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 - General Problems - Page 471: 55

Answer

$0.14N$ ,rightward

Work Step by Step

There will be a rightward force$F_1$ on $Q_1$ due to $Q_2$ as they are oppositely charged. There will be another force $F_2$towards left on $Q_1$ due to electric field. Let the rightward force be positive. Thus, $F_{net}=F_1+F_2$ $\implies F_{net}=\frac{KQ_1Q_2}{x^2}-Q_1E$ We plug in the known values to obtain: $F_{net}=\frac{8.99\times 10^9(6.7\times 10^{-6}\times 1.8\times 10^{-6})}{(0.47)^2}-6.7\times 10^{-6}\times 53000=0.14N$
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