Griffiths Chapter 4: induced and permanent dipoles, polarization P, bound surface and volume charges, the electric displacement D and Gauss's law with free charge, linear dielectrics, dielectric capacitors, and energy in dielectric systems.
Subject: Electrodynamics (Griffiths) · 60 slides · diagram-first lesson
Open the interactive version of this deck · Homework for this lesson
Title
Griffiths, Chapter 4
Polarization, bound charge, the displacement D, and linear dielectrics
Objectives
Chapters 2 and 3 assumed everything happened in vacuum with a few conductors. Real space is full of insulating matter, and it responds.
The organising idea: matter is a sea of tiny dipoles. Everything here follows from what a field does to them, and what they then do back.
Here is the gap this deck closes. Chapters 2 and 3 assumed everything happened in vacuum, with a few conductors sitting in otherwise empty space. Real apparatus is full of insulating matter — plastic, glass, oil, air, water — and every one of those responds to a field.
The organising idea fits in a sentence: matter is a sea of tiny dipoles. Everything here follows from what a field does to those dipoles, and what they then do back to the field.
The arc runs outward from the single atom. First what a field does to one atom or molecule, then how to describe the average over enormous numbers of them, then what field that averaged description produces, and finally the shortcut that makes real problems solvable.
Section
Section 1
Concept
Figure (svg): Left: a conductor with electrons flowing freely to the surface. Right: a dielectric where each atom's charge shifts slightly but nothing travels.
In an insulator every electron is bound to a particular atom or molecule. A field cannot make them travel — but it can make them shift, by a distance far smaller than an atom.
Tiny as those shifts are, there are around 10^29 atoms per cubic metre, and their combined effect is large enough to change a capacitor's capacitance by a factor of 80.
Get this contrast straight first, because conductor intuitions carried into dielectric problems produce systematically wrong answers.
In a conductor, some electrons are free to travel through the whole material. In an insulator, every electron is bound to a particular atom or molecule and can only shift slightly within it.
So a field cannot make charge flow in a dielectric — but it can make charge shift, by a distance far smaller than an atom. The picture is the contrast: charge travelling all the way to the surface on the left, charge stretching but staying home on the right.
Those tiny shifts matter because of sheer numbers. There are something like ten to the twenty-nine atoms in a cubic metre, and their combined effect can change a capacitor's capacitance by a factor of eighty.
Concept
Figure (svg): An animation of an atom whose nucleus is pulled off-centre and back as an applied field switches on and off, stretching it into a dipole.
\[ \mathbf{p} = \alpha\mathbf{E} \]
atomic polarizability — The constant alpha relating an atom's induced dipole moment to the field that induced it. Bigger, floppier atoms have larger alpha — caesium polarises about 50 times more easily than helium.
The relationship is linear because the displacement is minuscule. Push much harder and the atom ionises, and the linear model collapses — but ordinary fields are nowhere near that.
This is the single-atom picture, and the animation makes the mechanism visible: the nucleus is pulled one way, the electron cloud the other, and the atom stretches.
The result is an induced dipole moment proportional to the applied field, with the constant of proportionality called the atomic polarizability. Bigger, floppier atoms have a larger one — caesium polarises roughly fifty times more easily than helium.
The proportionality deserves a word. It holds because the displacement is minuscule compared with the size of the atom, so the restoring force is effectively linear — exactly like a spring under a small load.
Push much harder and the atom ionises, and the linear model collapses entirely. Ordinary laboratory fields are nowhere near that threshold, which is why the linear description works as well as it does.
Worked example
Griffiths' Example 4.1. A point nucleus of charge +q inside a uniform spherical electron cloud of charge −q and radius a.
Figure (svg): A nucleus displaced a small distance d from the centre of a uniform negative sphere, with the restoring field marked.
Write the field inside a uniformly charged sphere at distance d from its centre, in newtons per coulomb
Why: This is the Deck 2 result: inside a uniform ball of charge the field grows linearly with distance from the centre.
\[ E_e = \frac{1}{4\pi\varepsilon_0}\frac{qd}{a^3} \]
Set the external field equal to this restoring field at equilibrium
Why: The nucleus stops moving when the outward push of the applied field exactly balances the inward pull of the cloud it has been displaced from.
\[ E = \frac{1}{4\pi\varepsilon_0}\frac{qd}{a^3} \quad\Rightarrow\quad qd = 4\pi\varepsilon_0 a^3 E \]
Recognise the left-hand side as the induced dipole moment, in coulomb metres
Why: Charge q separated by distance d is exactly a dipole moment p = qd.
\[ \mathbf{p} = 4\pi\varepsilon_0 a^3\,\mathbf{E} \quad\Rightarrow\quad \alpha = 4\pi\varepsilon_0 a^3 \]
Verify the result against physical expectations, in coulomb metres squared per volt
Why: Alpha depends only on the atom's VOLUME — a bigger atom polarises more easily, exactly as measured. Putting in a typical atomic radius of 0.05 nm gives roughly 10⁻⁴¹ C m²/V, within an order of magnitude of the measured value for hydrogen. Not bad for a model with one parameter.
A satisfying calculation, because a crude model gets within an order of magnitude of the measured answer.
The model is a point nucleus inside a uniform sphere of negative charge. When a field displaces the nucleus, it feels a restoring pull from the cloud it has moved away from — and the field inside a uniformly charged sphere is a result already derived back in Deck 2.
Equilibrium is where the external push balances that internal pull. Solve for the displacement, then recognise charge times displacement as the induced dipole moment.
The striking part is the result: the polarizability depends only on the atom's volume. Bigger atom, easier to polarise — which is exactly what is measured. Put in a typical atomic radius and the number lands within an order of magnitude of the experimental value for hydrogen, which is remarkable for a model with one parameter and no quantum mechanics in it.
Concept
Some molecules are born with a dipole moment. Water is the standard example — its two hydrogens sit on one side of the oxygen.
Figure (svg): An animation of three misaligned polar molecules swinging round until their dipole arrows line up with a steady applied field.
\[ \mathbf{N} = \mathbf{p}\times\mathbf{E} \]
The torque is zero when p is parallel to E — that is the stable alignment. Antiparallel is also torque-free, but unstable, like a pendulum balanced upside down.
Alignment is never complete: thermal motion constantly knocks molecules out of line. Raise the temperature and a material's polarisation drops, which is why dielectric constants are temperature-dependent.
This is the second mechanism, and it is physically quite different from stretching.
Some molecules are born with a dipole moment, because their charge is not symmetrically arranged. Water is the standard example — the two hydrogens sit on one side of the oxygen, at about a hundred and five degrees.
A field does not need to create a dipole here; one already exists. What the field does is torque it into alignment, which the animation shows for three molecules starting at different angles.
The torque is zero when the dipole is parallel to the field, which is the stable alignment, and also when antiparallel — which is unstable, like a pendulum balanced upside down.
One crucial caveat: alignment is never complete, because thermal motion constantly knocks molecules out of line. That is why polar materials have temperature-dependent dielectric constants, and it is the main way this mechanism differs from stretching.
Concept
In a uniform field the two ends of a dipole feel equal and opposite forces: torque, but no net force.
Figure (svg): Left: a dipole in a uniform field with balanced forces. Right: the same dipole in a converging field where the near end feels a stronger pull.
\[ \mathbf{F} = (\mathbf{p}\cdot\nabla)\mathbf{E} \]
In a non-uniform field the ends are in different field strengths and the forces no longer cancel. Once aligned, a dipole is always pulled toward the region of stronger field.
This is why a charged comb picks up neutral scraps of paper: the comb polarises them, then pulls the induced dipoles into its own strong near field.
Keep torque and force firmly separate here, because they routinely get merged.
In a uniform field, the two ends of a dipole feel equal and opposite forces. There is a torque, which rotates it, but no net force, so it does not translate.
