Deduce the expression for the potential energy of a system of two point charges and `q_(2)` brought from infinity to the points with positions, `r_(1)` and `r_(2)` respectively, in presence of external electric field E.

Explanation

By definition, electric potential energy of any charge placed in the region of electric field is equal to the work done in bringing charge 9 from infinity to that point and given by U =qv. where. V is the electric potential (as potential at infinity is assumed to be zero) where, the charge q is placed. m
Now, considering the electric potentials at positions `r_(1)` and `r_(2)` as `V_(1)` and `V_(2)` respectively. Therefore, total potential energy of the system of two charges `q_(1)`, and `q_(2)` placed at points with position vectors, `r_(1)` and `r_(2)` in the region of Eis given by U = work done in bringing charge q, from infinite to that position in E is equal to the work done for charged `q_(2)`.from infinite to that position in E + work done to that of charge `q_(2)` at these positions in presence of `q_(1)` .
`i.e U = q_(1) V_(1) +q_(2) V_(2)= (1)/(4 pi epsilon_(0)) . (q_(1)q_(2))/(|r_(2)-r_(1)|) `