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: 31


The object has a charge of $+1.2\times 10^{-8}~C$

Work Step by Step

We can find the magnitude of the charge that creates a 270,000 N/C electric field at a point 2.0 cm away: $E = \frac{k~q}{r^2}$ $q = \frac{E~r^2}{k}$ $q = \frac{(270,000~N/C)(0.020~m)^2}{9.0\times 10^9~N~m^2/C^2}$ $q = 1.2\times 10^{-8}~C$ An electric field points away from a positive charge. Therefore, the object has a charge of $+1.2\times 10^{-8}~C$.
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