In the circuit shown here $C_1 = 6\,\mu F$,$C_2 = 3\,\mu F$ 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 finally on $C_2$ in $\mu C$?

The circuit shows two capacitors $A$ and $B$ of capacitances $C$ and $2C$ respectively. When they are fully charged,the cell is removed and the capacitors are connected with their plates of opposite polarities touching each other. Then:
$(a)$ Charge on $A$ is $\frac{4CE}{9}$
$(b)$ Charge on $B$ is $\frac{8CE}{9}$
$(c)$ Loss of energy in this process is $\frac{CE^2}{3}$
The correct statement$(s)$ is/are:

$A$ capacitor $C_{1}$ of capacitance $5\,\mu F$ is charged to a potential of $30\,V$ using a battery. The battery is then removed and the charged capacitor is connected to an uncharged capacitor $C_{2}$ of capacitance $10\,\mu F$ as shown in the figure. When the switch is closed,charge flows between the capacitors. At equilibrium,the charge on the capacitor $C_{2}$ is . . . . . . $\mu C$.

Two metallic spheres of radii $1 \, cm$ and $3 \, cm$ are given,with charges of $-1 \times 10^{-2} \, C$ and $5 \times 10^{-2} \, C$ respectively. If these are connected by a conducting wire,the final charge on the bigger sphere is

$A$ capacitor of capacitance $C$ and potential $V$ has energy $E$. It is connected to another capacitor of capacitance $2C$ and potential $2V$. Then the loss of energy is $\frac{x}{3} E$,where $x$ is . . . . . . .

Two spheres $A$ and $B$ of radius $4\,cm$ and $6\,cm$ are given charges of $80\,\mu C$ and $40\,\mu C$ respectively. If they are connected by a fine wire,the amount of charge flowing from one to the other is