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Question

An axon is the relatively long tail-like part of a neuron, or nerve cell. The outer surface of the axon membrane (dielectric constant = 5, thickness = 1 10-8m) is charged positively, and the inner portion is charged negatively. Thus, the membrane is a kind of capacitor. Assuming that an axon can be treated like a parallel plate capacitor with a plate area of 5.8 10-6 m2, what is its capacitance?

Answer 1

C=eA/d

C is the capacitance in farads, F
e is the permittivity of the insulator used (or ε0 for a vacuum)
A is the area of each plane electrode, measured in square metres
d is the separation between the electrodes, measured in metres


e0= 8.8541878176 × 10−12 F/m (C2/Jm),
e=er e0
er- relative permeability or relative dialectric constant (http://en.wikipedia.org/wiki/Dielectric_constant)

Now
C=er e0 A / d
Finally

C=5.0 x 8.85e−12 x 5.8e-6 /1e-8 =
C=25.7 e-9 F = 25.7 nF