Fields and charges (I)

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. 馃檪

Original post:

My previous posts focused mainly on photons, so this one should be focused more on matter-particles, things that have a mass聽and a聽charge.聽However, I will use it more as an opportunity to talk about fields聽and聽present some results from聽electrostatics聽using our new vector differential operators (see my posts on vector analysis).

Before I do so, let me note something that is obvious but… Well… Think about it: photons carry the electromagnetic force, but have no electric charge themselves. Likewise, electromagnetic fields have energy and are caused聽by charges,聽but so they also聽carry聽no charge. So…聽Fields聽act聽on a charge, and photons聽interact聽with electrons, but it’s only matter-particles (notably the electron and the proton, which is made of quarks) that actually carry electric charge. Does that make sense? It should. 馃檪

Another thing I want to remind you of, before jumping into it all head first, are the basic units and relations that are valid always, regardless of what we are talking about. They are represented below:

Capture

Let me recapitulate the main points:

  • The speed of light is always the same, regardless of the reference frame (inertial or moving), and nothing can travel faster than light (except mathematical points, such as the phase velocity of a wavefunction).
  • This universal rule is the basis of relativity theory and the mass-energy equivalence relation E = mc2.
  • The constant speed of light also allows us to redefine the units of time and/or聽distance such that c = 1. For example, if we re-define the unit of distance as the distance traveled by light in one second, or the unit of time as the time light needs to travel one meter, then c = 1.
  • Newton’s laws of motion define a force as the product of a mass and its acceleration: F = ma. Hence, mass is a measure of inertia, and the unit of force is 1 newton (N) = 1 kg路m/s2.
  • The momentum of an object is the product of its mass and its velocity: p = mv. Hence, its unit is 1 kg路m/s = 1 N路s. Therefore, the concept of momentum combines force (N) as well as time (s).
  • Energy is defined in terms of work: 1 Joule (J) is the聽work done when applying a force of one newton over a distance of one meter: 1 J = 1 N路m. Hence, the concept of energy combines force (N) and distance (m).
  • Relativity theory establishes the relativistic energy-momentum relation pc = Ev/c, which can also be written as聽E2聽= p2c2聽+ m02c4, with m0聽the rest聽mass of an object (i.e. its mass when the object would be at rest, relative to the observer, of course). These equations reduce to m = E and E2= p2+ m02聽when choosing time and/or distance units such that聽c = 1. The mass聽m聽is the total mass of the object, including its inertial mass as well as the equivalent mass of its kinetic energy.
  • The relationships above establish (a) energy and time and (b) momentum and position as聽complementary聽variables and, hence, the Uncertainty Principle can be expressed in terms of both. The Uncertainty Principle, as well as the Planck-Einstein relation and the聽de Broglie聽relation (not shown on the diagram), establish a聽quantum of action, h, whose dimension combines force, distance and time (h 鈮埪6.626脳10鈭34聽N路m路s). This quantum of action (Wirkung)聽can be defined in various ways, as it pops up in more than one fundamental relation, but one of the more obvious approaches is to define h as the proportionality constant between the energy of a photon (i.e. the ‘light particle’) and its frequency: h = E/谓.

Note that we talked about forces and energy above, but we didn’t say anything about the origin聽of these forces. That’s what we are going to do now, even if we’ll limit ourselves to the electromagnetic force only.

Electrostatics

According to Wikipedia, electrostatics deals with the phenomena and properties of stationary or slow-moving electric charges with no acceleration. Feynman usually uses the term when talking about stationary聽charges only. If a current is involved (i.e. slow-moving charges with no acceleration), the term聽magnetostatics is preferred. However, the distinction does not matter all that much because聽 鈥 remarkably!聽鈥 with stationary charges and steady currents, the electric and magnetic fields (E and B) can be analyzed as separate fields: there is no interconnection whatsoever!聽That shows, mathematically, as a neat separation between (1) Maxwell’s first and second equation and (2) Maxwell’s third and fourth equation:

  1. Electrostatics: (i)聽鈭団E= 蟻/蔚0聽and (ii) E= 0.
  2. Magnetostatics: (iii) c2鈭嚸B= j/蔚0聽and (iv) B= 0.

Electrostatics: The 蟻 in equation (i) is the so-called charge density, which describes the distribution of electric charges in space: 蟻 = 蟻(x, y, z). To put it simply: 蟻 is the 鈥榓mount of charge鈥 (which we鈥檒l denote by聽螖q) per unit volume聽at a given point. As for 蔚0, that’s a constant which ensures all units are 鈥榗ompatible鈥. Equation (i) basically says we have some flux of E, the exact amount of which is determined by the charge density 蟻 or, more in general, by the charge distribution in space. As for equation (ii), i.e. E= 0, we can sort of forget about that. It means the聽curl聽of E is zero: everywhere, and always. So there鈥檚 no circulation of E. Hence, E is a so-called聽curl-free field, in this case at least, i.e. when only stationary charges and steady currents are involved.

Magnetostatics: The聽j in (iii) represents a steady current indeed, causing some聽circulation of聽B. The聽c2聽factor is related to the fact that magnetism is actually only a relativistic effect of electricity, but I can’t dwell on that here. I鈥檒l just refer you to what Feynman writes about this in his Lectures, and warmly recommend to read it. Oh… Equation (iv),聽B= 0, means that the divergence聽of B is zero: everywhere, and always. So there鈥檚 no flux of B. None. So B is a divergence-free field.

Because of the neat separation, we’ll just forget about B and talk about E only.

The electric potential

OK. Let’s try to go through the motions as quickly as we can. As mentioned in my introduction, energy is defined in terms of work done. So we should just multiply the force and the distance, right? 1 Joule = 1 newton 脳 1 meter, right? Well… Yes and no. In discussions like this, we talk potential聽energy, i.e. energy stored in the system, so to say. That means that we’re looking at work done against the force, like when we carry a bucket of water up to the third floor or, to use a somewhat more scientific description of what’s going on, when we are separating聽two masses. Because we’re doing work against聽the force, we put聽a minus sign in front of our integral:

formula 1

Now, the electromagnetic force works pretty much like gravity, except that, when discussing gravity, we only have positive ‘charges’ (the mass of some object is always positive). In electromagnetics, we have positive as well as negative charge, and please note that two like charges repel (that’s not the case with gravity). Hence, doing聽work against the electromagnetic force may involve bringing like charges together or, alternatively, separating opposite charges.聽We can’t say. Fortunately, when it comes to the math of it, it doesn’t matter: we will have the same minus sign in front of our integral. The point is: we’re doing work against聽the force, and so that’s what the minus sign stands for. So it has nothing to do with the specifics of the law of attraction and repulsion in this case (electromagnetism as opposed to gravity) and/or the fact that electrons carry negative charge. No.

Let’s get back to the integral. Just in case you forgot,聽the integral sign 鈭 stands for an S: the S of summa, i.e. sum in Latin, and we’re using these integrals because we’re adding an infinite number of infinitesimally small contributions to the total effort here indeed.聽You should recognize it, because it’s a general formula for energy or work. It is, once again, a so-called聽line integral, so it’s a bit different than the聽鈭f(x)dx stuff you learned from high school. Not very different, but different nevertheless. What’s different is that we聽have a vector dot product F鈥s after the integral sign here, so that’s not聽like f(x)dx. In case you forgot, that f(x)dx product represents the surface of an infinitesimally rectangle, as shown below: we make the base of the rectangle smaller and smaller, so dx becomes an infinitesimal indeed. And then we add them all up and get the area under the curve. If f(x) is negative, then the contributions will be negative.

420px-Integral_approximationsIntegral_example

But so we don’t have little rectangles here. We have two vectors, F and ds, and their vector dot product,聽F鈥ds, which will give you… Well… I am tempted to write: the tangential component of the force along the path, but that’s not quite correct: if ds was a unit vector, it would be true鈥攂ecause then it’s just like that h鈥n product I introduced in our first vector calculus class. However, ds is聽not聽a unit vector: it’s an infinitesimal vector, and, hence, if we write the tangential component of the force along the path as Ft, then聽F鈥s聽=聽|F||ds|cos胃 = F路cos胃路ds =聽Ft路ds. So this聽F鈥ds is a tangential component over an infinitesimally small segment of the curve. In short, it’s an infinitesimally small contribution to the total amount of work done indeed. You can make sense of this by looking at the geometrical representation of the situation below.

illustration 1

I am just saying this so you know what that integral stands for. Note that we’re聽not聽adding arrows once again, like we did when calculating amplitudes or so. It’s all聽much聽more straightforward really: a vector dot product is a scalar, so it’s just some real number鈥攋ust like any component of a vector (tangential, normal, in the direction of one of the coordinates axes, or in whatever direction) is not a vector but a real number. Hence, W is also just some real number. It can be positive or negative because… Well… When we’d be going down聽the stairs聽with our bucket of water, our minus sign doesn’t disappear. Indeed, our convention to put that minus sign there should obviously not聽depend on what聽point a and b we’re talking about, so we may actually be going along聽the direction of the force聽when going from a to b.

As a matter of fact, you should note that’s actually the situation which is depicted above. So then we get a negative聽number for W. Does that make sense? Of course聽it聽does: we’re obviously not doing any work here as we’re moving along the direction, so we’re surely聽not adding any (potential) energy to the system. On the contrary, we’re taking energy聽out of the system. Hence, we are reducing its (potential) energy and, hence, we should聽have a negative聽value for W indeed. So, just think of the minus sign being there to ensure we add potential energy to the system when going against the force, and reducing it when going with the force.

OK. You get this. You probably also know we’ll re-define W as a difference in potential between two points, which we’ll write as 桅(b) 鈥 桅(a). Now that should remind you of your high school integral聽鈭f(x)dx once again. For a definite integral over a line segment [a, b], you’d have to find the antiderivative of f(x), which you’d write as F(x), and then you’d take the difference F(b) 鈥 F(a) too. Now, you may or may not remember that this antiderivative was actually a family聽of functions F(x) + k, and k could be any constant 鈥 5/9, 6蟺, 3.6脳10124, 0.86, whatever! 鈥 because such constant vanishes when taking the derivative.

Here we have the same, we can define an infinite number of functions 桅(r) + k, of which the gradient聽will yield… Stop! I am going too fast here. First, we need to re-write that W function above in order to ensure we’re calculating stuff in terms of the unit聽charge,聽so we write:

unit chage

Huh?聽Well… Yes. I am using the聽definition聽of the field聽E聽here really: E is the force (F) when putting a聽unit charge in the field. Hence, if we want the work done per unit charge, i.e. W(unit), then we have to integrate the聽vector dot product E路ds聽over the path from a to b. But so now you see what I want to do. It makes the聽comparison with our high school integral complete. Instead of taking a derivative in regard to one variable only, i.e. dF(x)/dx) = f(x), we have a function 桅 here not in one but in three variables: 桅 = 桅(x, y, z) = 桅(r) and, therefore, we have to take the vector derivative聽(or聽gradient聽as it’s called) of 桅聽to get E:

桅(x, y, z) = (鈭偽/鈭倄, 鈭偽/鈭倅, 鈭偽/鈭倆) = 鈥E(x, y, z)

But so it’s the same principle as what you learned how to use to solve your high school integral.聽Now, you’ll usually see the expression above written as:

E =聽鈥

Why so short? Well… We all just love these mysterious abbreviations, don’t we? 馃檪 Jokes aside, it’s true some of those vector equations pack an awful lot of information. Just take Feynman’s advice here: “If it helps to write out the components to be sure you understand what’s going on, just do it. There is nothing inelegant about that. In fact, there is often a certain cleverness in doing just that.” So… Let’s move on.

I should mention that we can only apply this more sophisticated version of the ‘high school trick’ because聽桅 and E are like temperature (T) and heat flow (h): they are fields. T is a scalar field and h is a vector field, and so that’s why we can and should apply our new trick: if we have the scalar field, we can derive the vector field.聽In case you want more details, I’ll just refer you to our first聽vector calculus class. Indeed, our so-called First Theorem聽in vector calculus was just about the more sophisticated version of the ‘high school trick’: if we have some聽scalar field 蠄聽(like temperature or potential, for example: just substitute the聽聽in the equation below for T or 桅), then we’ll always find that:

First theorem

The聽螕 here is the curve between point 1 and 2, so that’s the path along which we’re going, and must represent some vector field.

Let’s go back to our W integral. I should mention that it doesn’t matter聽what path we take: we’ll always get the same value for W, regardless of what path we take. That’s why the illustration above showed two possible paths: it doesn’t matter which one we take. Again, that’s only because E is a vector field. To be precise, the electrostatic field is a so-called conservative vector field, which means that we can’t get energy out of the field by first carrying some charge along one path, and then carrying it back along another. You’ll probably find that’s obvious, 聽and it is. Just note it somewhere in the back of your mind.

So we’re done. We should just substitute E for聽桅, shouldn’t we? Well… Yes. For minus , that is. Another minus sign. Why? Well… It makes that W(unit) integral come out alright. Indeed, we want a formula like W =聽桅(b) 鈥 桅(a), not like 桅(a) 鈥 桅(b). Look at it. We could, indeed, define E聽as the (positive) gradient of some scalar field 蠄聽= 鈥撐, and so we could write聽E = , but then we’d find that W = 鈥揫(b) 鈥 (a)] = (a) 鈥 (b).