In a non-uniform field the two ends sit in different field strengths, so the forces no longer cancel. Once aligned, the dipole gets pulled toward the region of stronger field — always.
That single result explains a whole family of everyday observations: a charged comb picking up paper, a balloon sticking to a wall, a stream of water bending toward a charged rod. In each case a neutral object is polarised first and then pulled into the stronger field near the source.
Concept
Tracking 10^29 individual dipoles is hopeless. Average them instead.
\[ \mathbf{P} \equiv \text{dipole moment per unit volume} \quad \text{(C/m}^2\text{)} \]
Figure (svg): A block of material with many small aligned dipole arrows inside it, and a single large arrow beside it labelled P.
P is a field: it can vary from point to point, and it points along the average dipole alignment. Its units are coulombs per square metre — the same as a surface charge density, which is a hint about what comes next.
Chapter 4's central question is now sharp: given P, what field does the material produce?
The move here is from microscopic to macroscopic, and it is a modelling decision worth naming as one.
Tracking ten to the twenty-nine individual dipoles is hopeless, so instead we average: the polarization P is the dipole moment per unit volume.
P is a field in its own right. It can vary from place to place, it points along the average dipole alignment, and its magnitude says how strongly the material is polarised.
Look at the units — coulombs per square metre, the same as a surface charge density. That is not a coincidence, and it is a strong hint about what is coming in the next section.
With P defined, the central question of the chapter becomes sharp: given P, what field does the material produce?
Intuition
'Polarisation' covers two physically distinct processes, and materials differ in which one dominates.
| mechanism | what moves | temperature dependence | typical materials |
|---|---|---|---|
| stretching (induced) | the electron cloud, within each atom | almost none | noble gases, oils, most solids |
| aligning (orientational) | whole molecules, rotating | strong — heat fights alignment | water, HCl, polar liquids |
The stretching mechanism is nearly temperature-independent because it is an electrostatic tug-of-war inside an atom, and thermal energy is far too small to matter there.
Alignment is a different story: thermal collisions constantly knock molecules out of line, so a polar material's dielectric constant falls as it gets hotter. Water's drops from about 88 near freezing to about 55 near boiling.
This table prevents a real confusion, because 'polarisation' names two physically different processes.
Stretching is an electrostatic tug-of-war inside a single atom. Thermal energy is far too small to interfere with it, so it is essentially temperature-independent.
Aligning is a competition between the field's torque and thermal agitation, so it depends strongly on temperature. Heat the material and the alignment degrades.
The concrete evidence is water: its dielectric constant falls from about eighty-eight near freezing to about fifty-five near boiling. Nothing about the individual molecules changed — only how well the field can keep them lined up against the thermal buffeting.
Estimation
A field of 10⁶ V/m — near the breakdown limit of air — is applied to a hydrogen-like atom with polarizability α ≈ 10⁻⁴¹ C·m²/V.
Predict first
Roughly how far does the nucleus shift relative to the electron cloud?
Correct: About 10⁻¹⁷ m — far smaller than the atom
Why: p = αE = (10⁻⁴¹)(10⁶) = 10⁻³⁵ C·m. Dividing by the elementary charge, d = p/e = 10⁻³⁵/1.6 × 10⁻¹⁹ ≈ 6 × 10⁻¹⁷ m. An atom is about 10⁻¹⁰ m across, so the displacement is roughly a millionth of the atom's size — which is exactly why the linear model p = αE works so well, and why the cloud can be assumed to keep its shape.
Tiny displacements, enormous numbers of atoms. That is the whole story of dielectrics in one sentence.
Worth doing, because this estimate justifies every approximation made so far.
Take a near-breakdown field and a typical polarizability, compute the induced dipole moment, and divide by the elementary charge to get an effective displacement.
The answer is around ten to the minus seventeen metres, against an atomic size of about ten to the minus ten. The displacement is roughly a millionth of the atom's width.
Which is why the linear relation holds so well, and why the model could assume the electron cloud keeps its spherical shape. The perturbation is genuinely tiny. Tiny displacements and astronomical numbers of atoms — that is the whole story of dielectrics in one line.
