Two charge $ + \,q$ and $ – \,q$ are situated at a certain distance. At the point exactly midway between them

Electric field at a point $(x, y, z)$ is represented by $\vec E = 2x\hat i + {y^2}\hat j$ if potential at $(0,0,0)$ is $2\, volt$ find potential at $(1, 1, 1)$

Two thin wire rings each having a radius $R$ are placed at a distance $d$ apart with their axes coinciding. The charges on the two rings are $ + q$ and $ – q$. The potential difference between the centres of the two rings is

The electric potential $V(x, y, z)$ for a planar charge distribution is given by: 

$V\left( {x,y,z} \right) = \left\{ {\begin{array}{*{20}{c}}
{0\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,for\,x\, < \, – d}\\
{ – {V_0}{{\left( {1 + \frac{x}{d}} \right)}^2}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,for\, – \,d\, \le x < 0}\\
{ – {V_0}\left( {1 + 2\frac{x}{d}} \right)\,\,\,\,\,\,\,\,\,\,\,for\,0\, \le x < d}\\
{ – 3{V_0}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,for\,x\, > \,d}
\end{array}} \right.$

where $-V_0$ is the potential at the origin and $d$ is a distance. Graph of electric field as a function of position is given as

A charge $ + q$ is fixed at each of the points $x = {x_0},\,x = 3{x_0},\,x = 5{x_0}$….. $\infty$, on the $x – $axis and a charge $ – q$ is fixed at each of the points $x = 2{x_0},\,x = 4{x_0},x = 6{x_0}$,….. $\infty$. 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