College Physics (7th Edition)

Published by Pearson
ISBN 10: 0-32160-183-1
ISBN 13: 978-0-32160-183-4

Chapter 16 - Electric Potential, Energy, and Capacitance - Learning Path Questions and Exercises - Exercises - Page 593: 36



Work Step by Step

Given/known: $\kappa=1$ (as air is in between the plates), $A=0.525\,m^{2},$ $d=2.15\,mm=2.15\times10^{-3}\,m$ and $\varepsilon_{0}=8.854\times10^{-12}\,F/m$ Recall: $C=\frac{\kappa\varepsilon_{0}A}{d}$ Result: $C=\frac{1\times8.854\times10^{-12}\,F/m\times0.525\,m^{2}}{2.15\times10^{-3}\,m}=2.16\times10^{-9}\,F$
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