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 782: 18.51a

Answer

$C = 1.14 \times 10^{-10} F$

Work Step by Step

Given: Consider a parallel-plate capacitor having an area of 3225 square millimeters (5 $in^2$), a plate separation of 1 mm (0.04 in.), and a material having a dielectric constant of 3.5 positioned between the plates. Required: capacitance of this capacitor Solution: Using a combination of Equations 18.26 and 18.27: $C = \frac{εA}{l} = \frac{ε_{r}ε_{0}A}{l} = \frac{(4)(8.85 \times 10^{-12} F/m)(3225 mm^{2})(1 m^{2}/10^{6}mm^{2})}{(1 \times 10^{-3} m)} = 1.14 \times 10^{-10} F$
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