Chapter 29 - Potential and Field - Exercises and Problems - Page 865: 61

\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& 3.33& 33.3&10\\ \hline \rm Capacitor \;2& 3.33& 66.7&20\\ \hline \end{array}

First, we need to find the charge stored in each capacitor when they are connected individually to the 10-V battery. $$Q_1=\Delta V_1C_1=(10)(10\mu)=\bf100\;\rm \mu C$$ Note that the voltage of each capacitor when connected to the battery is 10 V. $$Q_2=\Delta V_2C_2=(10)(20\mu)=\bf200\;\rm \mu C$$ This means that the second capacitor carries more charge than the first, This means that, when we connect the two capacitors as seen in the figure below, the negative charge from the negative plate of $C_2$ will move toward the positive plate of $C_1$ until they reach an equilibrium point. And yet the negative plate of $C_2$ remains negative while the positive plate in $C_1$ is now negative. Since the plates of the two capacitors are connected positive-from$C_2$ to the negative from $C_1$, then $$ Q_1'+Q_2'=-100+200={\bf 100\mu} $$ Thus, $$Q_1'=100\mu-Q_2'\tag 1$$ And since they are connected in parallel, $$\Delta V_1'=\Delta V_2'$$ Thus, $$\dfrac{Q_1'}{C_1}=\dfrac{Q_2'}{C_2}$$ Plug from (1), $$\dfrac{100\mu-Q_2'}{C_1}=\dfrac{Q_2'}{C_2}$$ Plug the known; $$\dfrac{100\mu-Q_2'}{10 \color{red}{\bf\not}\mu}=\dfrac{Q_2'}{20 \color{red}{\bf\not}\mu}$$ So $$200\mu-2Q_2'=Q_2'$$ $$Q_2'=\color{red}{\bf 66.7}\;\rm \mu F$$ And hence, $$Q_1'=\color{red}{\bf 33.3}\;\rm \mu F$$ $$\Delta V_1'=\dfrac{Q_1'}{C_1}=\dfrac{33.3}{10}=\color{red}{\bf 3.33}\;\rm V$$ $$\Delta V_2'=\dfrac{Q_2'}{C_2}=\dfrac{66.7}{20}=\color{red}{\bf 3.33}\;\rm V$$ \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& 3.33& 33.3&10\\ \hline \rm Capacitor \;2& 3.33& 66.7&20\\ \hline \end{array}