Physics for Scientists and Engineers: A Strategic Approach with Modern Physics (3rd Edition)

Published by Pearson
ISBN 10: 0321740904
ISBN 13: 978-0-32174-090-8

Chapter 26 - The Electric Field - Exercises and Problems: 20

Answer

The magnitude of the electric field strength is $6.1\times 10^5~N/C$ and the electric field is directed downward.

Work Step by Step

We can find the charge on the bead. $Q = (1.0\times 10^{10})(-1.6\times 10^{-19}~C)$ $Q = -1.6\times 10^{-9}~C$ For the bead to be suspended in the air, the force (directed upward) from the electric field must be equal in magnitude to the bead's weight. We can find the magnitude of the electric field strength. $E~Q = mg$ $E = \frac{mg}{Q}$ $E = \frac{(1.0\times 10^{-4}~kg)(9.80~m/s^2)}{1.6\times 10^{-9}~C}$ $E = 6.1\times 10^5~N/C$ Since the bead has a negative charge, the electric field must be directed downward for the electrostatic force on the bead to be directed upward. The magnitude of the electric field strength is $6.1\times 10^5~N/C$ and the electric field is directed downward.
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