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: 13a

Answer

$q=3.3nC$

Work Step by Step

We know that $V=K\frac{q}{r}$ This can be written as: $q=\frac{V r}{K}$ We plug in the known values to obtain: $q=\frac{200\times 0.15}{9\times 10^9}$ $q=3.3\times 10^{-9}$ $q=3.3nC$
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