In the given circuit,the charge $Q_2$ on the $2 \ \mu F$ capacitor changes as $C$ is varied from $1 \ \mu F$ to $3 \ \mu F$. $Q_2$ as a function of '$C$' is given properly by:

$64$ identical mercury drops,each charged to a potential of $10\, V$,are combined to form a single larger drop. The potential of the new larger drop is........$V$.

$A$ series combination of $n_1$ capacitors,each of value $C_1$,is charged by a source of potential difference $4V$. When another parallel combination of $n_2$ capacitors,each of value $C_2$,is charged by a source of potential difference $V$,it has the same total energy stored in it as the first combination has. The value of $C_2$ in terms of $C_1$ is:

Two parallel plate capacitors with capacitances $C$ and $2C$ are connected in parallel. They are charged to a potential difference $V$. Now,if the battery is disconnected and a dielectric of dielectric constant $K$ is inserted between the plates of the capacitor with capacitance $C$,find the potential difference across each capacitor.

$A$ particle of charge $Q$ and mass $M$ moves in a circular path of radius $R$ in a uniform magnetic field of magnitude $B$. The same particle now moves with the same speed in a circular path of same radius $R$ in the space between the cylindrical electrodes of a cylindrical capacitor. The radius of the inner electrode is $R/2$ while that of the outer electrode is $3R/2$. Then the potential difference between the capacitor electrodes must be

Four equal capacitors are connected to a $10 \, V$ battery as shown in the figure. The potentials of $A$ and $B$ are: