91Ó°ÊÓ

We need to find the charge on and the potential difference across each capacitor.

The common capacitance of the capacitor $C_1$and $C_2$will be

$C_{12}=C_1+C_2=16\mathrm{μ}F$

The total capacitance of the capacitor is

$C=\frac{C_{12}{C}_3}{C_{12}+C_3}=\frac{16·{10}^{-6}·2·{10}^{-6}}{8·{10}^{-6}}=16/9\mathrm{μ}F$

This means that the total charge on the system which means, in the capacitor $C_3$and distributed among $C_1$and $C_2$will be given as

$q=C∆V=16/9·9=16\mathrm{μ}C$

Among the two capacitors connected in parallel, this charge will be in a ratio of $3:1$, the ratio of capacitances. Since the potential difference is the same for both capacitors. So,

$$\begin{gathered} q_1=4\mathrm{μ}C \\ {q}_2=12\mathrm{μ}C \end{gathered}$$

The potential differences can be found knowing that the charge is the same for both $C_{12}$and $C_3$. We'll have

$C_{12}∆V_{12}=q=C_3∆V_3⇒∆V_{12}=\frac{C_3}{C_{12}}∆V_3$

It means the potential difference across the two first capacitors will be $8$times the potential difference of the capacitor $C_3$. That is

$$\begin{gathered} ∆V_{12}=8V, \\ ∆V_3=1V \end{gathered}$$