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: 18.55b


$V = 139.07 V$

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: Voltage required if a vacuum were used Solution: Using the combination of Equations 18.24, 18.26 and18.27: $C = \frac{Q}{V} = ε\frac{A}{l} = ε_{r}ε_{0}\frac{A}{l}$ Solving for V if a vacuum was used: $V = \frac{Ql}{ε_{0}A} = \frac{(2.0 \times 10^{-10} C )(4 \times 10^{-3} m)}{(8.85 \times 10^{-12} F/m)(650 mm^{2})(1 m^{2}/10^{6}mm^{2})} = 139.07 V$
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