Capacitors

This post briefly explores the properties of capacitors. Why? Well… Just because they’re an element in electric circuits, and so we should try to fully understand how they function so we can understand how electric circuits work. Indeed, we’ll look at some interesting DC and AC circuits in the very near future. 🙂

Feynman introduces condensers − now referred to as capacitors – right from the start, as he explains Maxwell’s fourth equation, which is written as c2×B =  ∂E/∂t + j0 in differential form, but easier to read when integrating over a surface S bounded by a curve C:

The ∂E/∂t term implies that changing electric fields produce magnetic effects (i.e. some circulation of B, i.e. the c2×B on the left-hand side). We need this term because, without it, there could be no currents in circuits that are not complete loops, like the circuit below, which is just a circuit with a capacitor made of two flat plates. The capacitor is charged by a current that flows toward one plate and away from the other. It looks messy because of the complicated drawing: we have a curve C around one of the wires defining two surfaces: S1 is a surface that just fills the loop and, hence, crosses the wire, while S2 is a bowl-shaped surface which passes between the plates of the capacitor (so it does not cross the wire).

If we look at C and S1 only, then the circulation of B around C is explained by the current through the wire, so that’s the j0 term in Maxwell’s equation, which is probably how you understood magnetism during your high-school time. However, no current goes through the S2 surface, so if we look at C and S2 only, we need the ∂E/∂t to explain the magnetic field. Indeed, as Feynman points out, changing the location of an imaginary surface should not change a real magnetic field! 🙂

Let’s look at those charged sheets. For a single sheet of charge, we found two opposite fields of magnitude E = (1/2)·σ/ε0. Now, it is easy to see that we can superimpose the solutions for two parallel sheets with equal and opposite charge densities +σ and −σ, so we get:

between the sheets = σ/ε0 and E outside = 0

Now, actual capacitors are not made of some infinitely thin sheet of charge: they are made of some conductor and, hence, we get that shielding effect and we’re talking surface charge densities +σ and −σ, so the actual picture is more like the one below. Having said that, the formula above is still correct: E is σ/ε0 between the plates, and zero everywhere else (except at the edge, but I’ll talk about that later).

We’re now ready to tackle the first property of a capacitor, and that is its capacity. In fact, the correct term is capacitance, but that sounds rather strange, doesn’t it?

The capacity of a capacitor

We know the two plates are both equipotentials but with different potential, obviously! If we denote these two potentials as Φ1 and Φrespectively, we can define their difference Φ1 − Φ2 as the voltage between the two plates. It’s unit is the same as the unit for potential which, as you may or may not remember, is potential energy per unit charge, so that’s newton·meter/coulomb. [In honor of the guy who invented the first battery, 1 N·m/C is usually referred to as one volt, which – quite annoyingly – is also abbreviated as V, even if the voltage and the volt are two very different things: the volt is the unit of voltage.]

Now, it’s easy to see that the voltage, or potential difference, is the amount of work that’s required to carry one unit charge from one plate to the other. To be precise, because the coulomb is a huge unit − it’s equivalent to the combined charge of some 6.241×1018 protons − we should say that the voltage is the work per unit charge required to carry a small charge from one plate to the other. Hence, if d is the distance between the two plates (as shown in the illustration above), we can write:

Q is the total charge on each plate (so it’s positive on one, and negative on the other), A is the area of each plate, and is the separation between the two plates. What the equation says is that the voltage is proportional to the charge, and the constant of proportionality is d over ε0A. Now, the proportionality between V and Q is there for any two conductors in space (provided we have a plus charge on one, and a minus charge on the other, and so we assume there are no other charges around). Why? It’s just the logic of the superposition of fields: we double the charges, so we double the fields, and so the work done in carrying a unit charge from one point to the other is also doubled! So that’s why the potential difference between any two points is proportional to the charges.

Now, the constant of proportionality is called the capacity or capacitance of the system. In fact, it’s defined as C = Q/V. [Again, it’s a bit of a nuisance the symbol (C) is the same as the symbol that is used for the unit of charge, but don’t worry about it.] To put it simply, the capacitance is the ability of a body to store electric charge. For our parallel-plate condenser, it is equal to C =  ε0A/d. Its unit is coulomb/volt, obviously, but – again in honor of some other guy – it’s referred to as the farad: 1 F = 1 C/V.

To build a fairly high-capacity condenser, one could put waxed paper between sheets of aluminium and roll it up. Sealed in plastic, that made a typical radio-type condenser. The principle used today is still the same. In order to reduce the risk of breakdown (which occurs when the field strength becomes so large that it pulls electrons from the dielectric between the plates, thus causing conduction), higher capacity is generally better, so the voltage developed across the condenser will be smaller. Condensers used to be fairly big, but modern capacitors are actually as small as other computer card components. It’s all interesting stuff, but I won’t elaborate on it here, because I’d rather focus on the physics and the math behind the engineering in this blog. 🙂

Onward! Let’s move to the next thing. Before we do so, however, let me quickly give you the formula for the capacity of a charged sphere (for a parallel-plate capacitor, it’s C = ε0A/d, as noted above): C = 4πε0a. You’ll wonder: where’s the ‘other’ conductor here? Well… When this formula is used, it assumes some imaginary sphere of infinite radius with opposite charge −Q.

The energy of a capacitor

I talked about the energy of fields in various places, most notably my posts on fields and charges. The idea behind is quite simple: if there’s some distribution of charges in space, then we always have some energy in the system, because a certain amount of work was required to bring the charges together. [For the concept of energy itself, please see my post on energy and potential.] Remember that simple formula, and the equally simple illustration:

Also remember what we wrote above: the voltage is the work per unit charge required to carry a small charge from one plate to the other. Now, when charging a conductor, what’s happening is that charge gets transferred from one plate to another indeed, and the work required to transfer a small charge dQ is, obviously, equal to V·dQ. Hence, the change in energy is dU = V·dQ. Now, because V = Q/C, we get dU = (Q/C)·dQ, and integrating this from zero charge to some final charge Q, we get:

U = (1/2)·Q2/C = (1/2)·C·V2

Note how the capacity C, or its inverse 1/C, appears as a a constant of proportionality in both equations. It’s the charge, or the voltage, that’s the variable really, and the formulas say the energy is proportional to the square of the charge, or the voltage. Finally, also note that we immediately get the energy of a charged sphere by substituting C for 4πε0a (see the capacity formula in the previous section):

Now, Feynman applies this energy formula to an interesting range of practical problems, but I’ll refer you to him for that: just click on the link and check it out. 🙂

OK… Next thing. The next thing is to look at the dielectric material inside capacitors.

Dielectrics

You know the dielectric inside a capacitor increases its capacity. In case you wonder what I am talking about: the dielectric is the waxed paper inside of that old-fashioned radio-type condenser, or the oxide layer on the metal foil used in more recent designs. However, before analyzing dielectric, let’s first look at what happens when putting another conductor in-between the plates of our parallel-plate condenser, as shown below.

As a matter of fact, the neutral conductor will also increase the capacitance of our condenser. Now how does that work? It’s because of the induced charges. As I explained in my post on how shielding works, the induced charges reduce the field inside of the conductor to zero. So there is no field inside the (neutral) conductor. The field in the rest of the space is still what it was: σ/ε0, so that’s the surface density of charge (σ) divided by ε0. However, the distance over which we have to integrate to get the potential difference (i.e. the voltage V) is reduced: it’s no longer d but d minus b, as there’s no work involved in moving a charge across a zero field. Hence, instead of writing V = E·d = σ·d/ε0, we now write V = σ·(d−b)/ε0. Hence, the capacity C = Q/V = ε0A/is now equal to C = Q/V = ε0A/(d−b), which we prefer to write as:

Now, because 0 < 1 − b/d < 1, we have a factor (1 − b/d)−1 that is greater than 1. So our capacitor will have greater capacity which, remembering our C = Q/V and U = (1/2)·C·V2, formulas, implies (a) that it will store more charge at the same potential difference (i.e. voltage) and, hence, (a) that it will also store more energy at the same voltage.

Having said that, it’s easy to see that, if there’s air in-between, the risk of the capacitor breaking down will be much more significant. Hence, the use of conducting material to increase the capacitance of a capacitor is not recommended. [The question of how a breakdown actually occurs in a vacuum is an interesting one: the vacuum is expected to undergo electrical breakdown at or near the so-called Schwinger limit. If you want to know more about it, you can read the Wikipedia article on this.]

So what happens when we put a dielectric in-between. It’s illustrated below. The field is reduced but it is not zero, so the positive charge on the surface of the dielectric (look at the gaussian surface S shown by the broken lines) is less than the negative charge on the conductor: in the illustration below, it’s a 1 to 2 ratio.

But what’s happening really? What’s the reality behind? Good question. The illustration above is just a mathematical explanation. It doesn’t tell us anything − nothing at all, really − on the physics of the situation. As Feynman writes:

“The experimental fact is that if we put a piece of insulating material like lucite or glass between the plates, we find that the capacitance is larger. That means, of course, that the voltage is lower for the same charge. But the voltage difference is the integral of the electric field across the capacitor; so we must conclude that inside the capacitor, the electric field is reduced even though the charges on the plates remain unchanged. Now how can that be? Gauss’ Law tells us that the flux of the electric field is directly related to the enclosed charge. Consider the gaussian surface S shown by broken lines. Since the electric field is reduced with the dielectric present, we conclude that the net charge inside the surface must be lower than it would be without the material. There is only one possible conclusion, and that is that there must be positive charges on the surface of the dielectric. Since the field is reduced but is not zero, we would expect this positive charge to be smaller than the negative charge on the conductor. So the phenomena can be explained if we could understand in some way that when a dielectric material is placed in an electric field there is positive charge induced on one surface and negative charge induced on the other.”

Now that’s a mathematical model indeed, based on the formula for the work involved in transferring charge from one plate to the other:

W = ∫ F·ds = ∫qE·d= q·∫E·ds = qV

If your physics classes in high school were any good, you’ve probably seen the illustration above. Having said that, the physical model behind is more complicated, and so let’s have a look at that now.

The key to the whole analysis is the assumption that, inside a dielectric, we have lots of little atomic or molecular dipoles. Feynman presents an atomic model (shown below) but we could also think of highly polar molecules, like water, for instance. [Note, however, that, with water, we’d have a high risk of electrical breakdown once again.]

The micro-model doesn’t matter very much. The whole analysis hinges on the concept of a dipole moment per unit volume. We’ve introduced the concept of the dipole moment tout court in a previous post, but let me remind you: the dipole moment is the product of the distance between two equal but opposite charges q+ and q.

Now, because we’re using the d symbol for the distance between our plates, we’ll use δ for the distance between the two charges. Also note that we usually write the dipole moment as a vector so we keep track of its direction and we can use it in vector equations. To make a long story: p = qδ and, using boldface for vectors, p = qδ. [Please do note that δ is a vector going from the negative to the positive charge, otherwise you won’t understand a thing of what follows.]

As mentioned above, we can have atomic or molecular or whatever other type of dipoles, but what we’re interested in is the dipole moment per unit volume, which we write as:

P = Nqδ, with N the number of dipoles per unit volume.

For rather obvious reasons, P is also often referred to as the polarization vector. […] OK. We’re all set now. We should distinguish two possibilities:

1. P is uniform, i.e. constant, across our sheet of material.
2. P is not uniform, i.e. P varies across the dielectric.

So let’s do the first case first.

1. Uniform P

This assumption gives us the mathematical model of the dielectric almost immediately. Indeed, when everything is said and done, what’s going on here is that the positive/negative charges inside the dielectric have just moved in/out over that distance δ, so at the surface, they have also moved in/out over the very same distance. So the image is effectively the image below, which is equivalent to that mathematical of a dielectric we presented above.

Of course, no analysis is complete without formulas, so let’s see what we need and what we get.

The first thing we need is the surface density of the polarization charge induced on the surface, which was denoted by σpol, as opposed to σfree, which is the surface density on the plates of our capacitor (the subscript ‘free’ refers to the fact that the electrons are supposed to be able to move freely, which is not the case in our dielectric). Now, if A is the area of our surface slabs, and if, for each of the dipoles, we have that q charge, then the illustration above tells us that the total charge in the tiny negative surface slab will be equal to Q = A·δ·q·N. Hence, the surface charge density σpol = Q/A = A·δ·q·N/A = N·δ·q. But N·δ·q is also the definition of P! Hence, σpol = P. [Note that σpol is positive on one side, and negative on the other, of course!]

Now that we have σpol, we can use our E = σ/ε0 formula and add the fields from the dielectric and the capacitor plates respectively. Just think about that gaussian surface S, for example. The field there, taking into account that σpol and σfree have opposite signs, is equal to:

Using our σpol = P identity, we can also write this as E = (σfree−P)/ε0. But what’s P? Well… It’s a property of the material obviously, but then it’s also related to the electric field, of course! For larger E, we can reasonably assume that δ will be larger too (assuming some grid of atoms or molecules, we should obviously not assume a change in N or q) and, hence, dP/dE is supposed to be positive. In fact, it turns out that the relation between E and P is pretty linear, and so we can define some constant of proportionality and write E ≈ kP. Moreover, because the E and P vectors have the same direction, we can actually write E ≈ kP. Now, for historic reasons, we’ll write our k as k = ε0·χ, so we’re singling out our ε0 constant once more and – as usual – we add some gravitas to the analysis by using one of those Greek capital letters (χ is chi). So we have P = ε0·χ·E, and our equation above becomes:

Now, remembering that V = E·d and that the total charge on our capacitor is equal to Q = σfree·A, we get the formula which you may or may not know from your high school physics classes:

So… As Feynman puts it: “We have explained the observed facts. When a parallel-plate capacitor is filled with a dielectric, the capacitance is increased by the factor 1+χ.” The table below gives the values for various materials. As you can see, water’d be a great dielectric… if it wouldn’t be so conducive. 🙂

As for the assumption of linearity between E and P, there’s stuff on the Web on non-linear relationships too, but you can google that yourself. 🙂 Let’s now analyze the second case.

