Two capacitors of capacitances $3 \mu F$ and $6 \mu F$ are charged to a potential of $12 V$ each. They are now connected to each other,with the positive plate of each joined to the negative plate of the other. The potential difference across each will be $volt$.

$A$ capacitor of capacity $C_1$ is charged up to $V$ volt and then connected to an uncharged capacitor of capacity $C_2$. The final potential difference across each will be:

Two isolated conducting spheres $S_{1}$ and $S_{2}$ of radius $\frac{2}{3} R$ and $\frac{1}{3} R$ have $12\, \mu C$ and $-3\, \mu C$ charges,respectively,and are at a large distance from each other. They are now connected by a conducting wire. $A$ long time after this is done,the charges on $S_{1}$ and $S_{2}$ are respectively:

$A$ capacitor of capacitance $2 \mu F$ is charged with the help of a $60 \ V$ battery. After disconnecting the battery,if this capacitor is connected in parallel with another uncharged capacitor of capacitance $1 \mu F$,then the potential difference across the plates of the $2 \mu F$ capacitor is: (in $V$)

$A$ 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 . . . . . . .

Two identical capacitors have same capacitance $C$. One of them is charged to the potential $V$ and other to the potential $2V$. The negative ends of both are connected together. When the positive ends are also joined together,the decrease in energy of the combined system is: