Electricity and Magnetism

Follow the field from Coulomb to Maxwell: charge and current, induction, and the four equations that turn out to describe light itself.

Charge and Coulomb's law

Rub a plastic rod on wool and it will pick up scraps of paper against the pull of the whole Earth, which is the first hint that the electric force is not a small correction to anything.

This course assumes the mechanics in Classical Mechanics (Newton's laws, work and energy, conservative forces, circular motion and torque) and the calculus in Calculus (derivatives, definite integrals, and separation of variables). It assumes no vector calculus at all: every field equation in it, including the four that Maxwell assembled, is written and used in integral form. What it does assume is a willingness to take a force law and follow it, because almost everything in the next twelve lessons is a consequence of the one law derived here.

Two kinds of charge

The oldest recorded fact in the subject is that amber, rubbed, attracts light objects, which is where the word electricity comes from: the Greek for amber is elektron. For two thousand years that was the whole of it. The useful step was Stephen Gray's, in 1729, when he showed the attracting property could be conducted along a moist thread for hundreds of feet, and so was something transferable rather than a property of a rubbed surface. Materials divided at once into conductors, which pass it, and insulators, which do not.

Charles du Fay found the crucial complication in 1733. Two glass rods, each rubbed with silk, repel each other. Two amber rods, each rubbed with fur, also repel each other. But a rubbed glass rod and a rubbed amber rod attract. There is no way to fit that into a single quantity that objects have more or less of, so du Fay proposed two electricities, vitreous and resinous, with like repelling like and unlike attracting.

Benjamin Franklin recast it around 1750 as a single fluid present in every body in a natural amount. Rubbing transfers some from one body to the other, so one ends with a surplus, which he called positive, and the other with a deficit, which he called negative. Franklin's picture is closer to the truth than du Fay's, since what actually moves is electrons, and it has one famous cost. Franklin guessed that the fluid flowed from the glass into the silk, so he called the glass negative and the silk positive, and the guess was backwards. The carrier of ordinary electric current in a metal is the electron, which by Franklin's convention carries a negative charge, so conventional current is drawn flowing the way positive charge would flow, which is opposite to the actual drift of the electrons. Every circuit diagram ever drawn still carries that error, harmlessly, because nothing physical depends on which sign is called positive.

What survives from all of this is one experimental rule: like charges repel, unlike charges attract, and nothing else in the subject has ever been observed. There is no third sign.

Conserved, and quantised

Two further facts, neither of them obvious, make charge a useful quantity rather than a description of rubbed rods.

The first is conservation. The total charge of an isolated system never changes. Rubbing does not create charge, it separates it: the wool ends up as positive as the rod is negative, and if you weigh the pair together the net charge is what it was. This survives into contexts Franklin never imagined. A neutron decays into a proton, an electron and an antineutrino, and the charges before and after are both zero. A gamma ray of sufficient energy converts into an electron and a positron, never into one alone. Charge conservation is tested harder than almost any law in physics: if the electron could decay into neutral particles, charge would vanish, and the Borexino detector's search for that decay puts the electron's lifetime above about 10²⁸ years, which is eighteen orders of magnitude longer than the age of the universe.

The second is quantisation. Charge comes in integer multiples of a fixed amount. Robert Millikan's oil drop experiment, run from 1909 and refined to 1913, sprayed a fine mist of oil between two horizontal plates and balanced individual droplets against gravity by adjusting the field. The charge on each droplet came out as a multiple of a single unit, and no droplet ever showed a fraction of it. His published value was $1.592 \times 10^{- 19}$ coulombs, low by 0.64 per cent against the modern figure, and the reason is instructive: he used a value for the viscosity of air that was slightly wrong, and the error propagated straight into $e$ . Since the 2019 redefinition of SI units the elementary charge is not measured at all but fixed by definition, at exactly

e = 1.602176634 \times 10^{- 19} C

and the coulomb is now defined as the charge of $1 / e$ elementary charges. Quarks carry $\pm e / 3$ and $\pm 2 e / 3$ , but no experiment has ever isolated one, so every free object carries an integer multiple of $e$ .

The quantum is small enough that it is invisible in ordinary electrostatics. A charge of one microcoulomb, easily produced by rubbing, is about $6 \times 10^{12}$ elementary charges, so treating charge as a smooth continuous fluid costs nothing at laboratory scales, exactly as treating water as continuous costs nothing when the vessel holds $10^{25}$ molecules.

Coulomb's measurement

Knowing that charges push and pull says nothing about how hard. Several people guessed an inverse square by analogy with gravity, and Joseph Priestley argued for it in 1767 from an experiment of Franklin's that this course reaches in the third lesson. The measurement that settled it was Charles Augustin de Coulomb's, published in 1785.

His instrument was a torsion balance, and the reason it worked is that it converts an unmeasurably small force into a measurable angle. A fine silver wire hangs from a fixed head and carries a light horizontal needle with a small pith ball at one end, all sealed inside a glass cylinder to keep out draughts. A second, fixed ball is brought in and both are charged, and the repulsion swings the needle round until the twist in the wire balances it. Coulomb calibrated the wire separately by hanging known small torques on it, so the deflection angle read directly as a force. The trick, which is the whole art of the thing, is that the restoring torque of a thin wire can be made almost arbitrarily weak by making the wire long and fine, so a force of a few micronewtons becomes a swing of tens of degrees.

His published numbers deserve to be looked at rather than summarised. Starting with the balls 36 degrees apart, he halved the separation to 18 degrees and found the torsion needed rose from 36 units to 144, a factor of exactly four for a halving of distance. Then he pressed to 8.5 degrees, where an inverse square predicts $36 \times (36 / 8.5)^{2} = 646$ units, and measured 575.5, about 11 per cent low. He attributed the shortfall to charge leaking away from the balls during the minutes the reading took, which was almost certainly right, and it is worth noticing that the law was announced on the strength of two good points and one poor one. The law is believed today because of the far more sensitive experiments described at the end of this lesson, not because of Coulomb's data.

The law

For two point charges $q_{1}$ and $q_{2}$ separated by a distance $r$ , the force each exerts on the other has magnitude

F = k \frac{| q_{1} q_{2} |}{r^{2}}

and is directed along the line joining them, repulsive if the charges have the same sign and attractive if they do not. The two forces are equal in magnitude and opposite in direction, which is Newton's third law and not an extra assumption: it comes out of the symmetry of the expression under swapping the labels.

"Point charge" is doing real work in that sentence. The law as written is exact only for charges of no size, and applies to spheres of finite size only because of a theorem proved in the third lesson, the electrical twin of the shell theorem that lets a planet be treated as a point.

The constant depends on the units. In SI,

k = 8.9875 \times 10^{9} {N m² C}^{- 2}

which is usually written instead as $k = 1 / (4 π ε_{0})$ , with $ε_{0} = 8.8542 \times 10^{- 12}$ farads per metre called the permittivity of free space. Moving the $4 π$ into the constant looks like gratuitous ugliness at this stage, and at this stage it is. The payoff arrives in the third lesson, where the $4 π$ cancels the surface area of a sphere and leaves Gauss's law clean. Physics chose to put the awkwardness in the force law rather than in the field equations, on the grounds that the field equations get used more.

Example. A charge of $+ 3.0$ μC and a charge of $- 5.0$ μC sit 20 cm apart in air. What force does each feel?

$F = k | q_{1} q_{2} | / r^{2} = (8.9876 \times 10^{9}) (3.0 \times 10^{- 6}) (5.0 \times 10^{- 6}) / (0.20)^{2}$ . The numerator is $8.9876 \times 10^{9} \times 1.5 \times 10^{- 11} = 0.1348$ , and dividing by $0.04$ gives $F = 3.37$ N. The signs are opposite, so it is an attraction, and both charges feel 3.37 N pulling them together.

Now you. A charge of $+ 2.0$ μC and a charge of $+ 7.0$ μC sit 15 cm apart. What force does each feel?

Answer

$F = (8.9876 \times 10^{9}) (2.0 \times 10^{- 6}) (7.0 \times 10^{- 6}) / (0.15)^{2} = (8.9876 \times 10^{9}) (1.4 \times 10^{- 11}) / 0.0225 = 5.59$ N, repulsive, since both charges are positive.

Superposition

Coulomb's law covers two charges, and the world contains more. The additional experimental fact, and it is a fact rather than a deduction, is superposition: the force on a charge from several others is the vector sum of the forces each would exert alone, unaffected by the presence of the rest. A third charge placed between two others does not screen them or weaken their interaction.

That is not guaranteed. A theory in which the force between two charges depended on what else was nearby would be perfectly consistent, and gravity in general relativity is exactly such a theory, which is why gravitational fields do not simply add. Electromagnetism is linear to the limits of measurement, and every calculation in this course leans on it.

In practice superposition means resolving into components and adding. On a line it means keeping track of signs.

Example. Three charges sit on the $x$ axis: $q_{1} = + 5.0$ nC at $x = 0$ , $q_{2} = + 3.0$ nC at $x = 4.0$ cm, and $q_{3} = - 8.0$ nC at $x = 10.0$ cm. What is the net force on $q_{2}$ ?

Take rightwards as positive. The charge $q_{1}$ is positive and 4.0 cm to the left, so it pushes $q_{2}$ rightwards: $F_{1} = (8.9876 \times 10^{9}) (5.0 \times 10^{- 9}) (3.0 \times 10^{- 9}) / (0.040)^{2} = 8.43 \times 10^{- 5}$ N, positive. The charge $q_{3}$ is negative and 6.0 cm to the right, so it pulls $q_{2}$ rightwards too: $F_{3} = (8.9876 \times 10^{9}) (3.0 \times 10^{- 9}) (8.0 \times 10^{- 9}) / (0.060)^{2} = 5.99 \times 10^{- 5}$ N, also positive. Both point the same way, so the net force is $1.44 \times 10^{- 4}$ N in the $+ x$ direction.

Now you. Three charges sit on the $x$ axis: $q_{1} = + 6.0$ nC at $x = 0$ , $q_{2} = + 2.0$ nC at $x = 5.0$ cm, and $q_{3} = + 9.0$ nC at $x = 10.0$ cm. What is the net force on $q_{2}$ ?

Answer

Both neighbours are positive, so they push $q_{2}$ in opposite directions. From $q_{1}$ : $(8.9876 \times 10^{9}) (6.0 \times 10^{- 9}) (2.0 \times 10^{- 9}) / (0.050)^{2} = 4.31 \times 10^{- 5}$ N to the right. From $q_{3}$ : $(8.9876 \times 10^{9}) (2.0 \times 10^{- 9}) (9.0 \times 10^{- 9}) / (0.050)^{2} = 6.47 \times 10^{- 5}$ N to the left. The net is $2.16 \times 10^{- 5}$ N in the $- x$ direction, towards $q_{1}$ .

How strong is it, really

Put two charges of one coulomb each a metre apart and the force between them is $8.99 \times 10^{9}$ N, the weight of nine hundred thousand tonnes. That tells you immediately that a coulomb is an enormous amount of static charge, and that the situation described never occurs: the charges would tear themselves off the objects holding them long before you assembled them.

A better feel comes from asking how little charge separation is needed to matter. One gram of copper contains $6.022 \times 10^{23} / 63.546 = 9.48 \times 10^{21}$ atoms, each with 29 electrons, so $2.75 \times 10^{23}$ electrons in total. Strip away one electron in every $10^{9}$ and the sphere is left with $2.75 \times 10^{14}$ elementary charges, which is $4.40 \times 10^{- 5}$ C. Two such spheres a metre apart repel with $k q^{2} / r^{2} = 17.4$ N, about the weight of a 1.8 kg brick, from removing a billionth of the available electrons from a gram of metal. Matter is neutral to a staggering precision because it has no choice: the smallest imbalance produces forces that undo it.

The comparison with gravity is the sharpest statement of this. In a hydrogen atom, with the electron at the Bohr radius of 52.9 pm, the electrical attraction is

F_{e} = \frac{k e^{2}}{r^{2}} = 8.24 \times 10^{- 8} N

while the gravitational attraction between the same two particles is $G m_{e} m_{p} / r^{2} = 3.63 \times 10^{- 47}$ N. The ratio is $2.27 \times 10^{39}$ , and note that it does not depend on $r$ , since both fall as the inverse square: it is $k e^{2} / (G m_{e} m_{p})$ , a pure number built from constants. Gravity is weaker than electricity by thirty-nine orders of magnitude.

Which raises the obvious question of why gravity, and not electricity, shapes the solar system. The answer is the two signs. Gravity has only one, so mass accumulates and its effect grows without limit; charge has two, so any large accumulation attracts its own neutraliser and is cancelled. The Sun holds about $10^{57}$ protons and very nearly exactly $10^{57}$ electrons, and the residue is what matters. Astronomy is gravitational not because gravity is strong but because electricity is so strong that it is never left over.

How well do we know the exponent

Writing $F \propto r^{- 2}$ commits to the exponent being exactly 2, and that is a strong claim about a measured quantity. Coulomb's own data would support $r^{- 2.06}$ as happily as $r^{- 2}$ .

The good bound does not come from measuring forces at all, but from a null experiment described in the third lesson: an inverse square law, and only an inverse square law, implies that the electric field inside a hollow closed conductor is exactly zero. Any charge placed inside must then move entirely to the outside surface. Henry Cavendish did this in 1773 with concentric spheres and, finding no charge inside, concluded the exponent was $2 \pm 0.02$ . Maxwell repeated it with better instruments in 1873 and tightened the bound to about one part in twenty thousand. The modern version, done by Williams, Faller and Hill in 1971 with a five-shell apparatus and a lock-in amplifier, writes the law as $F \propto r^{- (2 + q)}$ and reports $| q | < 3 \times 10^{- 16}$ .

What is being tested is more than a curve fit. In modern theory the exponent is exactly 2 if and only if the photon has exactly zero rest mass, and a massive photon would give a force falling off exponentially with a range set by that mass. The null result therefore converts into a bound on the photon mass, currently below about $10^{- 18}$ electronvolts in mass-energy, which is roughly $10^{- 54}$ kilograms. This is one of the places where a laboratory measurement in a basement constrains a fundamental constant of the universe.

Example. Two identical positive charges are fixed on the $x$ axis, $+ 9.0$ μC at $x = 0$ and $+ 4.0$ μC at $x = 1.00$ m. Where on the line between them does a test charge feel no net force?

Let the point be at distance $r$ from the 9 μC charge, so it is $1.00 - r$ from the other. Setting the magnitudes equal, $k (9) q / r^{2} = k (4) q / (1 - r)^{2}$ , and both $k$ and the test charge $q$ cancel, which is why the answer does not depend on either. Taking the square root of both sides gives $3 / r = 2 / (1 - r)$ , so $3 - 3 r = 2 r$ and $r = 0.60$ m. The null point sits closer to the smaller charge, as it must.

Now you. Two positive charges are fixed on the $x$ axis, $+ 16$ μC at $x = 0$ and $+ 4.0$ μC at $x = 1.50$ m. Where between them is the net force on a test charge zero?

Answer

$16 / r^{2} = 4 / (1.5 - r)^{2}$ , so $4 / r = 2 / (1.5 - r)$ , giving $6 - 4 r = 2 r$ and $r = 1.00$ m from the 16 μC charge, that is 0.50 m from the 4.0 μC charge.

What the law does not say

Coulomb's law is a complete account of what two charges do to each other, and it contains a silence at its centre. Nothing in $F = k q_{1} q_{2} / r^{2}$ says how the second charge learns that the first one is there, or across what, or how quickly. Written as it stands, the force depends on the separation now, which means that moving one charge would change the force on the other instantaneously, at any distance. That is action at a distance, and Newton, who had the same problem with gravity, called the idea so great an absurdity that no competent thinker could fall into it, while continuing to use it because it worked.

The repair, developed by Faraday in the 1830s and made mathematical by Maxwell, is to split the interaction into two steps. A charge fills the space around it with a condition, and a second charge responds to the condition where it sits, knowing nothing about what produced it. The condition is the electric field, and it will turn out to have energy, to carry momentum, to take a definite time to propagate, and eventually to exist entirely on its own with no charge anywhere near it. That last case is light. Building the field is the next lesson.

The electric field

Coulomb's law says what two charges do to each other and leaves an embarrassment behind it, which is that neither charge has any way of knowing the other is there.

The previous lesson ended on that point. This one takes Michael Faraday's way out, which was to insist that the space between the charges is not empty but is in a state, and that the state is the real object of study. The move looks at first like bookkeeping, and for static charges it is. It stops being bookkeeping in the ninth lesson, where a field turns up with no charge responsible for it, and it stops being anything but the whole subject in the twelfth.

Dividing out the test charge

Put a charge $q_{0}$ at some point near a fixed charge $Q$ and it feels a force $F = k Q q_{0} / r^{2}$ . Double $q_{0}$ and the force doubles. Halve it and the force halves. The force therefore contains a factor that has nothing to do with $Q$ or with the geometry, and dividing it out leaves something that depends only on the source and the point:

E = \frac{F}{q_{0}}

That is the electric field, and its units are newtons per coulomb, later shown to be the same thing as volts per metre. It is a vector at every point of space, which is what makes it a field rather than a number: a temperature field assigns a scalar to every point, and this assigns an arrow.

Two cautions come with the definition. The first is that $q_{0}$ has to be small. A real test charge repels the charges on the conductors producing the field, redistributing them and changing the very thing being measured, so the definition is properly $E = lim_{q_{0} \to 0} F / q_{0}$ . The limit is a fiction, since charge is quantised and cannot go below $e$ , but it is a harmless one: laboratory charges are $10^{12}$ elementary charges or more, so there is plenty of room to be small without being zero.

The second is that the field is defined by the force on a positive charge. Place a negative charge in a field and the force is opposite to $E$ , which is a permanent minor nuisance and the source of a great many sign errors.

For a point charge $Q$ the field follows at once by dividing Coulomb's law by $q_{0}$ :

E = \frac{k Q}{r^{2}}

pointing away from $Q$ if $Q$ is positive and towards it if negative. A charge of 1 nC produces 8.99 N/C at a metre. The proton in a hydrogen atom produces $k e / r^{2} = 5.15 \times 10^{11}$ N/C at the Bohr radius of 52.9 pm, which is why atomic physics needs no external fields to be interesting: nothing built in a laboratory comes within five orders of magnitude of that.

What the field is for

At this stage the field is a repackaging, and it is worth being honest that nothing has been explained yet. $E = F / q_{0}$ gives the same answers as Coulomb's law because it was derived from Coulomb's law by dividing.

The reason to prefer it is that the two-step story it tells, source makes field, field pushes charge, survives when the one-step story does not. If the source charge is jiggled, the force on a distant charge does not change instantly; the change propagates outward at a finite speed, and during the transit the field near the distant charge is what it was, still pushing. Something has to hold the information in the interval, and that something is the field. By the end of this course the field will be carrying energy and momentum in its own right, and a light wave will be a piece of field that has detached from its source entirely and no longer cares where it came from.

Superposition, and the picture

Fields add, because forces add. The field of several charges is the vector sum of the fields each would produce alone, and the fact that the sum is unaffected by the presence of the other charges is the linearity noted in the previous lesson.

Faraday's own way of seeing a field was lines of force: curves drawn everywhere tangent to $E$ , starting on positive charges and ending on negative ones. The convention that makes them quantitative is that the number of lines through a small area perpendicular to them is proportional to the field strength there, so lines crowd where the field is strong. Around an isolated point charge the lines are radial spokes, and the number crossing a sphere of radius $r$ is fixed while the sphere's area grows as $r^{2}$ , so the density falls as $1 / r^{2}$ , which is the field. The picture has the inverse square built into the geometry of three dimensions.

Two things the lines cannot do are worth stating, because both are commonly assumed. They never cross: the field at a point has one direction, so two tangents cannot pass through the same place. And they are not trajectories. A charge released in a field accelerates along the field, but it then has velocity, and its subsequent path curves away from the line, exactly as a projectile does not follow the direction of gravity.

Example. A charge $q_{1} = + 4.0$ nC sits at $x = 0$ and $q_{2} = - 6.0$ nC at $x = 8.0$ cm. What is the field at the midpoint, $x = 4.0$ cm?

