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 - Page 715: 74a

Answer

$W=-24J$

Work Step by Step

We can find the initial potential energy of the charges as: $U_i=\frac{Kq_1q_2}{b}+\frac{Kq_1q_3}{a}+\frac{Kq_2q_3}{a}$ We plug in the known values to obtain: $U_i=\frac{9\times 10^9\times 10\times 10^{-6}\times(-20\times 10^{-6})}{0.06}+\frac{9\times 10^9\times 10\times 10^{-6}\times(30\times 10^{-6})}{0.10}+\frac{9\times 10^9\times (-20\times 10^{-6})\times(30\times 10^{-6})}{0.10}=-57J$ Now the final potential energy is $U_f=\frac{Kq_1q_2}{a}+\frac{Kq_1q_3}{a}+\frac{Kq_2q_3}{b}$ We plug in the known values to obtain: $U_f=\frac{9\times 10^9\times 10\times 10^{-6}\times(-20\times 10^{-6})}{0.10}+\frac{9\times 10^9\times 10\times 10^{-6}\times(30\times 10^{-6})}{0.10}+\frac{9\times 10^9\times (-20\times 10^{-6})\times(30\times 10^{-6})}{0.06}=-81J$ Now work done $W$ can be determined as: $W=U_f-U_i=-81-(-57)=-24J$
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