Derive the formula for the electric potential energy of system of three charges.

Explanation

Let charge `q_(1),q_(2)` and `q_(3)` are bring from infinity distance to at the point `P_(1), P_(2)` and `P_(3)`, located at distance `r_(1),r_(2)` and `r_(3)` respectively.
All three charges are brought as shown in figure.

To bring `q_(1)` first from infinity to `P_(1)`, the work done `W_(1)=0`
because there is no external force to bring `q_(1)` to `P_(1)`
Electric potential at `P_(2)` due to charge `q_(1)`
`V_(1)= (kq_(1))/(r_(12))`
Now work done to bring charge `q_(2)` at point `P_(2)`
`W_(2)=V_(1)xxq_(2)`
`:. W_(2)=(kq_(1)q_(2))/(r_(12))`
Electric potential at Pz due to charge `q_(1) + q_(2)`
`V_(2)=(kq_(1))/(r_(13))+(kq_(2))/(r_(23))`
`:.` Work done to bring charge `q_(3)` to `P_(3)`
`W_(3)` potential at `P_(3)` due to `q_(1)+q_(2)xxq_(3)` charge
`=k[(q_(1))/(r_(13))+(q_(2))/(r_(23))]xxq_(3)`
`=k[(q_(1)q_(3))/(r_(13))+(q_(2)q_(3))/(r_(23))]`
`:.` otal potential energy of charges `q_(1)+q_(2)+q_(3)`
`U=W_(1)+W_(2)+W_(3)`
`[ because` Electric force is conservative and hence W U]
`:. U=k[0+(q_(1)q_(2))/(r_(12))+(q_(1)q_(3))/(r_(13))+(q_(2)q_(3))/(r_(23))]`
`:. U=k[(q_(1)q_(2))/(r_(12))+(q_(1)q_(3))/(r_(13))+(q_(2)q_(3))/(r_(23))]`
Equation (3) is true for any sign of `q_(1)` and `q_(2 )`
If `q_(1) q_(2) gt 0 ` potential energy will be positive. And if `q_(1)q_(2) lt 0` potential energy will be negative.