You’ll say: so what? Well… Nothing much. It’s just that our field vectors would point from lower聽to higher聽values of聽, so they would be flowing uphill, so to say. Now,聽we don’t want that in physics. Why? It just doesn’t look good. We want our field vectors聽to be directed from higher potential to lower potential, always. Just think of it: heat (h) flows from higher temperature (T) to lower, and Newton’s apple falls from greater to lower height. Likewise, when putting a unit charge in the field, we want to see it move from higher to lower electric potential. Now, we can’t change the direction of E, because that’s the direction of the force and Nature doesn’t care about our conventions and so we can’t choose the direction of the force. But we can choose our convention. So that’s why we put a minus sign in front of 桅 when writing E = 鈥桅. It makes everything come out alright. 馃檪聽That’s why we also have a minus sign in the differential heat flow equation: h =聽鈥T.

So now we have the easy W(unit) =聽桅(b) 鈥 桅(a) formula that we wanted all along. Now, note that, when we say a unit charge, we mean a plus one聽charge. Yes: +1. So that’s the charge of the proton (it’s denoted by e) so you should stop thinking about moving electrons around! [I am saying this because I used to confuse myself by doing that. You end up with the same formulas for W and 桅 but it just takes you longer to get there, so let me save you some time here. :-)]

But… Yes? In reality, it’s electrons going through a wire, isn’t? Not protons. Yes. But it doesn’t matter. Units are units in physics, and they’re always +1, for whatever (time, distance, charge, mass, spin, etcetera). Always.聽For whatever. Also note that in laboratory experiments, or particle accelerators, we often use protons instead of electrons, so there’s nothing weird about it. Finally, and most fundamentally, if we have a 鈥揺 charge moving through a neutral wire in one direction, then that’s exactly the same as a +e charge moving in the other way.

Just to make sure you get the point, let’s look at that illustration once again. We already said that we have F and, hence, E聽pointing from a to b and we’ll be reducing the potential energy of the system when moving our unit charge from a to b, so W was some negative value. Now, taking into account we want field lines to point from higher to lower potential, 桅(a) should be larger聽than聽桅(b), and so… Well.. Yes. It all makes sense: we have a聽negative difference 桅(b)聽鈥撀犖(a) = W(unit), which amounts, of course, to the reduction in potential energy.

The last thing we need to take care of now, is the reference point. Indeed, any聽桅(r) + k function will do, so which one do we take? The approach here is to take a聽reference point聽P0聽at infinity. What’s infinity? Well… Hard to say. It’s a place that’s very far away from all of the charges we’ve got lying around here. Very far away indeed. So far away we can say there is nothing there really. No charges whatsoever. 馃檪 Something like that. 馃檪 In any case. I need to move on. So聽桅(P0) is zero聽and so we can finally jot down the grand result for the聽electric potential 桅(P) (aka as the electrostatic or electric field potential):

potential

So now we can calculate all potentials, i.e. when we know where the charges are at least. I’ve shown an example below. As you can see, besides having zero potential at infinity, we will usually also have one or more equipotentialsurfaces with zero potential. One could say these zero potential lines sort of ‘separate’ the positive and negative space. That’s not a very scientifically accurate description but you know what I mean.

potential

Let me make a few final notes about the units. First, let me, once again, note that our unit charge is plus聽one, and it will flow from positive to negative potential indeed, as shown below, even if we know that, in an actual electric circuit, and so now I am talking about a copper wire or something similar,聽that means the (free) electrons will move in the other direction.

1280px-Current_notationIf you’re smart (and you are), you’ll say: what about the right-hand rule for the magnetic force? Well… We’re not discussing the magnetic force here but, because you insist, rest assured it comes out alright. Look at the illustration below of the magnetic force on a wire with a current, which is a pretty standard one.

terminalSo we have a given聽B, because of the bar magnet, and then v, the velocity vector for the… Electrons? No. You need to be consistent. It’s the velocity vector for the unit charges, which are positive聽(+e). Now just calculate the force F = qvB = evB using the right-hand rule for the vector cross product, as illustrated below. So v is the thumb and B is the index finger in this case.聽All you need to do is tilt your hand, and it comes out alright.

507px-Right_hand_rule_cross_product

But… We know聽it’s electrons going the other way.聽Well… If you insist. But then you have to put a minus sign in front of the q, because we’re talking minus聽e (鈥揺). So now v is in the other direction and so vB is in the other direction indeed, but our force F = qvB =聽鈥揺vB聽is not. Fortunately not, because physical reality should not depend on our conventions. 馃檪 So… What’s the conclusion. Nothing. You may or may not want to remember that, when we say that our current聽j current flows in this or that direction, we actually might be talking electrons (with charge聽minus聽one) flowing in the opposite direction, but then it doesn’t matter. In addition, as mentioned above, in laboratory experiments or accelerators, we may actually be talking protons instead of electrons, so don’t assume electromagnetism is the business of electrons only.

