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

The capacity of a parallel plate capacitor increases with the

$A$ parallel plate air capacitor is charged to a potential difference of $V$ volts. After disconnecting the charging battery,the distance between the plates of the capacitor is increased using an insulating handle. As a result,the potential difference between the plates:

The plates of a parallel plate capacitor of capacity $50\,\mu F$ are charged to a potential of $100\;V$ and then disconnected from the battery. The distance between the plates is then doubled. How much work is done in doing so?

If $Q$ is the charge on the plates of a capacitor of capacitance $C$,$V$ is the potential difference between the plates,$A$ is the area of each plate,and $d$ is the distance between the plates,then the force of attraction between the plates is:

Two parallel plate capacitors have their plate areas $100 \,cm^2$ and $500 \,cm^2$ respectively. If they have the same charge and potential,and the distance between the plates of the first capacitor is $0.5 \,mm$,then the distance between the plates of the second capacitor is ........ $cm$.