An uncharged parallel plate capacitor having a dielectric of constant $K$ is connected to a similar air-cored parallel capacitor charged to a potential $V$. The two share the charge and the common potential is $V'$. The dielectric constant $K$ is

$A$ 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 of capacity $4 \, \mu F$ is charged to $80 \, V$ and another capacitor of capacity $6 \, \mu F$ is charged to $30 \, V$. When they are connected, the energy lost by the $4 \, \mu F$ capacitor is: (in $ \, mJ$)

$A$ $2\,\mu F$ capacitor has a potential difference of $200\;V$ across its terminals. It is disconnected from the battery and then another uncharged capacitor is connected in parallel to it. If the potential difference becomes $20\;V$,the capacitance of the second capacitor is.......$\mu F$.

$A$ 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:

Two capacitors having capacitance $C_{1}$ and $C_{2}$ respectively are connected as shown in the figure. Initially,capacitor $C_{1}$ is charged to a potential difference $V$ volt by a battery. The battery is then removed and the charged capacitor $C_{1}$ is now connected to uncharged capacitor $C_{2}$ by closing the switch $S$. The amount of charge on the capacitor $C_{2}$ after equilibrium is