Each is 4.0 cm from the point. The positive charge pushes a test charge away from itself, that is in the $+ x$ direction, with $E_{1} = (8.9876 \times 10^{9}) (4.0 \times 10^{- 9}) / (0.040)^{2} = 2.25 \times 10^{4}$ N/C. The negative charge pulls a test charge towards itself, which from the midpoint is also the $+ x$ direction, with $E_{2} = (8.9876 \times 10^{9}) (6.0 \times 10^{- 9}) / (0.040)^{2} = 3.37 \times 10^{4}$ N/C. They point the same way, so $E = 5.62 \times 10^{4}$ N/C in the $+ x$ direction.

Now you. A charge $q_{1} = + 3.0$ nC sits at $x = 0$ and $q_{2} = + 5.0$ nC at $x = 10.0$ cm. What is the field at the midpoint, $x = 5.0$ cm?

Answer

Both are positive, so their fields at the midpoint oppose. From $q_{1}$ : $(8.9876 \times 10^{9}) (3.0 \times 10^{- 9}) / (0.050)^{2} = 1.08 \times 10^{4}$ N/C in $+ x$ . From $q_{2}$ : $(8.9876 \times 10^{9}) (5.0 \times 10^{- 9}) / (0.050)^{2} = 1.80 \times 10^{4}$ N/C in $- x$ . The net field is $7.19 \times 10^{3}$ N/C in the $- x$ direction, towards the smaller charge.

Continuous charge, and the integral it forces

Real charged objects are not points. A charged rod holds something like $10^{13}$ elementary charges spread along it, and adding $10^{13}$ vectors is not a plan. The standard move is to go the other way and treat the charge as a continuum, cut it into pieces $d q$ small enough to count as points, and integrate.

Three densities cover the cases. A line carries $λ$ coulombs per metre, so $d q = λ d s$ . A surface carries $σ$ coulombs per square metre, so $d q = σ d A$ . A volume carries $ρ$ coulombs per cubic metre, so $d q = ρ d V$ . In each case the contribution of one piece to the field at the point of interest is $d E = k d q / r^{2}$ directed from the piece to the point, and the total is the vector sum, which means integrating the components separately.

The component bookkeeping is the part that makes these integrals painful, and the part that symmetry usually rescues. The uniformly charged ring is the cleanest illustration.

Take a ring of radius $a$ carrying total charge $Q$ spread evenly, and ask for the field at a point on its axis a distance $x$ from the centre. Every piece $d q$ of the ring is the same distance $\sqrt{x^{2} + a^{2}}$ from that point, so each contributes $k d q / (x^{2} + a^{2})$ . The directions differ, but for each piece there is a piece diametrically opposite whose contribution has the same axial component and the opposite transverse one, so the transverse components cancel in pairs and only the axial survives. The axial fraction of each contribution is $\cos θ = x / \sqrt{x^{2} + a^{2}}$ , so

E = \int \frac{k d q}{x^{2} + a^{2}} \cdot \frac{x}{\sqrt{x^{2} + a^{2}}} = \frac{k Q x}{(x^{2} + a^{2})^{3 / 2}}

where everything except $d q$ came out of the integral because it is the same for every piece, and $\int d q = Q$ .

Read the answer at its two extremes, which is always worth doing. At the centre, $x = 0$ , the field is zero, as symmetry demands: every piece is opposed by the piece across the ring. Far away, $x ≫ a$ , the $a^{2}$ is negligible and $E \to k Q / x^{2}$ , which is a point charge, as it must be. In between the field rises, peaks, and falls, and differentiating shows the peak sits at $x = a / \sqrt{2}$ .

Example. A ring of radius 5.0 cm carries 20 nC spread uniformly. What is the field on its axis 12 cm from the centre?

$x^{2} + a^{2} = (0.12)^{2} + (0.05)^{2} = 0.0144 + 0.0025 = 0.0169$ m², whose square root is exactly 0.13 m, so $(x^{2} + a^{2})^{3 / 2} = {0.13}^{3} = 2.197 \times 10^{- 3}$ . Then $E = (8.9876 \times 10^{9}) (20 \times 10^{- 9}) (0.12) / (2.197 \times 10^{- 3}) = 9.82 \times 10^{3}$ N/C, directed along the axis away from the ring.

Now you. A ring of radius 4.0 cm carries 15 nC. What is the field on its axis 3.0 cm from the centre?

Answer

$x^{2} + a^{2} = 0.0009 + 0.0016 = 0.0025$ , so the square root is 0.050 and the cube of that is $1.25 \times 10^{- 4}$ . Then $E = (8.9876 \times 10^{9}) (15 \times 10^{- 9}) (0.030) / (1.25 \times 10^{- 4}) = 3.24 \times 10^{4}$ N/C.

The line of charge, and a first infinity

Now a straight rod of length $2 L$ carrying $λ$ per metre, with the field wanted at a perpendicular distance $y$ from its midpoint. Put the rod along the $z$ axis from $- L$ to $+ L$ . A piece at height $z$ is $\sqrt{y^{2} + z^{2}}$ away and contributes $k λ d z / (y^{2} + z^{2})$ , of which the fraction $y / \sqrt{y^{2} + z^{2}}$ points along the perpendicular. Components along the rod cancel between $+ z$ and $- z$ , so

E = \int_{- L}^{L} \frac{k λ y d z}{(y^{2} + z^{2})^{3 / 2}} = \frac{2 k λ L}{y \sqrt{y^{2} + L^{2}}}

using the standard integral $\int d z / (y^{2} + z^{2})^{3 / 2} = z / (y^{2} \sqrt{y^{2} + z^{2}})$ , which can be checked by differentiating the right side.

Two limits again. For $y ≫ L$ the rod looks like a point of charge $2 λ L$ and the expression collapses to $k (2 λ L) / y^{2}$ . For $L ≫ y$ , meaning very close to a long rod, $\sqrt{y^{2} + L^{2}} \to L$ and

E = \frac{2 k λ}{y} = \frac{λ}{2 π ε_{0} y}

which falls off as $1 / y$ rather than $1 / y^{2}$ . That is not a failure of Coulomb's law. It is what happens when moving away from a line brings more of the line into view at a shallow angle, partly compensating the loss. The same logic will shortly give a field that does not fall off at all.

An infinite rod does not exist, and the result is still the one to use in practice, because "infinite" here means only $L ≫ y$ . At 1 cm from a 1 m rod the finite formula and the infinite one differ by about 0.02 per cent.

The dipole

Two equal and opposite charges $+ q$ and $- q$ separated by a small distance $d$ form an electric dipole, the most important arrangement in the subject after the point charge, because a neutral molecule with its charge slightly off-centre is one. Water is the standard example, with a dipole moment of $6.2 \times 10^{- 30}$ coulomb metres, and essentially all of the chemistry of water follows from it.

Define the dipole moment $p = q d$ , treated as a vector pointing from the negative charge to the positive one. On the axis, at a distance $r$ from the centre with $r ≫ d$ , the two fields nearly cancel. Writing them out, $E = k q / (r - d / 2)^{2} - k q / (r + d / 2)^{2}$ , and expanding each denominator to first order in $d / r$ gives

E = \frac{2 k p}{r^{3}}

The cancellation costs a power of $r$ : a dipole's field dies as the inverse cube, faster than a point charge's, because at a distance the two charges are increasingly well disguised as nothing. That single fact explains why neutral matter can be handled at all. If molecules produced $1 / r^{2}$ fields, every object in a room would be electrically screaming at every other.

Put a dipole in a uniform external field $E$ and the two forces, $+ q E$ and $- q E$ , cancel exactly, so there is no net force. There is a net torque, since the forces act at different places, of magnitude

τ = p E \sin θ

with $θ$ the angle between $p$ and $E$ . The torque turns the dipole into alignment with the field, which is why a charged rod attracts an uncharged scrap of paper: the field polarises the paper, aligning and slightly separating its molecular charges, and then, because the field is not uniform near the rod, the induced near end feels a stronger pull than the far end repels. The attraction of neutral matter to charge is a second-order effect requiring a non-uniform field, and it is the reason the very first observation in the subject, amber picking up chaff, took two thousand years to explain.

Motion in a field

Once $E$ is known, mechanics takes over: the force on a charge is $q E$ , and the rest is Newton's second law. In a uniform field the acceleration is constant, so the constant-acceleration results from mechanics apply unchanged, and a charged particle fired across a uniform field follows a parabola for exactly the reason a projectile does.

Example. An electron starts from rest and is accelerated through 1.0 cm by a uniform field of $1.50 \times 10^{4}$ N/C. How fast is it going?

The force is $e E = (1.602 \times 10^{- 19}) (1.50 \times 10^{4}) = 2.40 \times 10^{- 15}$ N, so the acceleration is $a = e E / m_{e} = 2.40 \times 10^{- 15} / 9.109 \times 10^{- 31} = 2.64 \times 10^{15}$ m/s². With $v^{2} = 2 a d$ and $d = 0.010$ m, $v^{2} = 5.28 \times 10^{13}$ and $v = 7.26 \times 10^{6}$ m/s. That is 2.4 per cent of the speed of light, so treating it non-relativistically is good to about a tenth of a per cent, and a slightly larger field would make that excuse fail.

Now you. An electron starts from rest and is accelerated through 2.5 cm by a uniform field of $8.0 \times 10^{3}$ N/C. How fast is it going?

Answer

$a = e E / m_{e} = (1.602 \times 10^{- 19}) (8.0 \times 10^{3}) / (9.109 \times 10^{- 31}) = 1.41 \times 10^{15}$ m/s². Then $v^{2} = 2 a d = 2 (1.41 \times 10^{15}) (0.025) = 7.04 \times 10^{13}$ , so $v = 8.39 \times 10^{6}$ m/s.

Gravity does not appear in that calculation, and the omission is not laziness. The field needed to hold an electron up against its own weight is $m_{e} g / e = 5.6 \times 10^{- 11}$ N/C, which is eleven orders of magnitude below any field a laboratory can avoid producing by accident. In charged-particle problems, gravity is noise.

Where the integrals run out

The method of this lesson is complete in principle: any static charge distribution, cut into pieces, integrated. In practice it collapses almost immediately. The ring was easy because every piece sat at the same distance. Move off the axis and the same ring gives an integral with no expression in elementary functions. A uniformly charged sphere, the single most important case in the subject, can be done this way, but it takes a page of unpleasant work with the cosine rule.

What rescues it is that the difficulty is in the wrong place. The integral is hard because it tracks where the field points, and yet the ring, the sphere and the infinite line all have enough symmetry that the direction of the field is obvious before any calculation begins. A method that used the symmetry first and asked only for the magnitude would be doing a fraction of the work.

Such a method exists, it takes a page to derive, it converts Coulomb's inverse square into a statement about closed surfaces, and it solves the sphere in three lines. It is Gauss's law, and it is next.

Flux and Gauss's law

The integrals at the end of the previous lesson were hard for a reason that has nothing to do with the physics, which is that they tracked the direction of a field whose direction symmetry had already settled.

This lesson builds the tool that exploits that. It needs the field of a point charge, $E = k Q / r^{2}$ , and nothing else from what has gone before. The result is the first of the four equations that will eventually be called Maxwell's, and it is worth saying now that it is not a new law: it is Coulomb's law rewritten, and every step of the rewriting is reversible.

Flux

Imagine the field as something streaming through space, and ask how much of it crosses a given surface. For a uniform field $E$ crossing a flat area $A$ face on, the natural measure is the product $E A$ . Tilt the surface by an angle $θ$ away from face on and the field crosses a smaller effective area, $A \cos θ$ , so the measure becomes $E A \cos θ$ . The quantity

Φ_{E} = E A \cos θ

is the electric flux, in units of newton metres squared per coulomb. Nothing is actually flowing; the name is inherited from fluid mechanics, where the same integral counts real litres per second, and the analogy is close enough to be useful and loose enough to be worth distrusting.

For a curved surface in a varying field, chop it into patches small enough to be flat and uniform, and add:

Φ_{E} = \int E \cos θ d A

The angle $θ$ is measured from the normal to the patch, which for a closed surface is taken to point outwards by convention. With that convention flux is signed: field leaving the surface counts positive, field entering counts negative, and a field that goes in one side and out the other contributes zero net.

From Coulomb to Gauss

Put a point charge $q$ at the centre of a sphere of radius $r$ . Two features make this the easiest surface in physics: the field has the same magnitude $k q / r^{2}$ everywhere on it, and it is everywhere perpendicular to it, so $\cos θ = 1$ . The flux is then the field times the area:

Φ_{E} = \frac{k q}{r^{2}} \times 4 π r^{2} = 4 π k q = \frac{q}{ε_{0}}

using $k = 1 / (4 π ε_{0})$ . This is the payoff promised in the first lesson for carrying $4 π$ around in Coulomb's law: the $r^{2}$ of the inverse square cancels against the $r^{2}$ of the sphere's area, and the $4 π$ cancels too, leaving a result with no geometry in it whatsoever. The flux does not depend on the radius. Doubling the sphere quarters the field and quadruples the area.

That cancellation is a fact about three dimensions and the exponent 2, and it is the whole content of what follows. Now generalise in two steps.

First, the surface need not be a sphere. Take any closed surface around the charge and imagine a narrow cone of field lines spreading out from the charge. It crosses the sphere in a patch and the odd surface in some other patch, further away by a factor $s$ and tilted by some angle. Being further reduces the field by $s^{2}$ ; the larger patch that the cone cuts, and the tilt, together increase the effective area by exactly $s^{2}$ . The flux through the two patches is identical. Since every part of the odd surface is covered by some such cone, the total flux is the same as for the sphere.

Second, a charge outside the surface contributes nothing. A cone of lines from an external charge pierces the closed surface an even number of times, entering and leaving alternately, and the entering and leaving contributions have opposite signs and equal magnitudes by the same argument as before. They cancel in pairs.

Put the two together, add up over however many charges there are (superposition again), and the result is Gauss's law:

Φ_{E} = \int E \cos θ d A = \frac{q_{enc}}{ε_{0}}

The flux through any closed surface equals the charge enclosed divided by $ε_{0}$ . Charges outside the surface still contribute to $E$ at every point of it; what they do not contribute to is the total.

Example. A charge of $3.0$ μC sits at the centre of a cube. What is the flux through one face?

Gauss's law gives the total flux out of the whole cube as $q / ε_{0} = 3.0 \times 10^{- 6} / 8.854 \times 10^{- 12} = 3.39 \times 10^{5}$ N m² C⁻¹. The cube has six faces and the charge sits symmetrically, so each face takes an equal share: $Φ = 5.65 \times 10^{4}$ N m² C⁻¹. Doing this by integrating the field over a square face is a genuinely unpleasant calculation, and symmetry has just replaced it with a division by six.

Now you. A charge of $8.0$ μC sits at the centre of a cube. What is the flux through one face?

Answer

Total flux is $q / ε_{0} = 8.0 \times 10^{- 6} / 8.854 \times 10^{- 12} = 9.04 \times 10^{5}$ N m² C⁻¹, and one sixth of that is $1.51 \times 10^{5}$ N m² C⁻¹.

Using it: three classic distributions

Gauss's law is always true and only sometimes useful. It becomes a method when the symmetry is good enough that $E$ can be pulled out of the integral, which requires a surface on which the field magnitude is constant and the angle is either 0 or 90 degrees. Three symmetries qualify: spherical, cylindrical and planar. Learning to see which one applies is most of the skill.

A uniformly charged sphere. Take total charge $Q$ spread evenly through a ball of radius $R$ . Outside it, choose a spherical surface of radius $r > R$ . By symmetry the field is radial and constant in magnitude on that surface, so $E \times 4 π r^{2} = Q / ε_{0}$ and

E = \frac{Q}{4 π ε_{0} r^{2}} = \frac{k Q}{r^{2}}

Outside, the ball is indistinguishable from a point charge at its centre. That is the electrical version of the shell theorem from mechanics, which Newton needed a geometrical tour de force to prove; here it takes two lines, and it is the theorem that licensed treating charged spheres as points in the first lesson.

Inside, at radius $r < R$ , the enclosed charge is only the fraction $r^{3} / R^{3}$ of the total, so $E \times 4 π r^{2} = Q r^{3} / (ε_{0} R^{3})$ and $E = k Q r / R^{3}$ . The field rises linearly from zero at the centre to $k Q / R^{2}$ at the surface, then falls as the inverse square outside. Both expressions agree at $r = R$ , which is the check worth doing every time.

A long line. For a line with $λ$ coulombs per metre, take a cylinder of radius $r$ and length $L$ coaxial with it. No flux passes through the flat ends, since the field is radial and runs parallel to them; the curved side has area $2 π r L$ with the field perpendicular to it everywhere. So $E (2 π r L) = λ L / ε_{0}$ , and the $L$ cancels:

E = \frac{λ}{2 π ε_{0} r}

which is the result the previous lesson obtained by integrating over the rod and taking a limit. Compare the labour.

A large flat sheet. For a sheet carrying $σ$ coulombs per square metre, take a small cylinder poking through it, with flat ends of area $A$ parallel to the sheet on either side. By symmetry the field points straight out of the sheet on both sides, so nothing crosses the curved wall and each end contributes $E A$ . The enclosed charge is $σ A$ , so $2 E A = σ A / ε_{0}$ and

E = \frac{σ}{2 ε_{0}}

with no $r$ in it at all. The field of an infinite sheet does not fall off with distance. This is the limiting case of the pattern noticed for the line: the more of the source stays in view as you retreat, the slower the decline, and a plane keeps all of it in view. It is also the reason a parallel plate capacitor, two such sheets with opposite charge, has a uniform field between its plates, which is the subject of the fifth lesson.

Example. A ball of radius 10.0 cm carries 50 nC spread uniformly through its volume. Find the field at 5.0 cm and at 20.0 cm from the centre.

At 5.0 cm we are inside, so $E = k Q r / R^{3} = (8.9876 \times 10^{9}) (50 \times 10^{- 9}) (0.050) / (0.100)^{3} = 2.25 \times 10^{4}$ N/C. At 20.0 cm we are outside, so the ball counts as a point: $E = k Q / r^{2} = (8.9876 \times 10^{9}) (50 \times 10^{- 9}) / (0.200)^{2} = 1.12 \times 10^{4}$ N/C. The field at the surface, from either formula, is $4.49 \times 10^{4}$ N/C, and it is the maximum.

Now you. A ball of radius 6.0 cm carries 24 nC spread uniformly. Find the field at 3.0 cm and at 12.0 cm from the centre.

Answer

Inside: $E = k Q r / R^{3} = (8.9876 \times 10^{9}) (24 \times 10^{- 9}) (0.030) / (0.060)^{3} = 3.00 \times 10^{4}$ N/C. Outside: $E = k Q / r^{2} = (8.9876 \times 10^{9}) (24 \times 10^{- 9}) / (0.120)^{2} = 1.50 \times 10^{4}$ N/C.

What a conductor does

A conductor contains charges free to move through it. Wait long enough after any disturbance and the charges stop moving, which is electrostatic equilibrium, and that single condition forces several consequences that Gauss's law makes quick.

If charges have stopped moving, the field inside the conducting material must be zero. Any field there would push the free charges, and they would still be moving. This is not a property of the metal but a consequence of equilibrium: the charges rearrange themselves precisely until their own field cancels whatever was applied, and the arrangement that achieves it is what equilibrium means.

Now take any closed surface drawn entirely within the metal. The field is zero everywhere on it, so the flux is zero, so the enclosed charge is zero. Since the surface can be drawn anywhere, all excess charge on a conductor sits on its surface. Shrink the surface to hug the outer boundary and the same argument says any net charge lives in a layer at most a few atoms deep.

Just outside the surface, take a small flat pillbox with one end in the metal and one end just outside. The inner end contributes nothing since the field there is zero, the walls contribute nothing since the field just outside a conductor is perpendicular to it (any parallel component would drive a surface current), and the outer end contributes $E A$ . So

E = \frac{σ}{ε_{0}}

just outside a conductor, twice the field of an isolated sheet with the same charge density, because a conductor's charge produces field on one side only.

