Physics for Scientists and Engineers: A Strategic Approach with Modern Physics (4th Edition)

Published by Pearson
ISBN 10: 0133942651
ISBN 13: 978-0-13394-265-1

Chapter 37 - The Foundation of Modern Physics - Exercises and Problems - Page 1082: 11


$r = 5.21\times 10^{-7}~m$

Work Step by Step

The force exerted upward on the droplet by the electric field must be equal in magnitude to the weight of the droplet. We can find the mass of the droplet: $mg = E~q$ $mg = \frac{\Delta V~q}{d}$ $m = \frac{\Delta V~q}{g~d}$ $m = \frac{(25~V)(15)(1.6\times 10^{-19}~C)}{(9.8~m/s^2)~(12\times 10^{-3}~m)}$ $m = 5.1\times 10^{-16}~kg$ We can find the radius of the droplet: $\rho = \frac{m}{V}$ $V = \frac{m}{\rho}$ $\frac{4}{3}~\pi~r^3 = \frac{m}{\rho}$ $r^3 = \frac{3m}{4\pi~\rho}$ $r = \sqrt[3] {\frac{3m}{4\pi~\rho}}$ $r = \sqrt[3] {\frac{(3)(5.1\times 10^{-16}~kg)}{(4\pi)(860~kg/m^3)}}$ $r = 5.21\times 10^{-7}~m$
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