If the charge on each plate of a parallel plate capacitor is $Q$ and the magnitude of the electric field between the plates is $E$,then the force on each plate of the parallel plate capacitor will be:

$A$ parallel plate capacitor is made by stacking $n$ equally spaced plates connected alternately. If the capacitance between any two plates is $C$, then the resultant capacitance is:

When the distance between the two plates of a parallel plate capacitor is $6\, cm$,its capacitance is $15\ \mu F$. If the distance between the plates is reduced to $2\, cm$,the capacitance of the parallel plate capacitor becomes ... $\mu F$.

If the plates of a parallel plate capacitor connected to a battery are moved close to each other,then
$A$. the charge stored in it,increases.
$B$. the energy stored in it,decreases.
$C$. its capacitance increases.
$D$. the ratio of charge to its potential remains the same.
$E$. the product of charge and voltage increases.
Choose the most appropriate answer from the options given below:

$A$ parallel plate capacitor has $1\,\mu F$ capacitance. One of its two plates is given $+2\,\mu C$ charge and the other plate,$+4\,\mu C$ charge. The potential difference developed across the capacitor is......$V$

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: