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Chapter 17: Problem 103

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

Expert verified
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

01

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 = \(5 \times 10^{-12} m^2\)
02

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 \(8.854 \times 10^{-12} \mathrm{F/m}\).
03

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: $$ C= 5 \times (8.854 \times 10^{-12} \mathrm{F/m}) \times \frac{5 \times 10^{-12} m^2}{4.4 \times 10^{-9} m} $$
04

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 \(5.01 \times 10^{-14} \mathrm{F}\).

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Most popular questions from this chapter

A parallel plate capacitor has a capacitance of \(1.20 \mathrm{nF}\) There is a charge of \(0.80 \mu \mathrm{C}\) on each plate. How much work must be done by an external agent to double the plate separation while keeping the charge constant?
An alpha particle (charge \(+2 e\) ) moves through a potential difference \(\Delta V=-0.50 \mathrm{kV} .\) Its initial kinetic energy is $1.20 \times 10^{-16} \mathrm{J} .$ What is its final kinetic energy?
In \(1911,\) Ernest Rutherford discovered the nucleus of the atom by observing the scattering of helium nuclei from gold nuclei. If a helium nucleus with a mass of \(6.68 \times 10^{-27} \mathrm{kg},\) a charge of \(+2 e,\) and an initial velocity of \(1.50 \times 10^{7} \mathrm{m} / \mathrm{s}\) is projected head-on toward a gold nucleus with a charge of \(+79 e\), how close will the helium atom come to the gold nucleus before it stops and turns around? (Assume the gold nucleus is held in place by other gold atoms and does not move.)
A parallel plate capacitor is connected to a \(12-\mathrm{V}\) battery. While the battery remains connected, the plates are pushed together so the spacing is decreased. What is the effect on (a) the potential difference between the plates? (b) the electric field between the plates? (c) the magnitude of charge on the plates?
A 6.2 -cm by 2.2 -cm parallel plate capacitor has the plates separated by a distance of \(2.0 \mathrm{mm} .\) (a) When \(4.0 \times 10^{-11} \mathrm{C}\) of charge is placed on this capacitor, what is the electric field between the plates? (b) If a dielectric with dielectric constant of 5.5 is placed between the plates while the charge on the capacitor stays the same, what is the electric field in the dielectric?
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