Deduce the expression for the potential energy of a system of two point charges `q_1` and `q_2` brought from infinity to the points `vecr_1` and `vecr_2` respectively in the presence of external electric field `vecE`.

Explanation

To deduce the potential energy of a system of two point charges `q_(1)` and `q_(2)` being brought from infinity to the points `vecr_(1)` and `vecr_(2)` respectively in an external electric field `vecE_(1)`, first we calculate the work done in bringing the charge `q_1` from infinity of `vecr_1` . Work done in this step is `W_(1) = q_(1) V (vecr_(1))` , where `V (vecr_(1))` is the electrostatic potential at `r_1` .
Now we consider the work done in bringing `q_2` to `vecr_2` . In this step work `W_(2)` is done against the electric field and work `W_(12)` is done against the field due to charge `q_(1)` . Obviously `W_(2) = q_(2) V (vecr_(2))` where `V (vecr_(2))` is the electrostatic potential at `vecr_(2)` and `W_(12) = (q_(1) q_(2))/(4 pi in_(0) . r_(12))` , where `r_(12) = |vecr_(1) - vecr_(2)|` = distance between `q_(1)` and `q_(2)`
`therefore` Total work done in bringing `q_(2)` to `vecr_(2) = W_(2) + W_(12) = q_(2) V (vecr_(2)) + (q_(1) q_(2))/(4pi in_(0) . r_(12))`
`therefore` Potential energy of the system U = The total work done in assembling the configuration
`U = W_(1) + W_(2) + W_(12) = q_(1) V (r_(1)) + q_(2) V (r_(2)) + (q_(1) q_(2))/(4 pi in_(0) . r_(12))`