To conclude this disproportionately long introduction (we’re finally ready to talk more difficult stuff), I should just make a note on the units. Electric potential is measured in volts, as you know. However, it’s obvious from all that I wrote above that it’s the聽difference聽in potential that matters really. From the聽definition above, it should be measured in the same unit as our unit for energy, or for work, so that’s the聽joule. To be precise, it should be measured in聽joule聽per unit charge. But here we have one of the very few inconsistencies in physics when it comes to units. The proton is said to be the unit charge (e), but its actual value is measured in聽coulomb聽(C). To be precise: +1 e =聽1.602176565(35)脳10鈭19 C. So we do聽not聽measure voltage 鈥 sorry, potential difference 馃檪聽鈥 in joule but in joule per coulomb (J/C).

Now, we usually use another term for the joule/coulomb unit. You guessed it (because I said it): it’s the volt聽(V). One volt is one joule/coulomb: 1 V = 1 J/C. That’s not fair, you’ll say. You’re right, but so the proton charge e is not a so-called SI unit. Is the Coulomb an SI unit? Yes. It’s derived from the ampere聽(A) which, believe it or not, is actually an SI base聽unit. One ampere is聽6.241脳1018 electrons (i.e. one coulomb) per second. You may wonder how the ampere (or the coulomb) can be a base unit. Can they be expressed in terms of kilogram, meter and second, like all other base units. The answer is yes but, as you can imagine, it’s a bit of a complex description and so I’ll refer you to the Web for that.

The Poisson equation

I started this post by saying that I’d talk about fields and聽present some results from聽electrostatics聽using our ‘new’ vector differential operators, so it’s about time I do that. The first equation is a simple one. Using our聽E = 鈥桅 formula, we can re-write the聽鈭団E= 蟻/蔚0聽equation as:

鈭団E =鈭団⑩垏= 鈭2=聽鈥撓/蔚0

This is a so-called Poisson equation. The聽鈭2聽operator is referred to as the Laplacian and is sometimes also written as 螖, but I don’t like that because it’s also the symbol for the total differential, and that’s definitely not the same thing. The formula for the Laplacian is given below. Note that it acts on a scalar field (i.e. the potential function聽桅 in this case).

LaplacianAs Feynman notes: “The entire subject of electrostatics is merely the study of the solutions of this one equation.” However, I should note that this doesn’t prevent Feynman from devoting at least a dozen of his Lectures聽on it, and they’re not the easiest ones to read. [In case you’d doubt this statement, just have a look at his lecture on electric dipoles, for example.]聽In short: don’t think the ‘study of this one equation’ is easy. All I’ll do is just note some of the most fundamental results of this ‘study’.

Also note that聽鈭団 is one of our ‘new’ vector differential operators indeed: it’s the vector dot product of our del operator () with E. That’s something聽very聽different than, let’s say,聽桅. A little dot and some bold-face type make an enormous difference here. 馃檪 You may or may remember that we referred to the聽鈭団 operator as the divergence (div) operator (see my post on that).

Gauss’ Law

Gauss’ Law is not to be confused with Gauss’ Theorem, about which I wrote elsewhere. It gives the flux of E through a closed surface S,聽any聽closed surface S really, as the聽sum of all charges inside the surface聽divided by the electric constant聽蔚0聽(but then you know that constant is just there to make the units come out alright).

Gauss' Law

The derivation of Gauss’ Law is a bit lengthy, which is why I won’t reproduce it here, but you should note its derivation is based, mainly, on the fact that (a) surface areas are proportional to r2聽(so if we double the distance from the source, the surface area will quadruple), and (b) the magnitude of E is given by an inverse-square law, so it decreases as 1/r2.聽That explains why, if the surface S describes a sphere, the number we get from Gauss’ Law is independent of the radius of the sphere. The diagram below (credit goes to Wikipedia) illustrates the idea.

flux

The diagram can be used to show how a field and its flux can be represented. Indeed, the lines represent the flux of E emanating from a charge. Now, the total number of flux lines depends on the charge but is constant with increasing distance because the force is radial and spherically symmetric. A greater density of flux lines (lines per unit area) means a stronger field, with the density of flux lines (i.e. the magnitude of E) following an inverse-square law indeed, because the surface area of a sphere increases with the square of the radius. Hence, in Gauss’ Law, the two effect cancel out: the two factors vary with distance, but their product is a constant.

