Fundamentals of Physics Extended (10th Edition)

Published by Wiley
ISBN 10: 1-11823-072-8
ISBN 13: 978-1-11823-072-5

Chapter 24 - Electric Potential - Problems - Page 715: 75



Work Step by Step

We know that $E=K\frac{Q}{R^2}$ The above formula can be written as; $Q=\frac{ER^2}{K}$ where $R=6400\times 10^3m$. $R$ is the radius of the Earth. We plug in the known values to obtain: $Q=\frac{100\times (6400\times 10^3)^2}{9\times 10^9}$ $Q=4.5511\times 10^5C$ Now we can determine electric potential as; $V=K\frac{Q}{R}$ We plug in the known values to obtain: $V=9\times10^9\times \frac{4.5511\times10^5}{6400\times 10^3}$ $V=6.4\times10^8V$
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