$A$ $20\,\mu F$ capacitor is charged to a potential of $500\,V$ and then connected in parallel to a $10\,\mu F$ capacitor charged to $200\,V$. What is the common potential in $V$?

$A$ 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

$A$ $60\; pF$ capacitor is fully charged by a $20\; V$ supply. It is then disconnected from the supply and connected to another uncharged $60\; pF$ capacitor in parallel. The electrostatic energy that is lost in this process by the time the charge is redistributed between them is (in $nJ$):

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

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

$A$ capacitor having capacitance $C$ is charged to a voltage $V$. It is then removed and connected in parallel with another identical capacitor which is uncharged. The new charge on each capacitor is now