In the circuit diagram shown in the adjoining figure,the resultant capacitance between $P$ and $Q$ is ........ $\mu F$.

$A$ parallel combination of two capacitors of capacities $2C$ and $C$ is connected across a $5 \text{ V}$ battery. When they are fully charged,the charges and energies stored in them are $Q_1, Q_2$ and $E_1, E_2$ respectively. Then $\frac{E_1-E_2}{Q_1-Q_2}$ in $\text{J/C}$ is (capacity is in Farad,charge in Coulomb and energy in $\text{J}$)

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:

Two capacitances of capacity $C_1$ and $C_2$ are connected in series and a potential difference $V$ is applied across the combination. The potential difference across $C_1$ will be:

What is the equivalent capacitance of the system of capacitors between $A$ and $B$?

In the circuit shown,a potential difference of $30\, V$ is applied across $AB$. The potential difference between the points $M$ and $N$ is....$V$