Electric circuits (1): the circuit elements

OK. No escape. It’s part of physics. I am not going to go into the nitty-gritty of it all (because this is a blog about physics, not about engineering) but it’s good to review the basics, which are, essentially, Kirchoff’s rules. Just for the record, Gustav Kirchhoff was a German genius who formulated these circuit laws while he was still a student, when he was like 20 years old or so. He did it as a seminar exercise 170 years ago, and then turned it into doctoral dissertation. Makes me think of that Dire Straits song—That’s the way you do it—Them guys ain’t dumb. 🙂

So this post is, in essence, just an ‘explanation’ of Feynman’s presentation of Kirchoff’s rules, so I am writing this post basically for myself, so as to ensure I am not missing anything. To be frank, Feynman’s use of notation when working with complex numbers is confusing at times and so, yes, I’ll do some ‘re-writing’ here. The nice thing about Feynman’s presentation of electrical circuits is that he sticks to Maxwell’s Laws when describing all ideal circuit elements, so he keeps using line integrals of the electric field E around closed paths (that’s what a circuit is, indeed) to describe the so-called passive circuit elements, and he also recapitulates the idea of the electromotive force when discussing the so-called active circuit element, so that’s the generator. That’s nice, because it links it all with what we’ve learned so far, i.e. the fundamentals as expressed in Maxwell’s set of equations. Having said that, I won’t make that link here in this post, because I feel it makes the whole approach rather heavy.

OK. Let’s go for it. Let’s first recall the concept of impedance.

The impedance concept

There are three ideal (passive) circuit elements: the resistor, the capacitor and the inductor. Real circuit elements usually combine characteristics of all of them, even if they are designed to work like ideal circuit elements. Collectively, these ideal (passive) circuit elements are referred to as impedances, because… Well… Because they have some impedance. In fact, you should note that, if we reserve the terms ending with -ance for the property of the circuit elements, and those ending on -or for the objects themselves, then we should call them impedors. However, that term does not seem to have caught on.

You already know what impedance is. I explained it before, notably in my post on the intricacies related to self- and mutual inductance. Impedance basically extends the concept of resistance, as we know it from direct current (DC) circuits, to alternating current (AC) circuits. To put it simply, when AC currents are involved – so when the flow of charge periodically changes reverses direction – then it’s likely that, because of the properties of the circuit, the current signal will lag the voltage signal, and so we’ll have some phase difference telling us by how much. So, resistance is just a simple real number R – it’s the ratio between (1) the voltage that is being applied across the resistor and (2) the current through it, so we write R = V/I – and it’s got a magnitude only, but impedance is a ‘number’ that has both a magnitude as well as phase, so it’s a complex number, or a vector.

In engineering, such ‘numbers’ with a magnitude as well as a phase are referred to as phasors. A phasor represents voltages, currents and impedances as a phase vector (note the bold italics: they explain how we got the pha-sor term). It’s just a rotating vector really. So a phasor has a varying magnitude (A) and phase (φ) , which is determined by (1) some maximum magnitude A0, (2) some angular frequency ω and (3) some initial phase (θ). So we can write the amplitude A as:

A = A(φ) = A0·cos(φ) = A0·cos(ωt + θ)

As usual, Wikipedia has a nice animation for it:


In case you wonder why I am using a cosine rather than a sine function, the answer is that it doesn’t matter: the sine and the cosine are the same function except for a π/2 phase difference: just rotate the animation above by 90 degrees, or think about the formula: sinφ = cos(φ−π/2). 🙂

So A = A0·cos(ωt + θ) is the amplitude. It could be the voltage, or the current, or whatever real variable. The phase vector itself is represented by a complex number, i.e. a two-dimensional number, so to speak, which we can write as all of the following:

A = A0·eiφ = A0·cosφ + i·A0·sinφ = A0·cos(ωt+θ) + i·A0·sin(ωt+θ)

= A0·ei(ωt+θ) = A0·eiθ·eiωt = A0·eiωt with A= A0·eiθ

That’s just Euler’s formula, and I am afraid I have to refer you to my page on the essentials if you don’t get this. I know what you are thinking: why do we need the vector notation? Why can’t we just be happy with the A = A0·cos(ωt+θ) formula? The truthful answer is: it’s just to simplify calculations: it’s easier to work with exponentials than with cosines or sines. For example, writing ei(ωt + θ) = eiθ·eiωt is easier than writing cos(ωt + θ) = … […] Well? […] Hmm… 🙂

See! You’re stuck already. You’d have to use the cos(α+β) = cosα·cosβ − sinα·sinβ formula: you’d get the same results (just do it for the simple calculation of the impedance below) but it takes a lot more time, and it’s easier to make mistake. Having said why complex number notation is great, I also need to warn you. There are a few things you have to watch out for. One of these things is notation. The other is the kind of mathematical operations we can do: it’s usually alright but we need to watch out with the i2 = –1 thing when multiplying complex numbers. However, I won’t talk about that here because it would only confuse you even more. 🙂

Just for the notation, let me note that Feynman would write Aas A0 with the little hat or caret symbol (∧) on top of it, so as to indicate the complex coefficient is not a variable. So he writes Aas Â0 = A0·eiθ. However, I find that confusing and, hence, I prefer using bold-type for any complex number, variable or not. The disadvantage is that we need to remember that the coefficient in front of the exponential is not a variable: it’s a complex number alright, but not a variable. Indeed, do look at that A= A0·eiθ equality carefully: Ais a specific complex number that captures the initial phase θ. So it’s not the magnitude of the phasor itself, i.e. |A| = A0. In fact, magnitude, amplitude, phase… We’re using a lot confusing terminology here, and so that’s why you need to ‘get’ the math.

The impedance is not a variable either. It’s some constant. Having said that, this constant will depend on the angular frequency ω. So… Well… Just think about this as you continue to read. 🙂 So the impedance is some number, just like resistance, but it’s a complex number. We’ll denote it by Z and, using Euler’s formula once again, we’ll write it as:

Z = |Z|eiθ = V/I = |V|ei(ωt + θV)/|I|ei(ωt + θI= [|V|/|I|]·ei(θ− θI)

So, as you can see, it is, literally, some complex ratio, just like R = V/I was some real ratio: it is a complex ratio because it has a magnitude and a direction, obviously. Also please do note that, as I mentioned already, the impedance is, in general, some function of the frequency ω, as evidenced by the ωt term in the exponential, but so we’re not looking at ω as a variable: V and I are variables and, as such, they depend on ω, but so you should look at ω as some parameter. I know I should, perhaps, not be so explicit on what’s going on, but I want to make sure you understand.

So what’s going on? The illustration below (credit goes to Wikipedia, once again) explains. It’s a pretty generic view of a very simple AC circuit. So we don’t care what the impedance is: it might be an inductor or a capacitor, or a combination of both, but we don’t care: we just call it an impedance, or an impedor if you want. 🙂 The point is: if we apply an alternating current, then the current and the voltage will both go up and down, but the current signal will lag the voltage signal, and some phase factor θ tells us by how much, so θ will be the phase difference.


Now, we’re dividing one complex number by another in that Z = V/I formula above, and dividing one complex number by another is not all that straightforward, so let me re-write that formula for Z above as:

V = IZ = I∗|Z|eiθ

Now, while that V = IZ formula resembles the V = I·R formula, you should note the bold-face type for V and I, and the ∗ symbol I am using here for multiplication. The bold-face for V and I implies they’re vectors, or complex numbers. As for 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. [It’s obviously not a vector dot product either, because a vector dot product yields a real number, not some other vector.]

Now we write V and I as you’d expect us to write them:

  • = |V|ei(ωt + θV) = V0·ei(ωt + θV)
  • = |I|ei(ωt + θI= I0·ei(ωt + θI)

θV and θare, obviously, the so-called initial phase of the voltage and the current respectively. These ‘initial’ phases are not independent: we’re talking a phase difference really, between the voltage and the current signal, and it’s determined by the properties of the circuit. In fact, that’s the whole point here: the impedance is a property of the circuit and determines how the current signal varies as a function of the voltage signal. In fact, we’ll often choose the t = 0 point such that θand so then we need to find θI. […] OK. Let’s get on with it. Writing out all of the factors in the V = IZ = I∗|Z|eiθ equation yields:

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

Now, this equation must hold for all values of t, so we can equate the magnitudes and phases and, hence, the following equalities must hold:

  1. |V| = |I||Z| ⇔ |Z| = |V|/|I|
  2. ωt + θV =  ωt + θθ ⇔ θ = θV − θI


Of course, you’ll complain once again about those complex numbers: voltage and current are something real, isn’t it? And so what is really about this complex numbers? Well… I can just say what I said already. 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.

