Consider the situation shown in the figure. The switch $S$ is kept open for a long time and then closed. Find the charge that flows through the battery when $S$ is closed.

$A$ conducting body $1$ has some initial charge $Q$,and its capacitance is $C$. There are two other conducting bodies,$2$ and $3$,having capacitances $C_2 = 2C$ and $C_3 \rightarrow \infty$. Bodies $2$ and $3$ are initially uncharged. Body $2$ is touched with body $1$. Then,body $2$ is removed from body $1$ and touched with body $3$,and then removed. This process is repeated $N$ times. The charge on body $1$ at the end must be:

Using a battery,a $100 \ pF$ capacitor is charged to $60 \ V$ and then the battery is removed. After that,a second uncharged capacitor is connected to the first capacitor in parallel. If the final voltage across the second capacitor is $20 \ V$,its capacitance is: (in $pF$)

$A$ 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

$A$ capacitor of $2\,\mu F$ is charged as shown in the diagram. When the switch $S$ is turned to position $2,$ the percentage of its stored energy dissipated is ......$\%$

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