$A$ regular hexagon of side $6 \ cm$ has a charge of $2 \ \mu C$ at each of its vertices. What is the potential at the centre of the hexagon? $\left[\frac{1}{4 \pi \epsilon_0}=9 \times 10^9 \text{ SI units}\right]$

In the rectangle shown below,two corners have charges $q_1 = -5\,\mu C$ and $q_2 = +2.0\,\mu C$. The work done in moving a charge $q_0 = +3.0\,\mu C$ from $B$ to $A$ is ......... $J$. (Given: $1/4\pi\varepsilon_0 = 10^{10}\,N\cdot m^2/C^2$)

The work done in moving a charge of $20 \ C$ from point $A$ to point $B$ is $2 \ J$. What is the potential difference between the two points (in $V$)?

A charge $+q$ is fixed at each of the points $x = x_0, x = 3x_0, x = 5x_0, \dots, \infty$ on the $x$-axis, and a charge $-q$ is fixed at each of the points $x = 2x_0, x = 4x_0, x = 6x_0, \dots, \infty$. Here $x_0$ is a positive constant. Take the electric potential at a point due to a charge $Q$ at a distance $r$ from it to be $Q/(4\pi\varepsilon_0 r)$. Then, the potential at the origin due to the above system of charges is:

An infinite number of charges, each numerically equal to $q$ and of the same sign, are placed along the $x-$axis at $x = 1, 2, 4, 8, \dots \, \text{meters}$. The electric potential at $x = 0$ due to this set of charges is:

Three particles,each having a charge of $10\,\mu C$,are placed at the corners of an equilateral triangle of side $10\,cm$. The electrostatic potential energy of the system is.....$J$ (Given $\frac{1}{4\pi\varepsilon_0} = 9 \times 10^9\,N\cdot m^2/C^2$).