Question

An axon has the outer part of its membrane positively charged and the inner part negatively charged. The membrane has a thickness of \(4.4 \mathrm{nm}\) and a dielectric constant \(\kappa=5 .\) If we model the axon as a parallel plate capacitor whose area is \(5 \mu \mathrm{m}^{2},\) what is its capacitance?

Short answer

Answer: The capacitance of the axon modeled as a parallel plate capacitor is approximately \(5.01 \times 10^{-14} \mathrm{F}\).

Step-by-step solution

Write down the given values

Given the problem, we have the following values: - Thickness of the membrane (distance between the plates), d = \(4.4 \times 10^{-9} m\) - Dielectric constant, \(\kappa = 5\) - Area of the capacitor, A = \(5 \times 10^{-12} m^2\)

Use the vacuum permittivity constant

In order to determine the capacitance, we need to use the vacuum permittivity constant, \(\epsilon_0\). The value of vacuum permittivity is \(8.854 \times 10^{-12} \mathrm{F/m}\).

Calculate the capacitance using the formula

Now, we can use the formula for the capacitance of a parallel plate capacitor with a dielectric: $$ C=\kappa \epsilon_0\frac{A}{d} $$ Substituting the known values into the formula: $$ C= 5 \times (8.854 \times 10^{-12} \mathrm{F/m}) \times \frac{5 \times 10^{-12} m^2}{4.4 \times 10^{-9} m} $$

Solve for capacitance

Performing the multiplication and division, we get: $$ C \approx 5.01 \times 10^{-14} \mathrm{F} $$ The capacitance of the axon modeled as a parallel plate capacitor is approximately \(5.01 \times 10^{-14} \mathrm{F}\).