$A$ particle has a mass $400$ times that of an electron and a charge double that of an electron. It is accelerated by a potential difference of $5 \ V$. If the particle was initially at rest,what will be its final kinetic energy in $eV$?

Three equal electric charges,each of magnitude $q$,are placed at the vertices of an equilateral triangle of side length $L$. The electrostatic potential energy of the system is:

Four electric charges $+q, +q, -q$ and $-q$ are placed in order at the corners of a square of side $2r$. The electric potential at a point $P$ midway between the two negative charges is:

An electron is accelerated through a potential difference of $2.5 \ V$. The energy gained by the electron is . . . . . . . (Charge of an electron = $1.6 \times 10^{-19} \ C$)

The electric charges $+2q$, $+2q$, $-2q$ and $-2q$ are placed at the corners of a square of side $2L$ as shown in the figure. The electric potential at point $A$, midway between the two charges $+2q$ and $+2q$, is $(\epsilon_0 = \text{permittivity of free space})$

An electric charge $10^{-6} \mu C$ is placed at the origin $(0,0) \text{ m}$ of an $X-Y$ coordinate system. Two points $P$ and $Q$ are situated at $(\sqrt{3}, \sqrt{3}) \text{ m}$ and $(\sqrt{6}, 0) \text{ m}$ respectively. The potential difference between the points $P$ and $Q$ will be: