Lecture 4

Energy of a system of charges

Charges in electric fields accelerate, gaining kinetic energy. Where does this energy come from?

The answer is ‘from potential energy’.
Just as we talk about gravitational potential energy (when an object gains energy accelerating downwards), we can also talk about electrical potential energy.
When an electric field does work on a charge, energy is being transferred from the electrical potential energy to kinetic energy.
Similarly, if a charge is somehow pushed against the electrical field (i.e. a positive charge is pushed backwards along the field lines, or a negative charge is pushed forwards along the field lines), that charge does work on the field and the work is ‘stored’ as electrical potential energy.
The work done to move a particle between two points is independent of the path taken between those points. We will see a proof of this in the lecture. It is also proved in Section 23.1 of Y&F (see the discussion starting at equation 23.7 and working down to equation 23.10).¹
Because of this, if one moves a particle in a closed loop, back to where it started, the change in the electrical potential energy is zero. If one pushes a charge from A to B and it takes an amount of work $W$, if the charge then returns to A the field does the precise same amount of work $W$ on the charge itself. We call the electric field from static charges ‘conservative’ for this reason — it conserves energy.²

The electric potential energy of a single ‘test’ point charge of magnitude $q_0$ in the presence of other charges $q_1$, $q_2$ etc is

$U = \frac{q_0}{4 \pi \epsilon_0}\left(\frac{q_1}{r_1} + \frac{q_2}{r_2} + \cdots \right)\,,$ where $r_1$, $r_2$ etc are the distances from the test charge to the other charges in turn. Important things to note:

Potential energy is a scalar quantity, and is proportional to $1/r$ for any pair of charges. Contrast this with force which is a vector quantity which has magnitude proportional to $1/r^2$.
Considering the potential energy from a pair of charges, $q_0$ and $q_1$, it will be positive if $q_0$ and $q_1$ have the same sign and negative if $q_0$ and $q_1$ have differing signs. In other words, the potential energy of a negative-positive pair of charges is negative, while the potential energy of a matched pair of charges is positive.
$U$ as derived above is the potential energy of the test charge due to the other charges, not the total potential energy of the whole set of charges. Just as the net force acting on a charge comes about from adding all the forces from the other charges, so the potential energy of a single charge comes about from adding up the effect that all the other charges are having on it.
The SI units of $U$ are ${\rm J}$ (Joules), as you might expect. (In lecture 2 we saw that $\epsilon_0$ has units of ${\rm C^2 N^{-1}\,m^{-2}}$. You can use this together with the expression for $U$ given above to check the units of $U$ come out right.)

Electrical potential

You’ll notice in the expression for $U$ given above that the test charge $q_0$ factors out from the expression. We can therefore define a potential energy per unit charge, also known as the electrostatic potential, electric potential or sometimes just potential. It is denoted $V$, and as it is the potential energy per unit charge we can immediately say $V \equiv U/q_0$. As such, the potential due to the charges $q_1$, $q_2$ etc is:

\[V = \frac{1}{4 \pi \epsilon_0} \left(\frac{q_1}{r_1} + \frac{q_2}{r_2} + \cdots \right)\,.\]

The SI units of potential are Volts, and by definition $1{\rm V} = 1\,{\rm J} / {\rm C}$, i.e. a Volt is a Joule per Coulomb.

In many ways the electrical potential is more fundamental than the potential energy itself. In some ways, it is even more fundamental than the electric field!³ But perhaps more immediately relevant, it can also make some calculations a lot simpler than working directly with the electric field itself, especially when conductors are involved (more about this in later lectures).

Once we know the potential, we can actually recover the electric field by taking a suitable derivative. Mathematically, we write

$\vec{E} = -\nabla V$ where $\nabla$ is a special type of derivative known as the ‘gradient’. It will be covered formally in MATH1551 Single Mathematics B but for now we just note that it is written

\[\nabla V \equiv \begin{bmatrix} \partial V \over {\partial x} \\ \partial V \over {\partial y} \\ \partial V \over {\partial z} \end{bmatrix}.\]

The fact that $\vec{E} = -\nabla V$ is justified as follows:

Change in potential energy is equal to minus the work done, which is equal to the integral of force times distance ($\Delta U = -\int \vec{F} \cdot \vec{ {\rm d} r}$).
Reversing this relationship, we can say force must be equal to the derivative of potential energy along the direction of travel ($\vec{F} = -\nabla U$).
But since electric field is force per unit charge ($\vec{E} = \vec{F}/q_0$), and electric potential is potential energy per unit charge ($V = U/q_0$), electric field must be equal to the derivative of the electric potential ($\vec{E} = -\nabla V$).

Alternatively, in the lecture we will just directly calculate the electric field from the potential for a single point charge, and show that it gives the right answer, giving an alternative (though less fundamental) justification for the formula $\vec{E} = -\nabla V$.

Calculating a potential from a field

By integrating $\vec{E} = - \nabla V$ along a line, or by working straight from the expression for work done ($\Delta U = -\int \vec{F} \cdot \vec{ {\rm d} l}$), you can also show that:

\[\Delta V = - \int_A^B \vec{E} \cdot \vec{ {\rm d} l},\]

where $\Delta V$ is the change in the potential going from point $A$ to $B$.

Note that since $V$ is measured in Volts, $\vec{E}$ itself must have SI units of ${\rm V\, m^{-1}}$. In lecture 2, I said that the SI electric field units were ${\rm N\,C^{-1}}$. As an exercise, you might like to show that these two ways of expressing the SI units are equivalent, i.e. ${\rm 1\, V\, m^{-1} = 1\, N\,C^{-1}}$.

The electron-volt (eV)

As discussed above, a Joule is by definition equal to a Coulomb-Volt. So, if we have a charge of 1 Coulomb and we move it through a potential difference of 1 Volt, it will gain 1 Joule of energy.

However, in atomic and particle physics, it is often more convenient to express things in terms of the energy that a single electron gains when it is moved up a potential difference of 1 Volt. This is called an electron-volt ($\rm eV$). Since the charge on an electron is $e \simeq -1.602 \times 10^{-19}\,{\rm C}$, we have that

\[1\,{\rm eV} = 1.602 \times 10^{-19}\,{\rm J}.\]

We will not use this unit much in this course, but it is worth knowing about.

Energy of a dipole

Above, I pointed out that the energy to make a dipole spin must come from potential energy. Section 21.7 of Y&F shows that $U = -\vec{p} \cdot \vec{E}$. In the lecture, I’ll outline a second way of deriving this result — by adding the potential energy associated with the two charges making up the dipole. While this derivation involves some mathematical manipulations that might not yet be completely obvious, the result (whatever way it is obtained) can be made intuitive, like the torque.

The lowest value of $U$ is $U=-\vert\vec{p}\vert \vert\vec{E}\vert$, and comes when the dipole aligns with the fieldline. In this configuration a dipole is stable because there is no energy available to spin it.
If the dipole is perpendicular to the fieldlines, $U=0$. Despite this, energy is available, because the minimum of potential energy is negative (as just discussed). So, the dipole will tend to start spinning.
If the dipole is anti-aligned with the fieldlines (i.e. the positive charge and negative charge are flipped relative to their most stable configuration), the potential energy is maximised ($U = +\vert\vec{p}\vert \vert\vec{E}\vert$).

Note how similar the expression for energy and torque is. The sensible units of torque are ${\rm N\,m}$, while the sensible units of energy are ${\rm J}$ but these are two ways of writing the same unit.

Visualising the potential

The electric potential can be visualised using contours, like those that show height on a map. On a map, contours show lines of equal height as seen from above. Here’s a quick animation to show the construction of contours on a hill with a dip in the top, just as an example.

In the case of electromagnetism, potential contours are a helpful complement to fieldlines and/or quiver plots.

On each line, the potential is constant
Like fieldlines, potential contours never touch or cross (why?)
Contours always cross fieldlines at right-angles (why?)

To get a feel for this, try playing with the examples below. First, let’s take a case with two positive charges. Try moving the charges closer and further apart, and see if you can understand how the potential contours are changing.

Now let’s take a case with a positive and a negative charge (i.e. a dipole). Positive potential contours are red; negative potential contours are blue. Again, have a play and see if you can understand what is happening.

Footnotes

Really, the full Coulomb’s law isn’t needed for this proof — only the fact that the force is radial between two charges. ↩
That said, if there is a time-varying electromagnetic field, then the electric part is no longer conservative. Only static electric fields are conservative. This is hugely important later, but won’t concern us too much just now. ↩
This is quite a remarkable claim, but it is justified by the fact that the electric potential is the first step on the way to understanding something called a gauge field theory. This is well beyond the level of the first year course, so don’t worry about it for now unless you want to. ↩