# Electric potential energy

#### Prerequisites

- Coulomb's law
- Energy of place -- potential energy
- What do I have to know about integrals?
- The dimensional analysis tool

The idea of potential energy is that by changing relative position of two objects, kinetic energy can be changed in a reversible way as a result of the forces between objects. When this happens, we can define a "potential" energy — an energy of place that can serve as a way to "store" energy. With electrical forces, consider two identical positive charges approaching each other in vacuum with equal and opposite velocities. As they approach, they each feel a force, slowing each other down. Eventually, they come to a stop (instantaneously), turn around, and are repelled back, speeding up. The kinetic energy in the system is drained by the forces, being replaced by the "energy of place" — by the fact that the charges are close to each other. The forces between the charges then push them back, restoring the kinetic energy of the charges. Since the "draining" of the KE was reversible, we expect that the work done by the electric force between the charges can be expressed as a potential energy.

## Work done by the electric force

The electric force is a bit more complicated than our other two examples (gravity and springs). Since the force between two charges falls off like $1/r^2$, we can't treat it as a constant (like we could with gravity) and we can't treat it using a simple average (like we could with a spring). We have to do the integral explicitly. In order not to make things too messy, we'll set it up as simply as possible physically so the integral looks like something you've seen in calculus. And we'll see that dimensional analysis gives us almost the entire result anyway.

Let's consider a physical system made up of two positive charges. We'll nail one down and consider the work done by the force of that charge, $Q_B$, on a second charge, $Q_A$, that's allowed to move. We'll put the fixed charge (labeled B) at the origin, and let the movable charge (labeled A) move along the $x$ axis. Charge B exerts an electrical force on charge A. How much work does that force do as charge A moves?

From Coulomb's law, and since the distance between the two charges is $x$, we can see that if we move the charge A out by a distance $Δx$, the work done will be

$$W = F^E_{B \rightarrow A} \Delta x = \bigg(\frac{k_C Q_A Q_B}{x^2}\bigg) \Delta x = k_C Q_A Q_B \bigg( \frac{\Delta x}{x^2} \bigg) $$

Since the magnitude of the force is changed continuously — and not linearly, we really have to do the integral. So we just make the $Δx$* *really small, call it $dx$, and integrate:

$$W = \int_{x_i}^{x_f} F^E_{B \rightarrow A} dx = \bigg(\frac{k_C Q_A Q_B}{x^2}\bigg) \Delta x = k_C Q_A Q_B \int_{x_i}^{x_f} \frac{dx}{x^2} $$

We are just adding up the little bits of work done by the electric force on charge A as it moves out from an initial position $x_i$ to a final position $x_f$. (Remember that the integral symbol is a stretched "s" standing for "sum".)

We only have to do the integral of $x^{-2}dx$*.* Our simple rules for the integrals of powers tell us that the integral of x^{n} is x^{n+1}/(n+1) so for n = -2 we get

$$\int \frac{dx}{x^2} = - \frac{1}{x} $$

When we put in our initial and final positions we get

$$\int_{x_i}^{x_f} \frac{dx}{x^2} = \bigg[ - \frac{1}{x}\bigg]_{x_i}^{x_f} = \bigg(- \frac{1}{x_f} \bigg) - \bigg(- \frac{1}{x_i} \bigg) = \frac{1}{x_i} - \frac{1}{x_f}$$

(Follow the signs through carefully! There are a lot of minus signs and more to come!)

## Electric potential energy

Putting the result of our integration into our definition of work, we can see that the work only depends on the starting and ending values, so we can define a potential energy, $U(x)$, that is a function of $x$:

$$W = \int_{x_i}^{x_f} F^E_{B \rightarrow A} dx = k_C Q_A Q_B \int_{x_i}^{x_f} \frac{dx}{x^2} = \frac{k_C Q_A Q_B}{x_i} - \frac{k_C Q_A Q_B}{x_f}$$

$$W = -\Delta U_{elec} = - \bigg[U_{elec}(x_f) - U_{elec}(x_i) \bigg] = U_{elec}(x_i) - U_{elec}(x_f)$$

By comparing the first and second lines, it's clear that we should identify our electric PE between two charges, Q_{A} and Q_{B} as

$$U_{elec}(x) = \frac{k_C Q_A Q_B}{x}$$

where we have been explicit to note that U_{elec} is a function of the separation of the two charges, $x$.

Now this is for the special case of one charge nailed down and the axis $x$. We can make this more general by realizing that it's not the coordinate that matters here — that was only to make the integral look simple. What matters physically is the distance between the two charges. It doesn't really matter that the other charge is nailed down. If they are both moving we have to pick a particular instant to consider, but at that instant the PE will still look the same.

We then have the general result for ** the electric PE between any two charges**:

$$U_{elec}(x) = \frac{k_C Q_A Q_B}{r_{AB}}$$

where $r_{AB}$ is the distance between the charges A and B.

## Dimensional analysis

We could have guessed this result from dimensional analysis. Since the electric force looks like $k_CQQ/r^2$ and since the work just multiplies by a distance, we might have guessed that we would get $k_CQQ/r$. We couldn't have told that there would be no extra constant out front, though, without having done the integral.

Joe Redish 11/3/11

Last Modified: May 27, 2019