13.7 Electric potential energy: potential difference  (Page 3/10)

Note that the energies calculated in the previous example are absolute values. The change in potential energy for the battery is negative, since it loses energy. These batteries, like many electrical systems, actually move negative charge—electrons in particular. The batteries repel electrons from their negative terminals (A) through whatever circuitry is involved and attract them to their positive terminals (B) as shown in [link] . The change in potential is $\mathrm{\Delta }V=V_{\text{B}}{\mathrm{–V}}_{\text{A}}=\text{+12 V}$ and the charge $q$ is negative, so that $\text{ΔPE}=q\mathrm{\Delta }V$ is negative, meaning the potential energy of the battery has decreased when $q$ has moved from A to B.

The electron volt

The energy per electron is very small in macroscopic situations like that in the previous example—a tiny fraction of a joule. But on a submicroscopic scale, such energy per particle (electron, proton, or ion) can be of great importance. For example, even a tiny fraction of a joule can be great enough for these particles to destroy organic molecules and harm living tissue. The particle may do its damage by direct collision, or it may create harmful x rays, which can also inflict damage. It is useful to have an energy unit related to submicroscopic effects. [link] shows a situation related to the definition of such an energy unit. An electron is accelerated between two charged metal plates as it might be in an old-model television tube or oscilloscope. The electron is given kinetic energy that is later converted to another form—light in the television tube, for example. (Note that downhill for the electron is uphill for a positive charge.) Since energy is related to voltage by $\text{ΔPE}=q\mathrm{\Delta }V,$ we can think of the joule as a coulomb-volt.