In the circuit shown,$C_1 = 6\,\mu F, C_2 = 3 \,\mu F$ and battery $B = 20\,V$. The switch $S_1$ is first closed. It is then opened and afterwards $S_2$ is closed. What is the charge (in $\mu C$) finally on $C_2$?

Two conducting spheres of radii $R_1$ and $R_2$ are charged with charges $Q_1$ and $Q_2$ respectively. On bringing them in contact,there is:

$A$ $20 \mu F$ capacitor is charged to $5 V$ and isolated. It is then connected in parallel with an uncharged $30 \mu F$ capacitor. The decrease in the energy of the system will be (in $\text{J}$)

Two identical capacitors have the same capacitance $C$. One of them is charged to potential $V_1$ and the other to $V_2$. The negative ends of the capacitors are connected together. When positive ends are also connected,the decrease in energy of the combined system is

$A$ $2\, \mu F$ capacitor $C_{1}$ is first charged to a potential difference of $10\, V$ using a battery. Then the battery is removed and the capacitor is connected to an uncharged capacitor $C_{2}$ of $8\, \mu F$. The charge in $C_{2}$ on equilibrium condition is $\ldots\, \mu C$. (Round off to the Nearest Integer)

Two capacitors of capacitance $3 \ \mu F$ and $6 \ \mu F$ are charged to a potential of $12 \ V$. They are then connected by joining their positive plates together and their negative plates together. What will be the potential difference across each capacitor in $V$?