Electric Potential

Sources and Targets

As we mentioned in the previous section, potential energy is stored in the relationships between charges, and so we need to calculate the potential energy between every pair of charges in order to find the total energy of the system. This quickly becomes unwieldy as the number of charges increases.

However, remember that only changes in potential energy are significant. If you have a bunch of charges and they're not moving with respect to each other, then it doesn't really matter what their potential energy is. In fact, you can choose \(P_\infty\) in such a way that the potential energy of those charges is whatever you want it to be. If one of those charges moves, then that will change the potential energy of the system, and that change is important. Even in that case, however, you only need consider the bonds between the stationary charges and the moving charge; you can still ignore the relationships that the stationary charges have with each other. In analogy with previous chapters, we'll call the stationary charges sources and the moving charges targets.

Assuming we only have one target, the potential energy of the system is $$P=k{q_{s1}q_t\over d_1}+k{q_{s2}q_t\over d_2}+\dots+P_\infty$$ where \(q_{s1}\), \(q_{s2}\), etc are the charges of the various source charges, and \(d_1\), \(d_2\), etc are the distances from those source charges to the target. If I factor out \(q_t\), I'm left with $$P=q_t\left[k{q_{s1}\over d_1}+k{q_{s2}\over d_2}+\dots+{P_\infty\over q_t}\right]$$ The part in brackets depends only on the source charges and the location of the target, and is called the electric potential at the location of the target.

Electric Potential

The electric potential (or just "potential") has the same relationship with electric potential energy, as the electric field does to the electric force. Like the electric field, it is created by the source charges, and is defined at every point in space, whether or not there is a target charge there to feel it.

The symbol of the electric potential is \(V\), and its value is the potential energy the system would have if a target charge \(q_t\) were placed at that location, divided by \(q_t\): $$V={P\over q_t} \qquad \hbox{or} \qquad P=q_tV$$ From the first equation, we see that the units of potential are joules per coulomb which is given the name of volt: \(1\u{J/C}=1\u{V}\).

There are three potential (ha ha) sources of confusion you may run into when dealing with the electric potential.

  1. The name: "Electric potential energy" and "electric potential" sound awfully similar, even though they're very different: not only do they have different units, but the potential energy is a property of a system of charges while the potential is a property of a location in space. So be careful, whenever you see the word "potential", to notice whether the word "energy" follows it or not.
  2. The symbol: A capital V is used as the variable for potential, and also as an abbreviation for volts, its unit. Thus you will run into expressions like "V=5V", where V means two different things. It's usually obvious in context what we're talking about, but be careful to read "5V" above as "5 volts" and not "5 times the potential".
  3. Differences: Just as with potential energy, the potential has an arbitrary baseline, so that only differences in potential matter. In fact, when electricians speak of "voltage", they're not referring to potential but the difference in potential between two points (which is called the potential difference or \(\Delta V\)). So for instance, a "9 volt battery" does not have a potential of 9 volts; rather, one of its terminals is 9 volts higher than the other. It is very rare to see a formula that involves V directly.

    Because potential differences are so important, it's useful to derive the following formula: $$\begin{eqnarray} P_f&=&q_tV_f\\ P_i&=&q_tV_i\\ \implies P_f-P_i&=&q_t(V_f-V_i)\\ \implies \Delta P&=&q_t\Delta V\\ \end{eqnarray}$$

    A positive source charge creates an electric potential. Location A is closer to the source, while location B is farther away. At which location is the electric potential larger?
    A) Point A
    B) Point B
    C) Both points are at the same potential.

    As was the case with electric fields, we ask "What Would a Proton Do?" In this case, suppose a proton was placed at point A, and then moved to point B. That proton would be moving away from the source charge, which is exactly what it wants to do: therefore the system is relaxing, and the potential energy is decreasing. Because \(P=q_tV\) and \(q_t\) is positive for a proton, that means the potential decreases from A to B as well: the potential at A is higher.

    On the other hand, suppose an electron moved from point A to point B. It would be moving from a higher to a lower potential, but its potential energy would increase. Why? For one, since the change in potential \(\Delta V\) is negative, then according to the formula \(\Delta P=q_t\Delta V\), the change in potential energy is the product of two negative numbers (negative \(q_t\) and negative \(\Delta V\)), or a positive number. Also, we note that when I move an electron away from a positive charge that is not what it wants to do, and so the system becomes "tenser".