# The electric potential

#### Prerequisites

We introduced the idea of an electric field,

$$\overrightarrow{E} = \frac{\overrightarrow{F}^{elec}_A}{q_A}$$

to be able to talk about the force that a "test" object with a charge $q_A$* would* feel if it were placed at a point $\overrightarrow{r}$ in a very general way, without actually specifying what the "test" object's charge is. Though the equation above contains the charge $q_A$*,* the electric field is independent of test charge sign or magnitude for a simple reason: The Coulomb Force law also is proportional to $q_A$, so it cancels out. The electric field does not change when $q_A$_{ }is changed. Note that the field is a *vector function* — a vector (with units Newtons/Coulomb) that is assigned to each point in space.

While this is a convenient definition in many ways (just as it was convenient for us in discussing gravity to introduce the *gravitational field, *$\overrightarrow{g}$), it has the same cumbersome difficulties that we had with electric force. Coulomb's law only holds for point charges. Furthermore, if we have lots of charges as in the example below we have to construct the force for each pair using Coulomb's law and add up the vectors, paying particular attention to separating the vector components

So let's look further at what we learned in studying Newton's Laws. We learned that using potential energy was often more convenient than forces, since potential energy is a scalar. Although potential energy could be positive or negative, it added just like numbers and unlike forces or fields, which add like vectors where we have to worry about taking components. But what if we wanted to get the force back out of the energy? We could calculate forces by taking the vector derivative (the gradient) of the potential energy as function of position.

Here is where we have to be careful! We calculated the electric field based on the force on a "test" particle. Potential energy on the other hand is generally defined as the energy of a system, not of an object. The electrical potential energy of a system is simply the sum of the potential energies of all pairs of charges. In the example on the right, the electric potential energy of charges A,B,C and D is the sum of all interactions six pairs of charges in total (AB, AC, AD, BC, BD, CD).

How can we adapt the concept of electrical potential energy to a test charge?

The answer is a bit tricky, so I've marked it as a dangerous bend: We calculate the **change **in the electrical potential energy of the system if a test charge is brought into the system as shown on the right. With the new test charge, we have four additional interactions to consider (test-A, test-B, test-C, test-D)

This way we can identify an electrical potential energy for each point in space:

*The electric potential energy field (at a point in space) is the change in potential energy of the system if a test charge were to be positioned at that point in space. *

While considering this, note that we really only think about an instant in frozen time and don't pay attention to any affect our test charge might have on the source charges.*

Suppose we have a system of (imagined as fixed in place) charges $q_1, q_2, ..., q_N$ at positions $\overrightarrow{r}_1, \overrightarrow{r}_2, ... \overrightarrow{r}_N$ respectively. This system will have an electrical energy

$$U^{elec}_{1...N} = \sum_{j = i+1,...,N} \sum_{i = 1...N} \frac{k_Cq_iq_j}{r_{ij}}$$

where $r_{ij}$ is the distance between charges i and j. (We sum over all i but only over j from i+1 up to N. This prevents us from counting each pair twice. Try it with N =3 to see how that works! Don't worry too much about it, though. We're going to drop this in a minute.)

If we now place a test charge $q_0$ down at a position $\overrightarrow{r}_0$ in the presence of others, the electrical energy that will be added to the system is the interaction energy of $q_0$ with each of the other charges:

$$U^{elec}_0 = \frac{k_Cq_0q_1}{r_{01}} + \frac{k_Cq_0q_2}{r_{02}} + ... \frac{k_Cq_0q_N}{r_{0N}} = \sum^N_{i = 1} \frac{k_Cq_0q_i}{r_{0i}} $$

where $r_{01}, r_{02}, ..., r_{0N}$ are the magnitudes of the distances between charge 0 and charges, 1, 2, ...N.

Just as in our analysis of the electric field, every term in this expression for the added electric potential energy is proportional to $q_0$, the value of the test charge. We can therefore factor it out and define the electric energy of the test charge placed at point $\overrightarrow{r}_0$ divided by the value of the test charge as ** the electric potential at $\overrightarrow{r}_0$**: (or sometimes

*the electrostatic potential*):

$$V(\overrightarrow{r}_0) = \frac{U^{elec}_0}{q_0} = \frac{k_Cq_1}{r_{01}} + \frac{k_Cq_2}{r_{02}} + ... \frac{k_Cq_N}{r_{0N}} = \sum^N_{i = 1} \frac{k_Cq_i}{r_{0i}} $$

We see that just as with the electric field, this does NOT depend on the charge of the test charge — just on the charges of the source charges and all their positions. The value also depends on $\overrightarrow{r}_0$ — where we are looking. This is a *scalar field* — the assignment of a value (possibly positive or negative) to every point in space. We therefore define:

*The electric potential field,* $V$,* at a point in space is the change in potential energy of the system if a test charge* $q_0$* were to be positioned at that point in space divided by the charge *$q_0$*. (Therefore,* $V$* is independent of *$q_0$*.) *

## A note on notation

In essentially all introductory physics texts, the electric potential is identified by the symbol *"*$V$*"*. In more advanced physics texts you sometimes see the symbol "$φ$" used. Regrettably, in biology the symbol "$ψ$" is often used for the electric potential. In our general discussions we will mostly use $V$ but in our biological examples will occasionally switch to $ψ$. Don't worry. "What's in a name? That which we call a rose by any other name would smell as sweet." (Wm. Shakespeare, Romeo & Juliet, II, ii, 1-2)

## Units

The units of the electric potential, $V$, is clearly energy divided by charge or Joules/Coulomb. This is given the name "Volt" after Alessandro Volta, one of the early researchers in electricity who helped developed the idea of voltage (which he called "electrical pressure".) An interesting note is that one of his devices for measuring the strength of a voltage was the nerve to the leg of a frog. How strongly the leg twitched was a measure of the voltage difference applied across the ends of the axon. An example of biology helping the development of physics!

1 Volt = 1 Joule/Coulomb.

Yes, this is the "volt" you are familiar with from buying batteries.

One other point to note about units is that since the electric force is the gradient of the potential energy, the electric field is the gradient of the electric potential. Dividing the equation for the force and energy of charge $q_0$,

$$F^{elec}_{q_0} = -\overrightarrow \nabla U^{elec}_{q_0} = -\bigg(\frac{\partial U^{elec}_{q_0}}{\partial x}\hat{i} + \frac{\partial U^{elec}_{q_0}}{\partial y}\hat{j} + \frac{\partial U^{elec}_{q_0}}{\partial z}\hat{k}\bigg) $$

by $q_0$ gives the equation (which looks simpler since $q_0$ drops out)

$$E = -\overrightarrow \nabla V = -\bigg(\frac{\partial V}{\partial x}\hat{i} + \frac{\partial V}{\partial y}\hat{j} + \frac{\partial V}{\partial z}\hat{k}\bigg) $$

Analyzing the units of the derivative ($dV/dx$ is just like $ΔV/Δ x$) we see that the units of E field can also be expressed as "volts/meter". This is actually more common than using "Newtons/Coulomb" for the electric field.

* In traditional physics textbooks this ignoring of the effect of the test charge on the source charges is often described by saying, "let the test charge be arbitrarily small." This is kind of a mathematical limit. But since we know that the smallest charge we can have is the charge on one electron, and since many of the situations we we consider will involve molecules or membranes when only a few charges are involved, this model is not really appropriate.

Joe Redish 2/14/12 & Wolfgang Losert 2/18/13

Last Modified: May 14, 2019