In a parallel plate capacitor,if $10^{12}$ electrons pass from one plate to another,a potential difference of $10 \,V$ is developed across the plates. The capacitance of the capacitor is

If the distance between the plates of a parallel plate capacitor of capacity $10 \mu F$ is doubled,then the new capacity will be: (in $\mu F$)

The capacity of a parallel plate capacitor is $C$. What will be its capacity when the separation between the plates is halved?

The plates of a parallel plate capacitor are pulled apart with a velocity $v$. If at any instant their mutual distance of separation is $d$,then the magnitude of the time rate of change of capacity depends on $d$ as follows:

Why is a $1$ $F$ unit so big in practice?

$A$ parallel plate capacitor has circular plates of $10\, cm$ radius separated by an air-gap of $1\, mm$. It is charged by connecting the plates to a $100\, V$ battery. The change in energy stored in the capacitor when the plates are moved to a distance of $1\, cm$ while remaining connected to the battery is: