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 23 - The Electric Field - Exercises and Problems: 3

Answer

The strength of the electric field is $7640~N/C$ and it is directed downward.

Work Step by Step

By symmetry, the horizontal components of the electric field from each charge cancel out. The net electric field is the sum of the vertical components of the electric field from each charge, which are directed downward. $E = 2~\frac{k~q}{r^2}$ $E = 2~\frac{(9.0\times 10^9~N~m^2/C^2)(3.0\times 10^{-9}~C)}{[(0.050~m)(\sqrt{2})]^2}~cos(45^{\circ})$ $E = 7640~N/C$ The strength of the electric field is $7640~N/C$ and it is directed downward.
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.