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 - Search and Learn - Page 472: 7

Answer

$5.71\times10^{13}C$.

Work Step by Step

The electrical force equals the gravitational attraction. $$F_E=F_G$$ $$k\frac{Q^2}{r^2}=G\frac{M_EM_m}{r^2}$$ $$Q=\sqrt{ G\frac{M_EM_m}{k}}$$ Use the data from the inside front cover of the book. $$Q=\sqrt{6.67\times10^{-11}\frac{(5.98\times10^{24}kg)(7.35\times10^{22}kg) }{8.988\times10^{9}}}=5.71\times10^{13}C$$
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