Two conducting spheres of radii $R_1$ and $R_2$ are charged with charges $Q_1$ and $Q_2$ respectively. When they are brought into contact with each other:

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

$A$ parallel plate capacitor of capacitance $C$ is charged to a potential $V$ by a battery. Another capacitor of capacitance $2C$ is charged to a potential $2V$ by another battery. After removing the batteries,they are connected in parallel such that the positive plate of one is connected to the negative plate of the other. Calculate the final energy of the system.

$A$ conducting sphere of radius $R$ carrying a charge $q$ is joined by a conducting wire to a conducting sphere of radius $2R$ carrying a charge $-2q$. The charge flowing between them will be:

Two capacitors $C_1$ and $C_2 = 2C_1$ are connected with a switch $S$ as shown in the figure. Initially,the switch is open and the charge on capacitor $C_1$ is $Q$. Now,when the switch is closed,the final charges on the capacitors are:

$A$ capacitor $P$ with capacitance $10 \times 10^{-6} \text{ F}$ is fully charged with a potential difference of $6.0 \text{ V}$ and disconnected from the battery. The charged capacitor $P$ is connected across another capacitor $Q$ with capacitance $20 \times 10^{-6} \text{ F}$. The charge on capacitor $Q$ when equilibrium is established will be $\alpha \times 10^{-5} \text{ C}$. (Assume capacitor $Q$ does not have any charge initially). The value of $\alpha$ is . . . . . . .