Two capacitors $C_1$ and $C_2 = 2C_1$ are connected in a circuit with a switch $S$ between them as shown in the figure. Initially,the switch is open and $C_1$ holds charge $Q$. The switch is closed. At steady state,the charge on each capacitor will be

$A$ capacitor of capacitance $10\,\mu F$ has been charged to $100\,V$. It is now connected to another uncharged capacitor in parallel. The common potential becomes $40\,V$. The capacitance of the other capacitor is......$\mu F$.

$A$ 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 $V$)

Two insulated metallic spheres of $3\,\mu F$ and $5\,\mu F$ capacitances are charged to $300\, V$ and $500\, V$ respectively. The energy loss,when they are connected by a wire,is (in $,J$)

Two identical capacitors are joined in parallel,charged to a potential $V$ and then separated and then connected in series,i.e.,the positive plate of one is connected to the negative plate of the other. What happens to the system?

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