Assume the length of an axon membrane of about 0.10 cm is excited by an action p

Question

Assume the length of an axon membrane of about 0.10 cm is excited by an action potential (length excited = nerve speed pulse duration = 50 m/s 2.0 ms = 10 cm). 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 = oA/d and Q = CV to investigate the charge as follows. Use typical values for a cylindrical axon of cell thickness d = 1.1 108 m, axon radius r = 1.2 101 m, and cell-wall dielectric constant = 2.9.

(a) How much energy does it take to restore the inner wall of the axon to 7.0 102 V from +3.0 102 V? (Assume that no energy is required to first restore the potential to 0 V from the excited potential of +3.0 102 V.)

A: __________ J

(b) Find the average current in the axon wall during this process. (Hint: The time required to return to the resting state is 3.0 ms.)

A: _________ A

A Hint From My Professor:

The capacitance C = 0 A/d , ( A=L/2Pi r)
a. The potential energy of the capacitor equation ( use V=7e-2volt)
b. the charge relation with the capacitance and the potential
first, for resting state, the charge (Qi) on the outer surface of the
membrane ( use V=7e-2)
second, the net positive charge (Qf) on this surface in the excited
state(use V=3e-2volt)
Then find delta Q ( change in charge)
Then use the equation relates I, Q, t to find I

Explanation / Answer

C = oA/d ; A= 2*pi*r*L; L=10cm =0.1 m

C= 1.76 nF

potential difference(dV)= 7.0 102

Energy= C*(dV)2/2 = 4.31 x 10-12 J

b) Qi= CVo ; Qf= C(dV)

charge flowed(Q)= Qf-Qi = C( dV-Vo)

average current=Q/t ; t-time

I= 5.867 x 10-8 A

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