Assume a length of axon membrane of amount 0.10 m is excited by an action potent

Question

Assume a length of axon membrane of amount 0.10 m is excited by an action potential (length excited = nerve speed times pulse duration = 50.m/s times 2.0 times 10^3 s = 0.10 m). In the resting state, the outer surface of the axon wall is charged positively with k^+ ions and the inner wall has an equal and opposite charge of negative organic ions, as shown in the figure below. Model the axon as a parallel-plate capacitor and take c = kepsilon_0A/d and Q = CdeltaV to investigate the charge as follows. Use typical values for a cylindrical axon of cell wall thickness d= 1.8 times 10^-8 m, axon radius r = 1.9 times 10^1 mum, and cell-wall dielectric constant k = 2.3.

Explanation / Answer

capacitance of a parallel plate capacitor is dielectric constant*epsilon*area/distance

here area=pi*r^2=pi*(1.9*10^1*10^(-6))^2=1.1341*10^(-9) m^2

capacitance=2.3*8.85*10^(-12)*1.1341*10^(-9)/(1.8*10^(-8))=1.2825 pF

part a:

potential difference=0.07 volts

then charge stored=potential difference*capacitance=1.2825*10^(-12)*0.07=8.9774*10^(-14) C

each ion has charge of one proton=1.6*10^(-19) C

then number of ions=total charge/charge on each ion=8.9774*10^(-14)/(1.6*10^(-19))=5.61*10^5

charge per area=charge/area

=8.9774*10^(-14)/(1.1341*10^(-9))=7.9158*10^(-5) C/m^2

1 e/A^2 (where A=1 angstrom=10^(-10) m)

=1.6*10^(-19) C/(10^(-20) m^2)

=16 C/m^2

hence the charge density present is very less.

part b:

total voltage difference increase=0.03-(-0.07)=0.1 V

hence charge flow=capacitance*potential difference=1.2825*10^(-12)*0.1=1.2825*10^(-13) C

number of Na+ ions=total charge/charge on one ion=1.2825*10^(-13)/(1.6*10^(-19))=8.0156*10^5

part c:
average current=charge/time

=1.2825*10^(-13)/(2*10^(-3))=6.4125*10^(-11) A

=6.4125*10^(-5) uA

part d:

energy required=charge flown^2/(2*capacitance)

=(1.2825*10^(-13))^2/(2*1.2825*10^(-12))=6.4125*10^(-15) J

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