Real world
Discussion prompt
A neutral object has no net charge. So why is the attraction never repulsion?
Every one of these is a neutral object being attracted. That should be impossible for a monopole — and it is exactly what the dipole force in a non-uniform field predicts.
Four things you have almost certainly done or watched.
The common thread is that a neutral object gets attracted in every case. That should be impossible for anything obeying a simple monopole rule, which is exactly why these observations need this chapter to explain them.
The prompt sharpens the logic. The induced dipole always lines up with the field, so its near end always carries the charge opposite to the source, and the near end always sits in the stronger field. Attraction always wins.
A permanent dipole could in principle be held backwards and repelled, but the torque would flip it into alignment immediately. Induced polarisation never gets the sign wrong in the first place.
Pattern
The last point is the one that will bite in Section 4. The material's own bound charge contributes to the field that polarises it — a self-consistency requirement that the susceptibility quietly absorbs.
This is the procedure for the microscopic half of the chapter, before bound charges enter.
Identify the mechanism, get the individual dipole moment, multiply by number density to get P, and keep torque and force distinct — torque aligns, force needs a gradient.
The last point matters most for what follows, and it is worth flagging as a warning: the field that counts is the total field inside the material, not the one applied from outside. The material's own bound charge contributes to the field that polarises it.
That self-consistency requirement is what makes the naive approach circular, and it gets quietly resolved by the susceptibility later in the deck.
Check
Try it before clicking.
Check your understanding
A neutral atom is placed in a uniform electric field. What happens to it?
Answer: A
Why: The field pulls the nucleus one way and the electron cloud the other, creating an induced dipole p = αE. But the field is uniform, so the two ends feel equal and opposite forces and there is no net force. A net force requires a field GRADIENT.
This checks whether torque and force have stayed separate.
A neutral atom in a uniform field acquires an induced dipole but feels no net force, because the field is identical at both ends.
Option D is the interesting wrong answer. An induced dipole is created already pointing along the field, so there is nothing left to rotate. Rotation applies only to permanent dipoles that started out misaligned, and conflating the two mechanisms is what makes that option tempting.
Option C is worth a sentence too — neutral means no net charge, not no response.
Section
Section 2
Prediction
Figure (svg): A dipole sitting at an angle to a uniform field, with its positive end up and to the right and the field pointing to the right.
Predict first
A dipole sits at 45 degrees to a uniform field. What happens to it?
Correct: It rotates to line up with the field, and does not translate
Why: The torque p × E is non-zero whenever p and E are not parallel, so the dipole rotates toward alignment. But the field is uniform, so the forces on its two ends are exactly equal and opposite and there is no net force. Rotation without translation. Add a field gradient and you would also get translation, toward the stronger field.
Once aligned, the torque vanishes and the dipole sits still. That aligned state is what a polarised material is made of, molecule by molecule.
A quick commitment, though the reasoning is worth stating even where the answer is straightforward.
The torque is the cross product of p with E, which is non-zero whenever the two are not parallel, so the dipole rotates toward alignment.
But the field is uniform, so the forces on the two ends are exactly equal and opposite and there is no translation. Rotation without translation.
Add a gradient and you would get translation as well, toward the stronger field. Separating those two effects cleanly is the whole point of the slide.
Once aligned the torque vanishes and the dipole sits still — and that aligned state is exactly what a polarised material is made of, molecule by molecule.
Concept
Rather than adding up 10^29 dipole fields, Griffiths shows the whole object is equivalent to a surface charge plus a volume charge.
\[ \sigma_b = \mathbf{P}\cdot\hat{\mathbf{n}}, \qquad \rho_b = -\nabla\cdot\mathbf{P} \]
Figure (svg): A polarised block with aligned dipoles: positive charge accumulating on the right face, negative on the left, and the interior neutral.
Look at the interior: each dipole's positive end sits right next to the next dipole's negative end, and they cancel. Only at the surfaces is there nothing to cancel against.