Now, if we describe the location of charges in terms of charge densities (蟻), then we can write Qint聽as:

Q int

Now, Gauss’ Law also applies to an infinitesimal cubical surface and, in one of my posts on vector calculus, I showed that the flux of E out of such cube is given by E路dV. At this point, it’s probably a good idea to remind you of what this ‘new’ vector differential operator聽鈥, i.e. our聽‘divergence’ operator,stands for: the divergence of E聽(i.e. 鈥 applied to E, so that’s E) represents the聽volume density of the flux of E out of an infinitesimal volume around a given point. Hence, it’s the flux per unit volume, as opposed to the flux out of the infinitesimal cube itself, which is the product of聽E聽and dV, i.e. E路dV.

So what? Well… Gauss’ Law applied to our infinitesimal volume gives us the following equality:

ES 1

That, in turn, simplifies to:

ES 2

So that’s Maxwell’s first equation once again, which is equivalent to our Poisson equation:聽E =聽鈭2=聽鈥撓/蔚0. So what are we doing here? Just listing equivalent formulas? Yes. I should also note they can be derived from Coulomb’s law of force, which is probably the one you learned in high school. So… Yes. It’s all consistent. But then that’s what we should expect, of course. 馃檪

The energy in a field

All these formulas look very abstract. It’s about time we use them for something. A lot of what’s written in Feynman’s聽Lectures聽on electrostatics is applied stuff indeed: it focuses, among other things, on calculating the potential in various circumstances and for various distributions of charge. Now, funnily enough, while that E =聽鈥撓/蔚0聽equation is equivalent to Coulomb’s law and, obviously, much more compact to write down, Coulomb’s law is easier to start with for basic calculations. Let me first write Coulomb’s law. You’ll probably recognize it from your high school days:

Coulomb's law

F1聽is the force on charge q1, and聽F2聽is the force on charge q2. Now, q1聽and q2. may attract or repel each other but, in both cases, the forces will be equal and opposite. [In case you wonder, yes, that’s basically the law of action and reaction.] The聽e12聽vector is the unit vector from q2聽to q1, not from q1聽to q2, as one might expect. That’s because we’re not聽talking gravity here: like charges do not attract but repel and, hence, we have to switch the order here. Having said that, that’s basically the only peculiar thing about the equation. All the rest is standard:

  1. The force is inversely聽proportional to the square of the distance and so we have an inverse-square law here indeed.
  2. The force is proportional to the charge(s).
  3. Finally, we have聽a proportionality constant,聽1/4蟺蔚0, which makes the units come out alright. You may wonder why it’s written the way it’s written, i.e. with that聽4蟺 factor, but that factor (4蟺 or 2蟺) actually disappears in a number of calculations, so then we will be left with just a聽1/蔚0聽or a 1/2蔚0聽factor. So don’t worry about it.

We want to calculate potentials and all that, so the first thing we’ll do is calculate the force on a unit charge. So we’ll divide that equation by聽q1, to calculate E(1) =聽F1/q1:

E 1

Piece of cake. But… What’s E(1) really? Well… It’s the force on the unit charge (+e), but so it doesn’t matter whether or not that unit charge is actually there, so it’s the field E聽caused by a charge q2. [If that doesn’t make sense to you, think again.] So we can drop the subscripts and just write:

E 3

What a relief, isn’t it? The simplest formula ever: the (magnitude) of the field as a simple function of the charge q聽and its distance (r) from the point that we’re looking at, which we’ll write as P = (x, y, z). But what origin are we using to measure x, y and z. Don’t be surprised: the origin is聽q.

Now that’s a formula we can use in the聽桅(P) integral. Indeed, the antiderivative is聽鈭(q/4蟺蔚0r2)dr. Now, we can bring q/4蟺蔚0 out and so we’re left with 鈭(1/r2)dr. Now 鈭(1/r2)dr聽is equal to 鈥1/r + k, and so the whole antiderivative is聽鈥q/4蟺蔚0r聽+ k. However, the minus sign cancels out with the minus sign in front of the 桅(P) = 桅(x, y, z)聽 integral, and so we get:

E 4

You should just do the integral to check this result. It’s the same integral but with聽P0 (infinity) as point a and P as point b in the integral, so we have 鈭 as start value and r as end value. The integral then yields 桅(P) 鈥 桅(P0) = 鈥q/4蟺蔚0[1/r 鈥撀1/鈭). [The k constant falls away when subtracting聽桅(P0) from 桅(P).] But 1/鈭 = 0, and we had a minus sign in front of the integral, which cancels the sign of聽鈥q/4蟺蔚0. So, yes, we get the wonderfully simple result above. Also please do quickly check if it makes sense in terms of sign: the unit charge is +e, so that’s a positive charge. Hence,聽桅(x, y, z) will be positive if the sign of q is also positive, but negative if q would happen to be negative. So that’s OK.

