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 22 - Electric Charges and Forces - Exercises and Problems - Page 625: 34


(a) $2.35\times 10^{-6}$ (b) $E = 4.26\times 10^7~N/C$ The electric field must be directed upward.

Work Step by Step

(a) We can find the ratio of the electric force to the bee's weight: $\frac{q~E}{mg} = \frac{(23\times 10^{-12}~C)(100~N/C)}{(0.10\times 10^{-3}~kg)(9.8~m/s^2)} = 2.35\times 10^{-6}$ (b) We can find the magnitude of the required electric field: $q~E = mg$ $E = \frac{mg}{q}$ $E = \frac{(0.10\times 10^{-3}~kg)(9.8~m/s^2)}{23\times 10^{-12}~C}$ $E = 4.26\times 10^7~N/C$ Since the charge on the bee is positive, and the direction of the force due to the electric field must be upward, the electric field must be directed upward.
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