OK. We’re done with impedance. We can now discuss the impedors, including resistors (for which we won’t have such lag or phase difference, but the concept of impedance applies nevertheless).

Before I start, however, you should think about what I’ve done above: I explained the concept of impedance, but I didn’t do much with it. The real-life problem will usually be that you get the voltage as a function of time, and then you’ll have to calculate the impedance of a circuit and, then, the current as a function of time. So I just showed the fundamental relations but, in real life, you won’t know what θ and θI could possibly be. Well… Let me correct that statement: we’ll give you formulas for θ as we discuss the various circuit elements and their impedance below, and so then you can use these formulas to calculate θI. 🙂


Let’s start with what seems to be the easiest thing: a resistor. A real resistor is actually not easy to understand, because it requires us to understand the properties of real materials. Indeed, it may or may not surprise you, but the linear relation between the voltage and the current for real materials is only approximate. Also, the way resistors dissipate energy is not easy to understand. Indeed, unlike inductors and capacitors, i.e. the other two passive components of an electrical circuit, a resistor does not store but dissipates energy, as shown below.


It’s a nice animation (credit for it has to go to Wikipedia once more), as it shows how energy is being used in an electric circuit. 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 charges move with the field that is being generated by the power source (which is not shown here). So, 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 so the animation shows how some load – i.e. the resistor in this case – will consume their energy, so they lose their push (as shown by the change in color from yellow to black). So power, i.e.energy per unit time, is supplied, and is then consumed.

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. Electric power is voltage times current, so P = V·I (or v·i, if I use the small letters that are used in the two animations below). Now, Ohm’s Law (I = V/R) says that, if we’d want to double the current, we’d need to double the voltage, and so we’re quadrupling the power then: P2 = V2·I= (2·V1)·(2·I1) = 4·V1·I= 22·P1. 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.

But back to the impedance: Ohm’s Law is the Z = V/I law for resistors, but we can simplify it because we know the voltage across the resistor and the current that’s going through are in phase. Hence, θV and θare identical and, therefore, the θ = θθin Z = |Z|eiθ is equal to zero and, hence, Z = |Z|. Now, |Z| = |V|/|I| = V0/I0. So the impedance is just some real number R = V0/I0, which we can also write as:

R = V0/I= (V0·ei(ωt + α))/(I0·ei(ωt + α)) = V(t)/I(t), with α = θV = θI

The equation above goes from R = V0/Ito R = V(t)/I(t) = V/I. It’s note the same thing: the second equation says that, at any point in time, the voltage and the current will be proportional to each other, with R or its reciprocal as the proportionality constant. In any case, we have our formula for Z here:

Z = R = V/I = V0/I0

So that’s simple. Before we move to the next, let me note that the resistance of a real resistor may depend on its temperature, so in real-life applications one will want to keep its temperature as stable as possible. That’s why real-life resistors have power ratings and recommended operating temperatures. The image below illustrates how so-called heat-sink resistors can be mounted on a heat sink with a simple spring clip so as to ensure the dissipated heat is transported away. These heat-sink resistors are rather small (10 by 15 mm only) but are rated for 35 watt – so that’s quite a lot for such small thing – if correctly mounted.


As mentioned, the linear relation between the voltage and the current is only approximate, and the observed relation is also there only for frequencies that are not ‘too high’ because, if the frequency becomes very high, the free electrons will start radiating energy away, as they produce electromagnetic radiation. So one always needs to look at the tolerances of real-life resistors, which may be ± 5%, ± 10%, or whatever. In any case… On to the next.

Capacitors (condensers)

We talked at length about capacitors (aka condensers) in our post explaining capacitance or, the more widely used term, capacity: the capacity of a capacitor is the observed proportionality between (1) the voltage (V) across and (2) the charge (Q) on the capacitor, so we wrote it as:

C = Q/V

Now, it’s easy to confuse the C here with the C for coulomb, which I’ll also use in a moment, and so… Well… Just don’t! 🙂 The meaning of the symbol is usually obvious from the context.

As for the explanation of this relation, it’s quite simple: a capacitor consists of two separate conductors in space, with positive charge on one, and an equal and opposite (i.e. negative) charge on the other. Now, the logic of the superposition of fields implies that, if we double the charges, we will also double the fields, and so the work one needs to do to carry a unit charge from one conductor to the other is also doubled! So that’s why the potential difference between the conductors is proportional to the charge.

The C = Q/V formula actually measures the ability of the capacitor to store electric charge and, therefore, to store energy, so that’s why the term capacity is really quite appropriate. I’ll let you google a few illustrations like the one below, that shows how a capacitor is actually being charged in a circuit. Usually, some resistance will be there in the circuit, so as to limit the current when it’s connected to the voltage source and, therefore, as you can see, the R times C factor (R·C) determines how fast or how slow the capacitor charges and/or discharges. Also note that the current is equal to the time rate of change of the charge: I = dQ/dt.


In the above-mentioned post, we also give a few formulas for the capacity of specific types of condensers. For example, for a parallel-plate condenser, the formula was C =  ε0A/d. We also mentioned its unit, which is is coulomb/volt, obviously, but – in honor of Michael Faraday, who gave us Faraday’s Law, and many other interesting formulas – it’s referred to as the farad: 1 F = 1 C/V. The C here is coulomb, of course. Sorry we have to use C to denote two different things but, as I mentioned, the meaning of the symbol is usually clear from the context.

We also talked about how dielectrics actually work in that post, but we did not talk about the impedance of a capacitor, so let’s do that now. The calculation is pretty straightforward. Its interpretation somewhat less so. But… Well… Let’s go for it.

It’s the current that’s charging the condenser (sorry I keep using both terms interchangeably), and we know that the current is the time rate of change of the charge (I = dQ/dt). Now, you’ll remember that, in general, we’d write a phasor A as A = A0·eiωt with A= A0·eiθ, so Ais a complex coefficient incorporating the initial phase, which we wrote as θand θfor the voltage and for the current respectively. So we’ll represent the voltage and the current now using that notation, so we write: V = V0·eiωt and I = I0·eiωt. So let’s now use that C = Q/V by re-writing it as Q = C·V and, because C is some constant, we can write:

I = dQ/dt = d(C·V)/dt = C·dV/dt

Now, what’s dV/dt? Oh… You’ll say: V is the magnitude of V, so it’s equal to |V| = |V0·eiωt| = |V0|·|eiωt| = |V0| = |V0·eiθ| = |V0|·|eiθ| = |V0| = V0. So… Well… What? V0 is some constant here! It’s the maximum amplitude of V, so… Well… It’s time derivative is zero: dV0/dt = 0.

Yes. Indeed. We did something very wrong here! You really need to watch out with this complex-number notation, and you need to think about what you’re doing. V is not the magnitude of V but its (varying) amplitude. So it’s the real voltage V that varies with time: it’s equal to V0·cos(ωt + θV), which is the real part of our phasor V. Huh? Yes. Just hang in for a while. I know it’s difficult and, frankly, Feynman doesn’t help us very much here. Let’s take one step back and so – you will see why I am doing this in a moment – let’s calculate the time derivative of our phasor V, instead of the time derivative of our real voltage V. So we calculate dV/dt, which is equal to:

dV/dtd(V0·eiωt)/dt = V0·d(eiωt)/dt = V0·(iω)·eiωt = iω·V0·eiωt = iω·V

Remarkable result, isn’t it? We take the time derivative of our phasor, and the result is the phasor itself multiplied with iω. Well… Yes. It’s a general property of exponentials, but still… Remarkable indeed! We’d get the same with I, but we don’t need that for the moment. What we do need to do is go from our I = C·dV/dt relation, which connects the real parts of I and V one to another, to the I = C·dV/dt relation, which relates the (complex) phasors. So we write:

 I = C·dV/dt ⇔ I = C·dV/dt

Can we do that? Just like that? We just replace I and V by I and V? Yes, we can. Why? Well… We know that I is the real part of I and so we can write I = Re(I)+ Im(Ii = I + Im(Ii, and then we can write the right-hand side of the equation as C·dV/dt = Re(C·dV/dt)+ Im(C·dV/dt)·i. Now, two complex numbers are equal if, and only if, their real and imaginary parts are the same, so… Well… Write it all out, if you want, using Euler’s formula, and you’ll see it all makes sense indeed.

