Electric circuits (2): Kirchoff’s rules, and the energy in a circuit

My previous post was long and tedious, and all it did was presenting the three (passive) circuit elements as well as the concept of impedance. It show the inner workings of these little devices are actually quite complicated. Fortunately, the conclusions were very neat and short: for all circuit elements, we have a very simple relationship between (a) the voltage across the terminals of the element (V) and (b) the current that’s going through the circuit element (I). We found they are always in some ratio, which is referred to as the impedance, which we denoted by Z:

ZV/I ⇔ V = I∗Z

So it’s a ‘simple’ ratio, indeed. But… Well… Simple and not simple. It’s a ratio of two complex numbers and, therefore, it’s a complex number itself. That’s why I use the ∗ symbol when re-writing the ZV/I formula as V = I∗Z, so it’s clear we’re talking a product of two complex numbers). This ‘complexity’ is best understood by thinking of the voltage and the current as phase vectors (or phasors as engineers call them). Indeed, instead of using the sinusoidal functions we are used to, so that’s

• V = V0·cos(ωt + θV),
• I = I0·cos(ωt + θI), and
• Z = Z0·cos(ωt + θ) = (V0/I0)·cos(ωt + θV − θI),

we preferred the complex or vector notation, writing:

• = |V|ei(ωt + θV) = V0·ei(ωt + θV)
• = |I|ei(ωt + θI= I0·ei(ωt + θI)
• Z = |ZIei(ωt + θ= Z0·ei(ωt + θ= (V0/I0ei(ωt + θV − θI)

For the three circuit elements, we found the following solution for in terms of the previously defined properties of the respective circuit elements, i.e. their resistance (R), capacitance (C), and inductance (L) respectively:

1. For a resistor, we have Z(resistor) = Z= R
2. For an capacitor, we have Z(capacitor) = Z= 1/iωC = –i/(ωC)
3. For an inductor, we have Z(inductance) = ZL= iωL

We also explained what these formulas meant, using graphs like the ones below:

1. The graph on the left-hand side gives you the ratio of the peak voltage and peak current for the three devices as a function of C, L, R and ω respectively.
2. The graph on the right-hand side shows you the relationship between the phase of the voltage and the current for a capacitor and for an inductor. [For a resistor, the phases are the same, so no need for a graph. Also note that the lag of the phase of the current vis-á-vis the voltage phase is 90 degrees for an inductor, while it’s 270 degrees for a capacitor (which amounts to the current leading the voltage with a 90° phase difference).]

The inner workings of our circuit elements are all wonderful and mysterious, and so we spent a lot of time writing about them. That’s finished now. The summary about describes all of them in very simple terms, relating the voltage and current phasor through the concept of impedance, which is just a ratio—albeit a complex ratio.

As the graphs above suggest, we can build all kinds of crazy circuits now, and the idea of resonance as we’ve learned it when studying the behavior of waves will be particularly relevant when discussing circuits that are designed to filter certain frequencies or, the opposite, to amplify some. We won’t go that far in this post, however, as I just want to explain the basic rules one needs to know when looking at a circuit, i.e. Kirchoff’s circuit laws. There are two of them:

1. Kirchoff’s Voltage Law (KCL): The sum of the voltage drops around any closed path is zero.

The principle is illustrated below. It doesn’t matter whether or not we have other circuits feeding into this one: Kirchoff’s Voltage Law (KCL) remains valid.

We can write this law using the concept of circulation once again or, what you’ll probably like more, just using plain summation:

2. Kirchoff’s Current Law (KCL): The sum of the currents into any node is zero.

This law is written and illustrated as follows:

This Law requires some definition of a node, of course. Feynman defines a node as any set of terminals such as a, b, c, d in the illustration above which are connected. So it’s a set of connected terminals.

Now, I’ll refer you Feynman for some practical examples. The circuit below is one of them. It looks complicated but it all boils down to solving a set of linear equations. So… Well… That’s it, really. We’re done! We should do the exercises, of course, but then we’re too lazy for that, I guess. 🙂 So we’re done!

Well… Almost. I also need to mention how one can reduce complicated circuits by combining parallel impedances, using the following formula:

And then another powerful idea is the idea of equivalent circuits. The rules for this are as follows:

1. Any two-terminal network of passive elements is equivalent to (and, hence, can be replaced by) an effective impedance (Zeff).
2. Any two-terminal network of passive elements is equivalent to (and, hence, can be replaced by) a generator in series with an impedance.

These two principles are illustrated below: (a) is equivalent to (b) in each diagram.

The related formulas are:

1. IƐ/Zeff
2. VƐeff InZeff

Last but not least, I need to say something about the energy in circuits. As we noted in our previous post, the impedance will consist of a real and an imaginary part. We write:

Z = R + i·X

This gives rise to the following powerful equivalence: any impedance is equivalent to a series combination of a pure resistance and a pure reactance, as illustrated below (the ≡ sign stands for equivalence):

Of course, because this post risks becoming too short 🙂 I need to insert some more formulas now. If Z = R + i·X is the impedance of the whole circuit, then the whole circuit can be summarized in the following equation:

Ɛ = IZ = I∗(R + i·X)

Now, if we bring the analysis back to the real parts of this equation, then we may write our current as I = I0·cos(ωt). This implies we chose a t = 0 point so θI = 0. [Note that this is somewhat different than what we usually do: we usually chose our t = 0 point such that θV = 0, but it doesn’t matter.] The real emf is then going to be the real part of Ɛ = IZ = I∗(R + i·X), so we’ll write it as Ɛ (no bold-face), and it’s going to be the real part of that expression above, which we can also write as:

Ɛ = IZ = I0·ei(ωt) ∗(R + i·X)

So Ɛ is the real part of this Ɛ and, you should check, it’s going to be equal to:

Ɛ = I0·R·cos(ωt) − I0·X·sin(ωt)

The two terms in this equation represent the voltage drops across the resistance R and the reactance X in that illustration above. […] Now that I think of it, in line with the -or and -ance convention for circuit elements and their properties, should we, perhaps, say resistor and reactor in this case? 🙂 […] OK. That’s a bad joke. [I don’t seem to have good ones, isn’t it?] 🙂

Jokes aside, we see that the voltage drop across the resistance is in phase with the current (because it’s a simple cosine function of ωt as well), while the voltage drop across the purely reactive part is out of phase with the current (as you know, the sine and cosine are the same function, but with a phase difference of π/2 indeed).

You’ll wonder where are we going with this, so let me wrap it all up. You know the power is the emf times the current, and so let’s integrate this thing over one cycle to get the average rate (and then I mean a time rate of change) of the energy that gets lost in the circuit. So we need to solve the following integral:

This may look like a monster, but if you look back at your notes from your math classes, you should be able to figure it out:

1. The first integral is (1/2)I02·R..
2. The second integral is zero.

So what? Well… Look at it! It means that the (average) energy loss in a circuit with impedance Z = R + i·X only depends on the real part of Z, which is equal to I02·R/2. That’s, of course, how we want it to be: ideal inductances and capacitors store energy when being powered, and give whatever they stored when ‘charging’ back to the circuit when the current reverses direction.

So it’s a nice result, because it’s consistent with everything. Hmm… Let’s double-check though… Is it also consistent with the power equation for a resistor which, remember, is written as: P = V·I = I·R·I = I2·R. […] Well… What about the 1/2 factor?

Well… Think about it. I is a sine or a cosine here, and so we want the average value of its square, so that’s 〈cos2(ωt)〉 = 1/2.

Done! 🙂

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An introduction to electric circuits

In my previous post,I introduced electric motors, generators and transformers. They all work because of Faraday’s flux rule: a changing magnetic flux will produce some circulation of the electric field. The formula for the flux rule is given below:

It is a wonderful thing, really, but not easy to grasp intuitively. It’s one of these equations where I should quote Feynman’s introduction to electromagnetism: “The laws of Newton were very simple to write down, but they had a lot of complicated consequences and it took us a long time to learn about them all. The laws of electromagnetism are not nearly as simple to write down, which means that the consequences are going to be more elaborate and it will take us quite a lot of time to figure them all out.”

Now, among Maxwell’s Laws, this is surely the most complicated one! However, that shouldn’t deter us. 🙂 Recalling Stokes’ Theorem helps to appreciate what the integral on the left-hand side represents:

We’ve got a line integral around some closed loop Γ on the left and, on the right, we’ve got a surface integral over some surface S whose boundary is Γ. The illustration below depicts the geometry of the situation. You know what it all means. If not, I am afraid I have to send you back to square one, i.e. my posts on vector analysis. Yep. Sorry. Can’t keep copying stuff and make my posts longer and longer. 🙂

To understand the flux rule, you should imagine that the loop Γ is some loop of electric wire, and then you just replace C by E, the electric field vector. The circulation of E, which is caused by the change in magnetic flux, is referred to as the electromotive force (emf), and it’s the tangential force (E·ds) per unit charge in the wire integrated over its entire length around the loop, which is denoted by Γ here, and which encloses a surface S.

