College Physics (4th Edition)

Published by McGraw-Hill Education
ISBN 10: 0073512141
ISBN 13: 978-0-07351-214-3

Chapter 18 - Problems - Page 700: 42


(a) $C = 11.0~\mu F$ (b) $Q = 2.64\times 10^{-4}~C$

Work Step by Step

(a) Since the capacitors are connected in parallel, we can add the capacitance of each capacitor together to find the equivalent capacitance: $C = 2.0~\mu F+6.0~\mu F+3.0~\mu F$ $C = 11.0~\mu F$ (b) We can find the charge on the $6.0~\mu F$ capacitor: $Q = C~V$ $Q = (6.0 \times 10^{-6} F)(44.0~V)$ $Q = 2.64\times 10^{-4}~C$
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