Also note that the potential聽鈥 which, remember, represents the amount of work to be done when bringing a unit charge (e) from infinity to some distance r from a charge q 鈥 is proportional to the charge of q. We also know that the force and, hence, the work is proportional to the charge that we are bringing in (that’s how we calculated the work per unit in the first place: by dividing the total amount of work by the charge). Hence, if we’d not聽bring some unit charge but some other charge聽q2, the work done would also be proportional to q2. Now, we need to make sure we understand what we’re writing and so let’s tidy up and re-label our first charge once again as q1, and the distance r as聽r12, because that’s what r is: the distance between the two charges. We then have another obvious but nice result: the work done in bringing two charges together from a large distance (infinity) is

U 1Now, one of the many nice properties of fields (scalar or vector fields) and the associated energies (because that’s what we are talking about here) is that we can simply add up contributions. For example, if we’d have many charges and we’d want to calculate the potential 桅 at a point which we call 1, we can use the same聽桅(r) =聽q/4蟺蔚0r formula which we had derived for one聽charge only, for all charges, and then we simply add the contributions of each to get the total potential:

P 1

Now that we’re here, I should, of course, also give the continuum version of this formula, i.e. the formula used when we’re talking charge densities rather than individual charges. The sum then becomes an infinite sum (i.e. an integral), and qj聽(note that j goes from 2 to n) becomes a variable which we write as聽蟻(2). We get:

U 2

Going back to the discrete situation, we get the same type of sum when bringing multiple pairs of charges聽qiand聽qj聽together. Hence, the total electrostatic energy U聽is the sum of the energies of all possible pairs of charges:

U 3It’s been a while since you’ve seen any diagram or so, so let me insert one just to reassure you it’s as simple as that indeed:

U system

Now, we have to be aware of the risk of double-counting, of course. We should not聽be adding qiqj/4蟺蔚0rij聽twice. That’s why we write ‘all pairs’ under the 鈭 summation sign, instead of the usual i, j subscripts. The continuum version of this equation below makes that 1/2 factor explicit:

U 4

Hmm… What kind of integral is that? It’s a so-called double integral because we have two variables here. Not easy. However, there’s a lucky break. We can use the continuum version of our formula for聽桅(1) to get rid of the 蟻(2) and dV2聽variables and reduce the whole thing to a more standard ‘single’ integral. Indeed, we can write:

U 5Now, because our point (2) no longer appears, we can actually write that more elegantly as:

U 6That looks nice, doesn’t it? But do we understand it? Just to make sure. Let me explain it.聽The potential energy of the charge聽dV is the product of this charge and the potential at the same point. The total energy is therefore the integral over聽dV, but then we are counting energies twice, so that’s why we need the 1/2 factor. Now, we can write this even more beautifully as:

U 7

Isn’t this wonderful? We have an expression for the energy of a field, not in terms of the charges or the charge distribution, but in terms of the field they produce.

I am pretty sure that, by now, you must be suffering from ‘formula overload’, so you probably are just gazing at this without even bothering to try to understand. Too bad, and you should take a break then or just go do something else, like biking or so. 馃檪

First, you should note that you know this EE expression already: EE is just the square of the magnitude of the field vector E, so EE =聽E2. That makes sense because we know, from what we know about waves, that the energy is always proportional to the square of an amplitude, and so we’re just writing the same here but with a little proportionality constant (蔚0).

OK, you’ll say. But you probably still wonder what use聽this formula could possibly have. What is that number we get from some integration over all space? So we associate the Universe with some number聽U聽and then what? Well… Isn’t that just聽nice? 馃檪 Jokes aside, we’re actually looking at that EE = E2聽product inside聽of the integral as representing an聽energy density聽(i.e. the energy per unit volume). We’ll denote that with a lower-case聽u聽symbol and so we write:

D 6

Just to make sure you ‘get’ what we’re talking about here: u is the energy density in the little cube dV in the rather simplistic (and, therefore, extremely useful) illustration below (which, just like most of what I write above, I got from Feynman).

Capture

Now that聽should make sense to you鈥擨 hope. 馃檪 In any case, if you’re still with me, and if you’re not聽all formula-ed out聽you may wonder how we get that 蔚0EE =聽蔚0E2聽expression from that聽蟻桅 expression. Of course, you know that E =聽鈥撯垏桅, and we also have the Poisson equation 鈭2=聽鈥撓/蔚0, but that doesn’t get you very far. It’s one of those examples where an easy-looking formula requires a lot of gymnastics. However, as the objective of this post is to do some of that, let me take you through the derivation.