Hollow out the conductor and put nothing inside, and Gauss's law applied to a surface in the metal surrounding the cavity gives zero enclosed charge, and a further argument rules out equal and opposite charges on the cavity wall: the field in the cavity is exactly zero regardless of what happens outside. That is the Faraday cage, and it is why a car is a reasonable place to be in a thunderstorm and why sensitive instruments live inside metal boxes. Faraday made the point in 1843 with an ice pail: a charged sphere lowered inside a metal container induced exactly its own charge on the outer surface, and touching it to the inside transferred the charge completely, leaving the sphere neutral.

Put a charge $q$ inside the cavity, however, and the enclosed-charge argument runs the other way: a surface in the metal encloses $q$ plus whatever is on the cavity wall, and the flux is zero, so the cavity wall must carry exactly $- q$ . Charge conservation then puts $+ q$ on the outer surface, and it distributes itself according to the outer shape alone, knowing nothing about where the charge sits inside.

Example. A charge of $+ 4.0$ nC sits at the centre of a hollow, uncharged conducting shell with inner radius 5.0 cm and outer radius 8.0 cm. What charge sits on each surface, and what is the field at 10.0 cm from the centre?

A Gaussian sphere drawn inside the metal, say at 6.5 cm, encloses the central charge plus the inner surface and must enclose zero net charge, so the inner surface carries $- 4.0$ nC. The shell as a whole is uncharged, so the outer surface carries $+ 4.0$ nC. At 10.0 cm the enclosed charge is $+ 4.0 - 4.0 + 4.0 = + 4.0$ nC, giving $E = k q / r^{2} = (8.9876 \times 10^{9}) (4.0 \times 10^{- 9}) / (0.100)^{2} = 3.60 \times 10^{3}$ N/C, pointing outwards. The shell is invisible from outside.

Now you. A charge of $- 6.0$ nC sits at the centre of a hollow conducting shell that itself carries a net charge of $+ 2.0$ nC. What charge sits on each surface, and what is the field at 15.0 cm from the centre?

Answer

The inner surface must cancel the central charge, so it carries $+ 6.0$ nC. The shell's total is $+ 2.0$ nC, so the outer surface carries $2.0 - 6.0 = - 4.0$ nC. At 15.0 cm the enclosed charge is $- 4.0$ nC, so $E = (8.9876 \times 10^{9}) (4.0 \times 10^{- 9}) / (0.150)^{2} = 1.60 \times 10^{3}$ N/C, pointing inwards.

The null experiment

The shielding result is the sharpest test of Coulomb's law there is, and the reason was flagged at the end of the first lesson. Every step from the inverse square to "the field inside a closed conductor is zero" used the exact cancellation of $r^{2}$ against $r^{2}$ . If the exponent were $2 + q$ for any $q$ other than zero, the cancellation would be imperfect, flux would depend on the shape and size of the surface, and a charge would appear inside a hollow conductor.

That converts a hard measurement into an easy one. Instead of measuring a force to one part in $10^{16}$ , which is impossible, charge a hollow conductor heavily and look inside for any charge at all, which needs only a sensitive electrometer and a good null. Cavendish did it in 1773 with concentric spheres and got $| q | < 0.02$ . Maxwell, who edited and published Cavendish's unpublished papers, repeated it in 1873 and reached about $5 \times 10^{- 5}$ . Williams, Faller and Hill in 1971 used five nested shells, an alternating potential of 10 kV and a lock-in amplifier, and reached $3 \times 10^{- 16}$ .

Null experiments are worth noticing as a technique. They convert a question about the value of a quantity into a question about whether something is exactly zero, and zero is the one number that can be measured to arbitrary precision, because the answer does not depend on knowing any calibration.

What Gauss's law does not do

The law is completely general and it is not a general method. Given an arbitrary blob of charge, Gauss's law is still true, and it gives one equation for a field that varies over the surface in an unknown way, which is one equation and infinitely many unknowns. Symmetry is what collapses the unknowns to one, and only three symmetries do it.

It is also worth being clear about what enclosed means. The flux depends only on the charge inside, and the field at each point of the surface depends on every charge in the universe. Bring an external charge near a Gaussian sphere and the field on the near side rises while the far side falls, in such a way that the total is untouched. Students often read Gauss's law as saying the outside charge has no effect, and it says something much narrower and much stranger than that.

The deeper limitation is that this lesson has produced a vector field from a vector integral, and the second lesson's complaint stands: adding vectors is laborious, and for a distribution with no symmetry neither method helps. There is a third route, which uses the fact that the electric force, being an inverse square central force, is conservative in exactly the sense mechanics gave that word. That makes the whole field derivable from a single scalar function, and scalars add without components. It is the potential, and it is next.

Electric potential

Adding vectors is laborious, and the previous two lessons did a great deal of it for fields that a single number at each point could have described just as well.

The escape is a result already proved in mechanics: a force directed along the line to a fixed centre, with a magnitude depending only on distance from it, is conservative. The work it does between two points is independent of the route taken, which means a potential energy exists. Coulomb's force is exactly of that form, so everything mechanics says about gravitational potential energy carries across, with two changes: the force can be repulsive as well as attractive, and it is convenient to divide out the charge being moved.

The work is path independent

Take a fixed charge $Q$ at the origin and carry a test charge $q_{0}$ from a point at radius $r_{a}$ to one at radius $r_{b}$ along any path at all. Break the path into steps. Each step splits into a piece directly along the radius and a piece perpendicular to it, running along a sphere of constant radius. The force is radial, so it does no work at all on the perpendicular pieces: only the radial pieces count, and what they add up to is the net change in $r$ , whatever wandering happened in between.

So the work reduces to a single integral over $r$ . For like charges the force on the test charge points outward with magnitude $k Q q_{0} / r^{2}$ , and the work it does as the radius goes from $r_{a}$ to $r_{b}$ is

W = \int_{r_{a}}^{r_{b}} \frac{k Q q_{0}}{r^{2}} d r = k Q q_{0} (\frac{1}{r_{a}} - \frac{1}{r_{b}})

The answer depends only on the endpoints. Define the potential energy in the usual way, as minus the work done by the force, with the zero placed at infinite separation where the force vanishes:

U = \frac{k Q q_{0}}{r}

For like charges $U$ is positive and falls as they separate, which says that bringing them together costs work and that they will fly apart if released. For unlike charges $U$ is negative, which says the pair is bound: energy must be supplied to separate them to infinity.

Divide by the test charge and what remains belongs to the source and the point, exactly as dividing force by charge gave the field. That is the electric potential:

V = \frac{U}{q_{0}} = \frac{k Q}{r}

measured in joules per coulomb, called volts. A charge of 1 nC produces 8.99 V at a metre. The volt is the unit almost every practical measurement in electricity is made in, and it is worth registering that it is an energy per charge, not a force, not a field, and not a quantity of anything.

Why a scalar is worth having

The gain is immediate. Potentials from several charges add as ordinary numbers:

V = k \sum_{i} \frac{q_{i}}{r_{i}}

with the sign of each charge carried along and no components, no angles and no resolution into axes. For a continuous distribution the sum becomes $V = k \int d q / r$ , which is a scalar integral where the second lesson needed a vector one.

The ring is the case to compare. Every piece of a ring of radius $a$ carrying total charge $Q$ is the same distance $\sqrt{x^{2} + a^{2}}$ from a point on the axis, so

V = \frac{k Q}{\sqrt{x^{2} + a^{2}}}

with no cancellation argument, no cosine, and no worry about which way anything points. Compare the work needed for the field in the second lesson. The direction was recoverable all along, as the next section shows.

Example. A charge of $+ 5.0$ nC sits at the origin and a charge of $- 3.0$ nC at $x = 4.0$ cm. What is the potential at the point 3.0 cm directly above the origin?

That point is 3.0 cm from the first charge and, by Pythagoras on the 3, 4, 5 triangle, 5.0 cm from the second. Potentials add as numbers: $V = k (5.0 \times 10^{- 9} / 0.030) + k (- 3.0 \times 10^{- 9} / 0.050) = (8.9876 \times 10^{9}) (1.667 \times 10^{- 7} - 6.00 \times 10^{- 8}) = 959$ V. No geometry beyond the two distances was needed, which is the entire point.

Now you. A charge of $+ 4.0$ nC sits at the origin and a charge of $+ 6.0$ nC at $x = 8.0$ cm. What is the potential at the point 6.0 cm directly above the origin?

Answer

The distances are 6.0 cm and, from the 6, 8, 10 triangle, 10.0 cm. So $V = (8.9876 \times 10^{9}) (4.0 \times 10^{- 9} / 0.060 + 6.0 \times 10^{- 9} / 0.100) = (8.9876 \times 10^{9}) (1.267 \times 10^{- 7}) = 1138$ V.

Getting the field back

Potential would be a dead end if the field could not be recovered from it. It can, by differentiating.

Move a test charge a small distance $d s$ in some direction. The work done by the field is $q_{0} E_{s} d s$ , where $E_{s}$ is the component of $E$ along that direction, and by definition that work is $- q_{0} d V$ . Cancelling $q_{0}$ ,

E_{s} = - \frac{d V}{d s}

The field component in any direction is minus the rate of change of potential in that direction. In three dimensions this gives the three components as three partial derivatives, the operation called the gradient, but nothing in this course needs more than the one-dimensional form applied along a chosen axis.

Test it on the ring. Differentiating $V = k Q (x^{2} + a^{2})^{- 1 / 2}$ with respect to $x$ gives $d V / d x = - k Q x (x^{2} + a^{2})^{- 3 / 2}$ , so $E_{x} = k Q x / (x^{2} + a^{2})^{3 / 2}$ , which is exactly what the vector integral produced in the second lesson. One scalar integral and one derivative have replaced a symmetry argument and a component integral.

Two consequences of $E_{s} = - d V / d s$ are worth stating on their own. First, the field points from high potential to low, because of the minus sign, and a positive charge released from rest moves downhill in potential exactly as a mass falls downhill in height. Second, take a direction in which $V$ does not change: the field component along it is zero. The surfaces on which $V$ is constant, called equipotentials, are therefore everywhere perpendicular to the field. Around a point charge they are concentric spheres, and around any conductor in equilibrium they include the conductor's own surface, since the field inside is zero and no work is needed to move a charge anywhere within it. A conductor in electrostatic equilibrium is an equipotential volume, which is why one wire can be said to be at one voltage.

The units also reconcile here: volts per metre is the same as newtons per coulomb, since a joule is a newton metre.

The electron volt

A charge $q$ moved through a potential difference $V$ gains energy $q V$ , and for particles the natural unit is what an elementary charge gains across one volt:

1 eV = 1.602 \times 10^{- 19} J

It is a unit of energy, not of voltage, and it makes atomic numbers legible. The potential energy of the electron and proton in a hydrogen atom, at the Bohr radius of 52.9 pm, is $U = - k e^{2} / r = - 4.36 \times 10^{- 18}$ J, which is $- 27.2$ eV. The electron's kinetic energy is half the magnitude of that, by the virial result that holds for any inverse square orbit, so the total is $- 13.6$ eV, and 13.6 eV is exactly the energy needed to pull the electron off. That is the ionisation energy of hydrogen, measured spectroscopically to seven figures, and it drops out of a potential energy and a fact about orbits.

Example. An alpha particle, charge $+ 2 e$ and mass $6.645 \times 10^{- 27}$ kg, is accelerated from rest through a potential difference of 1.00 kV. How fast is it moving?

The energy gained is $q V = 2 (1.602 \times 10^{- 19}) (1000) = 3.204 \times 10^{- 16}$ J, which is 2000 eV, and all of it appears as kinetic energy. From $\frac{1}{2} m v^{2} = 3.204 \times 10^{- 16}$ , $v^{2} = 2 (3.204 \times 10^{- 16}) / (6.645 \times 10^{- 27}) = 9.64 \times 10^{10}$ , so $v = 3.11 \times 10^{5}$ m/s. Notice that the shape of the accelerating field never entered: only the endpoints matter, which is what path independence buys.

Now you. A proton, mass $1.673 \times 10^{- 27}$ kg, is accelerated from rest through 2.50 kV. How fast is it moving?

Answer

Energy gained is $(1.602 \times 10^{- 19}) (2500) = 4.005 \times 10^{- 16}$ J. Then $v^{2} = 2 (4.005 \times 10^{- 16}) / (1.673 \times 10^{- 27}) = 4.79 \times 10^{11}$ , so $v = 6.92 \times 10^{5}$ m/s.

Potential of a charged conductor, and why points spark

A conducting sphere of radius $R$ carrying charge $Q$ looks, from outside, like a point charge at its centre, so its surface sits at $V = k Q / R$ , and every point inside sits at the same value, since the interior field is zero. The potential does not drop to zero at the centre; it is flat there, which is what a zero field means.

Now connect two conducting spheres of different radii by a long thin wire. They form one conductor, so they must end at one potential, and charge flows until $k Q_{1} / R_{1} = k Q_{2} / R_{2}$ , that is $Q_{1} / Q_{2} = R_{1} / R_{2}$ . The larger sphere takes proportionally more charge. But surface charge density is $Q / 4 π R^{2}$ , so

\frac{σ_{1}}{σ_{2}} = \frac{Q_{1}}{Q_{2}} \cdot \frac{R_{2}^{2}}{R_{1}^{2}} = \frac{R_{2}}{R_{1}}

and the smaller sphere ends up with the higher density, and therefore, since $E = σ / ε_{0}$ just outside a conductor, the stronger field at its surface.

This is the quantitative version of the observation that charge concentrates at sharp points. A sharp region behaves like a small sphere, so it carries a high density and a high local field, and if that field exceeds about $3 \times 10^{6}$ V/m the air ionises and charge leaks away as a corona discharge. It is why high voltage equipment is built with fat rounded conductors and no sharp edges, and why a lightning rod is pointed: it is designed to leak, bleeding charge into the air and providing a preferred path when it does not.

The same number bounds what any isolated conductor can hold. A sphere of radius 10 cm reaches breakdown at $V = E R = (3 \times 10^{6}) (0.10) = 3 \times 10^{5}$ V, so 300 kV is roughly the ceiling for a 10 cm ball in air regardless of how much charge you try to push onto it. Van de Graaff generators get past that only by being large, and by being run in pressurised gas.

Example. Two conducting spheres, of radii 8.0 cm and 2.0 cm, are joined by a long wire and given 20 nC in total. Find the charge on each and the field just outside each.

Equal potentials require $Q_{1} / Q_{2} = R_{1} / R_{2} = 4$ , and the charges must sum to 20 nC, so $Q_{1} = 16$ nC on the large sphere and $Q_{2} = 4.0$ nC on the small one. The fields are $E_{1} = k Q_{1} / R_{1}^{2} = (8.9876 \times 10^{9}) (16 \times 10^{- 9}) / (0.080)^{2} = 2.25 \times 10^{4}$ V/m and $E_{2} = (8.9876 \times 10^{9}) (4.0 \times 10^{- 9}) / (0.020)^{2} = 8.99 \times 10^{4}$ V/m. The small sphere holds a quarter of the charge and carries four times the field, and it is the one that will spark first.

Now you. Two conducting spheres, of radii 9.0 cm and 3.0 cm, are joined by a long wire and given 24 nC in total. Find the charge on each and the field just outside each.

Answer

$Q_{1} / Q_{2} = 3$ , so $Q_{1} = 18$ nC and $Q_{2} = 6.0$ nC. Then $E_{1} = (8.9876 \times 10^{9}) (18 \times 10^{- 9}) / (0.090)^{2} = 2.00 \times 10^{4}$ V/m and $E_{2} = (8.9876 \times 10^{9}) (6.0 \times 10^{- 9}) / (0.030)^{2} = 5.99 \times 10^{4}$ V/m, three times larger.

What potential is not

Three confusions are worth heading off, because each of them survives into later lessons and does damage there.

Potential is not potential energy. $V$ belongs to a point in space whether or not anything is there; $U = q V$ belongs to a charge sitting at that point. The distinction is the same one as between $E$ and $F$ .

Only differences are physical. The choice of zero at infinity is a convention, convenient because it makes $V = k Q / r$ look tidy, and useless for an infinite line or an infinite plane, where the potential at infinity diverges and the zero has to be put somewhere finite instead. Nothing measurable changes: every experiment reports a difference. When a circuit is said to be at 5 V it means 5 V above whatever node was called ground.

And a high potential is not by itself dangerous, nor a low one safe. Shuffling across a carpet routinely puts a person at 10 kV or more relative to the room, and the resulting spark is startling rather than lethal because the total charge, and so the energy $q V$ , is tiny. What harms is energy delivered, and that is a question about how much charge is available and how fast it moves, which is the subject of the next two lessons.

Assembling charge costs energy

One more question closes the electrostatics half of the course. Bringing charges together against their mutual repulsion takes work, and that work must be stored somewhere.

Bring the first charge $q_{1}$ in from infinity to empty space: no other charge exists yet, so no work is done. Bring $q_{2}$ to a distance $r_{12}$ from it, moving against a potential $k q_{1} / r_{12}$ , at a cost of $k q_{1} q_{2} / r_{12}$ . Bring $q_{3}$ in and it must be pushed against the potential of both, costing $k q_{1} q_{3} / r_{13} + k q_{2} q_{3} / r_{23}$ . The total energy of the assembly is the sum over all distinct pairs,

U = k \sum_{pairs} \frac{q_{i} q_{j}}{r_{i j}}

and it does not depend on the order in which they were brought, which is another consequence of path independence.

That is a real, recoverable energy: release the charges and it appears as kinetic energy. Where was it while they sat still? The expression above locates it in the pairs, which is not a place. There is another way to compute the same number that locates it in the space between the charges instead, at a density fixed by the local field strength and nothing else. Working out which description is right, and finding the device that makes the question practical, is the next lesson.

Capacitance and dielectrics

Assembling charge takes work, and the previous lesson left the question of where that work goes while the charges sit still.

Answering it needs a device rather than an argument, because the answer is a claim about energy density and energy density has to be measured somewhere. The device is the capacitor: two conductors, one holding $+ Q$ and the other $- Q$ , with a field between them. It is the simplest arrangement in which a definite amount of electrostatic energy is stored in a definite volume, and it turns the question into arithmetic.

Capacitance

Take any two conductors, isolated from everything else, and move charge $Q$ from one to the other, leaving $+ Q$ on one and $- Q$ on the other. A potential difference $V$ appears between them. Double the charge and, because the field everywhere doubles by superposition and the potential difference is a line integral of the field, $V$ doubles too. The ratio is therefore a constant of the geometry alone:

C = \frac{Q}{V}

This is the capacitance, in coulombs per volt, called farads. Note that $Q$ here means the charge on the positive conductor, not the net charge of the pair, which is zero.

The farad is an absurdly large unit, and seeing why is the fastest way to get a feel for the quantity. Take two parallel plates of area $A$ separated by $d$ , with $d$ small enough that the field between them is uniform and edge effects can be ignored. Gauss's law at the surface of a conductor gives $E = σ / ε_{0} = Q / (ε_{0} A)$ between the plates, and since the field is uniform the potential difference is simply $V = E d = Q d / (ε_{0} A)$ . So

C = \frac{ε_{0} A}{d}

Plates of one square metre a millimetre apart give $C = (8.854 \times 10^{- 12}) (1) / (10^{- 3}) = 8.85$ nF. To reach one farad at that spacing you would need $1.13 \times 10^{8}$ m² of plate, an area of 113 square kilometres. Real capacitors of a farad and more exist, and they get there by making $d$ molecular rather than by making $A$ absurd, which is the trick described at the end of this lesson.

An isolated conductor also has a capacitance, with the second conductor taken to be a sphere at infinity. For a sphere of radius $R$ at potential $k Q / R$ , this gives $C = R / k = 4 π ε_{0} R$ . The Earth, with $R = 6.37 \times 10^{6}$ m, comes out at 709 μF, which is why connecting something to ground is a good way to make its potential stop changing: it is being connected to a large capacitor.

Example. Two parallel plates of area 20 cm² are separated by 0.50 mm of air and connected to a 12 V battery. Find the capacitance, the charge on each plate and the field between them.

