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The Coulomb potential energy is the interaction potential energy of two point charges in vacuum. With separation $r>0$, electric charges $q_1,q_2$, $k=1/(4\pi {\epsilon }_0)$ and zero potential energy at infinite separation,This is mutual interaction energy; it does not include the divergent self-energies of ideal point charges. Unlike electric potential, which has units of volts, $U$ has SI units of joules. With charge 1 fixed and positive radial direction away from it, the conservative force in Coulomb's law givesThus the $1/r$ potential energy and the inverse-square law for force are two descriptions of the same interaction. ^[ref].

For $q_1=+2\mathrm{nC}$, $q_2=-3\mathrm{nC}$ and $r=0.20\mathrm{m}$, using the stipulated approximation $k=8.99\times 10^9\mathrm{N}{\mathrm{m}}^2/{\mathrm{C}}^2$,At $r=0.40\mathrm{m}$, $U=-1.3485\times 10^{-7}\mathrm{J}$. Separating the pair from $0.20$ to $0.40\mathrm{m}$ therefore requires external work $\Delta U=+1.3485\times 10^{-7}\mathrm{J}$ if the initial and final kinetic energies agree; the electric force does $-1.3485\times 10^{-7}\mathrm{J}$ of work. Separating them all the way to infinity requires $+2.697\times 10^{-7}\mathrm{J}$. The negative initial potential energy means that the pair lies below the chosen separated reference; it does not mean a negative kinetic energy.

Like-signed electric charges have positive $U$, which decreases as they separate, so their electric force is repulsive. Opposite-signed electric charges have negative $U$, which decreases as they approach, so their electric force is attractive. This attractive interaction supplies the Coulomb potential used in the hydrogen atom; determining its stationary quantum states also requires quantum mechanics. ^[ref]. Dividing $U$ by a nonzero test electric charge $q_2$ gives the source's electric potential ${\Phi }_1=k{q}_1/r$. More generally, the electrostatic potential energy of point charges sums each unordered charge pair once. This energy description is useful because work between two separations follows from endpoint values, without integrating a complicated trajectory.

Let $\rho =a/\sqrt{3}$ be the distance from the origin to each fixed charge, and denote their position vectors by ${\mathbf{R}}_j$. For a small planar displacement $\mathbf{r}$, the Taylor expansioncan be summed using the threefold symmetry relationsThe electric potential due to the fixed charges is consequentlyThe moving charge has electrostatic potential energy of point charges $U=q\Phi$. Because the quadratic coefficient is positive, the origin is a strict local minimum of $U$, and the equilibrium is stable.

Interpret the coordinates as a plane in ordinary three-dimensional electrostatics. Set $k_e=1/(4\pi {ϵ}_0)$, where ${ϵ}_0$ is the vacuum permittivity. The work required to bring the second point charge in quasistatically is its electric charge times the first charge's electric potential:The individual point-charge self-energies are excluded; this is the electrostatic potential energy of point charges relative to infinite separation.

By the superposition principle, for $\mathbf{r}=(x,y)$ and $\mathbf{d}=(d,0)$ the electric field isAt the first position both contributions point toward positive $x$, whereas at the second they oppose one another. ThusAt $(d/2,d)$ the two distances are $\sqrt{5}d/2$. The vertical electric field components cancel and the horizontal components add:Dimensional convention. The source's description of space as two-dimensional cannot literally mean two-dimensional electrostatics with the same inverse-square formula. In a genuinely two-dimensional Gauss's law, the circumference grows as $r$, giving an electric field proportional to $1/r$ and a logarithmic electric potential. For example, integrating $1/r$ from $d$ to infinity diverges, so the same finite reference potential energy at infinite separation does not exist. The displayed results use the conventional three-dimensional Coulomb's law restricted to the stated plane.