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 25 - The Electric Potential - Exercises and Problems - Page 709: 7


$U = 4.8\times 10^{-6}~J$

Work Step by Step

We can find the electric potential energy: $U = \frac{kq_1q_2}{r_{12}}+\frac{kq_1q_3}{r_{13}}+\frac{kq_2q_3}{r_{23}}$ $U = k~(\frac{q_1q_2}{r_{12}}+\frac{q_1q_3}{r_{13}}+\frac{q_2q_3}{r_{23}})$ $U = (9.0\times 10^9~N~m^2/C^2)~[\frac{(2.0\times 10^{-9}~C)(3.0\times 10^{-9}~C)}{0.030~m}+\frac{(2.0\times 10^{-9}~C)(3.0\times 10^{-9}~C)}{0.040~m}+\frac{(3.0\times 10^{-9}~C)(3.0\times 10^{-9}~C)}{0.050~m}~]$ $U = 4.8\times 10^{-6}~J$
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.