Physics: Principles with Applications (7th Edition)

Published by Pearson
ISBN 10: 0-32162-592-7
ISBN 13: 978-0-32162-592-2

Chapter 17 - Electric Potential - General Problems - Page 499: 81

Answer

1.8J

Work Step by Step

$V_A=k(\frac{38\mu C}{0.12m}+\frac{38\mu C}{0.24m})=4.75\times10^6J$ $V_B=k(\frac{38\mu C}{0.184m}+\frac{38\mu C}{0.278m})=3.43\times10^6J$ $W=qV_{ba}=(-1.5\mu C)(3.43\times10^6J-4.75\times10^6J)=1.8J$
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