This post summarizes two of what may well be Feynman’s most tedious Lectures. Their title is the same: the electric field “in various circumstances.” At first, I wanted to skip them, but then I found some unifying principle: the fields involved are all quite simple. In fact, except in chapter seven, it’s only about (a) the field of a single charge and (b) the field of a so-called dipole, i.e. the field of two opposite charges next to each other. Both are depicted below, and the dipole field can actually be derived by adding the fields of the two single charges.
So… In a way, these two Lectures are just a bunch of formulas repeating the same thing over and over again. The thing to remember is that a complicated but neutral mess of charges will also create a dipole field and, if that mess would not be neutral as a whole, then the field of our lump of charge will look like that of a point charge, provided we look at it from a large enough distance (i.e. a distance that is large relative to the separation of the elementary charges involved). So the situation we’re looking at, is the one depicted below, which is really quite general.
Before going into the nitty-gritty, it is probably good to review one of the points I made in my previous post: the field inside of a spherical shell of charge (like the one below) is zero everywhere, i.e. for any point P inside the shell.
This has nothing to do with the phenomenon of shielding, which is a consequence of free electrons re-arranging themselves so as to cancel the field inside. If we’d be able to build the cage below from protons only, so we’d have a fixed distribution of charges, the inside would not be shielded from the external electrical field. [Credit for the animation must go to Wikipedia.]
Because of the symmetry of the situation, however, the field inside a rectangular, fixed and uniform distribution of charges would also be zero. Let me quickly go over the math for the example of the spherical shell. The randomly chosen point P defines small cones extending to the surface of the sphere, with their apex at P and cutting out some surface area Δa. In the illustration above, we have two symmetrical cones defining two surfaces Δa1 and Δa2 respectively. It is easy to see that:
Δa2/Δa1 = r22/r12
Note that r22/r12 is equal to (r2/r1)2 but that (r2/r1)2 is not equal to r2/r1. The square matters, and the square of a ratio is different than the ratio itself! In fact, it’s because of the inverse square law that the fields cancel exactly. Indeed, if the surface of the sphere is uniformly charged (which is the key assumption here), then the charge Δq on each of the area elements will be proportional to the area, so Δq2/Δq1 = Δa2/Δa1. Now, Coulomb’s Law also says that the magnitudes of the fields produced at P by these two surface elements are in the ratio of:
Huh? Yes. E2/E1 = (Δa2/Δa1)·(r12/ r22) = (Δa2/Δa1)·(Δa1/Δa2) = 1, according to the above. So… Yes, the fields cancel exactly, and because all parts of the surface can be paired off in the same way, the total field at P is zero, indeed! But what if we’d put a charge with equal sign at the center? Logic dictates the shell would balance it at the center. Hence, Feynman’s statement that a charge in an electrostatic field in free space can only be in equilibrium if there are mechanical constraints − as illustrated below – is false, and – I should add – the whole argument that follows has no relevance whatsoever for the quantum-mechanical model of an atom. But that’s a somewhat separate story which I’ll touch upon at the end of this post. Let me get back to the dipole problem.
The model of a dipole is illustrated below. We have two opposite charges separated by a distance d. The so-called dipole moment is defined as p = q·d, and we also have an associated vector p, whose magnitude is p (so that’s the product of q and d) and whose direction is that of the dipole axis from −q to +q. We could also define a vector d and write p as p = q·d. Just think about it. I am sure you’ll figure it out. 🙂
Now, Feynman derives the formula for the dipole potential in various ways—first in an easy way, and then in a not-so-easy way. 🙂 The not-so-easy way is the most interesting—in this case, that is! He first notes the general formula for the potential of some point charge q at the origin at some point P = (x, y, z). You’ve seen that before: it’s Φ0 = q/r. [Forget about the constant of proportionality (I mean that 1/4πε0 factor in Coulomb’s Law) for a while. We can stick it back in at the end of the argument.] What it says, is that, while the field follows an inverse square law, the potential has a 1/r dependence only (so when you double the distance, you halve the potential). Now, if we’d move the charge q along the z-axis, up a distance Δz, then the potential at P will change a little, by, say ΔΦ+. How much exactly? Well, Feynman notes that “it is just the amount that the potential would change if we were to leave the charge at the origin and move P downward by the same distance Δz.” His illustration below, and the associated formula below, speak for themselves:
Now I’ll refer you to Feynman itself for the detail of the whole argument. The bottom line is that he gets the following formula for the dipole potential:
Φ = −p·∇φ0
We have a vector dot product here of that dipole vector we defined above (p) and the gradient of φ0, which is the potential of a unit point of charge: φ0 = 1/4πε0r. So what? Well… We can re-write this as:
Φ = −(1/4πε0)p·∇(1/r)
Isn’t that great? For point charges, we have a field that’s the gradient of a potential that has a 1/r dependence, but so… Well… Here we have the potential of a dipole that’s the gradient of… Well… Just a number that has a 1/r dependence. 🙂
It explains why the dipole field E = −∇Φ varies inversely not as the square but as the cube of the distance from a dipole. I could give you the formula for E but, again, I don’t want to copy all of Feynman here and so I’ll just assume you believe me. Let me just wrap up in this section with the graph of the electric field, and note how the field vector E can be analyzed as the sum of a transverse component (i.e. the component in the x-y plane) and its component along the dipole axis (i.e. the component along the z-axis).
The dipole field of a lump of charges
The only thing that’s left is to define the p vector for a lump (or a mess as Feynman calls it) of charges. Note that the lump should not be neutral: if it is, then it will look like a point charge from a distance. But if it’s not neutral, then its field will be a dipole field. So the same formula applies but p is defined as p = ∑qidi. I copy the illustration above below so you can see what is what. 🙂
So… Is that it? Well… Yes. And… Well… No. All of the above assumes we know the charge distribution from the start. If we do, then my little summary above pretty much covers the whole subject. 🙂 However, we’ll often be talking some conductor with some total charge Q, without being able to say where the charges are, exactly. All that we know is that they will be spread out on the surface in some way.
Now… Well… That’s not quite exact. We also know they will distribute themselves so that the potential of the surface is constant, and that helps us some practical problems at least. What problems? Well… The problem of finding the field of charged conductors, which is the second topic that Feynman deals with in his two Lectures on the field “in various circumstances.”
However, that story risks becoming as tedious as Feynman’s Lectures on it, and so I’d rather not copy him here. Just look at the following illustrations. The first one gives the field lines and equipotentials for two point charges once again. It highlights two equipotentials in particular: A and B. Now look at the second illustration: we have a curved conductor with a given potential near a point charge and – lo and behold! – the field looks the same: we replace A by the surface of our conductor and all the rest vanishes. In fact, the illustration we could just put an imaginary point charge q at a suitable point and get the same field.
Now that’s what’s referred to as the method of images, and it’s illustrated in the third graph, where we have an “image charge” indeed. We see the equipotential halfway between the two charges which, in this case, is grounded conducting sheet. Why grounded? Because the plane had zero potential in our dipole field, as it was halfway between the two charges indeed.
Well… It doesn’t matter all that much. This is, indeed, the really boring stuff one just has to grind through in order to understand the next thing, which is hopefully somewhat more exciting.
Because you’re interested in physics, you probably know a thing or two about those quadrupole magnets used to focus particles beams in accelerators. They’re also referred to as lenses. The illustration below is the field of a quadrupole electric field, but a quadrupole magnetic field looks the same.
The point is: these lenses focus in one direction and, hence, in an actual accelerator or cyclotron, the Q-magnets will be arranged so as to alternately focus horizontally and vertically. Why can’t we build magnets so as to focus electric or magnetically charged particles simultaneously in two directions?
Well… It would require a tube built of protons, or electrons, in a stable configuration. We can’t do that. Technology just isn’t ready for it: we’re not able to build stable tubes of protons, or of electrons. 🙂 So the so-called Theorem of Earnshaw is still valid. Earnshaw’s Theorem says just that: simultaneous focusing in two directions at once is impossible. It applies to classical inverse-square law forces, such as the electric and gravitational force, but also the magnetic forces created by permanent magnets.
However, the theorem is subject to constraints, and these constraints can be exploited to create very interesting exceptions, like magnetic levitation. I warmly recommend the link. 🙂