Electric Potential and Potential Energy

Just as a raised mass has gravitational potential energy that depends on its position, a charge has ELECTROSTATIC potential energy that depends on its position relative to other charges -- the stored energy representing the work an external agent must do against the electrostatic (Coulomb) force to bring the charges into a given configuration. Since a system always tends toward its own lowest-energy configuration, work always has to be supplied by an outside agent to move it AWAY from that natural configuration; equivalently, work done AGAINST the electrostatic force always shows up as an INCREASE in the system's potential energy: $\vec{F}\cdot d\vec{r}=dU$.

To make this concrete, fix a source charge $+Q$ at the origin O, and consider bringing a small test charge $+q_0$ from a point at distance $r_1$ to a point at distance $r_2$ from O, working against the Coulomb repulsion $\vec{F}_E=\dfrac{1}{4\pi\epsilon_0}\dfrac{Qq_0}{r^2}\hat{r}$ at every step. Integrating the work along this path gives $\Delta U=\displaystyle\int_{r_1}^{r_2}\vec{F}_E\cdot d\vec{r}=\dfrac{Qq_0}{4\pi\epsilon_0}\left(\dfrac{1}{r_1}-\dfrac{1}{r_2}\right)$ -- and crucially, because the integral only involves the START and END distances $r_1,r_2$, this change in PE is completely independent of whatever PATH was actually taken between them: the electrostatic force is CONSERVATIVE, exactly like gravity.

Since the force -- and hence the PE -- naturally goes to zero as the separation $r\to\infty$, it is conventional (and always allowed, since only PE DIFFERENCES are ever physically meaningful) to fix the zero of potential energy AT infinity. With that convention, the PE of two point charges $q_1,q_2$ a distance $r$ apart is simply $U(r)=\dfrac{1}{4\pi\epsilon_0}\dfrac{q_1q_2}{r}$. Its SI unit is the joule; the more convenient practical unit at the atomic scale is the electron-volt, $1\,\text{eV}=1.6\times10^{-19}$ J (the kinetic energy an electron gains crossing a 1 V potential difference), with $1\,\text{meV}=1.6\times10^{-22}$ J and $1\,\text{keV}=1.6\times10^{-16}$ J as its own sub- and super-multiples. …