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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 200.0 - \(\mu\) F capacitor is placed across a \(12.0-\mathrm{V}\) battery. When a switch is thrown, the battery is removed from the capacitor and the capacitor is connected across a heater that is immersed in $1.00 \mathrm{cm}^{3}$ of water. Assuming that all the energy from the capacitor is delivered to the water, what is the temperature change of the water?
A cell membrane has a surface area of \(1.0 \times 10^{-7} \mathrm{m}^{2},\) a dielectric constant of \(5.2,\) and a thickness of \(7.5 \mathrm{nm}\) The membrane acts like the dielectric in a parallel plate capacitor; a layer of positive ions on the outer surface and a layer of negative ions on the inner surface act as the capacitor plates. The potential difference between the "plates" is \(90.0 \mathrm{mV}\). (a) How much energy is stored in this capacitor? (b) How many positive ions are there on the outside of the membrane? Assume that all the ions are singly charged (charge +e).
A shark is able to detect the presence of electric fields as small as $1.0 \mu \mathrm{V} / \mathrm{m} .$ To get an idea of the magnitude of this field, suppose you have a parallel plate capacitor connected to a 1.5 - \(V\) battery. How far apart must the parallel plates be to have an electric field of $1.0 \mu \mathrm{V} / \mathrm{m}$ between the plates?
The potential difference across a cell membrane from outside to inside is initially at \(-90 \mathrm{mV}\) (when in its resting phase). When a stimulus is applied, Na" ions are allowed to move into the cell such that the potential changes to \(+20 \mathrm{mV}\) for a short amount of time. (a) If the membrane capacitance per unit area is $1 \mu \mathrm{F} / \mathrm{cm}^{2},\( how much charge moves through a membrane of area \)0.05 \mathrm{cm}^{2} ?\( (b) The charge on \)\mathrm{Na}^{+}\( is \)+e$ How many ions move through the membrane?
The nucleus of a helium atom contains two protons that are approximately 1 fm apart. How much work must be done by an external agent to bring the two protons from an infinite separation to a separation of \(1.0 \mathrm{fm} ?\)
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