In the following figure C_1=5 μ F , C_2=C_3=10 μ F and =20 ~V ....

Given that, $C_1=5 \mu \mathrm{F}, C_2=C_3=10 \mu \mathrm{F}$ and $E=20 \mathrm{~V}$ When the switch is connected to point $A$. Then, situation of the circuit is shown below.
Charge stored in $C_1$, Q=C_1 V_1=C_1 E=5 $\times$20 \mu $\mathrm{C}=$100 \mu $\mathrm{C}$When switch is connect to point $B$, then situation circuit is shown below.
The charge $Q_1$ will start to distribute through, $C_2$ and $C_3$ to make the potential difference $V$ across all the three capacitors $C_1, C_2$ and $C_3$. Now, common potential difference, V=$\frac{\text { Total charge }}{\text { Total capacity }}=\frac{C_1 V_1+C_2 V_2+C_3 V_3}{C_1+C_2+C_3}$where, $V_1=20 \mathrm{~V}$, but $V_2=V_3=0 \mathrm{~V}$ \therefore \quad V=$\frac{100}{5+10+10}=\frac{100}{25}=$4 $\mathrm{~V}$Now, charge on $C_3, q_3=C_3 V$$\begin{aligned}& =10 \times4 \mu \mathrm{C}\\ & =40 \mu \mathrm{C} \end{aligned}$$$