Now, you can go from the line integral to the surface integral by noting Maxwell’s Law: −∂B/∂t = ×E. In fact, it’s the same flux rule really, but in differential form. As for (×E)n, i.e. the component of ×E that is normal to the surface, you know that any vector multiplied with the normal unit vector will yield its normal component. In any case, if you’re reading this, you should already be acquainted with all of this. Let’s explore the concept of the electromotive force, and then apply it our first electric circuit. 🙂

Indeed, it’s now time for a small series on circuits, and so we’ll start right here and right now, but… Well… First things first. 🙂

The electromotive force: concept and units

The term ‘force’ in ‘electromotive force’ is actually somewhat misleading. There is a force involved, of course, but the emf is not a force. The emf is expressed in volts. That’s consistent with its definition as the circulation of E: a force times a distance amounts to work, or energy (one joule is one newton·meter), and because E is the force on a unit charge, the circulation of E is expressed in joule per coulomb, so that’s a voltage: 1 volt = 1 joule/coulomb. Hence, on the left-hand side of Faraday’s equation, we don’t have any dimension of time: it’s energy per unit charge, so it’s x joule per coulomb . Full stop.

On the right-hand side, however, we have the time rate of change of the magnetic flux. through the surface S. The magnetic flux is a surface integral, and so it’s a quantity expressed in [B]·m2, with [B] the measurement unit for the magnetic field strength. The time rate of change of the flux is then, of course, expressed in [B]·mper second, i.e. [B]·m2/s. Now what is the unit for the magnetic field strength B, which we denoted by [B]?

Well… [B] is a bit of a special unit: it is not measured as some force per unit charge, i.e. in newton per coulomb, like the electric field strength E. No. [B] is measured in (N/C)/(m/s). Why? Because the magnetic force is not F = qE but F = qv×B. Hence, so as to make the units come out alright, we need to express B in (N·s)/(C·m), which is a unit known as the tesla (1 T = N·s/C·m), so as to honor the Serbian-American genius Nikola Tesla. [I know it’s a bit of short and dumb answer, but the complete answer is quite complicated: it’s got to do with the relativity of the magnetic force, which I explained in another post: both the v in F = qv×B equation as well as the m/s unit in [B] should make you think: whose velocity? In which reference frame? But that’s something I can’t summarize in two lines, so just click the link if you want to know more. I need to get back to the lesson.]

Now that we’re talking units, I should note that the unit of flux also got a special name, the weber, so as to honor one of Germany’s most famous physicists, Wilhelm Eduard Weber: as you might expect, 1 Wb = 1 T·m2. But don’t worry about these strange names. Besides the units you know, like the joule and the newton, I’ll only use the volt, which got its name to honor some other physicist, Alessandro Volta, the inventor of the electrical battery. Or… Well… I might mention the watt as well at some point… 🙂

So how does it work? On one side, we have something expressed per second – so that’s per unit time – and on the other we have something that’s expressed per coulomb – so that’s per unit charge. The link between the two is the power, so that’s the time rate of doing work. It’s expressed in joule per second. So… Well… Yes. Here we go: in honor of yet another genius, James Watt, the unit of power got its own special name too: the watt. 🙂 In the argument below, I’ll show that the power that is being generated by a generator, and that is being consumed in the circuit (through resistive heating, for example, or whatever else taking energy out of the circuit) is equal to the emf times the current. For the moment, however, I’ll just assume you believe me. 🙂

We need to look at the whole circuit now, indeed, in which our little generator (i.e. our loop or coil of wire) is just one of the circuit elements. The units come out alright: the poweremf·current product is expressed in volt·coulomb/second = (joule/coulomb)·(coulomb/second) = joule/second. So, yes, it looks OK. But what’s going on really? How does it work, literally?

A short digression: on Ohm’s Law and electric power

Well… Let me first recall the basic concepts involved which, believe it or not, are probably easiest to explain by briefly recalling Ohm’s Law, which you’ll surely remember from your high-school physics classes. It’s quite simple really: we have some resistance in a little circuit, so that’s something that resists the passage of electric current, and then we also have a voltage source. Now, Ohm’s Law tells us that the ratio of (i) the voltage V across the resistance (so that’s between the two points marked as + and −) and (ii) the current I will be some constant. It’s the same as saying that V and I are inversely proportional to each other.  The constant of proportionality is referred to as the resistance itself and, while it’s often looked at as a property of the circuit itself, we may embody it in a circuit element itself: a resistor, as shown below.

So we write R = V/I, and the brief presentation above should remind you of the capacity of a capacitor, which was just another constant of proportionality. Indeed, instead of feeding a resistor (so all energy gets dissipated away), we could charge a capacitor with a voltage source, so that’s a energy storage device, and then we find that the ratio between (i) the charge on the capacitor and (ii) the voltage across the capacitor was a constant too, which we defined as the capacity of the capacitor, and so we wrote C = Q/V. So, yes, another constant of proportionality (there are many in electricity!).

In any case, the point is: to increase the current in the circuit above, you need to increase the voltage, but increasing both amounts to increasing the power that’s being consumed in the circuit, because the power is voltage times current indeed, so P = V·I (or v·i, if I use the small letters that are used in the two animations below). For example, if we’d want to double the current, we’d need to double the voltage, and so we’re quadrupling the power: (2·V)·(2·I) = 22·V·I. So we have a square-cube law for the power, which we get by substituting V for R·I or by substituting I for V/R, so we can write the power P as P = V2/R = I2·R. This square-cube law says exactly the same: if you want to double the voltage or the current, you’ll actually have to double both and, hence, you’ll quadruple the power. Now let’s look at the animations below (for which credit must go to Wikipedia).

They show how energy is being used in an electric circuit in  terms of power. [Note that the little moving pluses are in line with the convention that a current is defined as the movement of positive charges, so we write I = dQ/dt instead of I = −dQ/dt. That also explains the direction of the field line E, which has been added to show that the power source effectively moves charges against the field and, hence, against the electric force.] What we have here is that, on one side of the circuit, some generator or voltage source will create an emf pushing the charges, and then some load will consume their energy, so they lose their push. So power, i.e. energy per unit time, is supplied, and is then consumed.

Back to the emf…

Now, I mentioned that the emf is a ratio of two terms: the numerator is expressed in joule, and the denominator is expressed in coulomb. So you might think we’ve got some trade-off here—something like: if we double the energy of half of the individual charges, then we still get the same emf. Or vice versa: we could, perhaps, double the number of charges and load them with only half the energy. One thing is for sure: we can’t both.

Hmm… Well… Let’s have a look at this line of reasoning by writing it down more formally.

1. The time rate of change of the magnetic flux generates some emf, which we can and should think of as a property of the loop or the coil of wire in which it is being generated. Indeed, the magnetic flux through it depends on its orientation, its size, and its shape. So it’s really very much like the capacity of a capacitor or the resistance of a conductor. So we write: emf = Δ(flux)/Δt. [In fact, the induced emf tries to oppose the change in flux, so I should add the minus sign, but you get the idea.]
2. For a uniform magnetic field, the flux is equal to the field strength B times the surface area S. [To be precise, we need to take the normal component of B, so the flux is B·S = B·S·cosθ.]  So the flux can change because of a change in B or because of a change in S, or because of both.
3. The emf = Δ(flux)/Δt formula makes it clear that a very slow change in flux (i.e. the same Δ(flux) over a much larger Δt) will generate little emf. In contrast, a very fast change (i.e. the the same Δ(flux) over a much smaller Δt) will produce a lot of emf. So, in that sense, emf is not like the capacity or resistance, because it’s variable: it depends on Δ(flux), as well as on Δt. However, you should still think of it as a property of the loop or the ‘generator’ we’re talking about here.
4. Now, the power that is being produced or consumed in the circuit in which our ‘generator’ is just one of the elements, is equal to the emf times the current. The power is the time rate of change of the energy, and the energy is the work that’s being done in the circuit (which I’ll denote by ΔU), so we write: emf·current = ΔU/Δt.
5. Now, the current is equal to the time rate of change of the charge, so I = ΔQ/Δt. Hence, the emf is equal to emf = (ΔU/Δt)/I = (ΔU/Δt)/(ΔQ/Δt) = ΔU/ΔQ. From this, it follows that: emf = Δ(flux)/Δt = ΔU/ΔQ, which we can re-write as:

Δ(flux) = ΔU·Δt/ΔQ

What this says is the following. For a given amount of change in the magnetic flux (so we treat Δ(flux) as constant in the equation above), we could do more work on the same charge (ΔQ) – we could double ΔU by moving the same charge over a potential difference that’s twice as large, for example – but then Δt must be cut in half. So the same change in magnetic flux can do twice as much work if the change happens in half of the time.

