College Physics (4th Edition)

Published by McGraw-Hill Education
ISBN 10: 0073512141
ISBN 13: 978-0-07351-214-3

Chapter 17 - Problems - Page 654: 41


$q = 2 e$

Work Step by Step

We can find the magnitude of the charge on the drop: $F = q~E$ $F = q~(\frac{V}{d})$ $q = \frac{F~d}{V}$ $q = \frac{(9.6\times 10^{-16}~N)(0.16~m)}{480~J/C}$ $q = 3.2\times 10^{-19}~C$ We can express the magnitude of the charge in terms of $e$: $q = \frac{3.2\times 10^{-19}~C}{1.6\times 10^{-19}~C} = 2 e$
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