Fundamentals of Physics Extended (10th Edition)

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

Chapter 25 - Capacitance - Problems - Page 742: 34d

Answer

$q_1=3.33\times10^{-4}C$

Work Step by Step

As we know that $q_1=\frac{C_1C_2 V}{C_1+C_2}$ We plug in the known values to obtain: $q_1=\frac{1\times10^{-6}\times5\times10^{-6}\times100}{1\times10^{-6}+5\times10^{-6}}$ $q_1=3.33\times10^{-4}C$
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