Now, does that mean the current is being doubled? We’re talking the same ΔQ and half the Δt, so… Well? No. The Δt here measures the time of the flux change, so it’s not the dt in I = dQ/dt. For the current to change, we’d need to move the same charge faster, i.e. over a larger distance over the same time. We didn’t say we’d do that above: we only said we’d move the charge across a larger potential difference: we didn’t say we’d change the distance over which they are moved.

OK. That makes sense. But we’re not quite finished. Let’s first try something else, to then come back to where we are right now via some other way. 🙂 Can we change ΔQ? Here we need to look at the physics behind. What’s happening really is that the change in magnetic flux causes an induced current which consists of the free electrons in the Γ loop. So we have electrons moving in and out of our loop, and through the whole circuit really, but so there’s only so many free electrons per unit length in the wire. However, if we would effectively double the voltage, then their speed will effectively increase proportionally, so we’ll have more of them passing through per second. Now that effect surely impacts the current. It’s what we wrote above: all other things being the same, including the resistance, then we’ll also double the current as we double the voltage.

So where is that effect in the flux rule? The answer is: it isn’t there. The circulation of E around the loop is what it is: it’s some energy per unit charge. Not per unit time. So our flux rule gives us a voltage, which tells us that we’re going to have some push on the charges in the wire, but it doesn’t tell us anything about the current. To know the current, we must know the velocity of the moving charges, which we can calculate from the push if we also get some other information (such as the resistance involved, for instance), but so it’s not there in the formula of the flux rule. You’ll protest: there is a Δt on the right-hand side! Yes, that’s true. But it’s not the Δt in the v = Δs/Δt equation for our charges. Full stop.

Hmm… I may have lost you by now. If not, please continue reading. Let me drive the point home by asking another question. Think about the following: we can re-write that Δ(flux) = ΔU·Δt/ΔQ equation above as Δ(flux) = (ΔU/ΔQ)·Δt equation. Now, does that imply that, with the same change in flux, i.e. the same Δ(flux), and, importantly, for the same Δt, we could double both ΔU as well as ΔQ? I mean: (2·ΔU)/(2·ΔQ) = ΔU/ΔQ and so the equation holds, mathematically that is. […] Think about it.

You should shake your head now, and rightly so, because, while the Δ(flux) = (ΔU/ΔQ)·Δt equation suggests that would be possible, it’s totally counter-intuitive. We’re changing nothing in the real world (what happens there is the same change of flux in the same amount of time), but so we’d get twice the energy and twice the charge ?! Of course, we could also put a 3 there, or 20,000, or minus a million. So who decides on what we get? You get the point: it is, indeed, not possible. Again, what we can change is the speed of the free electrons, but not their number, and to change their speed, you’ll need to do more work, and so the reality is that we’re always looking at the same ΔQ, so if we want a larger ΔU, then we’ll need a larger change in flux, or we a shorter Δt during which that change in flux is happening.

So what can we do? We can change the physics of the situation. We can do so in many ways, like we could change the length of the loop, or its shape. One particularly interesting thing to do would be to increase the number of loops, so instead of one loop, we could have some coil with, say, N turns, so that’s N of these Γ loops. So what happens then? In fact, contrary to what you might expect, the ΔQ still doesn’t change as it moves into the coil and then from loop to loop to get out and then through the circuit: it’s still the same ΔQ. But the work that can be done by this current becomes much larger. In fact, two loops give us twice the emf of one loop, and N loops give us N times the emf of one loop. So then we can make the free electrons move faster, so they cover more distance in the same time (and you know work is force times distance), or we can move them across a larger potential difference over the same distance (and so then we move them against a larger force, so it also implies we’re doing more work). The first case is a larger current, while the second is a larger voltage. So what is it going to be?

Think about the physics of the situation once more: to make the charges move faster, you’ll need a larger force, so you’ll have a larger potential difference, i.e. a larger voltage. As for what happens to the current, I’ll explain that below. Before I do, let me talk some more basics.

In the exposé below, we’ll talk about power again, and also about load. What is load? Think about what it is in real life: when buying a battery for a big car, we’ll want a big battery, so we don’t look at the voltage only (they’re all 12-volt anyway). We’ll look at how many ampères it can deliver, and for how long. The starter motor in the car, for example, can suck up like 200 A, but for a very short time only, of course, as the car engine itself should kick in. So that’s why the capacity of batteries is expressed in ampère-hours.

Now, how do we get such large currents, such large loads? Well… Use Ohm’s Law: to get 200 A at 12 V, the resistance of the starter motor will have to as low as 0.06 ohm. So large currents are associated with very low resistance. Think practical: a 240-volt 60 watt light-bulb will suck in 0.25 A, and hence, its internal resistance, is about 960 Ω. Also think of what goes on in your house: we’ve got a lot of resistors in parallel consuming power there. The formula for the total resistance is 1/Rtotal = 1/R+ 1/R+ 1/R+ … So more appliances is less resistance, so that’s what draws in the larger current.

The point is: when looking at circuits, emf is one thing, but energy and power, i.e. the work done per second, are all that matters really. And so then we’re talking currents, but our flux rule does not say how much current our generator will produce: that depends on the load. OK. We really need to get back to the lesson now.

A circuit with an AC generator

The situation is depicted below. We’ve got a coil of wire of, let’s say, N turns of wire, and we’ll use it to generate an alternating current (AC) in a circuit.

The coil is really like the loop of wire in that primitive electric motor I introduced in my previous post, but so now we use the motor as a generator. To simplify the analysis, we assume we’ll rotate our coil of wire in a uniform magnetic field, as shown by the field lines B.

Now, our coil is not a loop, of course: the two ends of the coil are brought to external connections through some kind of sliding contacts, but that doesn’t change the flux rule: a changing magnetic flux will produce some emf and, therefore, some current in the coil.

OK. That’s clear enough. Let’s see what’s happening really. When we rotate our coil of wire, we change the magnetic flux through it. If S is the area of the coil, and θ is the angle between the magnetic field and the normal to the plane of the coil, then the flux through the coil will be equal to B·S·cosθ. Now, if we rotate the coil at a uniform angular velocity ω, then θ varies with time as θ = ω·t. Now, each turn of the coil will have an emf equal to the rate of change of the flux, i.e. d(B·S·cosθ)/dt. We’ve got N turns of wire, and so the total emf, which we’ll denote by Ɛ (yep, a new symbol), will be equal to:

Now, that’s just a nice sinusoidal function indeed, which will look like the graph below.

When no current is being drawn from the wire, this Ɛ will effectively be the potential difference between the two wires. What happens really is that the emf produces a current in the coil which pushes some charges out to the wire, and so then they’re stuck there for a while, and so there’s a potential difference between them, which we’ll denote by V, and that potential difference will be equal to Ɛ. It has to be equal to Ɛ because, if it were any different, we’d have an equalizing counter-current, of course. [It’s a fine point, so you should think about it.] So we can write:

So what happens when we do connect the wires to the circuit, so we’ve got that closed circuit depicted above (and below)?

Then we’ll have a current I going through the circuit, and Ohm’s Law then tells us that the ratio between (i) the voltage across the resistance in this circuit (we assume the connections between the generator and the resistor itself are perfect conductors) and (ii) the current will be some constant, so we have R = V/I and, therefore:

[To be fully complete, I should note that, when other circuit elements than resistors are involved, like capacitors and inductors, we’ll have a phase difference between the voltage and current functions, and so we should look at the impedance of the circuit, rather than its resistance. For more detail, see the addendum below this post.]

OK. Let’s now look at the power and energy involved.

Energy and power in the AC circuit

You’ll probably have many questions about the analysis above. You should. I do. The most remarkable thing, perhaps, is that this analysis suggests that the voltage doesn’t drop as we connect the generator to the circuit. It should. Why not? Why do the charges at both ends of the wire simply discharge through the circuit? In real life, there surely is such tendencysudden large changes in loading will effectively produce temporary changes in the voltage. But then it’s like Feynman writes: “The emf will continue to provide charge to the wires as current is drawn from them, attempting to keep the wires always at the same potential difference.”

So how much current is drawn from them? As I explained above, that depends not on the generator but on the circuit, and more in particular on the load, so that’s the resistor in this case. Again, the resistance is the (constant) ratio of the voltage and the current: R = V/I. So think about increasing or decreasing the resistance. If the voltage remains the same, it implies the current must decrease or increase accordingly, because R = V/I implies that I = V/R. So the current is inversely proportional to R, as I explained above when discussing car batteries and lamps and loads. 🙂

Now, I still have to prove that the power provided by our generator is effectively equal to P = Ɛ·I but, if it is, it implies the power that’s being delivered will be inversely proportional to R. Indeed, when Ɛ and/or V remain what they are as we insert a larger resistance in the circuit, then P = Ɛ·I = Ɛ2/R, and so the power that’s being delivered would be inversely proportional to R. To be clear, we’d have a relation between P and R like the one below.

This is somewhat weird. Why? Well… I also have to show you that the power that goes into moving our coil in the magnetic field, i.e. the rate of mechanical work required to rotate the coil against the magnetic forces, is equal to the electric power Ɛ·I, i.e. the rate at which electrical energy is being delivered by the emf of the generator. However, I’ll postpone that for a while and, hence, I’ll just ask you, once again, to take me on my word. 🙂 Now, if that’s true, so if the mechanical power equals the electric power, then that implies that a larger resistance will reduce the mechanical power we need to maintain the angular velocity ω. Think of a practical example: if we’d double the resistance (i.e. we halve the load), and if the voltage stays the same, then the current would be halved, and the power would also be halved. And let’s think about the limit situations: as the resistance goes to infinity, the power that’s being delivered goes to zero, as the current goes to zero, while if the resistance goes to zero, both the current as well as the power would go to infinity!

Well… We actually know that’s also true in real-life: actual generators consume more fuel when the load increases, so when they deliver more power, and much less fuel, so less power, when there’s no load at all. You’ll know that, at least when you’re living in a developing country with a lot of load shedding! 🙂 And the difference is huge: no or just a little load will only consume 10% of what you need when fully loading it. It’s totally in line with what I wrote on the relationship between the resistance and the current that it draws in. So, yes, it does make sense:

An emf does produce more current if the resistance in the circuit is low (so i.e. when the load is high), and the stronger currents do represent greater mechanical forces.

That’s a very remarkable thing. It means that, if we’d put a larger load on our little AC generator, it should require more mechanical work to keep the coil rotating at the same angular velocity ω. But… What changes? The change in flux is the same, the Δt is the same, and so what changes really? What changes is the current going through the coil, and it’s not a change in that ΔQ factor above, but a change in its velocity v.

Hmm… That all looks quite complicated, doesn’t it? It does, so let’s get back to the analysis of what we have here, so we’ll simply assume that we have some dynamic equilibrium obeying that formula above, and so I and R are what they are, and we relate them to Ɛ according to that equation above, i.e.:

Now let me prove those formulas on the power of our generator and in the circuit. We have all these charges in our coil that are receiving some energy. Now, the rate at which they receive energy is F·v.

Huh? Yes. Let me explain: the work that’s being done on a charge along some path is the line integral ∫ F·ds along this path. But the infinitesimal distance ds is equal to v·dt, as ds/dt = v (note that we write s and v as vectors, so the dot product with F gives us the component of F that is tangential to the path). So ∫ F·ds = ∫ (F·v)dt. So the time rate of change of the energy, which is the power, is F·v. Just take the time derivative of the integral. 🙂

Now let’s assume we have n moving charges per unit length of our coil (so that’s in line with what I wrote about ΔQ above), then the power being delivered to any element ds of the coil is (F·v)·n·ds, which can be written as: (F·ds)·n·v. [Why? Because v and ds have the same direction: the direction of both vectors is tangential to the wire, always.] Now all we need to do to find out how much power is being delivered to the circuit by our AC generator is integrate this expression over the coil, so we need to find:

However, the emf (Ɛ) is defined as the line integral ∫ E·ds line, taken around the entire coil, and = F/q, and the current I is equal to I = q·n·v. So the power from our little AC generator is indeed equal to:

Power = Ɛ·I

So that’s done. Now I need to make good on my other promise, and that is to show that Ɛ·I product is equal to the mechanical power that’s required to rotate the coil in the magnetic field. So how do we do that?

We know there’s going to be some torque because of the current in the coil. It’s formula is given by τ = μ×B. What magnetic field? Well… Let me refer you to my post on the magnetic dipole and its torque: it’s not the magnetic field caused by the current, but the external magnetic field, so that’s the B we’ve been talking about here all along. So… Well… I am not trying to fool you here. 🙂 However, the magnetic moment μ was not defined by that external field, but by the current in the coil and its area. Indeed, μ‘s magnitude was the current times the area, so that’s N·I·S in this case. Of course, we need to watch out because μ is a vector itself and so we need the angle between μ and B to calculate that vector cross product τ = μ×B. However, if you check how we defined the direction of μ, you’ll see it’s normal to the plane of the coil and, hence, the angle between μ and B is the very same θ = ω·t that we started our analysis with. So, to make a long story short, the magnitude of the torque τ is equal to:

τ = (N·I·S)·B·sinθ

Now, we know the torque is also equal to the work done per unit of distance traveled (around the axis of rotation, that is), so τ = dW/dθ. Now dθ = d(ω·t) = ω·dt. So we can now find the work done per unit of time, so that’s the power once more:

dW/dt = ω·τ = ω·(N·I·S)·B·sinθ

But so we found that Ɛ = N·S·B·ω·sinθ, so… Well… We find that:

dW/dt = Ɛ·I

Now, this equation doesn’t sort out our question as to how much power actually goes in and out of the circuit as we put some load on it, but it is what we promised to do: I showed that the mechanical work we’re doing on the coil is equal to the electric energy that’s being delivered to the circuit. 🙂

It’s all quite mysterious, isn’t it? It is. And we didn’t include other stuff that’s relevant here, such as the phenomenon of self-inductance: the varying current in the coil will actually produce its own magnetic field and, hence, in practice, we’d get some “back emf” in the circuit. This “back emf” is opposite to the current when it is increasing, and it is in the direction of the current when it is decreasing. In short, the self-inductance effect causes a current to have ‘inertia’: the inductive effects try to keep the flow constant, just as mechanical inertia tries to keep the velocity of an object constant. But… Well… I left that out. I’ll take about next time because…

[…] Well… It’s getting late in the day, and so I must assume this is sort of ‘OK enough’ as an introduction to what we’ll be busying ourselves with over the coming week. You take care, and I’ll talk to you again some day soon. 🙂

Perhaps one little note, on a question that might have popped up when you were reading all of the above: so how do actual generators keep the voltage up? Well… Most AC generators are, indeed, so-called constant speed devices. You can download some manuals from the Web, and you’ll find things like this: don’t operate at speeds above 4% of the rated speed, or more than 1% below the rated speed. Fortunately, the so-called engine governor will take car of that. 🙂

In one of my posts on oscillators, I explain the concept of impedance, which is the equivalent of resistance, but for AC circuits. Just like resistance, impedance also sort of measures the ‘opposition’ that a circuit presents to a current when a voltage is applied, but it’s a complex ratio, as opposed to R = V/I. It’s literally a complex ratio because the impedance has a magnitude and a direction, or a phase as it’s usually referred to. Hence, one will often write the impedance (denoted by Z) using Euler’s formula:

Z = |Z|eiθ

The illustration below (credit goes to Wikipedia, once again) explains what’s going on. It’s a pretty generic view of the same AC circuit. The truth is: if we apply an alternating current, then the current and the voltage will both go up and down, but the current signal will usually lag the voltage signal, and the phase factor θ tells us by how much. Hence, using complex-number notation, we write:

V = IZ = I∗|Z|eiθ

Now, while that resembles the V = R·I formula, you should note the bold-face type for V and I, and the ∗ symbol I am using here for multiplication. First the ∗ symbol: that’s to make it clear we’re not talking a vector cross product A×B here, but a product of two complex numbers. The bold-face for V and I implies they’re like vectors, or like complex numbers: so they have a phase too and, hence, we can write them as:

• = |V|ei(ωt + θV)
• = |I|ei(ωt + θI)

To be fully complete – you may skip all of this if you want, but it’s not that difficult, nor very long – it all works out as follows. We write:

IZ = |I|ei(ωt + θI)∗|Z|eiθ = |I||Z|ei(ωt + θ+ θ) = |V|ei(ωt + θV)

Now, this equation must hold for all t, so we can equate the magnitudes and phases and, hence, we get: |V| = |I||Z| and so we get the formula we need, i.e. the phase difference between our function for the voltage and our function for the current.

θ= θI + θ

Of course, you’ll say: voltage and current are something real, isn’t it? So what’s this about complex numbers? You’re right. I’ve used the complex notation only to simplify the calculus, so it’s only the real part of those complex-valued functions that counts.

