Find the equivalent capacitance of the system of capacitors between points $A$ and $B$ as shown in the figure.

Assertion $(A)$: Two condensers of same capacity are connected first in parallel and then in series. The ratio of resultant capacities in the two cases will be $4: 1$.
Reason $(R)$: In parallel,capacity increases and in series,capacity decreases.

Two capacitors of capacitance $4\,\mu F$ and $6\,\mu F$ are connected in series. $A$ potential difference of $500\;V$ is applied to the outer plates of the two-capacitor system. The charge on each capacitor is numerically:

Two parallel plate capacitors are connected in series. Each capacitor has a plate area $A$ and a separation $d$ between the plates. The dielectric constants of the media between their plates are $2$ and $4$. The separation between the plates of a single air capacitor of plate area $A$ which effectively replaces the combination is:

In the given capacitor network,$C_1 = 10\,\mu F$,$C_2 = 5\,\mu F$,and $C_3 = 4\,\mu F$. What is the resultant capacitance between $A$ and $B$ in $\mu F$?

Two condensers of capacity $0.3\,\mu F$ and $0.6\,\mu F$ respectively are connected in series. The combination is connected across a potential of $6\,V$. The ratio of energies stored by the condensers will be