Chapter 29 - Potential and Field - Exercises and Problems - Page 865: 59

\begin{array}{|c|c|c|c|} \hline {\bf Finally:} & \Delta V\;({\rm V})& Q\;({\rm \mu C})&C\;({\rm \mu F})\\ \hline \rm Capacitor \;1& 55.5& 833&15\\ \hline \rm Capacitor \;2& 33.35& 667&20\\ \hline \rm Capacitor \;3& 22.23&667&30\\ \hline \end{array}

${\bf Initially:}$ We have only one capacitor that is connected to the battery. So that $$\Delta V_1=V_B=\color{red}{\bf 100}\;\rm V$$ Hence, the charge stored on this capacitor is given by $$\Delta V_1=\dfrac{Q_1}{C_1}$$ Thus, $$Q_1=C_1\Delta V_1$$ $$Q_1=(15\times 10^{-6})(100)=\color{red}{\bf 1500}\;\rm \mu C$$ ${\bf Finally:}$ We have 3 capacitors, and the battery is not connected, as we see in the second figure below. This means that the charge in the $C_1$ will be distributed among the system. Let's find the equivalent of the two in-series capacitor. $$(C_{eq})_{23}=\left[ \dfrac{1}{C_2}+\dfrac{1}{C_3}\right]^{-1}$$ Plug the known; $$(C_{eq})_{23}=\left[ \dfrac{1}{20}+\dfrac{1}{30}\right]^{-1}=\bf 12\;\rm \mu F$$ $$ \Delta V_2C_2=\Delta V_3C_3\tag 1$$ where $$\Delta V_1= \Delta V_{(C_{eq})_{23}}$$ Hence, $$\dfrac{Q_1'}{C_1}= \dfrac{Q_{(C_{eq})_{23}}}{(C_{eq})_{23}}$$ $$\dfrac{Q_1'}{15}= \dfrac{Q_{(C_{eq})_{23}}}{12}$$ $$12Q_1' = 15Q_{(C_{eq})_{23}} $$ Therefore, $$Q_{(C_{eq})_{23}} = \dfrac{4}{5}Q_1'\tag 2$$ Note that the charge $Q_1$ is now divided to the charge $Q_1'$ in the $C_1$ plus the charge $Q_{eq}$ in the $(C_{eq})_{23}$. Hence, $$Q_1'+Q_{(C_{eq})_{23}}=1500$$ Plug from (2), $$Q_1'+ \dfrac{4}{5} Q_1'=1500$$ Hence, $$Q_1'=\dfrac{5}{9}(1500)=\color{red}{\bf 833}\;\rm\mu C$$ So that, $$Q_{(C_{eq})_{23}}=1500-833=\bf 667\;\rm \mu C$$ And since the two capacitors are in-series, $Q_2=Q_3$. Thus, $$Q_2=Q_3=Q_{(C_{eq})_{23}}=\color{red}{\bf 667}\;\rm \mu C$$ $$\Delta V_1=\dfrac{Q_1'}{C_1}=\dfrac{833}{15}=\color{red}{\bf 55.5}\;\rm V$$ $$\Delta V_2=\dfrac{Q_2 }{C_2}=\dfrac{667}{20}=\color{red}{\bf 33.35}\;\rm V$$ $$\Delta V_3=\dfrac{Q_3}{C_3}=\dfrac{667}{30}=\color{red}{\bf 22.23}\;\rm V$$ \begin{array}{|c|c|c|c|} \hline {\bf Finally:} & \Delta V\;({\rm V})& Q\;({\rm \mu C})&C\;({\rm \mu F})\\ \hline \rm Capacitor \;1& 55.5& 833&15\\ \hline \rm Capacitor \;2& 33.35& 667&20\\ \hline \rm Capacitor \;3& 22.23&667&30\\ \hline \end{array}