So what do we get? The I = C·dV/dt gives us:

I = C·dV/dt = C·(iω)·V

That implies that I/V = C·(iω) and, hence, we get – finally! – what we need to get:

Z = V/I = 1/(iωC)

This is a grand result and, while I am sorry I made you suffer for it, I think it did a good job here because, if you’d check Feynman on it, you’ll see he – or, more probably, his assistants, – just skate over this without bothering too much about mathematical rigor. OK. All that’s left now is to interpret this ‘number’ Z = 1/(iωC). It is a purely imaginary number, and it’s a constant indeed, albeit a complex constant. It can be re-written as:

Z = 1/(iωC) = i-1/(ωC) = –i/(ωC) = (1/ωC)·ei·π/2

[Sorry. I can’t be more explicit here. It’s just of the wonders of complex numbers: i-1 = –i. Just check one my posts on complex numbers for more detail.] Now, a –i factor corresponds to a rotation of minus 90 degrees, and so that gives you the true meaning of what’s usually said about a circuit with a capacitor: the voltage across the capacitor will lag the current with a phase difference equal to π/2, as shown below. Of course, as it’s the voltage driving the current, we should say it’s the current that is lagging with a phase difference of 3π/2, rather than stating it the other way around! Indeed, i-1 = –i = –1·i = i2·i = i3, so that amounts to three ‘turns’ of the phase in the counter-clockwise direction, which is the direction in which our ωt angle is ‘turning’.


It is a remarkable result, though. The illustration above assumes the maximum amplitude of the voltage and the current are the same, so |Z| = |V|/|I| = 1, but what if they are not the same? What are the real bits then? I can hear you, indeed: “To hell with the bold-face letters: what’s V and I? What’s the real thing?”

Well… V and I are the real bits of = |V|ei(ωt+θV) = V0·ei(ωt+θVand of= |I|ei(ωt+θI= I0·ei(ωt+θV−θ) = I0·ei(ωt−θ) = I0·ei(ωt+π/2respectively so, assuming θV = 0 (as mentioned above, that’s just a matter of choosing a convenient t = 0 point), we get:

  • V = V0·cos(ωt)
  • I = I0·cos(ωt + π/2)

So the π/2 phase difference is there (you need to watch out with the signs, of course: θ = −π/2, but so it’s the current that seems to lead here) but the V0/Iratio doesn’t have to be one, so the real voltage and current could look like something below, where the maximum amplitude of the current is only half of the maximum amplitude of the voltage.


So let’s analyze this quickly: the V0/Iratio is equal to |Z| = |V|/|I| = V0/I= 1/ωC = (1/ω)(1/C) (note that it’s not equal to V/I = V(t)/I(t), which is a ratio that doesn’t make sense because I(t) goes through zero as the current switches direction). So what? Well… It means the ratio is inversely proportional to both the frequency ω as well as the capacity C, as shown below. Think about this: if ω goes to zero, V0/Igoes to ∞, which means that, for a given voltage, the current must go to zero. That makes sense, because we’re talking DC current when ω → 0, and the capacitor charges itself and then that’s it: no more currents. Now, if C goes to zero, so we’re talking capacitors with hardly any capacity, we’ll also get tiny currents. Conversely, for large C, we’ll get huge currents, as the capacitor can take pretty much any charge you throw at it, so that makes for small V0/Iratios. The most interesting thing to consider is ω going to infinity, as the V0/Iratio is also quite small then. What happens? The capacitor doesn’t get the time to charge, and so it’s always in this state where it has large currents flowing in and out of it, as it can’t build the voltage that would counter the electromotive force that’s being supplied by the voltage source.

graph 6OK. That’s it. Le’s discuss the last (passive) element.


