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

$A$ capacitor is charged with a battery and the energy stored is $U$. After disconnecting the battery,another capacitor of the same capacity is connected in parallel to the first capacitor. Then the energy stored in each capacitor is

$A$ capacitor of capacity $C_1$ is charged to a potential of $V_0$. After disconnecting it from the battery,it is connected to a capacitor of capacity $C_2$ as shown in the figure. The ratio of the total energy before and after the connection of switch $S$ is:

$A$ $10\,\mu F$ 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?......$V$

$A$ 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 fully charged to a potential difference of $50\, V$. After removing the source voltage, it is connected to an uncharged capacitor in parallel. Now, the potential difference across them becomes $20\, V$. The capacitance of the second capacitor is $\dots \mu F$.