A nerve cell in its resting state has a membrane potential of -70 mV, meaning th
ID: 1264565 • Letter: A
Question
A nerve cell in its resting state has a membrane potential of -70 mV, meaning that the potential inside the cell is 70 mV less than the potential outside due to a layer of negative charge on the inner surface of the cell wall and a layer of positive charge on the outer surface. This effectively makes the cell wall a charged capacitor. When the nerve cell fires, sodium ions, Na+, flood through the cell wall to briefly switch the membrane potential to +40mV. Model the central body of a nerve cell-the soma-as a 50-?m-diameter sphere with a 7.0-nm-thick cell wall whose dielectric constant is 9.0. Because a cell's diameter is much larger than the wall thickness, it is reasonable to ignore the curvature of the cell and think of it as a parallel-plate capacitor.
How many sodium ions enter the cell as it fires?
Explanation / Answer
a) it is not reasonable.
beause for a parallel plate capacitor, C = A*epsilon/d
for spherical cpacitor, C = 4*pi*epsilon/(1/a - 1/b)
a --> inner radius
b --> outer radius
b) C = 4*pi*epsilon/(1/a - 1/b)
= 4*pi*8.854*10^-12/( 1/(25*10^-6) - 1/(28.5*10^-6) )
= 2.264*10^-14 F
Q = C/V
= 2.264*10^-14/(70*10^-3)
= 3.23*10^-13 C
no of ions, N = Q/e
= 3.23*10^-13/(1.6*10^-19)
= 2.02*10^6 ions
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