We’ve spoiled the party a bit with that illustration above, as it gives the phase difference for an inductor already:

Z = iωL = ωL·ei·π/2, with L the inductance of the coil

So, again assuming that θV = 0, we can calculate I as:

= |I|ei(ωt+θI= I0·ei(ωt+θV−θ) = I0·ei(ωt−θ) = I0·ei(ωt−π/2

Of course, you’ll want to relate this, once again, to the real voltage and the real current, so let’s write the real parts of our phasors:

  • V = V0·cos(ωt)
  • I = I0·cos(ωt − π/2)

Just to make sure you’re not falling asleep as you’re reading, I’ve made another graph of how things could look like. So now’s it’s the current signal that’s lagging the voltage signal with a phase difference equal to θ = π/2.


Also, to be fully complete, I should show you how the V0/Iratio now varies with L and ω. Indeed, here also we can write that |Z| = |V|/|I| = V0/I0, but so here we find that V0/I0 =  ωL, so we have a simple linear proportionality here! For example, for a given voltage V0, we’ll have smaller currents as ω increases, so that’s the opposite of what happens with our ideal capacitors. I’ll let you think about that… 🙂


Now how do we get that Z = iωL formula? In my post on inductance, I explained what an inductor is: a coil of wire, basically. Its defining characteristic is that a changing current will cause a changing magnetic field in it and, hence, some change in the flux of the magnetic field. Now, Faraday’s Law tells us that that will cause some circulation of the electric field in the coil, which amounts to an induced potential difference which is referred to as the electromotive force (emf). Now, it turns out that the induced emf is proportional to the change in current. So we’ve got another constant of proportionality here, so it’s like how we defined resistance, or capacitance. So, in many ways, the inductance is just another proportionality coefficient. If we denote it by L – the symbol is said to honor the Russian phyicist Heinrich Lenz, whom you know from Lenz’ Law – then we define it as:

L = −Ɛ/(dI/dt)

The dI/dt factor is, obviously, the time rate of change of the current, and the negative sign indicates that the emf opposes the change in current, so it will tend to cause an opposing current. However, the power of our voltage source will ensure the current does effectively change, so it will counter the ‘back emf’ that’s being generated by the inductor. To be precise, the voltage across the terminals of our inductor, which we denote by V, will be equal and opposite to Ɛ, so we write:

V = −Ɛ = L·(dI/dt)

Now, this very much resembles the I = C·dV/dt relation we had for capacitors, and it’s completely analogous indeed: we just need to switch the I and V, and C and L symbols. So we write:

 V = L·dI/dt⇔ V = L·dI/dt

Now, dI/dt is a similar time derivative as dV/dt. We calculate it as:

dI/dtd(I0·eiωt)/dt = I0·d(eiωt)/dt = I0·(iω)·eiωt = iω·I0·eiωt = iω·I

So we get what we want and have to get:

V = L·dI/dt = iωL·I

Now, Z = V/I, so ZiωL indeed!

Summary of conclusions

Let’s summarize what we found:

  1. For a resistor, we have Z(resistor) = Z= R = V/I = V0/I0
  2. For an capacitor, we have Z(capacitor) = Z= 1/(iωC) = –i/(ωC)
  3. For an inductor, we have Z(inductance) = ZL= iωL

Note that the impedance of capacitors decreases as frequency increases, while for inductors, it’s the other way around. We explained that by making you think of the currents: for a given voltage, we’ll have large currents for high frequencies, and, hence, a small V0/Iratio. Can you think of what happens with an inductor? It’s not so easy, so I’ll refer you to the addendum below for some more explanation.

Let me also note that, as you can see, the impedance of (ideal) inductors and capacitors is a pure imaginary number, so that’s a complex number which has no real part. In engineering, the imaginary part of the impedance is referred to as the reactance, so engineers will say that ideal capacitors and inductors have a purely imaginary reactive impedance

However, in real life, the impedance will usually have both a real as well as an imaginary part, so it will be some kind of mix, so to speak. The real part is referred to as the ‘resistance’ R, and the ‘imaginary’ part is referred to as the ‘reactance’ X. The formula for both is given below:

formula resistance and reactance

But here I have to end my post on circuit elements. It’s become quite long, so I’ll discuss Kirchoff’s rules in my next post.