$C = ε_{0} A / d = (8.854 \times 10^{- 12}) (20 \times 10^{- 4}) / (5.0 \times 10^{- 4}) = 3.54 \times 10^{- 11}$ F, that is 35.4 pF. The charge is $Q = C V = (3.54 \times 10^{- 11}) (12) = 4.25 \times 10^{- 10}$ C, or 425 pC. The field is uniform, so $E = V / d = 12 / (5.0 \times 10^{- 4}) = 2.4 \times 10^{4}$ V/m, comfortably below the 3 × 10⁶ V/m at which air breaks down.

Now you. Two parallel plates of area 50 cm² are separated by 1.0 mm of air and connected to a 9.0 V battery. Find the capacitance, the charge and the field.

Answer

$C = (8.854 \times 10^{- 12}) (50 \times 10^{- 4}) / (1.0 \times 10^{- 3}) = 4.43 \times 10^{- 11}$ F, or 44.3 pF. Then $Q = C V = 3.98 \times 10^{- 10}$ C, and $E = V / d = 9.0 / 10^{- 3} = 9.0 \times 10^{3}$ V/m.

Combinations

Two capacitors side by side, both connected across the same pair of nodes, are in parallel. Each sees the same $V$ , and the charges add, so $Q = Q_{1} + Q_{2} = (C_{1} + C_{2}) V$ and

C_{parallel} = C_{1} + C_{2}

which is exactly what putting two capacitors side by side does geometrically: it adds their plate areas.

Two capacitors joined end to end, so that the same wire carries charge from one to the next, are in series. Here the argument that matters is about the isolated conductor between them. It started neutral and nothing can reach it, so whatever charge $- Q$ is drawn to one side must leave $+ Q$ on the other: every capacitor in a series chain carries the same charge, regardless of its size. The voltages add, so $V = Q / C_{1} + Q / C_{2}$ and

\frac{1}{C_{series}} = \frac{1}{C_{1}} + \frac{1}{C_{2}}

The series combination is always smaller than either member, which again matches the geometry: putting two capacitors in series is like increasing the separation.

Notice that the rules are the reverse of the ones resistors will follow in the next lesson. It is not worth memorising either; both fall out in two lines from what is shared and what adds.

Example. A 4.0 μF and a 12 μF capacitor are connected in series, and that pair is connected in parallel with a 5.0 μF capacitor across a 24 V supply. Find the total capacitance and the voltage across the 4.0 μF capacitor.

The series pair gives $1 / C = 1 / 4 + 1 / 12 = 4 / 12$ , so $C = 3.0$ μF. In parallel with 5.0 μF the total is 8.0 μF. The series branch is across the full 24 V, so it carries $Q = (3.0 \times 10^{- 6}) (24) = 72$ μC, and that same 72 μC sits on the 4.0 μF capacitor. Its voltage is therefore $V = Q / C = 72 / 4.0 = 18$ V, leaving $72 / 12 = 6$ V across the other, which correctly sums to 24 V. The smaller capacitor takes the larger share of the voltage.

Now you. A 6.0 μF and a 3.0 μF capacitor in series are connected in parallel with a 4.0 μF capacitor across 30 V. Find the total capacitance and the voltage across the 3.0 μF capacitor.

Answer

Series: $1 / C = 1 / 6 + 1 / 3 = 1 / 2$ , so $C = 2.0$ μF, and the total is $2.0 + 4.0 = 6.0$ μF. The series branch carries $Q = (2.0 \times 10^{- 6}) (30) = 60$ μC, so the 3.0 μF capacitor has $V = 60 / 3.0 = 20$ V across it, and the 6.0 μF has 10 V.

The energy of a charged capacitor

Charging a capacitor is a matter of moving charge from one plate to the other, and it gets harder as it goes: the first electron crosses a zero potential difference, and the last crosses the full $V$ . If the capacitor holds $q$ at some moment, its voltage is $q / C$ , and moving a further $d q$ costs $(q / C) d q$ . Integrating from empty to $Q$ ,

U = \int_{0}^{Q} \frac{q}{C} d q = \frac{Q^{2}}{2 C} = \frac{1}{2} C V^{2} = \frac{1}{2} Q V

The factor of one half is the average voltage during the charging, and it is where the energy that a battery delivers but a capacitor does not store goes: charge a capacitor through a resistor from a battery of EMF $V$ and the battery supplies $Q V$ while the capacitor keeps $\frac{1}{2} Q V$ , the other half being dissipated in the resistance no matter how small that resistance is.

The numbers get large quickly. A photographic flash capacitor of 100 μF charged to 200 V holds $\frac{1}{2} (10^{- 4}) (200)^{2} = 2.0$ J, which is unremarkable until it is released in a millisecond and becomes two kilowatts. A defibrillator is the same idea scaled up.

Example. A defibrillator stores energy in a 32 μF capacitor charged to 5.0 kV, and delivers it over 4.0 ms. How much energy, and what average power?

$U = \frac{1}{2} C V^{2} = \frac{1}{2} (32 \times 10^{- 6}) (5.0 \times 10^{3})^{2} = \frac{1}{2} (32 \times 10^{- 6}) (2.5 \times 10^{7}) = 400$ J. Average power is energy over time, $400 / 0.0040 = 1.0 \times 10^{5}$ W. A hundred kilowatts from a device that runs off a small battery, because the battery spends seconds filling the capacitor and the capacitor empties in milliseconds. Capacitors are poor stores of energy and superb stores of power.

Now you. A capacitor of 100 μF is charged to 300 V and discharged in 2.0 ms. How much energy, and what average power?

Answer

$U = \frac{1}{2} (100 \times 10^{- 6}) (300)^{2} = \frac{1}{2} (10^{- 4}) (9.0 \times 10^{4}) = 4.5$ J, and the average power is $4.5 / 0.0020 = 2.25 \times 10^{3}$ W.

The energy is in the field

Now the question the previous lesson left open. Take the parallel plate result and rewrite it, substituting $C = ε_{0} A / d$ and $V = E d$ :

U = \frac{1}{2} C V^{2} = \frac{1}{2} \frac{ε_{0} A}{d} (E d)^{2} = \frac{1}{2} ε_{0} E^{2} \times (A d)

and $A d$ is precisely the volume between the plates, the region where the field is. Dividing,

u = \frac{1}{2} ε_{0} E^{2}

joules per cubic metre, wherever there is a field. Every reference to charge, plates or geometry has vanished; what is left refers only to the local field strength.

This can be read two ways. Conservatively, it is an identity: an algebraic rearrangement of $\frac{1}{2} C V^{2}$ that happens to have a suggestive form, proved only for a parallel plate capacitor. Read strongly, it says the energy is genuinely located in the field, spread through space at a density $\frac{1}{2} ε_{0} E^{2}$ , and that charges are irrelevant to the accounting.

Nothing in electrostatics can distinguish the two readings, because in electrostatics the charges and the field always come together. The strong reading is nonetheless correct, and the proof is in the twelfth lesson: an electromagnetic wave carries energy through empty space at exactly the density this formula gives, with no charge anywhere near it, and that energy can be absorbed and weighed. Take this formula as a promissory note.

Meanwhile it says something sobering about capacitors as batteries. Air breaks down at $3 \times 10^{6}$ V/m, so the largest energy density an air capacitor can reach is $\frac{1}{2} (8.854 \times 10^{- 12}) (3 \times 10^{6})^{2} = 39.8$ J/m³. Petrol stores about $3.4 \times 10^{10}$ J/m³. The chemical fuel wins by a factor near $10^{9}$ , and even commercial supercapacitors, which use tricks well beyond a plain dielectric, reach only a few per cent of a lithium ion cell's energy per kilogram. Capacitors are used where power density and cycle life matter, not where energy is wanted.

What a dielectric does

Fill the gap between the plates with an insulating material and the capacitance rises, by a factor $κ$ called the dielectric constant or relative permittivity of the material. Faraday measured this in 1837 and the definition is still his: $κ$ is the ratio of the capacitance with the material to the capacitance in vacuum.

Material	Dielectric constant $κ$	Breakdown field / MV m⁻¹
Vacuum	$1$ exactly	no breakdown
Air	$1.00059$	$3$
Paper	$3.7$	$16$
Mica	$5.4$	$100$
Water at 20 degrees Celsius	$80.4$	conducts, so unusable
Barium titanate	$1200$ and above	$8$

The mechanism is polarisation. Every molecule in the material develops, or already has, a dipole moment that partly lines up with the applied field. The dipoles inside the bulk cancel against their neighbours, but at the two faces of the slab there is an uncancelled layer of charge: negative at the face nearest the positive plate, positive at the other. That bound surface charge is of opposite sign to the free charge on the adjacent plate, so it partly cancels it, and the field inside the material drops to $E = E_{0} / κ$ .

From there everything follows. If the capacitor is held at fixed charge, the field drops by $κ$ , so the voltage drops by $κ$ , so $C = Q / V$ rises by $κ$ . If it is held at fixed voltage by a battery, the field is unchanged and more charge flows in, which again means $C$ rose by $κ$ . Either way

C = \frac{κ ε_{0} A}{d}

and the energy density becomes $u = \frac{1}{2} κ ε_{0} E^{2}$ .

The two columns of the table have to be read together, because they pull in opposite directions. Water's $κ$ of 80 comes from whole molecules rotating into alignment, which is a large effect and a slow one: it collapses above about 10 GHz, where the molecules can no longer keep up, and water is useless as a capacitor dielectric anyway because it conducts. Mica has a modest $κ$ and an excellent breakdown field, so a mica capacitor can be run at high voltage. Since the stored energy goes as $κ E^{2}$ and $E$ is capped by breakdown, a material with a high $κ$ and a poor breakdown field can easily be the worse choice.

Practical capacitors take the dielectric idea to its limit. An electrolytic capacitor grows an aluminium oxide layer, sometimes under 100 nm thick, on an etched foil whose roughness multiplies its effective area many times over. With $d$ that small, $C = κ ε_{0} A / d$ delivers millifarads from something the size of a thumb, which is how the 113 square kilometres of the opening estimate is avoided.

The limits of the electrostatic picture

Five lessons have now built a complete theory of charges at rest: a force law, a field, two ways of computing that field, an energy, and a device that stores it. Everything in it is exact, and everything in it assumes nothing moves.

That assumption is doing more work than it looks. A conductor was defined as a material in which charges are free to move, and then every result about conductors was obtained by waiting until they had stopped. The waiting time was never asked about. The capacitor was charged, and how the charge got there was never modelled. Even the sentence "connect it to a battery" is outside the theory: nothing so far explains what a battery is, or why charge would keep flowing through a wire instead of arranging itself once and stopping.

Charge in steady motion is a genuinely different regime, with its own quantities and its own laws, and it is also the regime in which every application of electricity lives. It is next.

Current and circuits

Every result in the first five lessons was obtained by waiting until the charges had stopped moving, and nothing in them explains why, in a wire connected to a battery, they never do.

That is the gap this lesson fills. It needs the field, the potential and Gauss's law from earlier in the course, and it will end with a differential equation of exactly the kind Differential Equations treats first, solved by separating the variables. It is also where the subject stops being about isolated charged spheres and starts being about the objects that electricity is actually used for.

Current

Current is the rate at which charge crosses a surface:

I = \frac{d Q}{d t}

measured in coulombs per second, called amperes. One ampere is a large current in laboratory terms and an ordinary one in domestic terms; a 100 W lamp on a 230 V supply draws about 0.43 A.

Two things about the definition are worth pausing on. First, current is not a vector, despite having a direction along the wire: it is a flux, a scalar counting how much crosses a surface, in the same way that the flow rate of a river in litres per second is a scalar while the velocity of the water is not. Second, the direction assigned to a current is the direction positive charge would move, which as the first lesson explained is opposite to the actual motion of the electrons in a metal. Nothing goes wrong as long as the convention is kept, because a positive charge moving right and a negative charge moving left transport charge the same way.

Spread the current over the cross section and the local quantity is the current density $J = I / A$ , in amperes per square metre, which unlike $I$ does have a direction and is the thing that appears in material laws.

How fast do the charges actually move

Suppose a wire of cross section $A$ contains $n$ mobile charge carriers per cubic metre, each of charge $q$ , drifting at an average speed $v_{d}$ . In a time $d t$ every carrier within a distance $v_{d} d t$ of a chosen cross section will pass it, so the charge crossing is $n q A v_{d} d t$ , and

I = n q v_{d} A

For copper the carrier count is not a free parameter. Copper has a density of 8960 kg/m³ and a molar mass of 63.546 g/mol, and contributes very nearly one conduction electron per atom, so

n = \frac{8960}{0.063546} \times 6.022 \times 10^{23} = 8.49 \times 10^{28} m^{- 3}

Now put 1 A through a wire of 1 mm² cross section, which is ordinary domestic wiring:

v_{d} = \frac{I}{n q A} = \frac{1}{(8.49 \times 10^{28}) (1.602 \times 10^{- 19}) (10^{- 6})} = 7.35 \times 10^{- 5} m/s

That is 0.074 millimetres per second. An electron entering one end of a one metre wire arrives at the other end nearly four hours later.

Three speeds are being confused whenever this result surprises someone, and they differ by ten orders of magnitude each. The drift speed is the tenth of a millimetre per second just computed. The actual speed of an individual conduction electron in copper, set by quantum mechanics rather than by temperature, is about $1.6 \times 10^{6}$ m/s, some twenty billion times faster, but it is directed randomly and averages to almost nothing: the drift is a barely perceptible bias on a violent random motion. And the speed at which a lamp comes on when the switch is thrown is the speed at which the electric field propagates down the wire, a large fraction of the speed of light, because the field does not have to wait for any particular electron to arrive. The wire is already full of charge everywhere along it, and all that has to travel is the instruction to move.

Example. A copper wire of cross section 2.0 mm² carries 5.0 A. What is the drift speed?

$v_{d} = I / (n q A) = 5.0 / [(8.49 \times 10^{28}) (1.602 \times 10^{- 19}) (2.0 \times 10^{- 6})] = 1.84 \times 10^{- 4}$ m/s, about a fifth of a millimetre per second.

Now you. A copper wire of cross section 1.5 mm² carries 3.0 A. What is the drift speed?

Answer

$v_{d} = 3.0 / [(8.49 \times 10^{28}) (1.602 \times 10^{- 19}) (1.5 \times 10^{- 6})] = 1.47 \times 10^{- 4}$ m/s.

Resistance, and why Ohm's law is not a law

Something must keep the drift going. Left alone, a moving charge in a conductor would be accelerated by the field without limit; instead it scatters off lattice vibrations and impurities every $10^{- 14}$ seconds or so, losing its acquired velocity and starting again. The result is a steady drift proportional to the field rather than a steady acceleration, and for a wide class of materials the proportionality is remarkably good:

J = σ E

with $σ$ the conductivity of the material, or equivalently $E = ρ J$ with $ρ = 1 / σ$ the resistivity. This is Ohm's law in its useful form, discovered by Georg Ohm in 1827 and published to general indifference.

For a uniform wire of length $L$ and cross section $A$ , substitute $E = V / L$ and $J = I / A$ :

V = (\frac{ρ L}{A}) I = I R, R = \frac{ρ L}{A}

which is the familiar form. The resistance $R$ , in ohms, is a property of a particular object; the resistivity $ρ$ , in ohm metres, is a property of a material, and it is the one to reason with.

The range is the widest of any common physical property. Copper is $1.68 \times 10^{- 8}$ Ω m at 20 degrees Celsius, nichrome about $1.10 \times 10^{- 6}$ , silicon around $10^{3}$ pure, and good glass $10^{12}$ or more: twenty orders of magnitude between the wire and its insulation, which is why electrical engineering is possible at all.

Now the honesty. $V = I R$ is not a law of nature, it is a description of the materials that happen to obey it. A diode does not: its current rises exponentially with voltage in one direction and hardly at all in the other. A filament lamp does not, because it heats up and copper's resistivity rises by about 0.39 per cent per kelvin, so the resistance of a lamp at 2500 K is roughly ten times its cold value. An ionised gas does not, and can have a resistance that falls as current rises, which is why a fluorescent tube needs a ballast to stop it destroying itself. A superconductor has $ρ$ exactly zero below its transition temperature, which no amount of extrapolating Ohm's law would predict. Ohmic behaviour is common, useful and contingent.

A last observation that catches people out. The field inside a current-carrying copper wire is $E = ρ J$ , and for 1 A in 1 mm² that is $(1.68 \times 10^{- 8}) (10^{6}) = 0.017$ V/m. The electrostatics of the earlier lessons dealt in fields of $10^{4}$ V/m and up. Conduction is what happens when a tiny field acts on an enormous number of very mobile charges.

Example. A kettle element is made from nichrome wire of cross section 0.50 mm² and length 10 m, resistivity $1.10 \times 10^{- 6}$ Ω m. What is its resistance, and what power does it draw at 230 V?

$R = ρ L / A = (1.10 \times 10^{- 6}) (10) / (0.50 \times 10^{- 6}) = 22$ Ω. The power is $P = V^{2} / R = (230)^{2} / 22 = 2.40 \times 10^{3}$ W, which is a normal kettle. The current is $V / R = 10.5$ A, and the whole 2.4 kW appears as heat in the wire, because in a purely resistive element every joule delivered is dissipated.

Now you. A heating element of nichrome, cross section 0.30 mm² and length 8.0 m, is run at 230 V. What is its resistance and its power?

Answer

$R = (1.10 \times 10^{- 6}) (8.0) / (0.30 \times 10^{- 6}) = 29.3$ Ω, and $P = (230)^{2} / 29.3 = 1.80 \times 10^{3}$ W.

EMF, and what a battery does

A resistor connected to nothing carries no current, so a circuit needs a device that pushes charge from low potential to high, against the electric field, in the way a pump raises water. That device supplies an electromotive force, symbol $E$ , which is a poorly chosen name since it is not a force at all but an energy per unit charge, measured in volts.

In a chemical cell the pushing is done by a reaction that is energetically favourable, so a zinc atom gives up electrons at one electrode and a reduction consumes them at the other, and the energy released per electron transferred fixes the EMF. That is why cell voltages are set by chemistry and cluster around one to two volts: they are, in effect, reaction energies in electronvolts.

A real source has resistance of its own, the internal resistance $r$ , so the voltage appearing at its terminals is less than its EMF whenever current flows:

V_{terminal} = E - I r

A 12 V car battery with $r = 0.050$ Ω driving a 2.0 Ω load carries $I = 12 / 2.05 = 5.85$ A, so its terminals sit at $12 - (5.85) (0.050) = 11.7$ V and the load receives $I^{2} R = 68.5$ W while 1.7 W is wasted inside the battery. Starting the engine draws several hundred amperes, and the same 0.050 Ω then drops several volts, which is why the headlights dim.

Power in any two-terminal element is $P = V I$ , since $V$ is energy per charge and $I$ is charge per time. For a resistor, substituting $V = I R$ gives the two other forms, $P = I^{2} R = V^{2} / R$ . The first is the one to use when the current is known, the second when the voltage is, and mixing them up is the commonest arithmetic error in the subject.

Two rules that solve any network

Gustav Kirchhoff published in 1845, at twenty-one, two statements that reduce any network of sources and resistors to simultaneous linear equations.

The junction rule says the currents entering any junction sum to the currents leaving. This is charge conservation together with the observation that charge does not accumulate anywhere in a steady circuit: if it did, the resulting field would immediately stop it.

The loop rule says that the sum of the potential changes around any closed loop is zero. This is the statement that the electrostatic potential is a function of position: go round a loop and return to the same point and you return to the same potential. It is exactly the path independence of the fourth lesson, and it is worth registering now that this is the one result in the whole lesson with an expiry date. The ninth lesson produces an electric field for which the loop sum is not zero, and Kirchhoff's second rule survives there only by being rewritten.

Applying them needs one discipline: choose a direction for each unknown current before starting, and let the algebra return a negative number if the choice was wrong. Guessing correctly is not required and not worth trying.

The RC circuit

Connect a battery of EMF $E$ , a resistor $R$ and an uncharged capacitor $C$ in series and something genuinely time-dependent finally happens. Let $q$ be the charge on the capacitor. The loop rule gives

E - I R - \frac{q}{C} = 0

and since $I = d q / d t$ , this is a first order differential equation:

R \frac{d q}{d t} = E - \frac{q}{C}

Separate the variables, integrate, and impose $q = 0$ at $t = 0$ :

q (t) = C E (1 - e^{- t / R C})

The current is the derivative, $I = (E / R) e^{- t / R C}$ : full at the instant of connection, when the empty capacitor offers no back voltage, and decaying to nothing as it fills. For discharge through the same resistor with no battery, the same equation without $E$ gives $q = Q_{0} e^{- t / R C}$ .

