The force acting on a charged particle placed between the plates of a charged parallel plate capacitor is $F$. If one plate of the capacitor is removed,then the force acting on the same particle will become:

$A$ 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:

Charges of $2Q$ and $-Q$ are placed on two plates of a parallel plate capacitor. If the capacitance of the capacitor is $C$,the potential difference between the plates is:

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

$A$ parallel plate capacitor is charged and then disconnected from the charging battery. If the plates are now moved further apart by pulling them by means of insulating handles,then