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 25 - Electric Charges and Forces - Exercises and Problems: 23

Answer

The electric field has a strength of $4.5\times 10^4~N/C$ directed toward the bead.

Work Step by Step

We can find the magnitude of the electric field 4.0 cm from the bead: $E = \frac{k~q}{r^2}$ $E = \frac{(9.0\times 10^9~N~m^2/C^2)(8.0\times 10^{-9}~C)}{(0.040~m)^2}$ $E = 4.5\times 10^4~N/C$ An electric field points toward a negative charge. Therefore, the electric field has a strength of $4.5\times 10^4~N/C$ directed toward the bead.
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.