$A$ capacitor of capacitance $4 \mu F$ is charged to a potential difference of $6 \ V$ with a battery. The battery is then removed and in its place another capacitor of capacitance $8 \mu F$ is introduced and the circuit is closed. The potential difference attained by each of the capacitors in $V$ is

$A$ 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 . . . . . . .

$A$ $10 \ \mu F$ capacitor is charged to a potential difference of $50 \ V$ and connected in parallel to another uncharged capacitor. The common potential difference becomes $20 \ V$. The capacitance of the second capacitor is ........ $\mu F$.

$A$ $0.2\,F$ capacitor is charged to $600\,V$ by a battery. On removing the battery,it is connected to another parallel plate capacitor of $1\,F$. The potential decreases to.........$V$.

$A$ 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 $V$)

$A$ $10\ \mu F$ capacitor is charged to a potential difference of $1000\ V$. The terminals of the charged capacitor are disconnected from the power supply and connected to the terminals of an uncharged $6\ \mu F$ capacitor. What is the final potential difference across each capacitor in $V$?