Consider the situation shown in the figure. Capacitor $A$ has a charge $q$ on it,whereas $B$ is uncharged. The charge appearing on capacitor $B$ a long time after the switch $S$ is closed is

$A$ $10 \ \mu F$ capacitor is charged to a potential difference of $50 \ V$ and connected in parallel to another uncharged capacitor. The common potential difference becomes $20 \ V$. The capacitance of the second capacitor is ........ $\mu F$.

$A$ capacitor $P$ with capacitance $10 \times 10^{-6} \text{ F}$ is fully charged with a potential difference of $6.0 \text{ V}$ and disconnected from the battery. The charged capacitor $P$ is connected across another capacitor $Q$ with capacitance $20 \times 10^{-6} \text{ F}$. The charge on capacitor $Q$ when equilibrium is established will be $\alpha \times 10^{-5} \text{ C}$. (Assume capacitor $Q$ does not have any charge initially). The value of $\alpha$ is . . . . . . .

Using a battery,a $100 \ pF$ capacitor is charged to $60 \ V$ and then the battery is removed. After that,a second uncharged capacitor is connected to the first capacitor in parallel. If the final voltage across the second capacitor is $20 \ V$,its capacitance is: (in $pF$)

$A$ $10\,\mu F$ capacitor and a $20\,\mu F$ capacitor are connected in series across a $200\,V$ supply line. The charged capacitors are then disconnected from the line and reconnected with their positive plates together and negative plates together,and no external voltage is applied. What is the potential difference across each capacitor?

Two capacitors $C_1$ and $C_2$ are connected in parallel. If a charge $Q$ is given to the combination,the charge gets shared. Then the ratio of charge on $C_1$ to charge on $C_2$ is . . . . . . .