How Is Capacitance Calculated for an Axon Treated as a Parallel Plate Capacitor?

[sheri1987]

Homework Statement

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 *10-8 m) 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 5.50 * 10-6 m2, what is its capacitance?

Homework Equations

C= EoA/d where Eo= permittivity of free space = 1/(4*Pi*k) k= 8.99E9
A = area d= separation of plate

The Attempt at a Solution

I first solved for the permittivity of free space and got that answer to be 8.85E-12, then I plugged that in for Eo, 5.50E-6 for the A, and I was not sure what to do for the d? Use the 5? or 1 E-8? Can anyone help?

 

[Shooting Star]

> C= EoA/d where Eo= permittivity of free space

C= 5*EoA/d, where 5 is the dielectric constant, d=1*10^-8 m.

 

[Moetasim]

I would suggest using the 1E-8 value for the separation of plates (d) in this problem. This value is given in the homework statement and is likely a more accurate representation of the distance between the positively and negatively charged portions of the axon membrane. The 5 value may be the plate area, but it is not specified in the homework statement. Therefore, using the 1E-8 value for d, the capacitance of the axon membrane would be calculated as follows:

C = (8.85E-12 * 5.50E-6) / (1E-8) = 4.86E-5 Farads.

It's important to note that this calculation assumes ideal conditions and does not take into account any other factors that may affect the capacitance of the axon membrane.