Two capacitors of $10 \ pF$ and $20 \ pF$ are connected to $200 \ V$ and $100 \ V$ sources respectively. If they are connected by a wire,what is the common potential of the capacitors (in $V$)?

$A$ capacitor of capacitance $C_1 = 10 \mu F$ is charged using a $9 \text{ V}$ battery. It is then removed from the battery and connected to another uncharged capacitor $C_2 = 20 \mu F$ as shown in the figure. The charge on $C_2$ after equilibrium is reached is:

$A$ capacitor of capacitance $C_{0}$ is charged to a potential $V_{0}$ and is connected with another capacitor of capacitance $C$ as shown. After closing the switch $S,$ the common potential across the two capacitors becomes $V$. The capacitance $C$ is given by

Two identical capacitors each of capacitance $5 \mu F$ are charged to potentials $2 kV$ and $1 kV$ respectively. Their negative ends are connected together. When the positive ends are also connected together,the loss of energy of the system is

$A$ capacitor of capacity $C_1$ is charged to potential $V_1$ and then disconnected. An uncharged capacitor of capacity $C_2$ is connected in parallel with $C_1$. The resultant potential $V_2$ 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 $nJ$):