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: 47

Answer

$-6.8\times10^{5}C$

Work Step by Step

Because of the problem’s spherical symmetry, outside the Earth the electric field strength can be calculated using equation 16–4a. Solve for the charge. $$E=k\frac{Q}{r^2}$$ $$Q=\frac{Er^2}{k}$$ $$Q=\frac{(150N/C)(6.38\times10^6m)^2}{8.988\times10^{9}(N \cdot m^2)/C^2}$$ Since the field points toward the ball, the charge must be negative. $$Q=-6.8\times10^{5}C $$
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