6 Calculating the potential
We saw in the previous section the importance of metals (conductors) as a way of creating points in space at constant potential. Hence, metal wires, plates etc. are the practical means by which we create electric fields. Before we look into methods for calculating the potential it is interesting to ask what we would like to do. From a point of view of design we would like to create an electric field of known form in some region of space. So we might ask what arrangement of wires, plates etc. will give me the desired field? With a little thought you realise this is a difficult question. To calculate a field precisely we need to define (via Poisson’s equation) the charge density over \(\textit{all space}\). This is not practical! So in fact what we wish to do, is to produce a field that is close enough for a local region of space. However, in the end this means there is not a unique placement of charges/electrodes that can produce the required field. In other words, asking the question ‘What is the placement of electrodes I need to create a specific field?’ is a challenge and not possible to solve by direct methods! Fortunately, for this course and many applications we ask the question, ‘what is the field produced by a certain arrangement of plates, wires etc. ?’ This is much clearer, we just need to solve Laplace’s equation for the region of space in which we are interested (assuming the any field from outside our ‘system’ is negligible). Hence solving Laplace’s equation is a key goal in solving problems in electrostatics. Here we will look at a few of the methods.
6.1 Laplace’s equation
As we saw in section 3, in the regions of space where there are no charges, Laplace’s equation must apply. Any conductors (metals) in the system will, by definition, be at constant potential. If there is no difference in potential between them, we have no electric field. If we have a potential difference between the metals then charges will accumulate on them and give rise to an electric field. However, to solve Laplace’s equation we don’t need to know what these charge densities are, we just need to know the potentials, as they set the \(\textit{boundary conditions}\) for the solution. This may seem odd as we often naturally try to describe the physics of what is happening in terms of position and movement of charges (for example on the plates of a capacitor)! However, if we want to know the charge density we can calculate it later.
Before we look at ways for calculating the potential I will quote two \(\textit{uniqueness}\) theorems here (for proofs see Griffiths 3.15 and 3.16.
First uniqueness theoerm
The solution to Laplace’s equation in some volume \(\tau\) is uniquely determined if \(V\) is specified on the boundary surface \(S\).
Second uniqueness theorem
In a volume \(\tau\) surrounded by conductors and containing a specified charge density \(\rho\), the electric field is uniquely determined if the total charge is given.
In a nutshell, these mean if I find a solution for \(V\) in my region of space then it is the only solution - I can’t figure out some other arrangement of charges or potentials that give the same result. With this in mind we’ll look at various ways in which we might calculate the potential.
If we know the charge distribution in our system we can always, in principle, calculate \(V(\vec{r})\) from the results shown in section 3.4 (equation (3.5)).
6.2 Calculating potential - separation of variables
Laplace’s equation is a well known form of \(\textit{partial differential equation}\) (PDE). You have learnt techniques for solving PDEs, including the Laplace equation in your mathematics course. You will probably come across many more as you progress in your studies. Hence, all the techniques, in particular the method of separation of variables, that you learnt or will learn can be applied. How to solve partial differential equations is not the aim of this course so we will not explore these methods further. Neither will I set questions purely about solving PDEs, but I will assume you have seen them in your mathematics course.
6.3 Calculating potential - numerical methods
For more complex problems (where analytical solutions to Laplace’s equations cannot be found) we can apply numerical methods. Again we will not go into any detail in this course but we can note some interesting results for which very simple numerical methods may be applied. To show you how this works, let’s first look at Laplace’s equation in 2D,
\[\begin{eqnarray*} \displaystyle \frac{\partial ^2 V}{\partial x^2}+\frac{\partial ^2 V}{\partial y^2}=0 \end{eqnarray*}\]
The solution can’t have any local maximum (hill) or minimum (dip) in \(V\). However, one result we can find (see Griffiths chapter 3) is that the potential \(V\) at some point \((x,y)\) is the average of the potential around that point,
\[\begin{eqnarray*} \displaystyle V(x,y) =\frac{1}{2\pi R} \oint_{circle}V\;dl \end{eqnarray*}\]
Using this result I will demonstrate, in the lectures, a very simple way of numerically calculating the potential in a closed system, for an arbitary arrangements of constant potential surfaces in 2D.
This idea can be extended to 3D (again I will not prove this) to state that the potential at some point \(r\) in space is just the average of the potential surrounding \(r\),
\[\begin{eqnarray*} \displaystyle V(r)=\frac{1}{4\pi R^2}\oint_{sphere} V\;da \end{eqnarray*}\]
It is very easy to extend the numerical method we used in 2D to 3D by creating a 3D grid of points and doing the same local averaging.
The method I will show is crude but it demonstrates the principle. However, there is software that is specifically designed to solve these type of problems that is both fast and accurate.
6.4 Calculating potential - the method of images
For some arrangements of metals, in particular with some grounded at zero potential, it is possible to calculate the potential for some relatively simple geoemtries. The ‘trick’ of the method of images is to replace, in the calculation, the grounded surface with some carefully placed charges (of opposite sign) that have the effect of producing exactly the same constant potential in space as on the metal. The difficulty is ‘guessing’ where to place these ‘image’ charges, especially if you don’t know in advance that a solution by this method is possible! The method is best shown by example, for which we will give a few in the lectures.