The product $R C$ has units of seconds and is the time constant $τ$ . After one time constant the discharge has fallen to $1 / e$ , about 37 per cent; after three, to 5 per cent; after five, to 0.7 per cent, which is the usual working definition of finished. Note that $τ$ says nothing about how much charge is involved, only how fast.

Example. A 100 μF capacitor charged to 12 V is discharged through 47 kΩ. What is the time constant, the voltage after 10 s, and the time to fall to 1.0 V?

$τ = R C = (47 \times 10^{3}) (100 \times 10^{- 6}) = 4.7$ s. Voltage follows the same exponential as charge, so $V = 12 e^{- 10 / 4.7} = 12 e^{- 2.128} = 12 \times 0.1191 = 1.43$ V. For the time to reach 1.0 V, invert: $t = τ \ln (12 / 1.0) = 4.7 \times 2.485 = 11.7$ s.

Now you. A 220 μF capacitor charged to 9.0 V is discharged through 22 kΩ. Find the time constant and the voltage after 8.0 s.

Answer

$τ = (22 \times 10^{3}) (220 \times 10^{- 6}) = 4.84$ s. Then $V = 9.0 e^{- 8.0 / 4.84} = 9.0 e^{- 1.653} = 9.0 \times 0.1915 = 1.72$ V.

Where the current picture is about to fail

Steady currents look like a closed subject: charge conserved, energy accounted for, networks solvable, and one equation with a time in it. They are not, and the reason is an experiment that nothing in this lesson would predict.

Run two long parallel wires side by side and pass a current through both. They push on each other, attracting when the currents run the same way and repelling when they oppose. The force is easily measurable and it is not electrostatic: both wires are electrically neutral, containing exactly as many protons as electrons, so Coulomb's law says the force between them should be zero.

Something about charge in motion produces a force that charge at rest does not. It cannot be described by a field that acts along the line between the objects, because the force turns out to act sideways, and it cannot be described by anything depending on position alone, because it depends on how fast the charges are moving. It needs a second field, with its own force law and its own rules, and that field occupies the next four lessons.

The magnetic field

Two neutral wires carrying currents push on each other, and nothing in the first six lessons predicts it.

The previous lesson ended there. The escape is to define a second field, and the definition has to be different in kind from the electric one, because the effect being described is different in kind: it depends on how fast the charge is moving, and it acts at right angles to the motion. A field defined as force per unit charge cannot express either of those. What follows assumes the vector geometry of Classical Mechanics, in particular the right hand rule for a cross product, and nothing else new.

Oersted's accident

The connection between electricity and magnetism was suspected for decades and demonstrated by Hans Christian Oersted in April 1820, during a lecture in Copenhagen. He placed a compass needle near a wire and closed a circuit, and the needle swung.

The reason this was startling, rather than merely new, is the direction it swung. Every force known in 1820 acted along the line joining two objects: gravity, Coulomb's force, contact forces, tension. The needle near the wire did not point towards the wire or away from it. It set itself across the wire, tangent to a circle drawn around it, and reversing the current reversed the deflection. Oersted's own report notes that the effect encircles the wire, and no force law of the existing kind can produce a circulation.

Within months André-Marie Ampère had shown that two currents attract or repel each other directly, with no magnet involved, and had proposed that all magnetism is due to circulating currents, which is essentially the modern view: the magnetism of iron comes from electrons whose spin and orbital motion constitute microscopic currents. Magnetism is not a separate substance. It is what electricity does when it moves.

Defining B by what it does

Since the effect is not a force per unit charge, the field cannot be defined as one. Instead it is defined by the whole force law it produces, and the law is the experimental result:

F = q v B \sin θ

where $θ$ is the angle between the velocity and the field, with the force perpendicular to both $v$ and $B$ , in the direction given by the right hand rule for $v \times B$ and reversed if the charge is negative. Compactly, $F = q v \times B$ .

Everything about $B$ is contained in that. Fire a test charge through a point in several directions: there is exactly one direction along which it feels no force at all, and that line is the direction of $B$ , with the sign fixed by the right hand rule applied to any other direction. Fire it perpendicular to that line and the force is maximal, at $F = q v B$ , which fixes the magnitude. The definition is operational and slightly awkward, and it is awkward because the phenomenon is.

The unit of $B$ is the tesla, one newton per ampere metre. It is a large unit. The Earth's field is about 50 μT, a refrigerator magnet a few millitesla, a clinical scanner 1.5 to 3 T, and the strongest steady laboratory field about 45 T. Fields above roughly 100 T can only be made in pulses that destroy the apparatus, because a field exerts a pressure on the currents making it, a point the tenth lesson quantifies. The older unit, the gauss, is $10^{- 4}$ T, and the Earth's field of half a gauss is why it survives in geophysics.

Magnetic force does no work

One consequence follows immediately and is worth isolating, because a great deal depends on it.

The force is always perpendicular to the velocity. Work is force times displacement along the direction of motion, and there is none. So

the magnetic force does no work, ever

and the speed of a charged particle in a magnetic field never changes. Only the direction does. Whatever a magnetic field is doing when a crane lifts scrap iron, it is not supplying energy directly through this force, and any explanation that says otherwise is wrong somewhere.

This is also why $B$ has no potential in the sense of the fourth lesson. A potential exists for a force that does path-independent work; a force that does no work at all cannot be got at that way.

Circular motion

Send a charge into a uniform field perpendicular to it. The force has constant magnitude $q v B$ , since $v$ never changes, and is always perpendicular to the motion. That is precisely the condition for uniform circular motion, so Newton's second law with the centripetal acceleration $v^{2} / r$ gives $q v B = m v^{2} / r$ and

r = \frac{m v}{q B}

The radius is proportional to the momentum. The period is $T = 2 π r / v = 2 π m / (q B)$ , so the cyclotron frequency

f = \frac{q B}{2 π m}

does not depend on the speed at all. A fast particle goes round a bigger circle in exactly the same time as a slow one, because the two effects cancel.

That cancellation is what Ernest Lawrence exploited in 1932. A cyclotron holds two hollow D-shaped electrodes in a uniform field, with an alternating voltage across the gap between them at the cyclotron frequency. A particle crosses the gap, gets a kick, spirals out to a larger radius, and arrives back at the gap in step with the alternating voltage because the period does not depend on how fast it is now going. Lawrence's first machines were centimetres across and reached a megaelectronvolt.

The scheme fails at high energy, and the failure is instructive. As the particle approaches the speed of light its effective mass rises, the period lengthens, and it drifts out of step with the driving voltage. The synchrocyclotron sweeps the driving frequency to follow it, and the synchrotron instead ramps the magnetic field to hold the radius fixed, which is why large accelerators are rings of fixed radius rather than spirals. An electron cyclotron frequency of 28.0 MHz per millitesla is a useful figure to carry, and a proton in 1.5 T circulates at 22.9 MHz.

Example. A proton moves at $2.0 \times 10^{6}$ m/s perpendicular to a 0.40 T field. What is the radius of its path, and how long does one orbit take?

$r = m v / (q B) = (1.673 \times 10^{- 27}) (2.0 \times 10^{6}) / [(1.602 \times 10^{- 19}) (0.40)] = 5.22 \times 10^{- 2}$ m, about 5.2 cm. The period is $T = 2 π m / (q B) = 2 π (1.673 \times 10^{- 27}) / [(1.602 \times 10^{- 19}) (0.40)] = 1.64 \times 10^{- 7}$ s, and it would be the same for a proton of any speed in that field.

Now you. An electron moves at $5.0 \times 10^{6}$ m/s perpendicular to a 0.25 T field. What is the radius of its path?

Answer

$r = m v / (q B) = (9.109 \times 10^{- 31}) (5.0 \times 10^{6}) / [(1.602 \times 10^{- 19}) (0.25)] = 1.14 \times 10^{- 4}$ m, about a tenth of a millimetre. The electron's small mass makes it curl very tightly.

Selecting and weighing

Cross an electric and a magnetic field at right angles and send charges through the region. The electric force $q E$ is fixed; the magnetic force $q v B$ grows with speed. Arrange them to oppose and there is exactly one speed at which they cancel:

v = \frac{E}{B}

and the charge, whatever it is and whatever its mass, passes through undeflected. Everything else is swept aside. This is a velocity selector, and its output does not depend on $q$ or $m$ , which is what makes it useful as a first stage.

Feed that beam into a second region with a uniform field $B'$ and the particles curve with $r = m v / (q B')$ , so measuring where a particle lands measures its mass to charge ratio:

\frac{m}{q} = \frac{B' r}{v}

That is a mass spectrometer, and it is how isotopes were discovered. J. J. Thomson found in 1913 that neon gave two parabolas, at masses 20 and 22, and Francis Aston built the instrument properly and by 1922 had identified over 200 isotopes with it. The same principle, refined, is now standard for identifying molecules by their exact mass.

Thomson's earlier and more famous use of crossed fields, in 1897, was the other way round: he measured $m / q$ for cathode rays, found it a thousand times smaller than for any ion, and concluded that the particles were far lighter than atoms. That was the discovery of the electron, and it came out of the force law of this lesson.

Example. A velocity selector uses $E = 2.4 \times 10^{5}$ V/m and $B = 0.30$ T. Ions of charge $+ e$ emerge and enter a 1.00 T field, curving on a radius of 33.2 cm. What is their mass, in atomic mass units?

The selected speed is $v = E / B = 2.4 \times 10^{5} / 0.30 = 8.0 \times 10^{5}$ m/s. Then $m = q B' r / v = (1.602 \times 10^{- 19}) (1.00) (0.332) / (8.0 \times 10^{5}) = 6.65 \times 10^{- 26}$ kg. Dividing by the atomic mass unit, $1.6605 \times 10^{- 27}$ kg, gives 40.0 u, so this is argon-40, which makes up 99.6 per cent of atmospheric argon.

Now you. In the same instrument, another singly charged ion curves on a radius of 29.0 cm. What is its mass in atomic mass units?

Answer

$m = q B' r / v = (1.602 \times 10^{- 19}) (1.00) (0.290) / (8.0 \times 10^{5}) = 5.81 \times 10^{- 26}$ kg, which divided by $1.6605 \times 10^{- 27}$ is 35.0 u: chlorine-35.

Force on a wire, and torque on a loop

A current is charge in motion, so a current-carrying wire in a field feels the sum of the forces on its carriers. Take a straight segment of length $L$ and cross section $A$ , with $n$ carriers per cubic metre each of charge $q$ drifting at $v_{d}$ . The segment contains $n A L$ carriers, each feeling $q v_{d} B \sin θ$ , so the total is $n A L q v_{d} B \sin θ$ . But $n q v_{d} A$ is exactly the current, from the previous lesson, so

F = I L B \sin θ

with the direction given by the right hand rule applied to the current direction and the field. Ten amperes through 20 cm of wire across a 0.50 T field gives 1.0 N, which is easy to feel and is the whole basis of the electric motor.

Now bend the wire into a rectangular loop of sides $a$ and $b$ , carrying current $I$ , sitting in a uniform field with the loop's plane containing $B$ . The two sides of length $a$ that run perpendicular to the field feel forces $I a B$ , in opposite directions since the current runs opposite ways in them, and they are separated by $b$ . The net force is zero, exactly as for the electric dipole in the second lesson, and the net torque is not:

τ = I a b B = I A B

with $A = a b$ the area. For $N$ turns it is $N$ times larger, and for a general angle a factor $\sin θ$ appears, with $θ$ measured between the field and the normal to the loop. Defining the magnetic moment $μ = N I A$ , a vector along the loop's normal by the right hand rule,

τ = μ B \sin θ

which is the same form as the electric dipole's $τ = p E \sin θ$ , and for the same reason: at a distance, a current loop and a bar magnet are indistinguishable, and both are magnetic dipoles. This is Ampère's insight made quantitative.

Everything rotational in electrical engineering is here. A motor is a coil in a field, with a commutator that reverses the current twice per revolution so the torque never changes sign. A moving-coil galvanometer balances this torque against a spring, so the deflection reads the current. A compass needle is a magnetic moment aligning with the Earth's field, and Oersted's needle was reporting the field of his wire in exactly this way.

Example. A coil of 50 turns and area 4.0 cm² carries 0.20 A in a 0.10 T field, with its plane parallel to the field. What is the torque on it?

The magnetic moment is $μ = N I A = (50) (0.20) (4.0 \times 10^{- 4}) = 4.0 \times 10^{- 3}$ A m². With the plane parallel to the field, the normal is perpendicular to it, so $\sin θ = 1$ and the torque is maximal: $τ = μ B = (4.0 \times 10^{- 3}) (0.10) = 4.0 \times 10^{- 4}$ N m.

Now you. A coil of 120 turns and area 6.0 cm² carries 0.15 A in a 0.25 T field, with its plane parallel to the field. What is the torque?

Answer

$μ = N I A = (120) (0.15) (6.0 \times 10^{- 4}) = 1.08 \times 10^{- 2}$ A m², and $τ = μ B = (1.08 \times 10^{- 2}) (0.25) = 2.7 \times 10^{- 3}$ N m.

The Hall effect, and the sign of the carriers

Nothing so far distinguishes positive charges drifting one way from negative charges drifting the other, since both give the same current. Edwin Hall found the experiment that does, in 1879, while a graduate student.

Pass a current along a flat strip and apply a field perpendicular to its face. The carriers, whatever their sign, are pushed sideways by $q v \times B$ , and they pile up on one edge until the transverse electric field they create cancels the magnetic push. A voltage then appears across the strip, and its sign depends on the sign of the carriers, because positive carriers moving one way and negative carriers moving the other are deflected to the same edge, arriving there with opposite charge.

Balancing the two forces gives $q E_{H} = q v_{d} B$ , and with $E_{H} = V_{H} / w$ across a strip of width $w$ and $I = n q v_{d} w t$ for thickness $t$ , the width cancels:

V_{H} = \frac{I B}{n q t}

which also measures $n$ , the carrier density, an otherwise awkward quantity. In copper the effect is minute: 5 A through a strip 0.10 mm thick in a 1.0 T field gives $V_{H} = 3.7$ μV. In a semiconductor with $n$ around $10^{21}$ m⁻³, eight orders of magnitude fewer carriers, the same numbers give hundreds of volts in principle and a comfortably measurable signal in practice, which is why Hall sensors are made of semiconductors.

The results were a puzzle for fifty years. In some metals, notably zinc and cadmium, the Hall voltage comes out with the sign for positive carriers, which is impossible if conduction is by electrons. The resolution needed quantum band theory and the concept of a hole, a missing electron in a nearly full band that responds to fields exactly as a positive particle would. Hall's experiment was correct and unexplainable for half a century, which is a useful reminder that a clean measurement can outrun the theory available to interpret it.

What is still missing

The force law now covers everything a magnetic field does to a moving charge, a current, or a loop. It says nothing whatsoever about what produces the field.

That is a genuine asymmetry with the electric case, where Coulomb's law gave the source and the force in one statement. Here the source law has to be found separately, and the previous lesson's observation, that two parallel currents attract, is the experiment that will fix it. Doing so is next, and it will turn up an equation with the same structure as Gauss's law and one flat contradiction with it.

Sources of magnetism

The previous lesson defined the magnetic field entirely by what it does to a moving charge, and never said what produces one.

Filling that gap needs an experimental input, exactly as Coulomb's law was an experimental input for electrostatics. It arrived within weeks of Oersted's demonstration: Jean-Baptiste Biot and Félix Savart measured the field around a long straight wire in October 1820 and found it fell off as the inverse of the distance, and Ampère worked out the general rule behind it over the following year. This lesson assumes the force law $F = q v B \sin θ$ and the right hand rule from the previous one.

The contribution of a current element

The rule Biot, Savart and Ampère arrived at splits the wire into short pieces and adds their contributions, in the same spirit as cutting a charge distribution into point charges. A piece of length $d l$ carrying current $I$ contributes, at a point a distance $r$ away,

d B = \frac{μ_{0}}{4 π} \frac{I d l \sin θ}{r^{2}}

where $θ$ is the angle between the current direction and the line to the point, and the direction of $d B$ is perpendicular to both, by the right hand rule. This is the Biot-Savart law.

Compare it with Coulomb's law term by term, because the differences are the whole character of magnetism. Both fall as the inverse square. Both have a constant out front. But Coulomb's field points along the line from the source to the point, while this one points at right angles to that line, and Coulomb's source is a scalar while this one is a piece of a vector, so the $\sin θ$ kills the contribution of any element pointing straight at the field point. The circulation that puzzled Oersted's audience is built into the cross product.

The constant $μ_{0}$ , the permeability of free space, was for seventy years exactly $4 π \times 10^{- 7}$ T m A⁻¹ by definition, because the ampere was defined by fixing it. Since the 2019 redefinition of SI the ampere is fixed through the elementary charge instead, so $μ_{0}$ is a measured quantity, currently $1.25663706212 \times 10^{- 6}$ , which differs from $4 π \times 10^{- 7}$ by about five parts in $10^{10}$ . For every calculation in this course the old exact value is used and the difference is invisible.

One caution: a current element on its own is not a physical object. Current has to flow in a complete circuit, so the Biot-Savart law is a rule for a term in an integral, never for a measurable field by itself.

The straight wire and the loop

Integrate the law along an infinite straight wire and the result, which is what Biot and Savart measured, is

B = \frac{μ_{0} I}{2 π r}

circling the wire, with the right hand rule giving the sense: thumb along the current, fingers curl the way the field goes. One ampere at one metre gives $2 \times 10^{- 7}$ T, which is a very small field, and shows how weak magnetism from ordinary currents is. Ten amperes at 2 cm, on the other hand, gives $1.0 \times 10^{- 4}$ T, twice the Earth's field, which is why a compass held near a household cable misbehaves.

Integrating around a circular loop of radius $R$ gives, at the centre,

B = \frac{μ_{0} I}{2 R}

and on the axis at distance $x$ , $B = μ_{0} I R^{2} / [2 (R^{2} + x^{2})^{3 / 2}]$ . Far away this becomes $μ_{0} μ / (2 π x^{3})$ with $μ = I A$ the magnetic moment of the previous lesson: an inverse cube, exactly like an electric dipole. A current loop seen from a distance is a magnetic dipole and nothing else, which is why a compass needle and a coil are interchangeable.

Fields from separate wires superpose, exactly as electric fields do, so a problem with several wires is a matter of getting each direction right and then adding.

Example. Two long parallel wires 10 cm apart carry 8.0 A and 12 A in the same direction. What is the field at the midpoint between them?

Each wire is 5.0 cm from the midpoint. From the first, $B_{1} = μ_{0} I / (2 π r) = (4 π \times 10^{- 7}) (8.0) / (2 π \times 0.050) = 3.2 \times 10^{- 5}$ T; from the second, $B_{2} = 4.8 \times 10^{- 5}$ T. Because the currents run the same way, their fields circle the two wires in the same rotational sense, so at a point between them the two fields point in opposite directions and subtract. The net field is $1.6 \times 10^{- 5}$ T, directed as the stronger wire dictates.

Now you. Two long parallel wires 8.0 cm apart carry 6.0 A and 10 A in opposite directions. What is the field at the midpoint?

Answer

Each wire is 4.0 cm away, giving $B_{1} = (4 π \times 10^{- 7}) (6.0) / (2 π \times 0.040) = 3.0 \times 10^{- 5}$ T and $B_{2} = 5.0 \times 10^{- 5}$ T. With the currents opposed the two fields point the same way between the wires, so they add: $8.0 \times 10^{- 5}$ T.

Two wires, and the old ampere

Now the experiment that forced this lesson. Take two long parallel wires a distance $d$ apart carrying $I_{1}$ and $I_{2}$ . The first produces a field $μ_{0} I_{1} / (2 π d)$ at the second, perpendicular to it, and the second, of length $L$ , feels $F = I_{2} L B$ from the previous lesson. So

\frac{F}{L} = \frac{μ_{0} I_{1} I_{2}}{2 π d}

Working through the directions with two right hand rules gives attraction for parallel currents and repulsion for antiparallel, which is the opposite of the intuition trained on magnets and charges, where like repels like.

Two wires a metre apart each carrying one ampere attract with $2 \times 10^{- 7}$ N per metre. That number is not a coincidence: from 1948 to 2019 it was the definition of the ampere, which is why $μ_{0}$ was exactly $4 π \times 10^{- 7}$ during that period. Defining a unit of current by a mechanical force is a striking choice, and it was made because force and length could be measured far more accurately than any property of a flowing charge. The modern definition fixes $e$ instead and lets the force be predicted.

Example. Two long parallel wires 25 cm apart carry 40 A and 60 A in the same direction. What force acts on a 3.0 m length of one of them?

$F / L = μ_{0} I_{1} I_{2} / (2 π d) = (4 π \times 10^{- 7}) (40) (60) / (2 π \times 0.25) = 1.92 \times 10^{- 3}$ N/m, so over 3.0 m the force is $5.76 \times 10^{- 3}$ N, attractive since the currents are parallel. Under six millinewtons from 40 and 60 amperes at a quarter of a metre: the magnetic force between currents is genuinely feeble, and everything practical about magnetism comes from stacking many turns or using iron.

Now you. Two long parallel wires 15 cm apart carry 30 A and 50 A in opposite directions. What force acts on a 2.0 m length of one of them?

Answer

$F / L = (4 π \times 10^{- 7}) (30) (50) / (2 π \times 0.15) = 2.0 \times 10^{- 3}$ N/m, so over 2.0 m the force is $4.0 \times 10^{- 3}$ N, repulsive because the currents oppose.

Ampère's law

The Biot-Savart law is to magnetism what the integral of Coulomb's law was to electrostatics: complete, and painful. The relief comes in the same shape as before, as a statement about a whole path rather than about individual sources.

Take the straight wire result and walk once round a circle of radius $r$ centred on the wire, in the direction the field points. The field has constant magnitude $μ_{0} I / (2 π r)$ everywhere on the circle and is everywhere tangent to it, so the product of field and path length is

B \times 2 π r = \frac{μ_{0} I}{2 π r} \times 2 π r = μ_{0} I

and the radius cancels, exactly as it did for flux through a sphere. Generalising, by the same style of argument that turned a sphere into any closed surface, gives Ampère's law:

\oint B \cos φ d l = μ_{0} I_{enc}

The sum of the field component along a closed path, all the way round, equals $μ_{0}$ times the current passing through any surface bounded by that path. Currents outside the loop contribute to $B$ at every point of it and contribute nothing to the total, exactly as external charges did for flux.

The quantity on the left is the circulation of the field, and it is the deep contrast with electrostatics. The circulation of the electrostatic field is zero round every closed path, which is the loop rule of the previous lesson and the reason a potential exists. The circulation of a magnetic field is not zero, so there is no magnetic potential. Electric field lines start and end on charges; magnetic field lines close on themselves.

Ampère's law, like Gauss's, is always true and useful only where symmetry lets $B$ come out of the integral. Two cases matter.

The solenoid. A long coil of $n$ turns per metre carrying $I$ has, inside it and far from the ends, a uniform field along the axis and almost nothing outside. Take a rectangular path with one side of length $L$ running along the axis inside, and the opposite side outside where the field is negligible; the two short sides are perpendicular to the field and contribute nothing. The path encloses $n L I$ of current, so $B L = μ_{0} n L I$ and

B = μ_{0} n I

with no reference to the coil's radius. A solenoid is the magnetic equivalent of the parallel plate capacitor: a way of making a uniform field in a defined region.

The toroid. Bend the solenoid into a doughnut and the field is entirely confined inside it, at $B = μ_{0} N I / (2 π r)$ for $N$ total turns. Nothing leaks, which is why transformer cores and fusion confinement devices are toroidal.

The solenoid formula also says why strong fields are hard. To reach 1 T needs $n I = 1 / μ_{0} = 7.96 \times 10^{5}$ ampere turns per metre, so a coil wound at 1000 turns per metre needs 796 A, and the $I^{2} R$ heating in ordinary copper at that current would destroy the coil in seconds. Every field above about 2 T is made either with superconducting wire, which has no resistance to heat it, or in a pulse short enough that the coil has not yet melted.

Example. A solenoid 40 cm long is wound with 800 turns and carries 2.5 A. What is the field inside it?

The turn density is $n = 800 / 0.40 = 2000$ turns per metre. Then $B = μ_{0} n I = (4 π \times 10^{- 7}) (2000) (2.5) = 6.28 \times 10^{- 3}$ T, about 6.3 mT, or 125 times the Earth's field.

Now you. A solenoid 25 cm long is wound with 1500 turns and carries 1.8 A. What is the field inside it?

Answer

$n = 1500 / 0.25 = 6000$ turns per metre, so $B = (4 π \times 10^{- 7}) (6000) (1.8) = 1.36 \times 10^{- 2}$ T, about 13.6 mT.

The law with nothing on the right

Gauss's law for the electric field said the flux out of a closed surface counts the charge inside. Ask the same question of the magnetic field and the answer is

\oint B \cos θ d A = 0

for every closed surface, always. This is the second of the four equations that will eventually be Maxwell's, and it is the only one with nothing on its right hand side.

What it asserts is that there is no magnetic charge. Cut a bar magnet in half and you do not get a north pole and a south pole; you get two shorter magnets, each with both. Every magnetic field line that enters a region leaves it, because field lines close on themselves rather than terminating.

There is no explanation for this within classical electromagnetism, which is worth saying plainly. Nothing in the theory forbids a magnetic monopole; the equations would accommodate one with a symmetrical extra term, and would arguably look better for it. Paul Dirac showed in 1931 that the existence of even a single monopole anywhere in the universe would force electric charge to be quantised, which is otherwise an unexplained fact. Searches have been thorough and unsuccessful: through moon rock, deep-sea sediment, accelerator debris and cosmic rays. Blas Cabrera's superconducting loop recorded exactly one candidate event, on 14 February 1982, of precisely the expected size, and in the decades of running since then, by that detector and much larger ones, nothing similar has ever appeared. The result is a bound rather than a discovery.

Where iron comes in

Ampère's proposal was that all magnetism is current, and it takes some believing when a bar magnet has no visible circuit in it. The currents are atomic: an electron has an intrinsic magnetic moment, close to one Bohr magneton, $μ_{B} = 9.274 \times 10^{- 24}$ J/T, and orbital motion contributes as well.

In most materials these moments point in random directions and cancel. Applying a field produces a small alignment, giving weak paramagnetism, or a small opposing response, giving weaker diamagnetism, and in both cases the effect disappears when the field is removed. In iron, cobalt, nickel and a few alloys, a quantum mechanical exchange interaction makes neighbouring moments prefer to be parallel, so they align spontaneously within regions called domains, and applying a modest field grows the favourably oriented domains at the expense of the others. That is ferromagnetism, and it multiplies an applied field by a factor of hundreds or thousands, which is why every motor and transformer has an iron core.

The magnitude is checkable. Iron has a density of 7874 kg/m³ and a molar mass of 55.845 g/mol, so it holds $8.49 \times 10^{28}$ atoms per cubic metre, and each contributes about 2.2 Bohr magnetons. If every one of them lined up, the magnetisation would be $M = (8.49 \times 10^{28}) (2.2) (9.274 \times 10^{- 24}) = 1.73 \times 10^{6}$ A/m, and the field it produces is $μ_{0} M = 2.18$ T. The measured saturation field of iron is 2.16 T. A count of atoms and a per-atom moment reproduce a bulk magnetic property to one per cent, which is about as direct a confirmation of Ampère's idea as could be asked for.

The limits are worth naming too. Heat a ferromagnet above its Curie temperature, 1043 K for iron, and thermal agitation destroys the alignment, leaving an ordinary paramagnet. And the response of iron is not linear: it saturates, it lags behind the applied field, and it retains magnetisation when the field is removed. That hysteresis is why permanent magnets exist and why transformer cores dissipate energy on every cycle.

The asymmetry that will not last

Two of the four field equations now exist: the flux of $E$ counts enclosed charge, and the flux of $B$ is always zero. Two more are needed, about circulation rather than flux, and one of them is already half written: the circulation of $E$ is zero, and the circulation of $B$ is $μ_{0} I_{enc}$ .

The set as it stands is a one-way street. Currents make magnetic fields; magnetic fields push currents around; and nothing at all connects a magnetic field back to an electric one. It is a natural thing to try, and several people tried it through the 1820s by putting a coil near a magnet and looking for a current, and every one of them found nothing. Faraday found out why in 1831, and the reason is the most consequential single fact in the subject.

Induction

Currents make magnetic fields, so the obvious experiment is to put a coil near a strong magnet and look for a current, and through the 1820s several capable people did exactly that and found nothing at all.

They found nothing because there was nothing to find. A steady magnetic field, however strong, drives no current in a stationary circuit. What Michael Faraday established on 29 August 1831, and Joseph Henry had found independently at Albany the year before but published later, is that the effect appears only while something is changing. This lesson assumes the magnetic force law and the field of a solenoid from the two previous lessons, and it is where the subject stops being a collection of static results.

Faraday's ring

Faraday's apparatus was a soft iron ring about 15 cm across with two separate coils wound on opposite sides of it, touching nowhere. One coil went to a battery, the other to a galvanometer some distance away. No current from the first coil could reach the second: the only connection was the iron.

Closing the battery circuit produced a momentary kick of the galvanometer needle, which then fell back to zero and stayed there while the current in the first coil ran steadily. Opening the circuit produced another kick, in the opposite direction. The effect was in the transitions, not the state.

Faraday spent the following months isolating what mattered, and the sequence in his diary is a model of experimental method. He replaced the battery coil with a permanent magnet thrust into a coil: same kick, and only while it moved. He moved the coil towards a stationary magnet instead: same kick again. He turned a copper disc between the poles of a magnet and drew a steady current from its rim to its axis, which was the first dynamo. By the end he had reduced everything to one quantity.

Flux, and Faraday's law

The quantity is magnetic flux, defined exactly as electric flux was:

Φ_{B} = B A \cos θ

for a uniform field through a flat area, and $\int B \cos θ d A$ in general, in units of tesla square metres, called webers. It counts the field lines threading the circuit.

Faraday's law is that the EMF induced in a circuit is the rate of change of the flux through it:

E = - N \frac{d Φ_{B}}{d t}

with $N$ the number of turns, since each turn is threaded separately. Only the rate matters. A large flux held constant produces nothing; a small flux changed quickly produces a lot.

That single statement covers every one of Faraday's arrangements, and the three ways of changing the flux are worth separating because they feel unrelated in the laboratory. $B$ can change, as when the current in a neighbouring coil is switched. $A$ can change, as when a circuit is stretched or a rod slides along rails. And $θ$ can change, as when a coil rotates, which is a generator. The flux does not care which.

Example. A coil of 200 turns encloses 25 cm² and sits in a field perpendicular to its plane. The field rises steadily from 0.10 T to 0.60 T over 0.20 s. What EMF appears?

The area and angle are fixed, so the flux change per turn is $Δ Φ = A Δ B = (25 \times 10^{- 4}) (0.50) = 1.25 \times 10^{- 3}$ Wb. The rate is $1.25 \times 10^{- 3} / 0.20 = 6.25 \times 10^{- 3}$ Wb/s, and with 200 turns the EMF is $200 \times 6.25 \times 10^{- 3} = 1.25$ V.

Now you. A coil of 150 turns encloses 40 cm² perpendicular to a field that falls steadily from 0.80 T to 0.20 T in 0.15 s. What EMF appears?

Answer

$Δ Φ = (40 \times 10^{- 4}) (0.60) = 2.4 \times 10^{- 3}$ Wb per turn, over 0.15 s giving $1.6 \times 10^{- 2}$ Wb/s, and with 150 turns, $E = 2.4$ V.

Where the EMF comes from, when the circuit moves

Faraday's law can be quoted, and it is more satisfying to derive the moving case from what is already known.

Take a conducting rod of length $L$ sliding at speed $v$ along two rails, all inside a uniform field $B$ perpendicular to the plane of the circuit. Every free charge in the rod is being carried along at $v$ , so each feels a magnetic force $q v B$ directed along the rod. The force per unit charge is $v B$ , and pushing charge from one end of a length $L$ to the other at that rate means an energy per unit charge of

E = B L v

which is an EMF by the definition of the sixth lesson, produced with no battery and no chemistry.

Check it against Faraday's law. The circuit's area grows at $L d x / d t = L v$ , so the flux grows at $B L v$ , and the two agree exactly. The rod is a source of EMF driving current round the circuit, and the magnetic force on the carriers is doing the driving.

There is a subtlety here that is worth not glossing over, because it is the entrance to relativity. The magnetic force does no work, as the seventh lesson insisted, so it cannot be the ultimate source of the energy that the circuit dissipates. It is not: the energy comes from whatever is pushing the rod. Once current flows, the current-carrying rod in the field feels a retarding force $B I L$ , and the agent pushing it must work against that. The magnetic force acts as an intermediary, redirecting the work of the pusher into the circuit, and the books balance exactly.

Lenz's law

The minus sign in Faraday's law is a separate physical statement, articulated by Heinrich Lenz in 1834: the induced current flows in the direction that opposes the change producing it. Push a magnet's north pole towards a loop and the loop's induced current makes a north pole facing the magnet, pushing back. Pull it away and the loop makes a south pole, pulling after it.

The justification is energy conservation, and the argument by contradiction is short. Suppose the current flowed the other way, aiding the change. Then pushing the magnet in would produce a force pulling it in further, accelerating it, producing more current, producing more force. Energy would appear from nothing, with the current in the loop and the kinetic energy of the magnet both increasing without any work being done. Lenz's law is therefore not an extra empirical fact but the only sign consistent with the first law of thermodynamics.

The practical consequences are large. Every generator resists being turned, and resists more when more current is drawn, which is why a power station burns more fuel when demand rises: the connection between the grid and the boiler is Lenz's law. Every motor generates a back EMF opposing the supply that drives it, which is why a motor draws a huge current at the instant of starting, when it is not yet turning and there is no back EMF, and a modest one at speed.

Example. A rod of length 0.25 m slides at 4.0 m/s along rails in a 0.50 T field, with the circuit closed by a 2.0 Ω resistance. Find the EMF, the current, the force needed to keep the rod moving, and check the power balance.

$E = B L v = (0.50) (0.25) (4.0) = 0.50$ V. The current is $I = E / R = 0.25$ A. The rod now carries current in a field, so it feels a retarding force $F = B I L = (0.50) (0.25) (0.25) = 3.13 \times 10^{- 2}$ N, and by Lenz's law it opposes the motion, so the same force must be applied to keep the speed constant. The mechanical power is $F v = (3.13 \times 10^{- 2}) (4.0) = 0.125$ W, and the electrical power dissipated is $E^{2} / R = 0.25 / 2.0 = 0.125$ W. They agree, as they must.

Now you. A rod of length 0.40 m slides at 2.5 m/s along rails in a 0.80 T field, with a 4.0 Ω resistance closing the circuit. Find the EMF, the current, the retarding force and the power.

Answer

$E = (0.80) (0.40) (2.5) = 0.80$ V, so $I = 0.80 / 4.0 = 0.20$ A. The force is $B I L = (0.80) (0.20) (0.40) = 6.4 \times 10^{- 2}$ N, and the mechanical power is $F v = 0.16$ W, matching $E^{2} / R = 0.64 / 4.0 = 0.16$ W.

The generator

Rotate a coil of $N$ turns and area $A$ at angular speed $ω$ in a uniform field $B$ . The angle between the field and the coil's normal is $θ = ω t$ , so the flux is $Φ = B A \cos ω t$ and

E = - N B A \frac{d}{d t} \cos ω t = N B A ω \sin ω t

an alternating EMF of peak value $N B A ω$ . This is the origin of alternating current, and of the fact that mains electricity is a sine wave: it is what a rotating machine naturally produces, and rotation is what a turbine naturally provides.

Example. A generator coil of 500 turns and area 100 cm² rotates at 50 revolutions per second in a 0.20 T field. What is the peak EMF, and what is the root mean square value?

$ω = 2 π f = 2 π (50) = 314.2$ rad/s. The peak EMF is $N B A ω = (500) (0.20) (0.0100) (314.2) = 314$ V. The rms value of a sine wave is the peak divided by $\sqrt{2}$ , so 222 V, which is roughly a European mains supply from a coil the size of a hand.

Now you. A coil of 250 turns and area 80 cm² rotates at 60 revolutions per second in a 0.35 T field. What is the peak EMF?

Answer

$ω = 2 π (60) = 377.0$ rad/s, so the peak EMF is $(250) (0.35) (80 \times 10^{- 4}) (377.0) = 264$ V.

Induction is not always wanted. A solid conductor moving in a non-uniform field, or sitting in a changing one, has induced currents circulating within its own bulk. These eddy currents obey Lenz's law and so oppose the motion, which is the whole of magnetic braking: drop a strong magnet down a copper pipe and it takes seconds to fall a metre, because the induced currents in the pipe wall produce a field that repels it from below and attracts it from above. The pipe is neither magnetic nor in contact with the magnet. In transformer cores the same currents are pure loss, and are suppressed by building the core from thin insulated laminations that break the circulating paths.

The mechanism problem

Two of Faraday's experiments give the same EMF and appear to have nothing in common.

Move the loop, hold the magnet still, and the explanation is the one derived above: charges in the moving conductor feel $q v \times B$ and are pushed round the circuit. No electric field is involved anywhere.

Move the magnet, hold the loop still, and that explanation is unavailable: the charges in the loop are not moving, so $q v \times B$ is zero. Something must still push them, and the only candidate is an electric field. So a changing magnetic field creates an electric field, in empty space, whether or not any conductor is there to reveal it.

Only relative motion matters, and the measured EMF is identical in the two cases, yet classical theory offers two entirely different mechanisms for it. Einstein opened his 1905 paper on special relativity with precisely this observation, noting that the asymmetry in the explanation has no counterpart in the phenomena, and taking that as evidence that the notion of absolute rest is empty. The two mechanisms are the same thing seen from two frames: what one observer calls a magnetic field, another moving relative to them calls a partly electric one. Electricity and magnetism are not two fields that interact but one field seen from different states of motion.

The death of potential

The second half of that argument has an immediate consequence, and it wrecks something built earlier in this course.

Written for the induced electric field, Faraday's law says

\oint E \cos φ d l = - \frac{d Φ_{B}}{d t}

The circulation of the electric field round a closed path is not zero. Take a loop of wire encircling a solenoid whose current is being ramped up, and carry a charge once round the loop: it returns to where it started having gained energy.

Every appearance of electric potential in this course assumed the opposite. The fourth lesson defined $V$ by the work done between two points, which is only a function of position if the work round any closed path is zero. Under a changing magnetic field it is not, so there is no potential function for an induced electric field, and asking for the voltage at a point in such a region is asking a question with no answer. Two voltmeters connected to the same two points on a loop encircling a changing flux, with their leads routed on opposite sides, read different values, and both are correct.

Kirchhoff's loop rule survives only by being rewritten: the sum of potential drops round a loop equals the induced EMF rather than zero, which is exactly how circuit analysis handles inductors. Within a circuit whose changing flux is confined to identifiable components, the old bookkeeping works. Outside such a circuit, potential is simply not available.

What comes next

Faraday's law says a circuit responds to a change in the flux through it, and says nothing about where that flux came from. In particular it does not exclude the circuit's own field.

A coil carrying a changing current therefore induces an EMF in itself, opposing its own change, and by Lenz's law that opposition acts to keep the current where it was. A circuit acquires something very like inertia. Working out how much, where the energy goes, and what happens when such a circuit is connected to a capacitor is the next lesson, and the answer to the last part is an oscillation whose frequency is built from $ε_{0}$ and $μ_{0}$ alone.

Inductance and magnetic energy

Faraday's law says a circuit responds to any change in the flux through it, and nothing in it excludes the flux the circuit produces itself.

