If the distance between the plates of a capacitor is halved and the area of the plates is doubled,what will be the new capacitance?

$A$ large cloud of size $5 \, km \times 5 \, km$ is hovering over the Earth at a height of $1 \, km$. Assuming the cloud and the Earth form a parallel plate capacitor with air as the dielectric,calculate the capacitance of the cloud-Earth system in $\mu F$.

Two parallel plate capacitors with different plate separations but the same capacitance are connected in series to a battery. Both capacitors are filled with air. The quantity that is $NOT$ the same for both the capacitors when they are fully charged is:

The plates of a parallel plate capacitor of capacity $C_1$ are moved closer together until they are half their original separation. The new capacitance $C_2$ is

In an insulated parallel-plate capacitor of capacitance $C$,the four surfaces have charges $Q_1, Q_2, Q_3$ and $Q_4$ as shown. The potential difference between the plates is

Four metal plates,each with surface area $A$ on one side,are placed with separation '$d$' as shown in the figure. The capacitance between $a$ and $b$ is ($\varepsilon_0$ = permittivity of free space).