Let’s do something with that Poisson equation first, so we’ll re-write it as 蟻=聽鈥撐02桅, and then we can substitute 蟻聽in the integral with the聽蟻桅 product. So we get:

U 8

Now, you should check out those fancy formulas with our new vector differential operators which we listed in our second class on vector calculus, but, unfortunately, none of them apply. So we have to write it all out and see what we get:

D 1

Now聽that聽looks horrendous and so you’ll surely think we won’t get anywhere with that. Well… Physicists don’t despair as easily as we do, it seems, and so they do substitute it in the integral which, of course, becomes an even more monstrous expression, because we now have two volume integrals instead of one! Indeed, we get:

D 2But if聽桅 is a vector field (it’s minus E, remember!), then 桅桅 is a vector field too, and we can then apply Gauss’ Theorem, which we mentioned in our first class on vector calculus, and which 鈥 mind you! 鈥撀爃as nothing to do with Gauss’ Law. Indeed, Gauss produced so much it’s difficult to keep track of it all. 馃檪 So let me remind you of this theorem. [I should also show why 桅桅 still yields聽a field, but I’ll assume you believe me.] Gauss’ Theorem聽basically shows how we can go from a volume integral to a surface integral:

Gauss Theorem-2If we apply this to the second integral in our U expression, we get:

D 4

So what? Where are we going with this? Relax. Be patient. What volume and surface are we talking about here? To make sure we have all charges and influences, we should integrate over all space and, hence, the surface goes to infinity. So we’re talking a (spherical) surface of enormous radius R whose center is the origin of our coordinate system. I know that sounds ridiculous but, from a math point of view, it is just the same like bringing a charge in from infinity, which is what we did to calculate the potential. So if we don’t difficulty with infinite line integrals, we should not have difficulty with infinite surface and infinite volumes. That’s all I can, so… Well… Let’s do it.

Let’s look at that product 桅桅鈥n in the surface integral. 桅 is a scalar and 桅 is a vector, and so… Well… 桅鈥n聽is a scalar too: it’s the normal component of 桅 = 鈥E.聽[Just to make sure, you should note that the way we define the normal unit vector n is聽such that 鈭囄︹n is some positive number indeed! So聽n will point in the same direction, more or less, as 鈭囄 = 鈥E. So the 胃 angle 聽between 鈭囄 =聽鈥揈 and n聽is surely less than 卤 90掳 and, hence, the cosine factor in the 鈭囄︹n聽= |鈭囄|n|cos胃 = |鈭囄cos胃 is positive, and so the whole vector dot product is positive.]

So, we have a product of two scalars here. 聽What happens with them if R goes to infinity? Well… The potential聽varies as 1/r as we’re going to infinity. That’s obvious from that 桅 = (q/4蟺蔚0)(1/r) formula: just think of q as some kind of average now, which works because we assume all charges are located within some finite distance, while we’re going to infinity. What about 桅鈥n?Well… Again assuming that we’re reasonably far away from the charges, we’re talking the聽density of flux lines here (i.e. the magnitude of E) which, as shown above, follows an inverse-square law, because the surface area of a sphere increases with the square of the radius. So 桅鈥n varies not as 1/r but as 1/r2. To make a long story short, the whole product 桅桅鈥n聽falls of as 1/r聽goes to infinity. Now, we shouldn’t forget we’re integrating a surface integral here, with r = R, and so it’s R going to infinity. So that surface integral聽has to go to zero when we include all space. The volume integral still stands however, so our formula for U now consists of one term only, i.e. the volume integral, and so we now have:

D 5

Done !

What’s left?

In electrostatics? Lots. Electric dipoles (like polar molecules), electrolytes, plasma oscillations, ionic crystals, electricity in the atmosphere (like lightning!), dielectrics and polarization (including condensers), ferroelectricity,… As soon as we try to apply our theory to matter, things become hugely complicated. But the theory works. Fortunately! 馃檪 I have to refer you to textbooks, though, in case you’d want to know more about it. [I am sure you don’t, but then one never knows.]

What I wanted to do is to give you some feel聽for those vector and field equations in the聽electrostatic聽case. We now need to bring magnetic field back into the picture and, most importantly, move to聽electrodynamics, in which the electric and magnetic field do not appear as completely separate things. No! In electrodynamics, they are fully interconnected through the time derivatives 鈭E/鈭倀 and 鈭B/鈭倀. That shows they’re part and parcel of the same thing really: electromagnetism.聽

But we’ll try to tackle that in future posts. Goodbye for now!

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