Total electric potential energy

[betelgeuse91]

Homework Statement

My textbook states that for point charges $q_1, q_2, ... ,$ where distance between $q_i$ and $q_j$ is $r_{ij},$ the total potential energy U is the sum $U = \dfrac{1}{4\pi\epsilon_0} \sum_{i<j} \dfrac{q_iq_j}{r_{ij}}$ and specifically mentions not to count them twice as (i,j) and (j,i), but I don't see why.

Homework Equations

Potential energy of a point charge $q_i$ is $U_i = \dfrac{q_i}{4\pi\epsilon_0}\sum_{i \neq j}\dfrac{q_j}{r_j}$

The Attempt at a Solution

Using the above equation, I think the total potential energy should be $U = \dfrac{1}{4\pi\epsilon_0} \sum \dfrac{q_iq_j}{r_{ij}}$, counting both cases (i,j) and (j,i). For instance, when there are two point charges $q_1$ and $q_2$, the potential energies of individuals is $U_1 = \dfrac{q_1}{4\pi\epsilon_0}\dfrac{q_2}{r_{12}}$, $U_2 = \dfrac{q_2}{4\pi\epsilon_0}\dfrac{q_1}{r_{21}}$ so that the total potential energy is the sum $U = U_1 + U_2 = \dfrac{1}{4\pi\epsilon_0} \sum_{1\leqslant i,j\leqslant2} \dfrac{q_iq_j}{r_{ij}}$ which counts both the cases (1,2) and (2,1).
Where am I wrong?

 

[betelgeuse91]

This condition should add to the last equation: $i\neq j$

 

[J Hann]

Suppose you have only 2 charges.
The work done in moving charge 1 into place is zero.
Then you can move charge 2 into place by doing work that depends on R12.
So you don't need R21, since you would be counting the work twice.
Further, if you add an additional charge then the additional work done
depends on R13 and R23 and you don't need to count R31 and R32.