In an isolated 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

$A$ parallel plate air capacitor is charged to a potential difference of $V \text{ volts}$. 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

$A$ parallel plate capacitor is charged by a battery and then the battery is disconnected. If the distance between the plates is increased using an insulated handle,then:

$A$ parallel-plate capacitor consists of a fixed plate and a movable plate that is allowed to slide in the direction parallel to the plates. Let $x$ be the distance of overlap,as shown in the figure. The separation between the plates is fixed. Assume that the plates are electrically isolated,so that their charges $\pm Q$ are constant. Force on the movable plate is proportional to

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 $\mu F$)

$A$ 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: