Positive charges repel each other, so if I place two positive
charges near one another and let ago, they will start moving, gaining
kinetic energy. Where does this energy come from? Each charge is being pushed by the electric force, and is moving in the same direction as it is being pushed, so the electric force is doing positive **work** on the charge.

But work is merely the *transfer* of energy; the energy still has to have a source, and in this case the source is the electric field. Just like gravity, the electric force is conservative, which means that it has an associated potential energy which falls and rises as the electric field does positive or negative work. We can think of potential energy as something like a storage tank for energy. When the electric field does positive work on a charge (as in the example above), the potential energy of the field decreases as the tank is depleted of energy. Conversely, the potential energy increases when the electric field steals energy by doing negative work (by slowing down a charge, for instance). Electric potential energy is a function of the *relative positions* of the charges in a system; it doesn't depend on how the charges got where they are, only where they are right now.

Interactive 7.1.1

If a charge starts moving spontaneously (that is, without any outside forces) due to the electric force, then it will move in the direction which *decreases* its potential energy. For a charge to start moving at all, it has to gain kinetic energy which must come from somewhere, and if there is nothing else pushing it, that energy must come from the electric potential energy, which drops as a result. One might say that the potential energy drops when the system does what it "wants" to do, or when the system is "relaxing", analogous to what happens in a spring whose potential energy decreases. Conversely, if the potential energy increases, then the system is "tensing": e.g. pulling two opposite charges away from each other will increase their potential energy because they are pulling back. Go through the example above again and see if this idea makes it easier to determine what the potential energy does.

Work done by the electric field depletes the potential energy by the same amount, so we can relate the work W done to a charge to the change in potential energy Δ*P*:

$$W_{\mathrm{by\,field}}=-\Delta P$$

where the change in potential energy is the final value minus the initial value. The work can be defined in terms of the force, but in general the force won't remain constant as the charge moves, so we need to break the charge's motion into a number of small displacements \(\vec dr\), find the work done by each, and then add all the small amounts of work together. In other words
$$ \Delta P=P_f-P_i=-\int \vec F\cdot d\vec r$$

Let's consider a specific case: a source charge
$$
\definecolor{r}{rgb}{1,0,0}
\definecolor{b}{rgb}{0,0,1}
\begin{eqnarray}
P_f-P_i &=& -\int_{x_i}^{x_f}F_x\,dx\cr
&=&-\int_{x_i}^{x_f}k{q_sq_t\over x^2}\,dx\cr
&=&-kq_sq_t\int_{x_i}^{x_f}x^{-2}\,dx\cr
&=&-kq_sq_t[-x^{-1}]_{x_i}^{x_f}\cr
\Rightarrow {\color{r}P_f}-{\color{b}P_i}&=&{\color{r} {kq_sq_t\over x_f}}-{\color b {kq_sq_t\over x_i}}\cr
\end{eqnarray}$$

Both sides of the final equation have a final value minus an initial value, and so one might suppose that we can match initial term with initial term, and final with final. This is too restrictive an assumption, however; the most we can assume from this result is that

$$P=k{q_sq_t\over d}+P_\infty$$

Here is a graph of the potential energy equation in two circumstances, as two charges move farther away from each other. If the charges have the same sign, so that *q _{s}q_{t}*>0, then the potential energy decreases as the charges separate; if the charges have opposite sign, the potential energy increases from negative infinity. In both cases, the potential energy approaches the same value

The variable *P _{∞}*, called the

Consider two 1µC charges as shown.

- Find the potential energy of the system if the two charges are 1 meter apart, and \(P_\infty=0\mathrm{J}\)?$$P=(9\ten9\u{Nm^2/C^2}){(10^{-6}\u{C})(10^{-6}\u{C})\over (1\u{m})}+0=9\u{mJ}$$
- If I separate the charges so they are now 2 meters apart, what is the change in their potential energy?The potential energy after they have been moved is $$P=(9\ten9\u{Nm^2/C^2}){(10^{-6}\u{C})(10^{-6}\u{C})\over (2\u{m})}+0=4.5\u{mJ}$$ and so the change in potential energy is $$\Delta P=P_f-P_i=4.5\u{mJ}-9\u{mJ}=-4.5\u{mJ}$$
- Repeat the same calculations if \(P_\infty=5\,\mathrm{mJ}\).The initial potential energy is $$P_i=(9\ten9\u{Nm^2/C^2}){(10^{-6}\u{C})(10^{-6}\u{C})\over (1\u{m})}+5\u{mJ}=14\u{mJ}$$ and similarly, the final potential energy is also 5mJ higher than before: $$P_f=(9\ten9\u{Nm^2/C^2}){(10^{-6}\u{C})(10^{-6}\u{C})\over (2\u{m})}+5=9.5\u{mJ}$$ The change in potential energy is thus $$\Delta P=P_f-P_i=9.5\u{mJ}-14\u{mJ}=-4.5\u{mJ}$$ which is exactly what it is when \(P_\infty=0\u{J}\).

Figure: Three different people calculate three different values for the gravitational potential energy of the red ball, and they're all right.

The choice of baseline *P _{∞}* is entirely up to you, and that mean that the potential energy of any given configuration of charges is also completely arbitrary: you might find that two charges have a potential energy of 1J or 0J or -15,000J. This doesn't matter, however, because only

Interactive 7.1.2

In the original configuration of the demo above, the total potential energy is \(-0.4{\,\rm mJ}\) where \(P_\infty=0{\,\rm J}\). How much work is required to assemble this configuration of charges, if the charges start far away from each other?

If the charges start far away from each other, then their initial potential energy is \(P_\infty=0{\,\rm J}\). Since the final potential energy is less than this, we don't have to do work on the charges to move them into place. In fact, we have to remove 0.4 mJ of energy from the system... by stopping them once they are in position, for example.