Five capacitors each of capacity $C$ are connected as shown in the figure. If their resultant capacity is $2 \mu F$,then the capacity of each capacitor is: (in $\mu F$)

In the circuit shown,the potential difference across the $4.5 \mu F$ capacitor 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:

To obtain $3\,\mu F$ capacitance from three capacitors of $2\,\mu F$ each,they will be arranged as:

The equivalent capacitance between points $A$ and $B$ of the circuit shown will be

Three parallel plate air capacitors are connected in parallel. Each capacitor has plate area $\frac{A}{3}$ and the separation between the plates is $d, 2d$ and $3d$ respectively. The equivalent capacity of the combination is ($\varepsilon_0 =$ absolute permittivity of free space).