Two points charges q₁, q₂, initially at infinity are brought one-by-one to points p₁ and p₂, in external Electric field specified by position vectors $\overrightarrow{{\mathrm{r}}_1} \mathrm{and} \overrightarrow{{\mathrm{r}}_2},$ relative to some origin. What is the potential energy of this charge configuration?

Given, two point charges q₁ and q₂ are brought from infinity to points P₁ and P₂ in the presence of an external field.

Let $\overrightarrow{\mathrm{E}}$ be the required external field. 

The potential energy of the system is equl to the total amount of work done in assembling the configuration. 

Therefore,
Work done on q₁ against the external field $\stackrel{\rightharpoonup }{\mathrm{E}}$= q₁.V($\stackrel{\mathbf{\rightharpoonup }}{{\mathbf{r}}_{\mathbf{1}}}$ )
Work done on q₂ against the external electric field $\stackrel{\rightharpoonup }{\mathrm{E}}$$= {\mathrm{q}}_2. \mathrm{V}\left(\overrightarrow{{\mathrm{r}}_2}\right)$

and,

Work done on q₂ against the field due to q₁₌ Work done on q₁ against the field due to q₂ which is given by, 

           $\frac{1}{4{\mathrm{\pi \epsilon }}_0}. \frac{{\mathrm{q}}_1 {\mathrm{q}}_2}{{\mathrm{r}}_{12}} = \frac{{\mathrm{q}}_1{\mathrm{q}}_2}{4{\mathrm{\pi \epsilon }}_0{\mathrm{r}}_{12}}$

where, r₁₂ is the distance between q₁ and q₂.

Using the superposition principle for fields, we add up the work done on q₂ against the two fields.
Therefore, work done in carrying q₂ to r₂ is
             ${\mathrm{q}}_2. \mathrm{V}\left(\overrightarrow{{\mathrm{r}}_2}\right) + \frac{{\mathrm{q}}_1 {\mathrm{q}}_2}{4{\mathrm{\pi \epsilon }}_0 {\mathrm{r}}_{12}}$

Hence, Potential energy of the system is given by,

         ${\mathrm{q}}_1.\mathrm{V}\left(\overrightarrow{{\mathrm{r}}_1}\right) + {\mathrm{q}}_2. \mathrm{V}\left(\overrightarrow{{\mathrm{r}}_2}\right) + \frac{{\mathrm{q}}_1 {\mathrm{q}}_2}{4{\mathrm{\pi \epsilon }}_0{\mathrm{r}}_{12}}$