Physics: Principles with Applications (7th Edition)

Published by Pearson
ISBN 10: 0-32162-592-7
ISBN 13: 978-0-32162-592-2

Chapter 16 - Electric Charge and Electric Field - Problems - Page 470: 41

Answer

$8.3\times10^{-10}C$

Work Step by Step

This situation is covered by equation 16–10. The separation distance is a red herring. $$E=\frac{Q/A}{\epsilon_o}$$ $$Q=\epsilon_oEA$$ $$=(8.85\times10^{-12}C^2/(N \cdot m^2))(130N/C)(0.85m)^2$$ $$=8.3\times10^{-10}C$$
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