Oh… And also note that, as mentioned above, we do not have such lag or phase difference when only resistors are involved. So we don’t need the concept of impedance in the analysis above. With this addendum, I just wanted to be as complete as I can be. 🙂

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Fields and charges (II)

Pre-script (dated 26 June 2020): This post has become less relevant (even irrelevant, perhaps) because my views on all things quantum-mechanical have evolved significantly as a result of my progression towards a more complete realist (classical) interpretation of quantum physics. In addition, some of the material was removed by a dark force (that also created problems with the layout, I see now). In any case, we recommend you read our recent papers. I keep blog posts like these mainly because I want to keep track of where I came from. I might review them one day, but I currently don’t have the time or energy for it. 🙂

Original post:

My previous posts was, perhaps, too full of formulas, without offering much reflection. Let me try to correct that here by tying up a few loose ends. The first loose end is about units. Indeed, I haven’t been very clear about that and so let me somewhat more precise on that now.

Note: In case you’re not interested in units, you can skip the first part of this post. However, please do look at the section on the electric constant εand, most importantly, the section on natural units—especially Planck units, as I will touch upon the topic of gauge coupling parameters there and, hence, on quantum mechanics. Also, the third and last part, on the theoretical contradictions inherent in the idea of point charges, may be of interest to you.]

The field energy integrals

When we wrote that down that u = ε0E2/2 formula for the energy density of an electric field (see my previous post on fields and charges for more details), we noted that the 1/2 factor was there to avoid double-counting. Indeed, those volume integrals we use to calculate the energy over all space (i.e. U = ∫(u)dV) count the energy that’s associated with a pair of charges (or, to be precise, charge elements) twice and, hence, they have a 1/2 factor in front. Indeed, as Feynman notes, there is no convenient way, unfortunately, of writing an integral that keeps track of the pairs so that each pair is counted just once. In fact, I’ll have to come back to that assumption of there being ‘pairs’ of charges later, as that’s another loose end in the theory.

Now, we also said that that εfactor in the second integral (i.e. the one with the vector dot product EE =|E||E|cos(0) = E2) is there to make the units come out alright. Now, when I say that, what does it mean really? I’ll explain. Let me first make a few obvious remarks:

1. Densities are always measures in terms per unit volume, so that’s the cubic meter (m3). That’s, obviously, an astronomical unit at the atomic or molecular scale.
2. Because of historical reasons, the conventional unit of charge is not the so-called elementary charge +e (i.e. the charge of a proton), but the coulomb. Hence, the charge density ρ is expressed in Coulomb per cubic meter (C/m3). The coulomb is a rather astronomical unit too—at the atomic or molecular scale at least: 1 e ≈ 1.6022×10−19 C. [I am rounding here to four digits after the decimal point.]
3. Energy is in joule (J) and that’s, once again, a rather astronomical unit at the lower end of the scales. Indeed, theoretical physicists prefer to use the electronvolt (eV), which is the energy gained (or lost) when an electron (so that’s a charge of –e, i.e. minus e) moves across a potential difference of one volt. But so we’ll stick to the joule as for now, not the eV, because the joule is the SI unit that’s used when defining most electrical units, such as the ampere, the watt and… Yes. The volt. Let’s start with that one.

The volt

The volt unit (V) measures both potential (energy) as well as potential difference (in both cases, we mean electric potential only, of course). Now, from all that you’ve read so far, it should be obvious that potential (energy) can only be measured with respect to some reference point. In physics, the reference point is infinity, which is so far away from all charges that there is no influence there. Hence, any charge we’d bring there (i.e. at infinity) will just stay where it is and not be attracted or repelled by anything. We say the potential there is zero: Φ(∞) = 0. The choice of that reference point allows us, then, to define positive or negative potential: the potential near positive charges will be positive and, vice versa, the potential near negative charges will be negative. Likewise, the potential difference between the positive and negative terminal of a battery will be positive.

So you should just note that we measure both potential as well as potential difference in volt and, hence, let’s now answer the question of what a volt really is. The answer is quite straightforward: the potential at some point r = (x, y, z) measures the work done when bringing one unit charge (i.e. +e) from infinity to that point. Hence, it’s only natural that we define one volt as one joule per unit charge:

1 volt = 1 joule/coulomb (1 V = 1 J/C).

Also note the following:

1. One joule is the energy energy transferred (or work done) when applying a force of one newton over a distance of one meter, so one volt can also be measured in newton·meter per coulomb: 1 V = 1 J/C = N·m/C.
2. One joule can also be written as 1 J = 1 V·C.

It’s quite easy to see why that energy = volt-coulomb product makes sense: higher voltage will be associated with higher energy, and the same goes for higher charge. Indeed, the so-called ‘static’ on our body is usually associated with potential differences of thousands of volts (I am not kidding), but the charges involved are extremely small, because the ability of our body to store electric charge is minimal (i.e. the capacitance (aka capacity) of our body). Hence, the shock involved in the discharge is usually quite small: it is measured in milli-joules (mJ), indeed.

The remark on ‘static’ brings me to another unit which I should mention in passing: the farad. It measures the capacitance (formerly known as the capacity) of a capacitor (formerly known as a condenser). A condenser consists, quite simply, of two separated conductors: it’s usually illustrated as consisting of two plates or of thin foils (e.g. aluminum foil) separated by an insulating film (e.g. waxed paper), but one can also discuss the capacity of a single body, like our human body, or a charged sphere. In both cases, however, the idea is the same: we have a ‘positive’ charge on one side (+q), and a ‘negative’ charge on the other (–q). In case of a single object, we imagine the ‘other’ charge to be some other large object (the Earth, for instance, but it can also be a car or whatever object that could potentially absorb the charge on our body) or, in case of the charged sphere, we could imagine some other sphere of much larger radius. The farad will then measure the capacity of one or both conductors to store charge.

Now, you may think we don’t need another unit here if that’s the definition: we could just express the capacity of a condensor in terms of its maximum ‘load’, couldn’t we? So that’s so many coulomb before the thing breaks down, when the waxed paper fails to separate the two opposite charges on the aluminium foil, for example. No. It’s not like that. It’s true we can not continue to increase the charge without consequences. However, what we want to measure with the farad is another relationship. Because of the opposite charges on both sides, there will be a potential difference, i.e. a voltage difference. Indeed, a capacitor is like a little battery in many ways: it will have two terminals. Now, it is fairly easy to show that the potential difference (i.e. the voltage) between the two plates will be proportional to the charge. Think of it as follows: if we double the charges, we’re doubling the fields, right? So then we need to do twice the amount of work to carry the unit charge (against the field) from one plate to the other. Now, because the distance is the same, that means the potential difference must be twice what it was.

Now, while we have a simple proportionality here between the voltage and the charge, the coefficient of proportionality will depend on the type of conductors, their shape, the distance and the type of insulator (aka dielectric) between them, and so on and so on. Now, what’s being measured in farad is that coefficient of proportionality, which we’ll denote by C(the proportionality coefficient for the charge), CV ((the proportionality coefficient for the voltage) or, because we should make a choice between the two, quite simply, as C. Indeed, we can either write (1) Q = CQV or, alternatively, V = CVQ, with C= 1/CV. As Feynman notes, “someone originally wrote the equation of proportionality as Q = CV”, so that’s what it will be: the capacitance (aka capacity) of a capacitor (aka condenser) is the ratio of the electric charge Q (on each conductor) to the potential difference V between the two conductors. So we know that’s a constant typical of the type of condenser we’re talking about. Indeed, the capacitance is the constant of proportionality defining the linear relationship between the charge and the voltage means doubling the voltage, and so we can write:

C = Q/V

Now, the charge is measured in coulomb, and the voltage is measured in volt, so the unit in which we will measure C is coulomb per volt (C/V), which is also known as the farad (F):

1 farad = 1 coulomb/volt (1 F = 1 C/V)

[Note the confusing use of the same symbol C for both the unit of charge (coulomb) as well as for the proportionality coefficient! I am sorrry about that, but so that’s convention!].

To be precise, I should add that the proportionality is generally there, but there are exceptions. More specifically, the way the charge builds up (and the way the field builds up, at the edges of the capacitor, for instance) may cause the capacitance to vary a little bit as it is being charged (or discharged). In that case, capacitance will be defined in terms of incremental changes: C = dQ/dV.

