Materials Science and Engineering: An Introduction

Published by Wiley
ISBN 10: 1118324579
ISBN 13: 978-1-11832-457-8

Chapter 18 - Electrical Properties - Questions and Problems - Page 781: 18.49


$l_{2} = 3.36 mm$

Work Step by Step

Given: A parallel-plate capacitor using a dielectric material having an $ε_{r}$ of 2.2 has a plate spacing of 2 mm (0.08 in.). Required: The new spacing between the plates if another material having a dielectric constant of 3.7 is used and the capacitance is to be unchanged Solution: Using a combination of Equations 18.26 and 18.27, we find: $C = \frac{εA}{l} = \frac{ε_{r}ε_{0}A}{l}$ Since $C_{1} = C_{2}$, let subscripts 1 and 2 be the initial and final state: $\frac{ε_{r1}ε_{0}A}{l_{1}} = \frac{ε_{r2}ε_{0}A}{l_{2}}$ $l_{2} = \frac{ε_{r2}l_{1}}{ε_{r1}} = \frac{(3.7)(2 mm)}{2.2} = 3.36 mm$
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