Question

The net charge difference across the membrane, just like the charge difference across the plates of a capacitor, is what leads to the voltage across the membrane.

How much excess charge (in picocoulombs, where 1 pc = 1x10^-12 C) must lie on either side of the membrane of an axon of length 1 cm to provide this potential difference (0.07 V) across the membrane? You may consider that a net positive charge with this value lies just outside the axon cell wall, and a negative charge with this value lies just inside cell wall, like the equal and opposite charges on capacitor plates.

Answer 1

Treating the axon as a capacitor, the charge on it (net charge difference across its membrane) is:

Q = CV

here, V = 0.07 volts

capacitance of a parallel plate capacitor is given by:

A = surface area of the cylindrical axon =

where r = radius of the axon and l = 1 cm = 0.01 m

d = thickness of the membrane on it and dielectric constant of the lipid bilayer of axon is approximately 6.

and so,

therefore, Q = CV = 194.71 pC.