Chapter 29 - Potential and Field - Exercises and Problems - Page 865: 54

\begin{array}{|c|c|c|c|} \hline & \Delta V\;({\rm V})& Q\;({\rm \mu C})&C\;({\rm \mu F})\\ \hline \rm Capacitor \;1& 1.0 & 4.0 &4.0 \\ \hline \rm Capacitor \;2& 1.0& 12 &12.0 \\ \hline \rm Capacitor \;3& 8&16 &2.0 \\ \hline \end{array}

First, we need to re-sketch this circuit and reduce it as we see in the figures below. We need to find the equivalent capacitance of the whole circuit, so we can find the total charge. The two capacitors, in the first circuit below, $C_1$ and $C_2$ are in parallel so their equivalent is given by $$(C_{eq})_{12 }=C_1+C_2=4+12$$ $$(C_{eq})_{12 }=\color{black}{\bf 16}\;\rm \mu F$$ The two capacitors, in the second circuit below, $(C_{eq})_{12}$ and $C_3$ are in series so their equivalent is given by $$(C_{eq})_{123}= \left[ \dfrac{1}{(C_{eq})_{12}}+\dfrac{1}{C_3} \right]^{-1}$$ Plug the known; $$(C_{eq})_{123}=\left[ \dfrac{1}{16}+\dfrac{1}{2} \right]^{-1}$$ $$(C_{eq})_{123 }=\color{black}{\bf \dfrac{16}{9}}\;\rm \mu F$$ Now the voltage of this equivalent capacitor $(C_{eq})_{123}$ is equal to that of the battery. Thus, $$\Delta V_{(C_{eq})_{123}}=V_B=\bf 9\;\rm V$$ where we know that $\Delta V_C=Q/C$ Hence, $$Q_{}=\Delta V_{(C_{eq})_{123}} (C_{eq})_{123}=(9)\left(\dfrac{16}{9}\right)=\bf 16\;\rm \mu C$$ In the second circuit below, $(C_{eq})_{12}$ and $C_3$ are in series, so $$Q_3=Q_{ }=Q_{(C_{eq})_{12}}=\color{red}{\bf 16}\;\rm \mu C$$ So the potential difference for $C_3$ is $$\Delta V_{3}=\dfrac{Q_3}{C_3}=\dfrac{16\times 10^{-6}}{2\times 10^{-6}}$$ $$\Delta V_{3}=\color{red}{\bf 8}\;\rm V$$ Hence, the potential difference for $(C_{eq})_{12}$ is given by $$\Delta V_{(C_{eq})_{12}}=V_B-\Delta V_3=9-8=\bf 1\;\rm V$$ And since the two capacitors $C_1$ and $C_2$ are in parallel, they must have the same potential difference. $$\Delta V_{1}=\Delta V_{2}=\color{red}{\bf 1}\;\rm V$$ Thus their charges are given by $$Q_2=\Delta V_{2}C_2=(1)(12\;\mu)=\color{red}{\bf 12}\;\rm \mu C$$ $$Q_1=\Delta V_{1}C_1=(1)(4\;\mu)=\color{red}{\bf 4}\;\rm \mu C$$ \begin{array}{|c|c|c|c|} \hline & \Delta V\;({\rm V})& Q\;({\rm \mu C})&C\;({\rm \mu F})\\ \hline \rm Capacitor \;1& 1.0 & 4.0 &4.0 \\ \hline \rm Capacitor \;2& 1.0& 12 &12.0 \\ \hline \rm Capacitor \;3& 8&16 &2.0 \\ \hline \end{array}