University Physics with Modern Physics (14th Edition)

Published by Pearson
ISBN 10: 0321973615
ISBN 13: 978-0-32197-361-0

Chapter 24 - Capacitance and Dielectrics - Problems - Exercises - Page 810: 24.19


(a) $Q_1 =156 \mathrm{~\mu C}$ and $Q_2 = 260 \mathrm{~\mu C}$ (b) $V_1 = V_2 = 52 \,\text{V}$

Work Step by Step

(a) As both capacitors are in parallel, therefore, they have the same potential even they have different capacitances. So the charge for each capacitor is calculated as next: For $C_1$ $$Q_1 = C_1 V = (3\mathrm{~\mu F} ) (52 \,\text{V}) = \boxed{156 \mathrm{~\mu C}}$$ For $C_2$ $$Q_2 = C_2 V = (5\mathrm{~\mu F} ) (52 \,\text{V}) = \boxed{260 \mathrm{~\mu C}}$$ (b) In parallel, the combination of the two capacitors has the same potential for each capacitor $$\boxed{V_1 = V_2 = 52 \,\text{V}}$$
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