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: 12


(a) $m = 1.9\times 10^{-15}~kg$ (b) $q = 1.024\times 10^{-17}~C$ (c) The droplet has a surplus of 64 electrons.

Work Step by Step

(a) We can find the mass of the droplet: $\frac{m}{V} = \rho$ $m = V~\rho$ $m = \frac{4}{3}~\pi~r^3~\rho$ $m = \frac{4}{3}~\pi~(0.80\times 10^{-6}~m)^3~(885~kg/m^3)$ $m = 1.9\times 10^{-15}~kg$ (b) 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 magnitude of the charge on the droplet: $E~q = mg$ $\frac{\Delta V~q}{d} = mg$ $q = \frac{mg~d}{\Delta V}$ $q = \frac{(1.9\times 10^{-15}~kg)(9.8~m/s^2)(11\times 10^{-3}~m)}{20~V}$ $q = 1.024\times 10^{-17}~C$ (c) Since the force on the droplet by the electric field is directed upward, the droplet must have a negative charge. Therefore, the droplet has a surplus of electrons. We can find the number of extra electrons: $n~(1.6\times 10^{-19}~C) = 1.024\times 10^{-17}~C$ $n = \frac{1.024\times 10^{-17}~C}{1.6\times 10^{-19}~C}$ $n = 64$ The droplet has a surplus of 64 electrons.
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