Electric Potential Energy and Potential

Unlike Coulomb's law, this formula is used with the signs of the charges, because the sign of the energy carries meaning. Like charges give a positive $E_e$: energy was put in to push them together, and they would fly apart if released. Unlike charges give a negative $E_e$: the pair has less energy than when separated, so work must be done to pull them apart, which is why they are said to be bound. Note that the $1/r$ here is not squared; energy falls off more slowly with distance than force does. (Take care not to confuse $E_e$, an energy in joules, with $E$, the field strength in N/C.)

The electric potential $V$ at a point is the electric potential energy per coulomb that a positive test charge would have there, $V=E_e/q$. Its unit is the volt, $1\text{V}=1\text{J/C}$. Around a single point charge, dividing $E_e$ by the test charge gives $V=kq/r$. Because potential is a scalar, the potential from several charges is found by simple addition with signs, no components needed. What matters physically is usually the potential difference $\Delta V$ between two points, since the work done moving a charge $q$ between them is $W=q\Delta V$.

Lines joining points of equal potential are called equipotential lines, and they always cross field lines at right angles; moving a charge along one takes no work. Between plates they are evenly spaced, which is what uniform means. A charge released in the gap feels a constant force $F=qE$, so it accelerates steadily. Rather than use kinematics, use energy: the electric potential energy it loses becomes kinetic energy, $|q|\Delta V=\frac{1}{2}m{v}^2$ for a charge starting from rest. This is exactly how the electron beam in an old television picture tube was produced, and how an X-ray tube still accelerates its electrons.

In the early twentieth century Robert Millikan sprayed tiny oil drops, some of which picked up electric charge, into the space between two horizontal parallel plates. By adjusting the potential difference he could make the electric force on a charged drop exactly balance its weight, so the drop hung still. With the field switched off he found the drop's mass from how quickly it drifted down through the air. At balance the forces are equal, so $qE=mg$, and with $E=\Delta V/d$ the charge on the drop is $q=mgd/\Delta V$.