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4. Electric Potential Energy

Energy in Electric Systems

Believe it or not, you already know a great deal about electric potential energy, which you studied extensively in Physics 7A. For instance, it takes energy to move two like charges closer together. We can model the process of moving charges closer together with the following energy interaction diagram below.

This process is represented mathematically as \(W = \Delta PE_{electric}\)

Now let’s imagine starting with a positive charge and a negative charge very far apart, and allowing them to come nearer. As the charges come together, their speed increases, so the kinetic energy of the charges also increases.  But as \(r\) decreases, \(PE_{electric}\) also decreases.  We see that the total energy of the too charges does not change:\[\Delta E_{\text{tot}} = 0 = \Delta P E_{\text{electric}} + \Delta K E\]

Like force, potential energy is an interaction and requires at least two charges.  It makes no sense to talk about the potential energy of a 45 µC charge unless you reference its position in a field created by other charges.  Just like it makes no sense ot talk about the gravitational potential energy of a 1 kg ball unless you also give its height above a reference level.  In electric systems, to have either a force or potential energy, two or more charges are required.

As both force and potential energy are interactions (that require at least two charges), one might expect them to be related in some way. Indeed, they are. Their relationship was studied in Physics 7A: the magnitude of the force is determined by how fast \(PE\) changes with position \(r\). \[ | \mathbf{F}_{\text{something on object}} | = \left| \dfrac{\text{d} PE}{\text{d} r} \right| \] Recall that graphically, evaluating the derivative at a certain location is equivalent to finding the slope of a \(PE\) vs \(r\) graph at that location. If the potential energy is changing rapidly, the graph will be steep.  That means the slope is big and the force at that spot is also large. If these ideas are unfamiliar to you, consult the Calculus Appendix of this volume or your introductory calculus text.

Example # 4

As studied in Physics 7A, the attraction between two atoms can be modeled as a Lennard-Jones interaction. Determine the force at a separation distance of a) 1.5 Angstroms and b) 4 Angstroms.

Solution

We are asked to evaluate the force at two different locations.  To determine the magnitude, we must draw tangent lines at each location (1.5 Å and 4 Å) and calculate each line's slope.  We must also determine the direction of the force (\(+r\) right or \(-r\) left).

a)  A tangent line must have the same slope as the original function. Taking a ruler and matching the slope, we find

The task now before us is to calculate the slope of this line. Between the 0.75 Å and 3 Å locations, the potential energy changes by 6 eV. The slope is rise over run, or \[\text{slope} \approx \dfrac{6 \text{ eV}}{(3Å - 0.75 Å)} = 2.67 \text{ eV/Å}\] While we have certainly determined the magnitude of the force, the units of force we are accustomed to are Newtons, not eV/Å. Before we call our work complete, we should convert to Newtons using these unit conversions:  \[1\text{ eV} = 1.6 \times 10^{-19} \text{ Joules,    } 1Å = 10^{-10} \text{ meters}\] \[\left( 2.67 \dfrac{\text{eV}}{Å} \right) \left( 1.60 \times 10^{-19} \dfrac{\text{J}}{\text{eV}} \right) \left( \dfrac{1 \text{ Å}}{10^{-10} \text{ m}} \right) = 4.3 \times 10^{-9} \text{ J/m} = 4.3 \times 10^{-9} \text{ N}\] As far as direction goes, at the 1.5 Å mark, the atoms are attracted. The
force is to the left. The force will always act to decrease the potential energy

b) At a separation of 4 Å, the potential energy graph is nearly flat. Thus, the slope approaches zero, and so does the force.  We can approximate: \[|\mathbf{F}| \propto \text{slope} \approx 0\]

Potential Energy of Two Charges

There is no single equation for potential energy. For instance, the potential energy of two atoms interacting (as in the Lennard-Jones interaction, above) is different than the equation for two single charges interacting. We have discussed the electric field created by a single charge, and the electric force between two charges. We now use this prior work, along with the relationship between force and potential energy, to determine the potential energy of two charges interacting. 

The magnitude of the force between two charges \(q\) and \(Q\) is \[|\mathbf{F}| =|kqQ/r^2|\]We know that the force is equal to the derivative of the potential energy with respect to position: \[|\mathbf{F}| = \left| \dfrac{\mathrm{d}PE}{\mathrm{d}r} \right| \]We would like to know the potential energy \(PE\) as a function of position \(r\).  That means we want a function of \(PE\) such that \[\dfrac{\mathrm{d}}{\mathrm{d}r}PE(r) = \dfrac{−kqQ}{r^2}\]We solve this by performing an integral, and find that \[PE = \dfrac{kqQ}{r} + \text{constant}\] We require that the derivative (or slope) of the potential energy with respect to position gives us force.  We find that the constant does not change the derivative (slope). It is up to us to determine the value of the constant, and in doing so determine the zero-point for potential energy.  This idea should be familiar from Physics 7A

It is convenient to describe charges incredibly far away as having zero potential energy. We can consider \(PE\) at very long distances mathematically by taking the limit of \(PE(r)\) as \(r \rightarrow \infty\) and finding \[0=PE = 0 + constant\]This leads us to the very useful conclusion \[\text{constant} = 0\] The zero-point for potential energy is 0 for charges separated by incredibly large distances.

With two signs, there are three different combinations of charges: both positive, both negative, one charge of each sign. However, when potential energy is concerned, you only need to consider two cases:  the charges are the same or the charges are different.  The graphs of potential energy between two charges for like and unlike charges are shown below.

For like charges, the potential energy is always positive,  that is because we need to put energy in the system to bring like charges closer together. Considering the total mechanical energy, (\(PE + KE\)), and knowing that kinetic energy is always positive in classical systems, the total mechanical energy must be positive as well. If the like charges are initially moving toward one another, energy transfers from \(KE\) to \(PE\) until finally all of the energy is in \(PE\), at which time the particles briefly stop, turn around, and move apart.

For unlike charges, there are two interesting cases. If the total mechanical energy is greater than zero (\(PE + KE >0\)), then the particles will have both kinetic energy and potential energy at all separation distances. If the total mechanical energy is less than 0, then the particles are confined to one another in a bound state.  At these energies, the particles lack sufficient energy to escape their electric attraction. A common example of this phenomenon is the hydrogen atom,  in which a negatively-charged electron is bound to a positively-charged proton (we'll explore this more in quantum mechanics).

Exercise

Note for either potential energy graph, the PE gets flat for large separation distances and steep for small separation distances. What can you conclude from this information?