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: 24.20


(a) $Q_1 = 60 \mathrm{~\mu C}$ and $Q_3 = 90 \mathrm{~\mu C}$ (b) $V_{ab} = 28 \,\text{V}$

Work Step by Step

(a) In parallel, both $C_1$ and $C_2$ have the same potential. So, we could get the next using the potential as $V = \dfrac{Q}{C}$ \begin{gather*} \dfrac{Q_1}{C_1} = \dfrac{Q_2}{C_2} \\ Q_1 = \dfrac{Q_2}{C_2}\,C_1 \\ Q_1 = \dfrac{30 \mathrm{~\mu C}}{3 \mathrm{~\mu F}}\,6 \mathrm{~\mu F}\\ Q_1 = 60 \mathrm{~\mu C} \end{gather*} The total charge of the combination $C_{12}$ is given by $$Q_{12} = Q_1 + Q_2 = 60 \mathrm{~\mu C} + 30 \mathrm{~\mu C} = 90 \mathrm{~\mu C}$$ $C_3$ and the combination $C_{12}$ are in series, where, in series both elements have the same charge, therefore, the charge of $C_3$ is $$Q_3 = \boxed{90 \mathrm{~\mu C}}$$ (b) The combination $C_{12}$ has a capacitance $$C_{12} = C_1 + C_2 = 6 \mathrm{~\mu F} + 3 \mathrm{~\mu F} = 9 \mathrm{~\mu F}$$ $C_{12}$ and $C_3$ are in series, which means they have the same charge and their equivalent capacitance is calculated by $$\dfrac{1}{C_{eq}} = \dfrac{1}{C_{12} } + \dfrac{1}{C_{3} } = \dfrac{1}{9\mathrm{~\mu F} } + \dfrac{1}{5 \mathrm{~\mu F} } = 0.311$$ Take the reciprocal $$C_{eq} = 3.21 \mathrm{~\mu F} $$ Now it is easy to find the potential between $a$ and $b$ as next $$V_{ab} = \dfrac{Q}{C_{eq}} = \dfrac{90 \mathrm{~\mu C}}{3.21 \mathrm{~\mu F} }= \boxed{28 \,\text{V}}$$
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