Let me conclude this section by giving you two formulas, which are also easily proved but so I will just give you the result:

1. The capacity of a parallel-plate condenser is C = ε0A/d. In this formula, we have, once again, that ubiquitous electric constant ε(think of it as just another coefficient of proportionality), and then A, i.e. the area of the plates, and d, i.e. the separation between the two plates.
2. The capacity of a charged sphere of radius r (so we’re talking the capacity of a single conductor here) is C = 4πε0r. This may remind you of the formula for the surface of a sphere (A = 4πr2), but note we’re not squaring the radius. It’s just a linear relationship with r.

I am not giving you these two formulas to show off or fill the pages, but because they’re so ubiquitous and you’ll need them. In fact, I’ll need the second formula in this post when talking about the other ‘loose end’ that I want to discuss.

Other electrical units

From your high school physics classes, you know the ampere and the watt, of course:

1. The ampere is the unit of current, so it measures the quantity of charge moving or circulating per second. Hence, one ampere is one coulomb per second: 1 A = 1 C/s.
2. The watt measures power. Power is the rate of energy conversion or transfer with respect to time. One watt is one joule per second: 1 W = 1 J/s = 1 N·m/s. Also note that we can write power as the product of current and voltage: 1 W = (1 A)·(1 V) = (1 C/s)·(1 J/C) = 1 J/s.

Now, because electromagnetism is such well-developed theory and, more importantly, because it has so many engineering and household applications, there are many other units out there, such as:

• The ohm (Ω): that’s the unit of electrical resistance. Let me quickly define it: the ohm is defined as the resistance between two points of a conductor when a (constant) potential difference (V) of one volt, applied to these points, produces a current (I) of one ampere. So resistance (R) is another proportionality coefficient: R = V/I, and 1 ohm (Ω) = 1 volt/ampere (V/A). [Again, note the (potential) confusion caused by the use of the same symbol (V) for voltage (i.e. the difference in potential) as well as its unit (volt).] Now, note that it’s often useful to write the relationship as V = R·I, so that gives the potential difference as the product of the resistance and the current.
• The weber (Wb) and the tesla (T): that’s the unit of magnetic flux (i.e. the strength of the magnetic field) and magnetic flux density (i.e. one tesla = one weber per square meter) respectively. So these have to do with the field vector B, rather than E. So we won’t talk about it here.
• The henry (H): that’s the unit of electromagnetic inductance. It’s also linked to the magnetic effect. Indeed, from Maxwell’s equations, we know that a changing electric current will cause the magnetic field to change. Now, a changing magnetic field causes circulation of E. Hence, we can make the unit charge go around in some loop (we’re talking circulation of E indeed, not flux). The related energy, or the work that’s done by a unit of charge as it travels (once) around that loop, is – quite confusingly! – referred to as electromotive force (emf). [The term is quite confusing because we’re not talking force but energy, i.e. work, and, as you know by now, energy is force times distance, so energy and force are related but not the same.] To ensure you know what we’re talking about, let me note that emf is measured in volts, so that’s in joule per coulomb: 1 V = 1 J/C. Back to the henry now. If the rate of change of current in a circuit (e.g. the armature winding of an electric motor) is one ampere per second, and the resulting electromotive force (remember: emf is energy per coulomb) is one volt, then the inductance of the circuit is one henry. Hence, 1 H = 1 V/(1 A/s) = 1 V·s/A.

The concept of impedance

You’ve probably heard about the so-called impedance of a circuit. That’s a complex concept, literally, because it’s a complex-valued ratio. I should refer you to the Web for more details, but let me try to summarize it because, while it’s complex, that doesn’t mean it’s complicated. 🙂 In fact, I think it’s rather easy to grasp after all you’ve gone through already. 🙂 So let’s give it a try.

When we have a simple direct current (DC), then we have a very straightforward definition of resistance (R), as mentioned above: it’s a simple ratio between the voltage (as measured in volt) and the current (as measured in ampere). Now, with alternating current (AC) circuits, it becomes more complicated, and so then it’s the concept of impedance that kicks in. Just like resistance, impedance also sort of measures the ‘opposition’ that a circuit presents to a current when a voltage is applied, but we have a complex ratio—literally: it’s a ratio with a magnitude and a direction, or a phase as it’s usually referred to. Hence, one will often write the impedance (denoted by Z) using Euler’s formula:

Z = |Z|eiθ

Now, if you don’t know anything about complex numbers, you should just skip all of what follows and go straight to the next section. However, if you do know what a complex number is (it’s an ‘arrow’, basically, and if θ is a variable, then it’s a rotating arrow, or a ‘stopwatch hand’, as Feynman calls it in his more popular Lectures on QED), then you may want to carry on reading.

The illustration below (credit goes to Wikipedia, once again) is, probably, the most generic view of an AC circuit that one can jot down. If we apply an alternating current, both the current as well as the voltage will go up and down. However, the current signal will lag the voltage signal, and the phase factor θ tells us by how much. Hence, using complex-number notation, we write:

V = IZ = I∗|Z|eiθ

Now, while that resembles the V = R·I formula I mentioned when discussing resistance, you should note the bold-face type for V and I, and the ∗ symbol I am using here for multiplication. First the ∗ symbol: that’s a convention Feynman adopts in the above-mentioned popular account of quantum mechanics. I like it, because it makes it very clear we’re not talking a vector cross product A×B here, but a product of two complex numbers. Now, that’s also why I write V and I in bold-face: they have a phase too and, hence, we can write them as:

• = |V|ei(ωt + θV)
• = |I|ei(ωt + θI)

This works out as follows:

IZ = |I|ei(ωt + θI)∗|Z|eiθ = |I||Z|ei(ωt + θ+ θ) = |V|ei(ωt + θV)

Indeed, because the equation must hold for all t, we can equate the magnitudes and phases and, hence, we get: |V| = |I||Z| and θ= θI + θ. But voltage and current is something real, isn’t it? Not some complex number? You’re right. The complex notation is used mainly to simplify the calculus, but it’s only the real part of those complex-valued functions that count. [In any case, because we limit ourselves to complex exponentials here, the imaginary part (which is the sine, as opposed to the real part, which is the cosine) is the same as the real part, but with a lag of its own (π/2 or 90 degrees, to be precise). Indeed: when writing Euler’s formula out (eiθ = cos(θ) + isin(θ), you should always remember that the sine and cosine function are basically the same function: they differ only in the phase, as is evident from the trigonometric identity sin(θ+π/) = cos(θ).]

Now, that should be more than enough in terms of an introduction to the units used in electromagnetic theory. Hence, let’s move on.

The electric constant ε0

Let’s now look at  that energy density formula once again. When looking at that u = ε0E2/2 formula, you may think that its unit should be the square of the unit in which we measure field strength. How do we measure field strength? It’s defined as the force on a unit charge (E = F/q), so it should be newton per coulomb (N/C). Because the coulomb can also be expressed in newton·meter/volt (1 V = 1 J/C = N·m/C and, hence, 1 C = 1 N·m/V), we can express field strength not only in newton/coulomb but also in volt per meter: 1 N/C = 1 N·V/N·m = 1 V/m. How do we get from N2/C2 and/or V2/mto J/m3?

Well… Let me first note there’s no issue in terms of units with that ρΦ formula in the first integral for U: [ρ]·[Φ] = (C/m3)·V = [(N·m/V)/m3)·V = (N·m)/m3 = J/m3. No problem whatsoever. It’s only that second expression for U, with the u = ε0E2/2 in the integrand, that triggers the question. Here, we just need to accept that we need that εfactor to make the units come out alright. Indeed, just like other physical constants (such as c, G, or h, for example), it has a dimension: its unit is either C2/N·m2 or, what amounts to the same, C/V·m. So the units come out alright indeed if, and only if, we multiply the N2/C2 and/or V2/m2 units with the dimension of ε0:

1. (N2/C2)·(C2/N·m2) = (N2·m)·(1/m3) = J/m3
2. (V2/m2)·(C/V·m) = V·C/m3 = (V·N·m/V)/m= N·m/m3 = J/m3

Done!

But so that’s the units only. The electric constant also has a numerical value:

ε0 = 8.854187817…×10−12 C/V·m ≈ 8.8542×10−12 C/V·m

This numerical value of εis as important as its unit to ensure both expressions for U yield the same result. Indeed, as you may or may not remember from the second of my two posts on vector calculus, if we have a curl-free field C (that means ×= 0 everywhere, which is the case when talking electrostatics only, as we are doing here), then we can always find some scalar field ψ such that C = ψ. But so here we have E = –ε0Φ, and so it’s not the minus sign that distinguishes the expression from the C = ψ expression, but the εfactor in front.

It’s just like the vector equation for heat flow: h = –κT. Indeed, we also have a constant of proportionality here, which is referred to as the thermal conductivity. Likewise, the electric constant εis also referred to as the permittivity of the vacuum (or of free space), for similar reasons obviously!

