Physics (10th Edition)

Published by Wiley
ISBN 10: 1118486897
ISBN 13: 978-1-11848-689-4

Chapter 19 - Electric Potential Energy and the Electric Potential - Problems - Page 539: 45


(a) 33 J (b) 8500 W

Work Step by Step

(a) Given: Capacitance $C=850\,\mu F,$ Potential difference $V=280\,V$ Recall: Energy $E=\frac{1}{2}CV^{2}$ Result: $E=\frac{1}{2}\times850\times10^{-6}\,F\times(280\,V)^{2}=33\,J$ (b) Given/known: $E=33\,J,$ Time $t=3.9\times10^{-3}\,s$ Recall: $Wattage=\frac{E}{t}$ Result: $Wattage=\frac{33\,J}{3.9\times10^{-3\,}s}=8500\,W$
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.