2. Non-uniform P

The analysis for non-uniform polarization is more general, and includes uniform polarization as a special case. To get going with it, Feynman uses an illustration (reproduced below) which is not so evident to interpret. Take your time to study it. The connects, once again, two equal but opposite charges. The P vector points in the same direction as the d vector, obviously, but has a different magnitude, because P is equal to P = Nqd. We also have the normal unit vector n here and an angle θ between the normal and P. Finally, the broken lines represent a tiny imaginary surface. To be precise, it represents, once again, an infinitesimal surface, or a surface element, as Feynman terms it.

Just take your time and think about it. If there’s no field across, then θ = π/2 and our surface disappears. If n and P point in the same direction, then θ = 0 and our surface becomes a tiny rectangle of height d. Feynman uses the illustration above to point out that the charge moved across any surface element is proportional to the component of P that is perpendicular to the surface. Hence, remembering what the vector dot product stands for, and remembering that both σpol as well as P are expressed per unit area, we can write:

σpol = P·n = |P|·|n|·cosθ = P·cosθ

So P·is the normal component of P, i.e. the component of P that’s perpendicular to our infinitesimal surface, and this component gives us the charge that moves across a surface element. [I know… The analysis is everything but easy here… But just hang in and try to get through it.]

Now, while the illustration above, and the formula, show us how some charge moves across the infinitesimal surface to create some surface polarization, it is obvious that it should not result in a net surface charge, because there are equal and opposite contributions from the dielectric on the two sides of the surface. However, having said that, the displacements of the charges do result in some tiny volume charge density, as illustrated below.

Now, I must admit Feynman does not make it easy to intuitively understand what’s going on because the various P vectors are chosen rather randomly, but you should be able to get the idea. P is not uniform indeed. Therefore, the electric field across our dielectric causes the P vectors to have different magnitudes and/or lengths. Now, as mentioned above, to get the total charge that is being displaced out of any volume bound by some surface S, we should look at the normal component of P over the surface S. To be precise, to get the total charge that is being displaced out of the volume V, we should integrate the outward normal component of P over the surface S. Of course, an equal excess charge of the opposite sign will be left behind. So, denoting the net charge inside V by ΔQpol, we write:

Now, you may or may not remember Gauss’ Theorem, which is related but not to be confused with Gauss’ Law (for more details, check one of my previous posts on vector analysis), according to which we can write:

[I know… You’re getting tired, but we’re almost there.] We can look at the net charge inside ΔQpol as an infinite sum of the (surface) charge densities σpol, but then added over the volume V. So we write:

Again, the integral above may not appear to be be very intuitive, but it actually is: we have a formula for the surface density for a surface element – so that’s something two-dimensional – and now we integrate over the volume, so the third spatial dimension comes in. Again, just let it sink in for a while, and you’ll see it all makes sense. In any case, the equalities above imply that:

and, therefore, that

σpol = −· P

You’ll say: so what? Well… It’s a nice result, really. Feynman summarizes it as follows:

“If there is a nonuniform polarization, its divergence gives the net density of charge appearing in the material. We emphasize that this is a perfectly real charge density; we call it “polarization charge” only to remind ourselves how it got there.”

Well… That says it all, I guess. To make sure you understand what’s written here: please note, once again, that the net charge over the whole of the dielectric is and remains zero, obviously!

The only question you may have is if non-uniform polarization is actually relevant. It is. You can google and you’re likely to get a lot of sites relating to multi-layered transducers and piezoelectric materials. 🙂 But, you’re right, that’s perhaps too advanced to talk about here.

Having said that, what I write above may look like too much nitty-gritty, but it isn’t: the formulas are pretty basic, and you need them if you want to advance in physics. In fact, Feynman uses these simple formulas in two more Lectures (Chapter 10 and 11 in Volume II, to be precise) to do some more analyses of real physics. However, as this blog is not meant to be a substitute for his Lectures, I’ll refer to him for further reading. At the very least, you have the basics here, and I hope it was interesting enough to induce you to look at the mentioned Lectures yourself. 🙂