Accelerating Charges and the Electron Volt

Joules are awkward for particles because the numbers come out tiny: 10⁻¹⁹ or smaller. Voltage, on the other hand, is something you set on a power supply and read off a dial. Since the work done on a charge q crossing a potential difference V is simply W = qV, physicists skip the joule conversion and just multiply charge by voltage. For the elementary charge, one electron volt (eV) is defined as the energy one electron gains crossing a one volt potential difference: 1 eV = 1.602x10⁻¹⁹ J.

Picture two parallel plates with a potential difference V between them. An electron released near the negative plate feels a force toward the positive plate and accelerates across the gap. By the work-energy theorem, the work done by the electric field equals the kinetic energy gained: eV = (1/2)mv², with no other forces and starting from rest. Solve for speed: v = √(2eV/m). Plug in V = 2000 volts, e = 1.602x10⁻¹⁹ C, and m = 9.11x10⁻³¹ kg, and you get v = √(2(1.602x10⁻¹⁹)(2000)/9.11x10⁻³¹) = about 2.65x10⁷ m/s, fast but still well under light speed.

If the particle carries charge nq instead of q, the energy gained becomes nqV, so the electron volt scales linearly with how many elementary charges the particle carries. A doubly charged ion crossing the same 2000 volts gains twice the energy of a single electron, though its larger mass usually means a smaller final speed. This same qV bookkeeping is what makes electric potential so useful across the unit: once you know the potential difference a charge crosses, you know its energy change without ever touching the field or the path it took.