Physics: Principles with Applications (7th Edition)

Published by Pearson
ISBN 10: 0-32162-592-7
ISBN 13: 978-0-32162-592-2

Chapter 18 - Electric Currents - Search and Learn - Page 525: 8

Answer

a) $1.7\times 10^{-8}\;\rm F$ b) $3.66\times10^{-5}\;\rm ion/m^3$

Work Step by Step

a) Let's assume that the shape of the axon is just like a cylinder. So, we can assume it is a cylindrical capacitor with two parallel plates. The capacitance of this kind of capacitor is given by $$C=\dfrac{k\varepsilon_0 A}{d}$$ The area of the plate inside this capacitor is the surface area of the cylinder which is given by $A=2\pi r h$ where $h$ is its height or length. Thus, $$C=\dfrac{2\pi k\varepsilon_0 rh }{d}=\dfrac{2\pi \cdot 3\cdot 8.85\times10^{-12}\cdot 10\times 10^{-6}\cdot 10\times10^{-2}}{10^{-8}}$$ $$C=\color{red}{\bf 1.7\times 10^{-8}}\;\rm F$$ b) We know that the potential difference change in the axon for one action is about 0.1 V. So we can calculate the change in charge during the action which is given by $$\Delta Q=C\Delta V=1.7\times 10^{-8}\cdot 0.1=\bf1.7\times 10^{-9} \;\rm C$$ Now we know that each $\rm Na^+$ loses an electron which means it is having a charge of $q=+1.6\times10^{-19}\;\rm C$, thus the number of ions moved is given by $$N=\dfrac{Q}{q}=\dfrac{1.7\times 10^{-9} }{1.6\times10^{-19}}=\bf 1.04\times10^{10}\;\rm Na^+$$ We know that the concentration of $\rm Na^+$ in the axon is about 15 mol/m$^3$ Thus, $$n=\dfrac{N}{V_{axon}}=\dfrac{N}{15N_AAh}$$ whereas $n$ is the number of ions per volume, and $N_A$ is Avogadro's number. Plugging the known; $$n=\dfrac{1.04\times10^{10}}{15\cdot 6.022\times10^{23}\cdot \pi (10\times10^{-6})^2\cdot 10\times10^{-2}}$$ $$n=\color{red}{\bf3.66\times10^{-5}}\;\rm Na^+/m^3$$
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