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

AI Thread Summary
Capacitance for an axon treated as a parallel plate capacitor can be calculated using the formula C = EoA/d, where Eo is the permittivity of free space, A is the area of the plates, and d is the separation between them. The dielectric constant of the axon membrane is 5, and its thickness is 1 * 10^-8 m. The area of the axon is given as 5.50 * 10^-6 m2. To find the capacitance, the correct values must be substituted into the formula, specifically using the dielectric constant to adjust Eo. The final capacitance calculation will provide insight into the electrical properties of the axon membrane.
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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?
 
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> C= EoA/d where Eo= permittivity of free space

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

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