Does the capacitance of a parallel plate capacitor depend on the potential difference $(V)$ between its plates?

Three plates $A, B$ and $C$ each of area $0.1 \ m^2$ are separated by $0.885 \ mm$ from each other as shown in the figure. A $10 \ V$ battery is used to charge the system. The energy stored in the system is

$A$ parallel plate capacitor is charged. Then the battery is removed. Now,if the plates are pulled apart,

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 $\mu F$.

$A$ parallel plate capacitor is to be designed with a voltage rating $1\; kV$,using a material of dielectric constant $3$ and dielectric strength about $10^{7}\; V\,m^{-1}$. (Dielectric strength is the maximum electric field a material can tolerate without breakdown,i.e.,without starting to conduct electricity through partial ionisation.) For safety,we should like the field never to exceed,say $10\%$ of the dielectric strength. What minimum area (in $cm^2$) of the plates is required to have a capacitance of $50\; pF$?