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Why is this? I am confused about it because W = Fd and both F and d have the same direction.

The potential energy U of the system is equal to the work we must do to assemble the system, bringing in each charge from an infinite distance.

Calculation:

Let's mentally build the system of Fig. 24-19, starting with one of the charges, say q1, in place and the others at infinity. Then we bring another one, say q2, in from infinity and put it in place. From Eq. 24-46 with d substituted for r, the potential energy U12 associated with the pair of charges q1 and q2 is

U12 = kq1q2/d

We then bring the last charge q3 in from infinity and put it in place. The work that we must do in this last step is equal to the sum of the work we must do to bring q3 near q1 and the work we must do to bring it near q2. From Eq. 24-46, with d substituted for r, that sum is

W13 + W23 = U13 + U23 = kq1q3/d + kq2q3/d

The total potential energy U of the three-charge system is the sum of the potential energies associated with the three pairs of charges. This sum (which is actually independent of the order in which the charges are brought together) is