That loophole is the whole of this lesson. It gives a circuit element with no static counterpart, an energy stored in a magnetic field, the transformer that made electrical distribution possible, and, at the end, a circuit that oscillates at a frequency assembled from $ε_{0}$ and $μ_{0}$ . It assumes Faraday's law and Lenz's law from the previous lesson, the solenoid field from the one before, and the technique of solving a first order differential equation by separating variables.

Self-inductance

Send a current $I$ through a coil and it produces a magnetic field, and that field threads the coil's own turns. Every contribution to the field is proportional to $I$ , so the total flux linkage is proportional to $I$ too, and the constant of proportionality is a property of the geometry:

N Φ_{B} = L I

$L$ is the self-inductance, in webers per ampere, called henries after Joseph Henry, who discovered self-induction in 1832. Combining this with Faraday's law,

E = - L \frac{d I}{d t}

The coil opposes changes in its own current. Increase the current and it pushes back; decrease it and it pushes forward, trying to keep the current going. This is not a metaphorical inertia. Put $L$ beside $m$ , $I$ beside $v$ , and $E$ beside force, and the equation is Newton's second law.

For a long solenoid the inductance is computable. The field inside is $B = μ_{0} n I$ from the eighth lesson, the flux through one turn is $μ_{0} n I A$ , and there are $n l$ turns in a length $l$ , so $N Φ = μ_{0} n^{2} A l I$ and

L = μ_{0} n^{2} A l = \frac{μ_{0} N^{2} A}{l}

The square on the turn count is why inductors are wound with many turns: doubling $N$ doubles both the field produced and the flux caught.

Example. A solenoid 25 cm long is wound with 500 turns over a cross section of 4.0 cm². What is its inductance, and what EMF appears across it if its current is changing at 200 A/s?

$L = μ_{0} N^{2} A / l = (4 π \times 10^{- 7}) (500)^{2} (4.0 \times 10^{- 4}) / (0.25) = 5.03 \times 10^{- 4}$ H, about half a millihenry. The induced EMF is $L d I / d t = (5.03 \times 10^{- 4}) (200) = 0.101$ V. A tenth of a volt from a rate of change that would take a large current to sustain, which is why inductors of any usefulness are wound on iron: the core multiplies $L$ by the relative permeability of the material, often by a factor of a thousand.

Now you. A solenoid 30 cm long with 800 turns over 6.0 cm² carries a current changing at 150 A/s. Find its inductance and the EMF across it.

Answer

$L = (4 π \times 10^{- 7}) (800)^{2} (6.0 \times 10^{- 4}) / (0.30) = 1.61 \times 10^{- 3}$ H, and the EMF is $(1.61 \times 10^{- 3}) (150) = 0.241$ V.

The LR circuit

Connect an inductor $L$ in series with a resistor $R$ across a source of EMF $E$ . The loop rule, in the amended form the previous lesson insisted on, gives

E - I R - L \frac{d I}{d t} = 0

which has exactly the structure of the RC circuit's equation. Separating variables and imposing $I = 0$ at the instant of connection,

I (t) = \frac{E}{R} (1 - e^{- t / τ}), τ = \frac{L}{R}

At the first instant the current is zero, because the inductor's back EMF exactly cancels the source: an inductor with no current in it behaves momentarily like a break in the circuit. After several time constants the current settles at $E / R$ and the inductor, with $d I / d t$ now zero, behaves like a plain piece of wire. The inductor's whole existence is in the transition.

Opening the switch is the dangerous case. The current is forced to fall abruptly, $d I / d t$ is enormous, and $L d I / d t$ can reach thousands of volts across the coil, which is where the spark at a switch comes from and how the ignition coil in a petrol engine makes its spark from a 12 V supply. Circuits driving inductive loads carry a diode across the coil for exactly this reason, giving the current somewhere to go while it dies away.

Example. A 0.50 H inductor in series with 25 Ω is connected to 12 V. Find the time constant, the final current, the current after 10 ms, and the energy finally stored.

$τ = L / R = 0.50 / 25 = 0.020$ s. The final current is $E / R = 12 / 25 = 0.48$ A. At $t = 0.010$ s, which is half a time constant, $I = 0.48 (1 - e^{- 0.5}) = 0.48 \times 0.3935 = 0.189$ A. The energy stored at full current is $\frac{1}{2} L I^{2} = \frac{1}{2} (0.50) (0.48)^{2} = 0.0576$ J, a result derived in the next section.

Now you. A 0.20 H inductor in series with 40 Ω is connected to 24 V. Find the time constant, the final current, the current after 2.5 ms, and the stored energy.

Answer

$τ = 0.20 / 40 = 5.0$ ms, and the final current is $24 / 40 = 0.60$ A. At 2.5 ms, half a time constant, $I = 0.60 (1 - e^{- 0.5}) = 0.236$ A. The stored energy is $\frac{1}{2} (0.20) (0.60)^{2} = 0.036$ J.

Energy in the magnetic field

Establishing a current in an inductor takes work, because the back EMF opposes the increase the whole way up. The source delivers power $E I$ against that back EMF, which is $L I d I / d t$ , so the total work done in raising the current from zero to $I$ is

U = \int_{0}^{I} L I' d I' = \frac{1}{2} L I^{2}

exactly parallel to the capacitor's $\frac{1}{2} C V^{2}$ , and for the same reason: the opposition grows as the quantity being established grows, so the average is half the final value.

Where is it? Run the same substitution as for the capacitor. For a solenoid, $L = μ_{0} n^{2} A l$ and $B = μ_{0} n I$ , so $I = B / (μ_{0} n)$ and

U = \frac{1}{2} μ_{0} n^{2} A l \frac{B^{2}}{μ_{0}^{2} n^{2}} = \frac{B^{2}}{2 μ_{0}} \times (A l)

with $A l$ the volume inside the solenoid, where the field is. Dividing,

u = \frac{B^{2}}{2 μ_{0}}

joules per cubic metre. The electric and magnetic energy densities are now both in hand, $\frac{1}{2} ε_{0} E^{2}$ and $B^{2} / 2 μ_{0}$ , and in the twelfth lesson a light wave will carry both, in equal amounts.

Comparing them says something practical. At 1 T, the magnetic density is $1 / (2 μ_{0}) = 3.98 \times 10^{5}$ J/m³, four orders of magnitude above the 39.8 J/m³ an air capacitor can reach at the breakdown field. Magnetic storage is the better of the two, and superconducting magnetic energy storage exists commercially on that basis, though it still loses to chemical fuel by a factor of $10^{5}$ .

The same quantity is also a pressure, since joules per cubic metre are newtons per square metre, and it is the pressure the field exerts on whatever carries the current making it. In the 45 T of the strongest steady laboratory magnet, $B^{2} / 2 μ_{0} = 8.06 \times 10^{8}$ Pa, about 8000 atmospheres, which is comparable to the yield strength of good steel. That is the real limit on magnetic fields: not the current, not the heat, but the fact that a strong enough magnet tears itself apart.

Mutual inductance and the transformer

Two coils sharing flux induce in each other, with $E_{2} = - M d I_{1} / d t$ , and the mutual inductance $M$ is the same in both directions, which is a small theorem worth believing rather than deriving here. That is Faraday's iron ring, and it is a transformer.

Wind $N_{p}$ turns and $N_{s}$ turns on a common core so that essentially all the flux is shared. Each turn on either side sees the same $d Φ / d t$ , so

\frac{V_{s}}{V_{p}} = \frac{N_{s}}{N_{p}}

and if the transformer is lossless, power in equals power out, so the currents go the other way: $I_{s} / I_{p} = N_{p} / N_{s}$ . A transformer trades voltage against current at fixed power, and it works only on alternating current, because a steady current produces no $d Φ / d t$ at all.

That constraint decided the war between direct and alternating current in the 1880s, and the reason is $I^{2} R$ . Send 500 MW down a line of 5 Ω resistance at 400 kV and the current is 1250 A, so the loss is $I^{2} R = 7.8$ MW, or 1.6 per cent. Send the same power at 132 kV and the current is 3788 A, so the loss is 71.7 MW, or 14.3 per cent. Nine times the loss for a third of the voltage, because the loss goes as the square. Long distance transmission demands high voltage, safe use demands low voltage, and only a transformer can convert between them. Edison's direct current system had no way to do it and lost on that point alone.

Real transformers fall short of the ideal in three named ways, and each is worth recognising because each is a term already met in this course. Not all the flux from one winding reaches the other, which is leakage and behaves as an unwanted series inductance. The windings have resistance, so the $I^{2} R$ that transmission avoids reappears inside the transformer, as copper loss. And the iron core is itself a conductor sitting in a changing flux, so eddy currents circulate in it and dissipate, which is why cores are built from thin laminations insulated from each other rather than from solid iron, and why the laminations are stacked along the flux direction so the circulating paths are broken. Add hysteresis loss, the energy spent driving the iron round its magnetisation loop on every cycle, and a large power transformer still manages better than 99 per cent efficiency, which is among the highest of any machine ever built.

The LC circuit

Now connect a charged capacitor across an inductor, with no resistance. The loop rule gives $q / C + L d I / d t = 0$ , and with $I = d q / d t$ ,

L \frac{d^{2} q}{d t^{2}} + \frac{q}{C} = 0

which is the equation of simple harmonic motion, identical in form to a mass on a spring with $L$ playing the mass and $1 / C$ the spring constant. Its solution oscillates at

ω = \frac{1}{\sqrt{L C}}

Physically the energy sloshes. It starts entirely electric, $Q^{2} / 2 C$ in the capacitor. As the capacitor discharges the current grows and the energy moves into the magnetic field of the inductor, and at the moment the capacitor is empty the current is maximal and all the energy is magnetic, $\frac{1}{2} L I^{2}$ . The inductor's inertia then keeps the current going, recharging the capacitor with the opposite polarity, and the cycle repeats. The total is constant, and equating the two extremes gives the peak current directly: $\frac{1}{2} L I_{0}^{2} = Q_{0}^{2} / 2 C$ , so $I_{0} = Q_{0} / \sqrt{L C} = ω Q_{0}$ , which is the same relation between amplitude and peak speed that a mass on a spring obeys.

No real circuit does this forever. Every inductor has resistance, so the equation acquires a damping term and becomes $L d^{2} q / d t^{2} + R d q / d t + q / C = 0$ , which is the damped oscillator that Differential Equations solves in full: oscillation with an amplitude decaying as $e^{- R t / 2 L}$ while $R$ is small, and no oscillation at all once $R$ exceeds $2 \sqrt{L / C}$ . The ratio of stored energy to energy lost per radian is the quality factor $Q = ω L / R$ , and a good radio tuning circuit reaches a few hundred, which is what lets it pick one station out of a band. That is the practical use of the whole section: an LC circuit responds strongly at one frequency and weakly at others, so it selects.

Example. An inductor of 10 mH is connected to a capacitor of 100 nF. At what frequency does the circuit oscillate?

$L C = (10^{- 2}) (10^{- 7}) = 10^{- 9}$ , whose square root is $3.162 \times 10^{- 5}$ . Then $ω = 1 / (3.162 \times 10^{- 5}) = 3.162 \times 10^{4}$ rad/s, and $f = ω / 2 π = 5.03 \times 10^{3}$ Hz, an audible tone.

Now you. An inductor of 2.5 mH is connected to a capacitor of 40 nF. What is the oscillation frequency?

Answer

$L C = (2.5 \times 10^{- 3}) (4.0 \times 10^{- 8}) = 1.0 \times 10^{- 10}$ , so $\sqrt{L C} = 1.0 \times 10^{- 5}$ and $ω = 10^{5}$ rad/s. Then $f = 10^{5} / 2 π = 1.59 \times 10^{4}$ Hz.

A hint worth taking seriously

Look at what $ω = 1 / \sqrt{L C}$ is made of.

The inductance of a solenoid is $μ_{0} n^{2} A l$ , proportional to $μ_{0}$ and otherwise pure geometry. The capacitance of a parallel plate capacitor is $ε_{0} A / d$ , proportional to $ε_{0}$ and otherwise pure geometry. So for any circuit built out of such elements, $L C$ is $ε_{0} μ_{0}$ multiplied by something with the dimensions of length squared, and

\frac{1}{\sqrt{ε_{0} μ_{0}}}

has the dimensions of a speed. Putting in the numbers, $ε_{0} μ_{0} = (8.854 \times 10^{- 12}) (1.2566 \times 10^{- 6}) = 1.113 \times 10^{- 17}$ , whose reciprocal square root is $3.00 \times 10^{8}$ m/s.

That number should be recognisable. It is the speed of light, and it has just fallen out of a capacitor and a coil, apparatus with no light in it anywhere. Wilhelm Weber and Rudolf Kohlrausch obtained it in 1856 by charging a capacitor, measuring its charge electrostatically with a torsion balance and then magnetically by discharging it through a galvanometer, and taking the ratio. They got $3.107 \times 10^{8}$ m/s, and Fizeau's 1849 measurement of the speed of light gave $3.15 \times 10^{8}$ m/s. Weber and Kohlrausch noted the closeness and did not know what to make of it.

Maxwell did. But before the connection can be made, one of the four field equations has to be repaired, because as it stands Ampère's law contradicts the conservation of charge. Finding the contradiction is the next lesson.

Maxwell's equations

Ampère's law says the circulation of the magnetic field round a closed path equals $μ_{0}$ times the current passing through it, and for a charging capacitor that instruction does not have a unique answer.

This lesson finds the contradiction, repairs it, and assembles the four equations that result. Everything it needs is already in hand: Gauss's law from the third lesson, the absence of monopoles and Ampère's law from the eighth, and Faraday's law from the ninth. The only new physics is one term, and it is the term that makes the twelfth lesson possible.

The surface that does not agree

Ampère's law is stated for a closed path, and the current it refers to is the current through a surface bounded by that path. Which surface is not specified, and for good reason: for a steady current it makes no difference. A wire that pierces one surface bounded by the loop pierces every other, because a steady current has nowhere to stop.

Now charge a capacitor. Current $I$ flows along the wire, arrives at one plate, and stops there, with an equal current leaving the other plate. Draw a circular path round the wire, a few centimetres from the capacitor, and apply Ampère's law.

Take the flat disc bounded by that circle. The wire pierces it, so the enclosed current is $I$ , and the law gives $\oint B d l = μ_{0} I$ .

Now take instead a surface shaped like a shopping bag, bounded by the same circle but bulging out and passing between the capacitor plates. No charge crosses the gap: that is what a capacitor is. So the enclosed current is zero, and the law gives $\oint B d l = 0$ .

The left hand side is identical in the two cases, since it is an integral round the same physical path in the same physical field. The right hand sides differ. Ampère's law, as it stands, is not merely incomplete but self-contradictory whenever the current is not steady.

The deeper way to say it is that Ampère's law implicitly asserts that current never accumulates anywhere, which is a statement about steady states masquerading as a law. Charge conservation permits accumulation and demands that when charge does build up in a region, the current in exceeds the current out by exactly the rate of build-up. A capacitor plate is precisely such a region.

The term that repairs it

Something must be passing through the shopping bag surface with the same effect as a current. Look at what is between the plates: charge is arriving on them, so the field between them is growing.

For a parallel plate capacitor of area $A$ carrying charge $Q$ , Gauss's law gives $E = Q / (ε_{0} A)$ , so the electric flux between the plates is $Φ_{E} = E A = Q / ε_{0}$ . Differentiate:

ε_{0} \frac{d Φ_{E}}{d t} = \frac{d Q}{d t} = I

Exactly $I$ . Not approximately, not proportionally: the rate of change of electric flux through the gap, multiplied by $ε_{0}$ , equals the conduction current in the wire, to the last digit, at every instant.

That is the fix. Define the displacement current

I_{d} = ε_{0} \frac{d Φ_{E}}{d t}

and write Ampère's law with it included:

\oint B \cos φ d l = μ_{0} (I_{enc} + ε_{0} \frac{d Φ_{E}}{d t})

Now the two surfaces agree. The flat disc has conduction current $I$ and no changing flux; the bulging one has no conduction current and displacement current $I$ ; the total is $μ_{0} I$ either way. The contradiction was not a paradox to be explained away, it was a missing term, and the requirement that the answer not depend on an arbitrary choice of surface determines that term uniquely.

Example. The plates of a capacitor have area 0.020 m², and the field between them is rising at $5.0 \times 10^{12}$ V m⁻¹ s⁻¹. What current is flowing in the wire?

The displacement current between the plates must equal the conduction current in the wire, and $I_{d} = ε_{0} d Φ_{E} / d t = ε_{0} A d E / d t$ since the area is fixed. So $I = (8.854 \times 10^{- 12}) (0.020) (5.0 \times 10^{12}) = 0.885$ A. The enormous rate of change of the field is what a modest current looks like when it is expressed this way, because $ε_{0}$ is very small.

Now you. The plates of a capacitor have area 0.035 m² and the field between them is rising at $3.0 \times 10^{12}$ V m⁻¹ s⁻¹. What current is flowing in the wire?

Answer

$I = ε_{0} A d E / d t = (8.854 \times 10^{- 12}) (0.035) (3.0 \times 10^{12}) = 0.930$ A.

James Clerk Maxwell added it in a paper of 1861 and 1862 titled On Physical Lines of Force, and the route he took there is worth knowing, because it is not the route above. He was working with an elaborate mechanical model of the ether, in which magnetic field lines were vortices in a fluid and electric currents were small idle wheels rolling between them, and the displacement current entered as the elastic deformation of the vortex material. Nobody believes the model. The term is correct, and it is a case where a physically wrong picture produced a physically right equation, which happens more often than tidy accounts of science admit.

The name is a fossil of the same model and is doubly unfortunate: nothing is displaced, and it is not a current. Nothing at all moves between the plates of a vacuum capacitor. It is a changing electric field, given a name that says a changing electric field produces a magnetic field exactly as a current does.

Is it real?

A reasonable objection is that the displacement current is a bookkeeping device inserted to make the equations behave, and that it makes no prediction of its own.

It does. The added term says there is a genuine magnetic field in the gap of a charging capacitor, where no charge is moving at all, circling the axis exactly as though the wire continued through. Applying the amended law to a circular path of radius $r$ inside a gap between plates of radius $R$ gives, for $r < R$ ,

B = \frac{μ_{0} I r}{2 π R^{2}}

rising linearly from zero on the axis to $μ_{0} I / (2 π R)$ at the rim, where it joins continuously onto the field of the wire outside. This has been measured, and it is there.

Example. A capacitor with circular plates of radius 4.0 cm is charged by a steady current of 0.50 A. How fast is the field between the plates changing, and what is the magnetic field 2.0 cm from the axis?

The plate area is $π (0.040)^{2} = 5.03 \times 10^{- 3}$ m². The displacement current must equal 0.50 A, and $I_{d} = ε_{0} A d E / d t$ , so $d E / d t = 0.50 / [(8.854 \times 10^{- 12}) (5.03 \times 10^{- 3})] = 1.12 \times 10^{13}$ V m⁻¹ s⁻¹. The magnetic field at 2.0 cm is $B = μ_{0} I r / (2 π R^{2}) = (4 π \times 10^{- 7}) (0.50) (0.020) / [2 π (0.040)^{2}] = 1.25 \times 10^{- 6}$ T, half the value at the rim, which is $2.50 \times 10^{- 6}$ T.

Now you. A capacitor with circular plates of radius 6.0 cm is charged at 1.2 A. Find the rate of change of the field between the plates, and the magnetic field 3.0 cm from the axis.

Answer

The area is $π (0.060)^{2} = 1.13 \times 10^{- 2}$ m², so $d E / d t = 1.2 / [(8.854 \times 10^{- 12}) (1.13 \times 10^{- 2})] = 1.20 \times 10^{13}$ V m⁻¹ s⁻¹. The magnetic field is $B = (4 π \times 10^{- 7}) (1.2) (0.030) / [2 π (0.060)^{2}] = 2.0 \times 10^{- 6}$ T.

The four equations

With the repair made, the whole of classical electromagnetism is four statements. Each is an integral over a closed surface or a closed path, each has been derived earlier in this course, and together they determine the fields completely once the charges and currents are given.

Gauss's law for electricity. The electric flux out of any closed surface counts the charge inside it.

