For the shown situation, in the steady state condition, what is the ratio of the charge stored in the first capacitor to the charge stored in the last $(n^{th})$ capacitor?

Two thin conducting shells of radii $R$ and $3R$ are shown in the figure. The outer shell carries a charge $+Q$ and the inner shell is neutral. The inner shell is earthed with the help of a switch $S$.

$A$ capacitor $A$ has a capacitance of $15\ \mu F$ when filled with a dielectric of constant $K = 15$. Another capacitor $B$ is air-filled and has a capacitance of $1\ \mu F$. Both are charged to $100\ V$. After charging, the dielectric is removed from capacitor $A$, and the two capacitors are connected in parallel. The common potential difference across them will be ... $V$.

The potential difference that must be applied across the series and parallel combination of $4$ identical capacitors is such that the energy stored in them becomes the same. The ratio of potential difference in series to parallel combination is

Two capacitors of capacities $1 \mu F$ and $C \mu F$ are connected in series and the combination is charged to a potential difference of $120 \ V$. If the charge on the combination is $80 \mu C$,the energy stored in the capacitor of capacity $C$ in $\mu J$ is