m
C = A*epsilon*D/d (here epsilon id permitivity of free space)
= 3.14*10^-7*8.854*10^-12*2/(5*10^-9)
= 1.11*10^-9 F <<<<<<<<<<<-------------Answer
Q = C*V
= 0.1*1.11*10^-9*
= 1.11*10^-10 C
no of unipositive charge carriers must cross the cell membrane,
N = Q/e
= 1.11*10^-10/(1.6*10^-19)
= 6.938*10^8 <<<<<<<<<<<-------------Answer
Related Questions
drjack9650@gmail.com
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.