A $10 \mu F$ capacitor is charged to $15 V$. It is next connected in series with an uncharged $5 \mu F$ capacitor. The series combination is finally connected across a $50 V$ battery, as shown in Figure. Find the new potential differences across the $5 \mu F$ and $10 \mu F$ capacitors.

The initial charge on the larger capacitor is

$$\begin{gathered} Q=C△V \\ \Rightarrow Q=\left(10 \mu F\right)\left(15 V\right)=150 \mu C \end{gathered}$$

An additional charge $q$ is pushed through the $50 V$ battery, giving the smaller capacitor charge $q$ and the larger charge $150+q$ (in $\mu C$).

$$\begin{gathered} \Rightarrow 50V=\frac{q}{5 \mu F}+\frac{150 \mu C+q}{10 \mu F} \\ \Rightarrow 500 \mu C=2q+150 \mu C+q \\ \Rightarrow q=117 \mu C \end{gathered}$$

Across the $5 \mu F$ capacitor $△V=\frac{q}{C}=\frac{117 \mu C}{5 \mu F }=23.3 v$

Across the $10 \mu F$ capacitor

$△V=\frac{150 \mu C+117 \mu C}{10 \mu F}=26.7 V$