Physics Technology Update (4th Edition)

Published by Pearson
ISBN 10: 0-32190-308-0
ISBN 13: 978-0-32190-308-2

Chapter 20 - Electric Potential and Electric Potential Energy - Problems and Conceptual Exercises - Page 720: 60

Answer

a) $A=0.0065m^2$ b) $V_{max}=7.8KV$

Work Step by Step

(a) We know that $A=\frac{Cd}{k\epsilon_{\circ}}$ We plug in the known values to obtain: $A=\sqrt{\frac{22\times 10^{-12}(2.6\times 10^{-3})}{(1.0059)(8.85\times 10^{-12})}}$ $A=0.0065m^2$ (b) As we know that $V_{max}=E_{max}d$ We plug in the known values to obtain: $V_{max}=(3.0\times 10^{6})(2.6\times 10^{-3})$ $V_{max}=7.8KV$
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