Answer
$\sum W=6.90\times10^{-18}J$
Work Step by Step
$d=1.0\times10^{-10}m$
$\sum W=3\times k\frac{q_1q_1}{d}$
$=3\times(8.99\times10^9\frac{N\cdot m^2}{C^2})\frac{(1.60\times10^{-19}C)^2}{1.0\times10^{-10}m}=6.90\times10^{-18}J$
Physics: Principles with Applications (7th Edition)
by Giancoli, Douglas C.
Published by
Pearson
ISBN 10:
0-32162-592-7
ISBN 13:
978-0-32162-592-2
Chapter 17 - Electric Potential - Problems - Page 496: 26
Update this answer!
You can help us out by revising, improving and updating this answer.
Update this answerAfter 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.
