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: 60a

Answer

$Q=12\mu C$

Work Step by Step

We know that: $W=\frac{KeQ}{R}$ This can be rearranged as: $Q=\frac{WR}{Ke}$ We plug in the known values to obtain: $Q=\frac{2.16\times 10^{-13}\times 0.08}{9\times 10^9(1.6\times 10^{-19})}=12\times 10^{-6}=12\mu C$
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