\oint E \cos θ d A = \frac{q_{enc}}{ε_{0}}

Gauss's law for magnetism. The magnetic flux out of any closed surface is zero, because there is no magnetic charge.

\oint B \cos θ d A = 0

Faraday's law. The circulation of the electric field round a closed path is minus the rate of change of the magnetic flux through it.

\oint E \cos φ d l = - \frac{d Φ_{B}}{d t}

The Ampère-Maxwell law. The circulation of the magnetic field round a closed path is $μ_{0}$ times the current through it, plus $μ_{0} ε_{0}$ times the rate of change of the electric flux.

\oint B \cos φ d l = μ_{0} I_{enc} + μ_{0} ε_{0} \frac{d Φ_{E}}{d t}

To these must be added the force law that connects the fields to matter, $F = q (E + v \times B)$ , usually called the Lorentz force, without which the equations describe fields that nothing can detect.

Two footnotes on the presentation. First, Maxwell did not write them like this. His 1865 paper A Dynamical Theory of the Electromagnetic Field gave twenty equations in twenty variables, including the potentials and written out component by component, and it is heavy going. The compact set above is Oliver Heaviside's, who reduced them to four in 1884 and 1885 using the vector notation he was inventing at the time, with Hertz arriving at much the same form independently. What everyone now calls Maxwell's equations is Heaviside's rewriting of Maxwell's physics, and the names attached to the four individual laws were largely settled by textbooks afterwards.

Second, these are the vacuum equations. Inside a material, the polarisation of the fifth lesson and the magnetisation of the eighth add their own bound charges and bound currents, and the usual dodge is to hide them by replacing $ε_{0}$ with $κ ε_{0}$ and $μ_{0}$ with the material's permeability. That works when the material responds linearly and instantly, which covers most of engineering and none of ferromagnetism, nonlinear optics or anything at high frequency. The vacuum form has no such caveats, which is one reason the rest of this course stays in vacuum.

The structure is worth reading as a table of what makes what. Charges make electric flux. Nothing makes magnetic flux. Changing magnetic flux makes electric circulation. Currents and changing electric flux make magnetic circulation. The pattern is nearly symmetric between the two fields, and the one asymmetry, the zero in the second equation, is exactly the absence of magnetic monopoles: put a magnetic charge density on the right of the second equation and a magnetic current on the right of the third, and the four become perfectly symmetric. Nature declined the offer.

Example. A long solenoid of radius 5.0 cm has its field increasing at 0.20 T/s. What induced electric field appears at 3.0 cm from the axis, and at 8.0 cm?

Use Faraday's law on a circle of radius $r$ centred on the axis, on which the induced field is tangential and uniform by symmetry, so the circulation is $E (2 π r)$ . Inside, at $r = 0.030$ m, the flux through the circle is $B π r^{2}$ , so $E (2 π r) = π r^{2} d B / d t$ and $E = (r / 2) d B / d t = (0.030 / 2) (0.20) = 3.0 \times 10^{- 3}$ V/m. Outside, at $r = 0.080$ m, the flux is only $B π R^{2}$ because there is no field beyond the solenoid, so $E = (R^{2} / 2 r) d B / d t = (0.0025 / 0.16) (0.20) = 3.13 \times 10^{- 3}$ V/m. Note that an electric field exists outside the solenoid where the magnetic field is zero, which is a genuinely non-local-looking result and entirely correct.

Now you. A long solenoid of radius 4.0 cm has its field increasing at 0.50 T/s. Find the induced electric field at 2.0 cm and at 10.0 cm from the axis.

Answer

Inside: $E = (r / 2) d B / d t = (0.020 / 2) (0.50) = 5.0 \times 10^{- 3}$ V/m. Outside: $E = (R^{2} / 2 r) d B / d t = (0.0016 / 0.20) (0.50) = 4.0 \times 10^{- 3}$ V/m.

What the set forbids

A good way to test whether the equations have been understood is to ask what each one rules out.

The first forbids a field diverging from nothing. Any region from which net flux emerges contains charge, in exact proportion, which is why a hollow conductor shields and why a charge cannot be quietly created in a box.

The second forbids an isolated north pole. Field lines have no ends, so any magnetic field line followed far enough returns to where it began.

The third forbids a static description of a changing world. It also, as the ninth lesson showed, forbids the electric potential from existing wherever a magnetic flux is changing.

The fourth, with its new term, forbids the current from vanishing. Combining it with the first gives charge conservation as a theorem rather than an assumption: the equations cannot be satisfied by a process that destroys charge. Ampère's law without the displacement current was, in this precise sense, incompatible with one of the best-tested facts in physics, and Maxwell's term is what reconciles them.

The equations with nothing in them

The last question is the one that makes this course's title too small for its subject.

Take the four equations and set the charge and the current to zero. Empty space, no sources, nothing anywhere. Two of the four become trivial: no flux of either kind out of any closed surface. The other two do not:

\oint E \cos φ d l = - \frac{d Φ_{B}}{d t}, \oint B \cos φ d l = μ_{0} ε_{0} \frac{d Φ_{E}}{d t}

Read them together. A changing magnetic field produces a circulating electric field. A changing electric field produces a circulating magnetic field. Neither needs a charge, a current or a wire. Each is the other's source, and the pair could in principle sustain each other indefinitely, propagating through vacuum with nothing material involved at any point.

Before the displacement current, that possibility did not exist: the second equation would have read zero, and a changing magnetic field would have produced an electric field that then did nothing. One term turned a one-way relationship into a loop.

Whether the loop actually closes, and at what speed the result travels, is a matter of solving the two equations rather than admiring them. That is the last lesson, and the constant $μ_{0} ε_{0}$ sitting in the second one has already given the answer away.

Electromagnetic waves

The previous lesson ended with two equations that describe empty space and refuse to be trivial, each field driving the other with no charge anywhere in sight.

This lesson solves them. The mathematics needs one new piece of notation, the partial derivative, introduced below in a sentence, and the result is that optics stops being a separate subject. The lesson also names the places where the classical theory runs out, because it does, and knowing where is what makes the rest of it trustworthy.

Guessing the shape of the answer

Solving a pair of field equations in full generality is a large task. Solving them for one particular shape of field is not, and if that shape works it is a solution, which is all that is needed to establish that such things exist.

So assume the simplest possible arrangement. Let the electric field point along $y$ and the magnetic field along $z$ , and let both depend only on the coordinate $x$ and on time. That describes a plane wave travelling along $x$ : at any instant the fields are uniform across every plane perpendicular to $x$ and vary only as you move along it.

The notation $\partial E / \partial x$ means the rate of change of $E$ with respect to $x$ at a fixed instant, and $\partial E / \partial t$ the rate of change with time at a fixed place. That is all a partial derivative is: an ordinary derivative with the other variable held still.

Two thin rectangles, and a wave equation

The strategy is to apply each of the two surviving equations to a rectangle so thin that the result becomes a statement about the fields at a point.

Draw the first rectangle in the $x y$ plane with one pair of sides of length $a$ running along $y$ , at positions $x$ and $x + Δ x$ , and the other pair, of length $Δ x$ , running along $x$ .

Take the circulation of $E$ round it. The electric field points along $y$ , so the two sides running along $x$ contribute nothing, since the field is perpendicular to them. The two sides of length $a$ contribute $E$ at their own position times $a$ , with opposite signs because the circuit traverses them in opposite directions, so the total is

a [E (x + Δ x) - E (x)] = a \frac{\partial E}{\partial x} Δ x

Now the flux of $B$ through the rectangle. The magnetic field points along $z$ , perpendicular to the rectangle, so the flux is $B$ times the area $a Δ x$ , and its rate of change is $(\partial B / \partial t) a Δ x$ .

Faraday's law equates the first to minus the second, and $a Δ x$ cancels from both sides:

\frac{\partial E}{\partial x} = - \frac{\partial B}{\partial t}

The area of the rectangle dropping out is what makes this a statement about the fields at a point rather than about a particular loop.

Now repeat with a rectangle in the $x z$ plane, sides of length $b$ along $z$ at $x$ and $x + Δ x$ . The magnetic field points along $z$ , so the same argument gives its circulation as $b (\partial B / \partial x) Δ x$ . The electric flux through this rectangle is $E$ times $b Δ x$ , since $E$ is perpendicular to it.

There are no charges and no currents, so the only term on the right is the displacement one, and

\frac{\partial B}{\partial x} = - ε_{0} μ_{0} \frac{\partial E}{\partial t}

The two minus signs are what the right hand rules give when the orientations are tracked consistently, and the reader who works them through and gets a different pair of signs has probably chosen the opposite direction of travel, which is equally valid.

That leaves two coupled equations, each relating one field to the other. Eliminate $B$ in the standard way: differentiate the first with respect to $x$ ,

\frac{\partial^{2} E}{\partial x^{2}} = - \frac{\partial}{\partial x} \frac{\partial B}{\partial t} = - \frac{\partial}{\partial t} \frac{\partial B}{\partial x}

where the order of the two derivatives can be swapped for any well behaved function. Substituting the second equation for $\partial B / \partial x$ , the two minus signs cancel:

\frac{\partial^{2} E}{\partial x^{2}} = ε_{0} μ_{0} \frac{\partial^{2} E}{\partial t^{2}}

That is the wave equation. Any function of the form $f (x - v t)$ satisfies it, provided $v^{2} = 1 / (ε_{0} μ_{0})$ : such a function is a fixed shape sliding along the $x$ axis at speed $v$ without changing, which is what a wave is. The identical elimination the other way gives the same equation for $B$ , so both fields travel together.

So electromagnetic waves exist, they are predicted by the four equations with nothing else assumed, and they travel at

v = \frac{1}{\sqrt{ε_{0} μ_{0}}}

The number

Put in the constants. $ε_{0} = 8.8542 \times 10^{- 12}$ F/m, measured with capacitors, and $μ_{0} = 1.2566 \times 10^{- 6}$ T m/A, measured with the force between current-carrying wires. Their product is $1.1127 \times 10^{- 17}$ , and

v = \frac{1}{\sqrt{1.1127 \times 10^{- 17}}} = 2.9979 \times 10^{8} m/s

That is the speed of light, obtained from two electrical measurements involving no light whatsoever.

Maxwell had the comparison in 1862. The electrical constants came from Weber and Kohlrausch's 1856 experiment, in which they measured a capacitor's charge twice, once electrostatically with a torsion balance and once magnetically by discharging it through a galvanometer, and took the ratio: 310,740,000 m/s, high by 3.7 per cent. The speed of light came from Fizeau's 1849 toothed wheel, 314,858,000 m/s, high by 5.0 per cent. Maxwell wrote that the velocity of transverse undulations in his hypothetical medium agreed so exactly with the velocity of light that we can scarcely avoid the inference that light consists in the transverse undulations of the same medium which is the cause of electric and magnetic phenomena.

Two independently measured constants of electricity, each known to a few per cent, combining to give a third quantity from an entirely different branch of physics: that is the strongest kind of evidence a physical theory can offer, because there was no adjustable parameter anywhere to make it come out right.

The confirmation took twenty-five years. Heinrich Hertz, between 1887 and 1888, built an oscillator from a spark gap between two metal spheres and a detector from a loop with a smaller gap, and saw sparks in the detector when the transmitter fired across the room. He measured the wavelength by finding the nodes of a standing wave reflected off a metal sheet, computed the speed, and got light's. He then showed the waves reflect, refract through a pitch prism, and can be polarised by a grid of wires. Asked what use it was, he said none whatsoever, and died in 1894 at thirty-six, seven years before Marconi sent a signal across the Atlantic.

What the wave looks like

The solution carries more information than its speed.

$E$ and $B$ are perpendicular to each other and both perpendicular to the direction of travel: the wave is transverse, which is why light can be polarised, and polarisation is direct evidence for the transverse character since a longitudinal wave has no orientation to filter.

They are in phase, peaking and vanishing together, and their magnitudes are locked. Substituting a travelling sine wave into either coupled equation gives

E = c B

so the fields are far from equal in size in SI units, which is a fact about the units rather than about the wave: the energy in the two is identical, as the next section shows.

Nothing in the derivation fixed the frequency. Any $f (x - c t)$ works, so the equations permit waves of every wavelength, all travelling at the same speed in vacuum, and the electromagnetic spectrum is one continuum. Radio at a hundred megahertz has a wavelength of three metres; a microwave oven at 2.45 GHz gives 12.2 cm, and the half-wavelength spacing of 6.1 cm between the nodes of the standing wave inside is why an oven has a turntable. Green light at 550 nm is $5.45 \times 10^{14}$ Hz. X-rays and gamma rays continue the same list. Radio and gamma rays differ by nineteen orders of magnitude in frequency and obey the identical four equations.

Example. A microwave oven operates at 2.45 GHz. What is the wavelength, and how far apart are the hot spots of the standing wave inside?

$λ = c / f = (2.998 \times 10^{8}) / (2.45 \times 10^{9}) = 0.122$ m, so 12.2 cm. A standing wave has antinodes every half wavelength, so the hot spots sit 6.1 cm apart, which is a substantial fraction of a dinner plate.

Now you. A mobile phone transmits at 1800 MHz. What is the wavelength?

Answer

$λ = c / f = (2.998 \times 10^{8}) / (1.80 \times 10^{9}) = 0.167$ m, about 16.7 cm. A quarter of that, 4.2 cm, is a typical internal antenna length.

Energy, and how much

The two energy densities from earlier lessons apply at every point of the wave: $\frac{1}{2} ε_{0} E^{2}$ from the fifth and $B^{2} / 2 μ_{0}$ from the tenth. Substituting $B = E / c$ and $c^{2} = 1 / (ε_{0} μ_{0})$ into the second gives $ε_{0} E^{2} / 2$ , identical to the first. A light wave carries exactly half its energy in the electric field and half in the magnetic, at every point and every instant.

The energy passing through unit area per second is the density times the speed, $u_{total} c = ε_{0} c E^{2}$ , usually written as the Poynting vector $S = E B / μ_{0}$ , which points in the direction of travel. For a sine wave the average of $E^{2}$ over a cycle is half the peak squared, so the intensity is $\frac{1}{2} ε_{0} c E_{0}^{2}$ , or equivalently $ε_{0} c E_{rms}^{2}$ .

This is the promissory note of the fifth lesson being redeemed. The energy density $\frac{1}{2} ε_{0} E^{2}$ was derived there from a capacitor and could have been dismissed as an algebraic rearrangement. Here the same expression describes energy in a region with no charge in it, travelling, arriving somewhere else, and warming a thermometer. The energy is genuinely in the field.

Example. Sunlight above the atmosphere delivers 1361 W/m². What are the electric and magnetic field strengths in it?

From $S = ε_{0} c E_{rms}^{2}$ , $E_{rms}^{2} = 1361 / [(8.854 \times 10^{- 12}) (2.998 \times 10^{8})] = 1361 / (2.654 \times 10^{- 3}) = 5.13 \times 10^{5}$ , so $E_{rms} = 716$ V/m and the peak is $\sqrt{2}$ times that, 1013 V/m. The magnetic field follows from $B = E / c$ : $B_{rms} = 716 / (2.998 \times 10^{8}) = 2.39 \times 10^{- 6}$ T, about five per cent of the Earth's field. Sunlight is a 700 V/m electric field oscillating $5 \times 10^{14}$ times a second, which is not how it feels.

Now you. A laser delivers 5000 W/m² onto a surface. What is the rms electric field in the beam?

Answer

$E_{rms}^{2} = 5000 / [(8.854 \times 10^{- 12}) (2.998 \times 10^{8})] = 5000 / (2.654 \times 10^{- 3}) = 1.88 \times 10^{6}$ , so $E_{rms} = 1.37 \times 10^{3}$ V/m.

Momentum and pressure

A wave that carries energy also carries momentum, at $p = U / c$ per unit energy, a result that comes out of Maxwell's equations and that relativity later makes inevitable for anything moving at $c$ . Absorbing a wave therefore delivers momentum, which is a force, which over an area is a pressure:

P = \frac{S}{c}

for an absorbing surface, and twice that for a perfect mirror, because the reflected wave carries momentum away in the opposite direction.

The numbers are tiny. Sunlight at 1361 W/m² pushes on a black surface with $1361 / (2.998 \times 10^{8}) = 4.54 \times 10^{- 6}$ Pa, ten orders of magnitude below atmospheric pressure. It is nonetheless real, it was measured by Lebedev in 1900 and by Nichols and Hull in 1901, and it is what shapes a comet's tail, which points away from the Sun regardless of which way the comet is going.

Example. A perfectly reflecting solar sail of 200 m² is deployed at the Earth's distance from the Sun on a 10 kg spacecraft. What thrust does it produce, and what speed change over a year of continuous thrust?

The pressure on a mirror is $2 S / c = 2 (1361) / (2.998 \times 10^{8}) = 9.08 \times 10^{- 6}$ Pa. Over 200 m² that is a force of $1.82 \times 10^{- 3}$ N, under two millinewtons, roughly the weight of a grain of rice. On 10 kg it gives an acceleration of $1.82 \times 10^{- 4}$ m/s², and over a year of $3.16 \times 10^{7}$ s the speed change is $5.7 \times 10^{3}$ m/s. A thrust too small to feel, applied for a year, beats a chemical rocket stage. The Japanese probe IKAROS demonstrated this in 2010 with a sail of about 196 m², measuring a thrust near 1.12 mN, below the ideal figure because a real sail is neither perfectly reflecting nor perfectly flat.

Now you. A perfectly absorbing sheet of 50 m² is held at the same distance from the Sun. What force does the sunlight exert on it?

Answer

An absorber feels $S / c = 1361 / (2.998 \times 10^{8}) = 4.54 \times 10^{- 6}$ Pa, so over 50 m² the force is $2.27 \times 10^{- 4}$ N, about a fifth of a millinewton.

Where this stops

The theory just completed is one of the most successful in physics, and it has edges, three of which were found within twenty years of Hertz.

There is no medium. Maxwell believed his waves were undulations of a material ether, and the whole nineteenth century framework assumed one. Michelson and Morley in 1887 looked for the Earth's motion through it, with an interferometer sensitive enough to detect a hundredth of the expected effect, and found nothing. The equations turn out not to need a medium: the fields are the wave. Einstein's 1905 paper, which opened with the induction problem of the ninth lesson, disposed of the ether and made the constancy of $c$ a postulate, and Maxwell's equations then turn out to be exactly relativistic already, which is why they needed no correction when mechanics did.

Energy is not continuous. The classical wave carries as little energy as you like, arriving smoothly. Shine dim ultraviolet on a metal and electrons come off immediately with an energy fixed by the frequency, not by the intensity, which is impossible on the classical picture and which Einstein explained in the same year by quantising the wave into photons of energy $h f$ . The classical theory works when the photon count is enormous, which for sunlight and radio it always is, and fails when it is small.

Accelerating charges radiate, which classically destroys the atom. An orbiting electron is accelerating, and the equations of this course say it must radiate, losing energy and spiralling into the nucleus in about $10^{- 11}$ seconds. Atoms exist. That contradiction is one of the doors into quantum mechanics, and Atoms and Elements goes through it.

None of this makes the four equations wrong. Quantum electrodynamics, the theory that replaces them at small scales, reproduces every result in this course in the limit of many photons, and it is the most precisely tested theory ever constructed, agreeing with measurement of the electron's magnetic moment to twelve significant figures. What has changed is the domain, not the content.

What was built

The course began with two charged rods and a torsion balance, and a force law that said what two charges do and could not say how.

Replacing the force with a field made the how expressible. Symmetry and energy gave two ways of computing that field, one closed surface at a time and one scalar function at a time. Charge in motion needed a second field, defined by a sideways velocity-dependent force, and produced by currents. Changing either field made the other, which gave induction, the generator and the whole electrical industry, and which cost the electric potential its existence. One missing term, forced by a capacitor and a choice of surface, completed the set and let the two fields sustain each other with no matter present at all.

Then the four equations, containing nothing but charges, currents and two constants measured with wire and glass, turned out to describe light. That was not a goal anyone had set. It is what the equations said when they were solved, and the fact that a theory can know more than the people who built it is the best argument there is for taking a derivation seriously enough to follow it to the end.

Electricity and Magnetism, from libre.university