What is the energy density of a parallel plate capacitor with area $A$ and charge $Q$?

Two protons $A$ and $B$ are placed in the space between the plates of a parallel plate capacitor charged up to $V$ volts (see figure). If the forces on the protons are $F_A$ and $F_B$ respectively,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

Five identical capacitor plates,each of area $A$,are arranged such that adjacent plates are at a distance $d$ apart. The plates are connected to a source of $emf$ $V$ as shown in the figure. Then the charges on plates $1$ and $4$ are,respectively:

Two identical thin metal plates have charges $q_{1}$ and $q_{2}$ respectively,such that $q_{1} > q_{2}$. The plates are brought close to each other to form a parallel plate capacitor of capacitance $C$. The potential difference between them is:

The area of each plate of a parallel plate capacitor is $20 \, cm^2$ and the separation between the plates is $2 \, mm$. If the dielectric strength of air is $3 \times 10^6 \, V/m$,the maximum possible value of the emf of the battery that can be connected across the plates of this capacitor and the corresponding charge on the plates are: