The ratio of potential difference that must be applied across parallel and series combination of two capacitors $C_1$ and $C_2$ with their capacitance in the ratio $1: 2$ so that energy stored in these two cases becomes same is

See the diagram. The area of each plate is $2.0 \,m^{2}$ and $d=2 \times 10^{-3} \,m$. $A$ charge of $8.85 \times 10^{-8} \,C$ is given to plate $Q$. Then the potential of $Q$ becomes (in $\,V$)

$A$ capacitor of capacitance $C = 900 \, pF$ is charged fully by a $100 \, V$ battery as shown in figure $(a)$. Then it is disconnected from the battery and connected to another uncharged capacitor of capacitance $C = 900 \, pF$ as shown in figure $(b)$. The electrostatic energy stored by the system in figure $(b)$ is $\dots \times 10^{-6} \, J$.

Four capacitors are connected in a circuit as shown in the figure. The effective capacitance in $\mu F$ between points $A$ and $B$ will be

The figure shows a diagonal symmetric arrangement of capacitors and a battery. The potentials of the points $B$ and $D$ are:

For the given circuit,find the charge on the $4\ \mu F$ capacitor in $\mu C$.