$A$ bullet of mass $2\,g$ has a charge of $2\,\mu C$. Through what potential difference must it be accelerated, starting from rest, to acquire a speed of $10\,m/s$?

Eight charges,each of magnitude $q$,are placed at the vertices of a cube placed in a vacuum. The electric potential at the centre of the cube due to this system of charges is . . . . . . . ($\varepsilon_0 = $ permittivity of vacuum,$a = $ length of each side of the cube.)

$A$ charge $+q$ is fixed at each of the points $x = x_0, x = 3x_0, x = 5x_0, \dots$ up to $\infty$ on the $X$-axis and a charge $-q$ is fixed at each of the points $x = 2x_0, x = 4x_0, x = 6x_0, \dots$ up to $\infty$. Here $x_0$ is a positive constant. Taking the potential at a point due to a charge $Q$ at a distance $r$ from it to be $\frac{Q}{4\pi\varepsilon_0 r}$,find the potential at the origin due to the above system of charges.

If four charges $+12 \text{ nC}, -20 \text{ nC}, +32 \text{ nC}$ and $-15 \text{ nC}$ are arranged at the four vertices of a square of side $\sqrt{2} \text{ m}$,then the net electric potential at the centre of the square due to these four charges is (in $\text{ V}$)

Calculate the electrostatic potential energy of a system of two charges $7\ \mu C$ and $-2\ \mu C$ placed at points $(-9\ cm, 0, 0)$ and $(9\ cm, 0, 0)$ respectively (in the absence of an external field).

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