One point to remember is that although we use image charges to calculate the potential (and hence field) in the region outside our conductor, there is no field in the region bounded by the conductor so care needs to be taken if you wish to calculate the energy and work done in creating the field. We will also mention this in lectures.
For a full discussion of the method you are referred to Griffiths, chapter …
6.5 Calculating potential - multipole expansions
Before it became straightforward to calculate potentials numerically, the use of \(\textit{multipole}\) expansions was a common way for calculating quite complex fields. We will not go into the details here (it is quite mathematically involved) but I will try and describe the idea. We will then look at the first term in the expansion, the dipole, that, as we will see, is a first point in discussing the effects of electric fields in matter and is also the fundamental way of describing magnetic fields (for which ‘magnetic’ charges do not exist).
Imagine that we have an object that has some arrangement of positive and negative charges within it. If there is a net charge on the object then, provided we go far enough away (so we can’t see its size), the field from the object will look more and more like that of a point charge; the field lines will radiate uniformly out into space. This is the first ‘pole’ , a monopole, of our expansion. Now suppose that the net charge on the object is zero but there is a simple separation of charge on it. At large distances this might seem as if there is no field. However, as we get nearer, we will start to see field lines emerging from the positive charges and looping back to return at the negative charge. Provided, we are far enough away and we have this simple charge separation we have the second term of our expansion - the dipole. However, we don’t have to have this dipole arrangement of charges and in the lectures we will give an example where we can have charge separation but arranged in a way that does not give rise to a dipole field at large distances - the quadrupole. The mathematical technique for characterising this hiearchy of charge distributions is known as the method of moments. For any charged object we can calculate these moments\(^*\)(monopole, dipole, quadrupole, octopole …) If we use the superposition you can see that the field from any charge distribution can be characterised by the sum of all these contributions. The potential for these moments falls off as \(1/r^n\) where \(n=2\) is the dipole, \(n=3\) the quadrupole etc., so at large distances we only need to use the first non-zero term in the expansion to get a good approximation of the field. As we come closer, we will need to include higher and higher order terms.
\(^*\) The concept of moments of functions occurs in other areas of physics, see for example, moments of inertia etc. for describing mass distributions.
6.6 The dipole potential
As explained earlier the electric (and magnetic) dipole play an important role in understanding the behaviour of electric and magnetic materials. We’ll therefore take a more detailed look at the electric dipole, as shown in the diagram below.
We consider the centre of the dipole (and the origin) to be the midpoint between the two charges. We can hence write the potentialat any point \(\vec{r}\) as \[\begin{equation*} V(\vec{r})=\frac{1}{4\pi\epsilon_0}\left (\frac{q}{r_+}-\frac{q}{r_-}\right) \end{equation*}\] We can use the cosine rule to show that,
\[\begin{equation*} r_\pm^2=r^2+\left(\frac{d}{2}\right)^2\mp rd \cos(\theta)=r^2\left(1\mp\frac{d}{r}\cos(\theta)+\frac{d^2}{4r^2}\right) \end{equation*}\]
So, \(\textit{provided}\) we are at large distances from the dipole we may neglect the last term to leave, \[\begin{equation*} \frac{1}{r_\pm}\simeq \frac{1}{r} \left( 1\mp \frac{d}{r}\cos \theta\right)^{-\frac{1}{2}} \end{equation*}\]
or, expanding the \((1\mp x)^{1/2}\) term further,
\[\begin{equation*} V_{dipole}(\vec{r}) = \frac{1}{4\pi\epsilon_0}\frac{qd\cos \theta}{r^2} \;\; ...(6.1) \end{equation*}\]
You should be careful to note how we have, in this formula, already defined the axis for taking \(\theta\) as along the length of the dipole. We note how this correctly gives a scalar quantity, as we require, for the potential. We can go further in describing the orientation of the axis of the dipole by defining a \(\textit{dipole moment}\) expressed as a vector,
\[\begin{equation*} \vec{p}=q \vec{d} \end{equation*}\] where \(\vec{d}\) is a vector pointing in the direction from the negative to the positive charge and \(d\) is the distance separating the charges. From equation (6.1) we can see we can express the term \(qd \cos \theta\) as a dot product and hence the dipole potential in the more concise form,
\[\begin{equation*} V_{dipole}(\vec{r})=\frac{1}{4\pi\epsilon_0}\frac{\vec{p} \cdot \hat{ \vec{r}}}{r^2} \;\; ... (6.2) \end{equation*}\]
We will show in the lectures, by using our relationship between the electric field and the potential that the electric field from the dipole mya be written in general form as,
\[\begin{equation*} \vec{E}_{dipole}(\vec{r})=\frac{1}{4\pi\epsilon_0}\frac{1}{r^3} \left [ 3(\vec{p}\cdot\hat{\vec{r}})\hat{\vec{r}}-\vec{p}\right] \;\; .... (6.3) \end{equation*}\]
You should take some time to look at this expression carefully and make sure you can relate it to the electric field lines you expect from the dipole (remember it is only vaild for large \(\vec{r}\). The dipole is the fundamental object through which we try to describe the effects of fields in matter.