$A$ 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 . . . . . . $\mu C$.

$A$ $2\,\mu F$ capacitor is charged to a potential of $10\,V$. Another $4\,\mu F$ capacitor is charged to a potential of $20\,V$. The two capacitors are then connected in a single loop,with the positive plate of one connected to the negative plate of the other. What heat is evolved in the circuit in $\mu J$?

$A$ 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 . . . . . . .

$A$ 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 capacitors $C_1$ and $C_2 = 2C_1$ are connected in a circuit with a switch 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 the capacitors will be:

$A$ capacitor of $10\,\mu F$ charged up to $250\,V$ is connected in parallel with another capacitor of $5\,\mu F$ charged up to $100\,V$. The common potential is.....$V$.