- #1

etotheipi

##U = qV_{1} - qV_{2} = q \times -\frac{Ed}{2}\cos{\theta} - q \times \frac{Ed}{2}\cos{\theta} = -E(qd)\cos{\theta} = -\vec{E} \cdot \vec{p}## where ##\vec{p} = q\vec{d}##.

My question is, why isn't the potential energy of the two charges ##U_{3} = -\frac{q^{2}}{4\pi\epsilon_{0}d^{2}}## also included in the summation above?

Since, for any collection of charged particles, the potential energies of subsystems can be added pairwise (i.e. ##-\Delta U = (W_{12} + W_{21}) + (W_{13} + W_{31}) + (W_{23} + W_{32}) = -\Delta U_{a} - \Delta U_{b} - \Delta U_{c}##. And to me, there seem to be three charged things at play here: the source (whatever it happens to be, e.g. perhaps an infinite charged sheet), and the two charges.

Consequently, the potential energy should be the sum of that of the source and +ve charge, the source and -ve charge, and the two charges. Although I think this might be incorrect since I haven't defined ##0V## to be at infinite distance!