Fundamentals of Physics Extended (10th Edition)

Published by Wiley
ISBN 10: 1-11823-072-8
ISBN 13: 978-1-11823-072-5

Chapter 24 - Electric Potential - Problems: 89

Answer

$W=2.30\times 10^{-30}J$

Work Step by Step

The initial electric energy is given as $U_i=\frac{Kq^2}{a}$ The final electric energy is given as $U_f=3\times \frac{Kq^2}{a}$ We know that $W=\Delta U=U_f-U_i$ We plug in the known values to obtain: $W=3\times \frac{Kq^2}{a}-\frac{Kq^2}{a}=2\times \frac{Kq^2}{a}$ We plug in the known values to obtain: $W=2\times \frac{9\times 10^9(1.6\times 10^{-19})^2}{2\times 10^{-6}}=2.30\times 10^{-30}J$
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.