Natural units

You may wonder whether we can’t find some better units, so we don’t need the rather horrendous 8.8542×10−12 C/V·m factor (I am rounding to four digits after the decimal point). The answer is: yes, it’s possible. In fact, there are several systems in which the electric constant (and the magnetic constant, which we’ll introduce later) reduce to 1. The best-known are the so-called Gaussian and Lorentz-Heaviside units respectively.

Gauss defined the unit of charge in what is now referred to as the statcoulomb (statC), which is also referred to as the franklin (Fr) and/or the electrostatic unit of charge (esu), but I’ll refer you to the Wikipedia article on it in case you’d want to find out more about it. You should just note the definition of this unit is problematic in other ways. Indeed, it’s not so easy to try to define ‘natural units’ in physics, because there are quite a few ‘fundamental’ relations and/or laws in physics and, hence, equating this or that constant to one usually has implications on other constants. In addition, one should note that many choices that made sense as ‘natural’ units in the 19th century seem to be arbitrary now. For example:

1. Why would we select the charge of the electron or the proton as the unit charge (+1 or –1) if we now assume that protons (and neutrons) consists of quarks, which have +2/3 or –1/3?
2. What unit would we choose as the unit for mass, knowing that, despite all of the simplification that took place as a result of the generalized acceptance of the quark model, we’re still stuck with quite a few elementary particles whose mass would be a ‘candidate’ for the unit mass? Do we chose the electron, the u quark, or the d quark?

Therefore, the approach to ‘natural units’ has not been to redefine mass or charge or temperature, but the physical constants themselves. Obvious candidates are, of course, c and ħ, i.e. the speed of light and Planck’s constant. [You may wonder why physicists would select ħ, rather than h, as a ‘natural’ unit, but I’ll let you think about that. The answer is not so difficult.] That can be done without too much difficulty indeed, and so one can equate some more physical constants with one. The next candidate is the so-called Boltzmann constant (kB). While this constant is not so well known, it does pop up in a great many equations, including those that led Planck to propose his quantum of action, i.e.(see my post on Planck’s constant). When we do that – so when we equate c, ħ and kB with one (ħ = kB = 1), we still have a great many choices, so we need to impose further constraints. The next is to equate the gravitational constant with one, so then we have ħ = kB = G = 1.

Now, it turns out that the ‘solution’ of this ‘set’ of four equations (ħ = kB = G = 1) does, effectively, lead to ‘new’ values for most of our SI base units, most notably length, time, mass and temperature. These ‘new’ units are referred to as Planck units. You can look up their values yourself, and I’ll let you appreciate the ‘naturalness’ of the new units yourself. They are rather weird. The Planck length and time are usually referred to as the smallest possible measurable units of length and time and, hence, they are related to the so-called limits of quantum theory. Likewise, the Planck temperature is a related limit in quantum theory: it’s the largest possible measurable unit of temperature. To be frank, it’s hard to imagine what the scale of the Planck length, time and temperature really means. In contrast, the scale of the Planck mass is something we actually can imagine – it is said to correspond to the mass of an eyebrow hair, or a flea egg – but, again, its physical significance is not so obvious: Nature’s maximum allowed mass for point-like particles, or the mass capable of holding a single elementary charge. That triggers the question: do point-like charges really exist? I’ll come back to that question. But first I’ll conclude this little digression on units by introducing the so-called fine-structure constant, of which you’ve surely heard before.

The fine-structure constant

I wrote that the ‘set’ of equations ħ = kB = G = 1 gave us Planck units for most of our SI base units. It turns out that these four equations do not lead to a ‘natural’ unit for electric charge. We need to equate a fifth constant with one to get that. That fifth constant is Coulomb’s constant (often denoted as ke) and, yes, it’s the constant that appears in Coulomb’s Law indeed, as well as in some other pretty fundamental equations in electromagnetics, such as the field caused by a point charge q: E = q/4πε0r2. Hence, ke = 1/4πε0. So if we equate kwith one, then ε0 will, obviously, be equal to ε= 1/4π.

To make a long story short, adding this fifth equation to our set of five also gives us a Planck charge, and I’ll give you its value: it’s about 1.8755×10−18 C. As I mentioned that the elementary charge is 1 e ≈ 1.6022×10−19 C, it’s easy to that the Planck charge corresponds to some 11.7 times the charge of the proton. In fact, let’s be somewhat more precise and round, once again, to four digits after the decimal point: the qP/e ratio is about 11.7062. Conversely, we can also say that the elementary charge as expressed in Planck units, is about 1/11.7062 ≈ 0.08542455. In fact, we’ll use that ratio in a moment in some other calculation, so please jot it down.

0.08542455? That’s a bit of a weird number, isn’t it? You’re right. And trying to write it in terms of the charge of a u or d quark doesn’t make it any better. Also, note that the first four significant digits (8542) correspond to the first four significant digits after the decimal point of our εconstant. So what’s the physical significance here? Some other limit of quantum theory?

Frankly, I did not find anything on that, but the obvious thing to do is to relate is to what is referred to as the fine-structure constant, which is denoted by α. This physical constant is dimensionless, and can be defined in various ways, but all of them are some kind of ratio of a bunch of these physical constants we’ve been talking about:

The only constants you have not seen before are μ0Rand, perhaps, ras well as m. However, these can be defined as a function of the constants that you did see before:

1. The μ0 constant is the so-called magnetic constant. It’s something similar as ε0 and it’s referred to as the magnetic permeability of the vacuum. So it’s just like the (electric) permittivity of the vacuum (i.e. the electric constant ε0) and the only reason why you haven’t heard of this before is because we haven’t discussed magnetic fields so far. In any case, you know that the electric and magnetic force are part and parcel of the same phenomenon (i.e. the electromagnetic interaction between charged particles) and, hence, they are closely related. To be precise, μ= 1/ε0c2. That shows the first and second expression for α are, effectively, fully equivalent.
2. Now, from the definition of ke = 1/4πε0, it’s easy to see how those two expressions are, in turn, equivalent with the third expression for α.
3. The Rconstant is the so-called von Klitzing constant, but don’t worry about it: it’s, quite simply, equal to Rh/e2. Hene, substituting (and don’t forget that h = 2πħ) will demonstrate the equivalence of the fourth expression for α.
4. Finally, the re factor is the classical electron radius, which is usually written as a function of me, i.e. the electron mass: re = e2/4πε0mec2. This very same equation implies that reme = e2/4πε0c2. So… Yes. It’s all the same really.

Let’s calculate its (rounded) value in the old units first, using the third expression:

• The econstant is (roughly) equal to (1.6022×10–19 C)= 2.5670×10–38 C2. Coulomb’s constant k= 1/4πεis about 8.9876×10N·m2/C2. Hence, the numerator e2k≈ 23.0715×10–29 N·m2.
• The (rounded) denominator is ħc = (1.05457×10–34 N·m·s)(2.998×108 m/s) = 3.162×10–26 N·m2.
• Hence, we get α = kee2/ħc ≈ 7.297×10–3 = 0.007297.

Note that this number is, effectively, dimensionless. Now, the interesting thing is that if we calculate α using Planck units, we get an econstant that is (roughly) equal to 0.08542455= … 0.007297! Now, because all of the other constants are equal to 1 in Planck’s system of units, that’s equal to α itself. So… Yes ! The two values for α are one and the same in the two systems of units and, of course, as you might have guessed, the fine-structure constant is effectively dimensionless because it does not depend on our units of measurement. So what does it correspond to?

Now that would take me a very long time to explain, but let me try to summarize what it’s all about. In my post on quantum electrodynamics (QED) – so that’s the theory of light and matter basically and, most importantly, how they interact – I wrote about the three basic events in that theory, and how they are associated with a probability amplitude, so that’s a complex number, or an ‘arrow’, as Feynman puts it: something with (a) a magnitude and (b) a direction. We had to take the absolute square of these amplitudes in order to calculate the probability (i.e. some real number between 0 and 1) of the event actually happening. These three basic events or actions were:

1. A photon travels from point A to B. To keep things simple and stupid, Feynman denoted this amplitude by P(A to B), and please note that the P stands for photon, not for probability. I should also note that we have an easy formula for P(A to B): it depends on the so-called space-time interval between the two points A and B, i.e. I = Δr– Δt= (x2–x1)2+(y2–y1)2+(z2–z1)– (t2–t1)2. Hence, the space-time interval takes both the distance in space as well as the ‘distance’ in time into account.
2. An electron travels from point A to B: this was denoted by E(A to B) because… Well… You guessed it: the of electron. The formula for E(A to B) was much more complicated, but the two key elements in the formula was some complex number j (see below), and some other (real) number n.
3. Finally, an electron could emit or absorb a photon, and the amplitude associated with this event was denoted by j, for junction.

