The outer surface of the axon membrane (dielectric constant = 5, thickness = 1 x

Question

The outer surface of the axon membrane (dielectric constant = 5, thickness = 1 x 10^-8 m) is charged positively, and the inner portion is charged negatively. Assuming the membrane acts like a parallel plate capacitor with a plate area of 5 x 10^-6 m^2, what is its capacitance in Farads? Epsilon = 8.85 x 10^-12 C^2/(N-m^2) The outer surface of the axon membrane (dielectric constant = 5, thickness = 1 x 10^-8 m) is charged positively, and the inner portion is charged negatively. Assuming the membrane acts like a parallel plate capacitor with a plate area of 5 x 10^-6 m^2, what is its capacitance in Farads? Epsilon = 8.85 x 10^-12 C^2/(N-m^2) The outer surface of the axon membrane (dielectric constant = 5, thickness = 1 x 10^-8 m) is charged positively, and the inner portion is charged negatively. Assuming the membrane acts like a parallel plate capacitor with a plate area of 5 x 10^-6 m^2, what is its capacitance in Farads? Epsilon = 8.85 x 10^-12 C^2/(N-m^2)

Explanation / Answer

Given that
A = 5*10^-6 m^2

k = 5

d = 1*10^-8 m

e0 = 8.854*10^-12

We ues capacitance formula

C = k*e0*A /d

= (5)*( 8.854*10^-12 )*(5*10^-6) / 1*10^-8

= 2.2135*10*-8

C = 22.14 nF

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