Addendum: Why is V = − Ɛ?

Inductors are not easy to understand—intuitively, that is. That’s why I spent so much time writing on them in my other post on them, to which I should be referring you here. But let me recapitulate the key points. The key idea is that we’re pumping energy into an inductor when applying a current and, as you know, the time rate of change is power: P = dW/dt, so we’re talking power here too, which is voltage times current: P = dW/dt = V·I. The illustration below shows what happens when an alternating current is applied to the circuit with the inductor. So the assumption is that the current goes in one and then in the other direction, so I > 0, and then I < 0, etcetera. We’re also assuming some nice sinusoidal curve for the current here (i.e. the blue curve), and so we get what we get for U (i.e. the red curve), which is the energy that’s stored in the inductor really, as it tries to resist the changing current: the energy goes up and down between zero and some maximum amplitude that’s determined by the maximum current.

power 2

So, yes, building up current requires energy from some external source, which is used to overcome the ‘back emf’ in the inductor, and that energy is stored in the inductor itself. [If you still wonder why it’s stored in the inductor, think about the other question: where else would it be stored?] How is stored? Look at the graph and think: it’s stored as kinetic energy of the charges, obviously. That explains why the energy is zero when the current is zero, and why the energy maxes out when the current maxes out. So, yes, it all makes sense! 🙂

Let me give another example. The graph below assumes the current builds up to some maximum. As it reaches its maximum, the stored energy will also max out. This example assumes direct current, so it’s a DC circuit: the current builds up, but then stabilizes at some maximum that we can find by applying Ohm’s Law to the resistance of the circuit: I = V/R. Resistance? But we were talking an ideal inductor? We are. If there’s no other resistance in the circuit, we’ll have a short-circuit, so the assumption is that we do have some resistance in the circuit and, therefore, we should also think of some energy loss to heat from the current in the resistance. If not, well… Your power source will obviously soon reach its limits. 🙂


So what’s going on then? We have some changing current in the coil but, obviously, some kind of inertia also: the coil itself opposes the change in current through the ‘back emf’. Now, it requires energy, or power, to overcome the inertia, so that’s the power that comes from our voltage source: it will offset the ‘back emf’, so we may effectively think of a little circuit with an inductor and a voltage source, as shown below.

circuit with coil

But why do we write V = − Ɛ? Our voltage source can have any voltage, can’t it? Yes. Sure. But so the coil will always provide an emf that’s exactly the opposite of this voltage. Think of it: we have some voltage that’s being applied across the terminals of the inductor, and so we’ll have some current. A current that’s changing. And it’s that current will generate an emf that’s equal to Ɛ = –L·(dI/dt). So don’t think of Ɛ as some constant: it’s the self-inductance coefficient L that’s constant, but I (and, hence, dI/dt) and V are variable.

The point is: we cannot have any potential difference in a perfect conductor, which is what the terminals are: any potential difference, i.e. any electric field really, would cause huge currents. In other words, the voltage V and the emf Ɛ have to cancel each other out, all of the time. If not, we’d have huge currents in the wires re-establishing the V = −Ɛ equality.

Let me use Feynman’s argument here. Perhaps that will work better. 🙂 Our ideal inductor is shown below: it’s shielded by some metal box so as to ensure it does not interact with the rest of the circuit. So we have some current I, which we assume to be an AC current, and we know some voltage is needed to cause that current, so that’s the potential difference V between the terminals.


The total circulation of E – around the whole circuit – can be written as the sum of two parts:

Formula circulaton

Now, we know circulation of E can only be caused by some changing magnetic field, which is what’s going on in the inductor:


So this change in the magnetic flux is what it causing the ‘back emf’, and so the integral on the left is, effectively, equal to Ɛ, not minus Ɛ but +Ɛ. Now, the second integral is equal to V, because that’s the voltage V between the two terminals a and b. So the whole integral is equal to 0 = Ɛ + V and, therefore, we have that:

V = − Ɛ = L·dI/dt

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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.

U 6

U 7

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 farad

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


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:

Fine-structure constant formula

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.”


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:

U 3

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):

U 4

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

U 6

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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