Physics: Principles with Applications (7th Edition)

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

Chapter 17 - Electric Potential - Search and Learn - Page 500: 6

Answer

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Work Step by Step

a. Let N electrons flow onto the plate. The charge on the capacitor is Q=Ne. However, we also know that Q=CV for a capacitor. $$Ne=CV$$ $$V=\frac{Ne}{C}$$ V is directly proportional to N, which was to be shown. b. We’re told that $\Delta V=1\;mV$. It is desired to resolve $\Delta N=1$. Solve for the required capacitance. $$V=\frac{Ne}{C}$$ $$\Delta V=\frac{e \Delta N}{C}$$ $$C=\frac{e \Delta N}{\Delta V }$$ $$C=\frac{(1.60\times10^{-19}C) (1)}{0.001 V }\approx 2\times10^{-16}F$$ c. Use the equation for the capacitance of a parallel-plate capacitor with a dielectric. $$C=\kappa\epsilon_o \frac{A}{d} =\kappa\epsilon_o \frac{\ell ^2}{d} $$ Solve for the side length of the square. $$\ell= \sqrt{\frac{Cd}{\kappa\epsilon_o }} $$ $$\ell= \sqrt{\frac{(1.60\times10^{-16}F) (100\times10^{-9}m)}{3(8.85\times10^{-12}C^2/(N\cdot m^2)) }} $$ $$\ell=7.76\times10^{-7}m=0.8\mu m$$
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