Question

When switch S is thrown to the left in figure, the plates of capacitor 1 acquire a potential difference $V_o$. Capacitors 2 and 3 are initially uncharged. The switch is now thrown to the right What are the final charges $q_1,q_2$ and $q_3$ on the capacitors?

Answer 1

$q_{total}=C_1{V}_o$
After switch is thrown towards right, $C_{23}$ and $C_1$ are in parallel. The common potential is
$V=\frac{\text{Total charge}}{\text{Total capacity}}$
$\frac{C_1{V}_o}{1+\left(\frac{C_2{C}_3}{C_2+C_3}\right)}$
Now, $q_{C_1}=C_{1}V=\frac{C_1^2{V}_o}{C_1+\left(\frac{C_2{C}_3}{C_2+C_3}\right)}$
This is the same result as given in the answer.
$q_{C_2}=q_{C_3}=C_{23}V$
$=\left(\frac{C_2{C}_3}{C_2+C_3}\right)\left[\frac{C_1{V}_o}{C_1+\frac{C_2{C}_3}{C_2+C_3}}\right]$