Coulomb potential energy ($U(r)=k{q}_1{q}_2/r$)

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.

A thin disk of radius $a>0$ in $z=0$ carries uniform surface charge density $\sigma =Q/(\pi a^2)$. Summing the Coulomb potential of concentric rings gives the displayed axis potential, with zero reference at infinity. Put $A=\sigma /(2{ϵ}_0)$, use spherical coordinates $z=r\cos \theta$, and let $P_2(t)=(3t^2-1)/2$. The near-center expansion, separately harmonic above and below the charged sheet, isThe absolute value represents the jump in normal electric field across the sheet; a single regular solid harmonic series cannot describe both half-spaces simultaneously. The constant and quadratic coefficients follow from the positive-axis Taylor series, and reflection symmetry supplies the other half-space. The far-field multipole expansion isThe monopole is the total charge, the dipole vanishes by reflection symmetry, and the next term is the axisymmetric electric quadrupole. These expansions converge for $r<a$ within each source-free half-space and for $r>a$ in the exterior, respectively; at a point of the sheet their potentials have continuous limiting values but their normal derivatives differ.

For an annulus of inner radius $a>0$, outer radius $b>a$ and uniform surface charge density $\sigma$, integrate circular rings of electric charge to get $V(x)=\sigma [\sqrt{x^2+b^2}-\sqrt{x^2+a^2}]/(2{ϵ}_0)$ with zero electric potential at infinity. Differentiation gives the displayed axial electric field. Its magnitude is greatest at $|x|=(ab{)}^{2/3}/\sqrt{a^{2/3}+b^{2/3}}$. Near the centre, $E_x=Qx/[2\pi {ϵ}_{0}ab(a+b)]+O(x^3)$, where $Q=\pi \sigma (b^2-a^2)$; an oppositely charged particle constrained to the axis consequently undergoes small-amplitude simple harmonic motion.