Now, that junction number j is about –0.1. To be somewhat more precise, I should say it’s about –0.08542455.

–0.08542455? That’s a bit of a weird number, isn’t it? Hey ! Didn’t we see this number somewhere else? We did, but before you scroll up, let’s first interpret this number. It looks like an ordinary (real) number, but it’s an amplitude alright, so you should interpret it as an arrow. Hence, it can be ‘combined’ (i.e. ‘added’ or ‘multiplied’) with other arrows. More in particular, when you multiply it with another arrow, it amounts to a shrink to a bit less than one-tenth (because its magnitude is about 0.085 = 8.5%), and half a turn (the minus sign amounts to a rotation of 180°). Now, in that post of mine, I wrote that I wouldn’t entertain you on the difficulties of calculating this number but… Well… We did see this number before indeed. Just scroll up to check it. We’ve got a very remarkable result here:

j ≈ –0.08542455 = –√0.007297 = –√α = –e expressed in Planck units

So we find that our junction number j or – as it’s better known – our coupling constant in quantum electrodynamics (aka as the gauge coupling parameter g) is equal to the (negative) square root of that fine-structure constant which, in turn, is equal to the charge of the electron expressed in the Planck unit for electric charge. Now that is a very deep and fundamental result which no one seems to be able to ‘explain’—in an ‘intuitive’ way, that is.

I should immediately add that, while we can’t explain it, intuitively, it does make sense. A lot of sense actually. Photons carry the electromagnetic force, and the electromagnetic field is caused by stationary and moving electric charges, so one would expect to find some relation between that junction number j, describing the amplitude to emit or absorb a photon, and the electric charge itself, but… An equality? Really?

Well… Yes. That’s what it is, and I look forward to trying to understand all of this better. For now, however, I should proceed with what I set out to do, and that is to tie up a few loose ends. This was one, and so let’s move to the next, which is about the assumption of point charges.

Note: More popular accounts of quantum theory say α itself is ‘the’ coupling constant, rather than its (negative) square –√α = j = –e (expressed in Planck units). That’s correct: g or j are, technically speaking, the (gauge) coupling parameter, not the coupling constant. But that’s a little technical detail which shouldn’t bother you. The result is still what it is: very remarkable! I should also note that it’s often the value of the reciprocal (1/α) that is specified, i.e. 1/0.007297 ≈ 137.036. But so now you know what this number actually stands for. 🙂

Do point charges exist?

Feynman’s Lectures on electrostatics are interesting, among other things, because, besides highlighting the precision and successes of the theory, he also doesn’t hesitate to point out the contradictions. He notes, for example, that “the idea of locating energy in the field is inconsistent with the assumption of the existence of point charges.”

Huh?

Yes. Let’s explore the point. We do assume point charges in classical physics indeed. The electric field caused by a point charge is, quite simply:

E = q/4πε0r2

Hence, the energy density u is ε0E2/2 = q2/32πε0r4. Now, we have that volume integral U = (ε0/2)∫EEdV = ∫(ε0E2/2)dV integral. As Feynman notes, nothing prevents us from taking a spherical shell for the volume element dV, instead of an infinitesimal cube. This spherical shell would have the charge q at its center, an inner radius equal to r, an infinitesimal thickness dr, and, finally, a surface area 4πr(that’s just the general formula for the surface area of a spherical shell, which I also noted above). Hence, its (infinitesimally small) volume is 4πr2dr, and our integral becomes:

To calculate this integral, we need to take the limit of –q2/8πε0r for (a) r tending to zero (r→0) and for (b) r tending to infinity (r→∞). The limit for r = ∞ is zero. That’s OK and consistent with the choice of our reference point for calculating the potential of a field. However, the limit for r = 0 is zero is infinity! Hence, that U = (ε0/2)∫EEdV basically says there’s an infinite amount of energy in the field of a point charge! How is that possible? It cannot be true, obviously.

So… Where did we do wrong?

Your first reaction may well be that this very particular approach (i.e. replacing our infinitesimal cubes by infinitesimal shells) to calculating our integral is fishy and, hence, not allowed. Maybe you’re right. Maybe not. It’s interesting to note that we run into similar problems when calculating the energy of a charged sphere. Indeed, we mentioned the formula for the capacity of a charged sphere: C = 4πε0r. Now, there’s a similarly easy formula for the energy of a charged sphere. Let’s look at how we charge a condenser:

• We know that the potential difference between two plates of a condenser represents the work we have to do, per unit charge, to transfer a charge (Q) from one plate to the other. Hence, we can write V = ΔU/ΔQ.
• We will, of course, want to do a differential analysis. Hence, we’ll transfer charges incrementally, one infinitesimal little charge dQ at the time, and re-write V as V = dU/dQ or, what amounts to the same: dU = V·dQ.
• Now, we’ve defined the capacitance of a condenser as C = Q/V. [Again, don’t be confused: C stands for capacity here, measured in coulomb per volt, not for the coulomb unit.] Hence, we can re-write dU as dU = Q·dQ/C.
• Now we have to integrate dU going from zero charge to the final charge Q. Just do a little bit of effort here and try it. You should get the following result: U = Q2/2C. [We could re-write this as U = (C2/V2)/2C =  C·V2/2, which is a form that may be more useful in some other context but not here.]
• Using that C = 4πε0r formula, we get our grand result. The energy of a charged sphere is:

U = Q2/8πε0r

From that formula, it’s obvious that, if the radius of our sphere goes to zero, its energy should also go to infinity! So it seems we can’t really pack a finite charge Q in one single point. Indeed, to do that, our formula says we need an infinite amount of energy. So what’s going on here?

Nothing much. You should, first of all, remember how we got that integral: see my previous post for the full derivation indeed. It’s not that difficult. We first assumed we had pairs of charges qi and qfor which we calculated the total electrostatic energy U as the sum of the energies of all possible pairs of charges:

And, then, we looked at a continuous distribution of charge. However, in essence, we still did the same: we counted the energy of interaction between infinitesimal charges situated at two different points (referred to as point 1 and 2 respectively), with a 1/2 factor in front so as to ensure we didn’t double-count (there’s no way to write an integral that keeps track of the pairs so that each pair is counted only once):

Now, we reduced this double integral by a clever substitution to something that looked a bit better:

Finally, some more mathematical tricks gave us that U = (ε0/2)∫EEdV integral.

In essence, what’s wrong in that integral above is that it actually includes the energy that’s needed to assemble the finite point charge q itself from an infinite number of infinitesimal parts. Now that energy is infinitely large. We just can’t do it: the energy required to construct a point charge is ∞.

Now that explains the physical significance of that Planck mass ! We said Nature has some kind of maximum allowable mass for point-like particles, or the mass capable of holding a single elementary charge. What’s going on is, as we try to pile more charge on top of the charge that’s already there, we add energy. Now, energy has an equivalent mass. Indeed, the Planck charge (q≈ 1.8755×10−18 C), the Planck length (l= 1.616×10−35 m), the Planck energy (1.956×109 J), and the Planck mass (2.1765×10−8 kg) are all related. Now things start making sense. Indeed, we said that the Planck mass is tiny but, still, it’s something we can imagine, like a flea’s egg or the mass of a hair of a eyebrow. The associated energy (E = mc2, so that’s (2.1765×10−8 kg)·(2.998×108 m/s)2 ≈ 19.56×108 kg·m2/s= 1.956×109 joule indeed.

Now, how much energy is that? Well… That’s about 2 giga-joule, obviously, but so what’s that in daily life? It’s about the energy you would get when burning 40 liter of fuel. It’s also likely to amount, more or less, to your home electricity consumption over a month. So it’s sizable, and so we’re packing all that energy into a Planck volume (lP≈ 4×10−105 m3). If we’d manage that, we’d be able to create tiny black holes, because that’s what that little Planck volume would become if we’d pack so much energy in it. So… Well… Here I just have to refer you to more learned writers than I am. As Wikipedia notes dryly: “The physical significance of the Planck length is a topic of theoretical research. Since the Planck length is so many orders of magnitude smaller than any current instrument could possibly measure, there is no way of examining it directly.”

So… Well… That’s it for now. The point to note is that we would not have any theoretical problems if we’d assume our ‘point charge’ is actually not a point charge but some small distribution of charge itself. You’ll say: Great! Problem solved!

Well… For now, yes. But Feynman rightly notes that assuming that our elementary charges do take up some space results in other difficulties of explanation. As we know, these difficulties are solved in quantum mechanics, but so we’re not supposed to know that when doing these classical analyses. 🙂

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