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: 18.55e

Answer

$P = 2.20 \times 10^{-7} C/m^{2}$

Work Step by Step

Given: A charge (Q) of $2.0 \times 10^{-10} C $ is to be stored on each plate of a parallel-plate capacitor having an area (A) of $650 mm^{2} (1.0 in^{2}) $ and a plate separation (l) of 4.0 mm (0.16 in.). Required: polarization in part a (V = 39.73 V) Solution: Using a combination of Equations 18.32 and 18.6: $P = ε_{0}(ε_{r} - 1)E = ε_{0}(ε_{r} - 1)\frac{V}{l} = \frac{(8.85 \times 10^{-12} F/m)(3.5-1)(39.73 V)}{(4 \times 10^{-3} m)}\\ = 2.20 \times 10^{-7} C/m^{2}$
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