Chapter 29 - Potential and Field - Exercises and Problems - Page 865: 55

\begin{array}{|c|c|c|c|} \hline & \Delta V\;({\rm V})& Q\;({\rm \mu C})&C\;({\rm \mu F})\\ \hline \rm Capacitor \;1& 9& 45&5\\ \hline \rm Capacitor \;2& 5.4& 21.6&4\\ \hline \rm Capacitor \;3& 3.6&21.6&6\\ \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, $C_2$ and $C_3$ are in series so their equivalent is given by $$(C_{eq})_{23}= \left[ \dfrac{1}{C_2}+\dfrac{1}{C_3} \right]^{-1}$$ Plug the known; $$(C_{eq})_{ 23}=\left[ \dfrac{1}{4}+\dfrac{1}{6} \right]^{-1}$$ $$(C_{eq})_{ 23 }=\color{black}{\bf 2.4}\;\rm \mu F$$ The two capacitors, in the second circuit below, $C_1$ and $(C_{eq})_{23}$ are in parallel so their equivalent is given by $$(C_{eq})_{123 }=C_1+(C_{eq})_{23}=5+2.4$$ $$(C_{eq})_{123 }=\color{black}{\bf 7.4}\;\rm \mu F$$ 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, $$Q_{}=\Delta V_{(C_{eq})_{123}} (C_{eq})_{123}=(9)(7.4 )=\bf 66.6\;\rm \mu C$$ In the second circuit below, $(C_{eq})_{23}$ and $C_1$ are in parallel, so they must have the same voltage as the battery. $$\Delta V_1=\Delta V_{12}=V_B=\color{red}{\bf 9}\;\rm V$$ So the charge stored in $C_1$ is given by $$Q_1=C_1\Delta V_1=(5)(9)=\color{red}{\bf 45}\;\rm \mu C$$ And hence, the charge stored in $(C_{eq})_{23}$ $$Q_{(C_{eq})_{23}}={(C_{eq})_{23}}\Delta V_{(C_{eq})_{23}}=(2.4)(9)= {\bf 21.6}\;\rm \mu C$$ We know that the two capacitors $C_2$ and $C_3$ are in series, so they must have the same stored charge as their equivalent. Hence, $$Q_2=Q_3=\color{red}{\bf 21.6}\;\rm \mu C$$ And their potentials are given by $$\Delta V_2=\dfrac{Q_2}{C_2}=\dfrac{21.6}{4}=\color{red}{\bf 5.4 }\;\rm V$$ $$\Delta V_3=\dfrac{Q_3}{C_3}=\dfrac{21.6}{6}=\color{red}{\bf 3.6}\;\rm V$$ Plug all of that into the table below. \begin{array}{|c|c|c|c|} \hline & \Delta V\;({\rm V})& Q\;({\rm \mu C})&C\;({\rm \mu F})\\ \hline \rm Capacitor \;1& 9& 45&5\\ \hline \rm Capacitor \;2& 5.4& 21.6&4\\ \hline \rm Capacitor \;3& 3.6&21.6&6\\ \hline \end{array}