Derive an expression for the potential at a distance from a point charge

Consider a positive point charge 'q' is placed at a point O. E be the electric field due to the charge. Let 'A' be the point at a distance 'x' from O.

Electric field due to +q at point 'A' is
$\mathrm{E}=\frac{1}{4\mathrm{\pi }{\in }_0}\frac{\mathrm{q}}{{\mathrm{x}}^2}$
If a charge q₀ is placed at a point point 'A' the force between two charges q and q₀ is
$\mathrm{F}=\frac{1}{4{\mathrm{\pi ϵ}}_0} \frac{{\mathrm{qq}}_0}{{\mathrm{x}}^2}\text{ along }\overrightarrow{\mathrm{OA}}$
Let the charge q₀ be moved towards +q from A to B through a distance 'dx' against eletric field

The work done in moving the charge
$\mathrm{d\omega }=-\mathrm{Fdx}$

[∴–ve sign force and displacement are opposite in direction]

The total amount of workdone in moving the test charge from $\infty$ to point P is
$\begin{array}{l}\mathrm{\omega }={\int }_{\mathrm{\infty }}^{\mathrm{r}} \mathrm{d\omega }={\int }_{\mathrm{\infty }}^{\mathrm{r}} -\mathrm{Fdx}={\int }_{\mathrm{\infty }}^{\mathrm{r}} \frac{1}{4\mathrm{\pi }{\in }_0}\frac{{\mathrm{qq}}_0}{{\mathrm{x}}^2}\mathrm{dx}\\ =\frac{-{\mathrm{qq}}_0}{4\mathrm{\pi }{\in }_0}{\int }_{\mathrm{\infty }}^{\mathrm{r}} {\mathrm{x}}^{-2}\mathrm{dx}=\frac{{\mathrm{qq}}_0}{4\mathrm{\pi }{\in }_0}{\left[\frac{1}{\mathrm{x}}\right]}_{\mathrm{\infty }}^{\mathrm{r}}\\ =\frac{{\mathrm{qq}}_0}{4\mathrm{\pi }{\in }_0}\left[\frac{1}{\mathrm{r}}-\frac{1}{\mathrm{\infty }}\right];\mathrm{w}=\frac{1}{4\mathrm{\pi }{\in }_0}\frac{{\mathrm{qq}}_0}{\mathrm{r}}\end{array}$
Now the potential at the point P is $\mathrm{v}=\frac{\mathrm{w}}{{\mathrm{q}}_0}$
$\mathrm{v}=\frac{1}{4\mathrm{\pi }{\in }_0}\frac{\mathrm{q}}{\mathrm{r}}$