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: 51

Answer

$5.2\times10^{-11}$m

Work Step by Step

For the force between charged particles we use $F = k\frac{Q_{1}Q_{2}}{r^{2}}$, where $k = 8.99\times10^{9}$. For circular motion we use $F = \frac{mv^{2}}{r}$. Combining these and solving for $r$ we get $r =k\frac{Q_{1}Q_{2}}{mv^{2}} $. We then substitute for all of our known values: $r = (8.99\times10^{9})\frac{(1.602\times10^{-19})(1.602\times10^{-19})}{(9.11\times10^{-31})(2.2\times10^{6})^{2}}$ $r = 5.2\times10^{-11}$m
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