If the polarisation is non-uniform, the cancellation is imperfect and a volume charge appears too — which is exactly what the divergence measures.
This is the central result of the chapter and what makes the subject tractable, so it deserves room.
Rather than adding up ten to the twenty-nine individual dipole fields, Griffiths shows the entire polarised object is exactly equivalent to a surface charge plus a volume charge.
The picture is the argument. Inside a uniformly polarised block, each dipole's positive end sits right next to its neighbour's negative end and they cancel in pairs, leaving the interior neutral. Only at the surfaces is there nothing left to cancel against, so charge accumulates there.
If the polarisation is non-uniform, the cancellation is imperfect and a volume charge appears too — and that imperfection is precisely what the divergence measures.
Intuition
Both formulas can be read straight off the picture, with no calculation.
Figure (svg): Left: dipoles hitting a surface end-on, leaving charge on it. Right: dipoles running parallel to a surface, leaving nothing.
The crucial point: these are real charges. They are made of real electrons and nuclei that really have shifted. 'Bound' only means they cannot travel through the material, not that they are fictitious.
Both formulas can be read straight off the picture, which is far better than memorising them.
For the surface charge: on a face perpendicular to P, every dipole in the surface layer sticks an end out and the full magnitude of P accumulates. On a face parallel to P, the dipoles run along the surface and nothing protrudes. That is exactly what a dot product with the outward normal gives — full value when parallel, zero when perpendicular.
For the volume charge: if the polarisation spreads outward, each little region sends more positive charge out than it receives, so it is left negative. Hence the minus sign in front of the divergence.
And here is the point that must not slide past: these are real charges. Real electrons and nuclei that really have moved. 'Bound' means they cannot travel through the material, not that they are imaginary.
Worked example
Griffiths' Example 4.2, and the single most reused result of the chapter.
Figure (svg): A sphere with uniform upward polarisation, showing positive bound charge on the top hemisphere and negative on the bottom, densest at the poles.
Find the volume bound charge, in coulombs per cubic metre
Why: P is uniform, so its divergence is zero and there is no bound charge in the interior at all.
\[ \rho_b = -\nabla\cdot\mathbf{P} = 0 \]
Find the surface bound charge, in coulombs per square metre
Why: On a sphere the outward normal is r-hat, and with P along z the dot product gives a cosine.
\[ \sigma_b = \mathbf{P}\cdot\hat{\mathbf{n}} = P\cos\theta \]
Recognise this as a problem already solved in Chapter 3
Why: A surface charge varying as cos theta on a sphere is exactly Example 3.9 — which the separation-of-variables machinery from the last deck handled.
\[ V(r,\theta) = \begin{cases} \dfrac{P}{3\varepsilon_0}r\cos\theta, & r<R \\[10pt] \dfrac{P}{3\varepsilon_0}\dfrac{R^3}{r^2}\cos\theta, & r>R\end{cases} \]
Take the gradient inside, in newtons per coulomb
Why: Since r cos theta is just z, the interior potential is linear in z and its gradient is a constant.
\[ \mathbf{E}_{\text{in}} = -\nabla V = -\frac{1}{3\varepsilon_0}\mathbf{P} \]
Verify both regions, in newtons per coulomb
Why: Inside, the field is perfectly UNIFORM and points OPPOSITE to the polarisation — a depolarising field, which is what you would expect from negative charge at the top and positive at the bottom of the interior. Outside, the potential is exactly that of a perfect dipole with p = (4/3)πR³P, which is the sphere's volume times its polarisation — the total dipole moment, as it must be.
The most reused result in the chapter, so the logic matters as much as the answer.
Uniform P means zero divergence, so there is no volume bound charge at all — everything happens on the surface. On a sphere the outward normal is r-hat, so the surface charge varies as the cosine of the polar angle: maximal at the poles, zero at the equator, positive on one hemisphere and negative on the other.
The elegant step is recognising that a cos-theta surface charge on a sphere is a problem already solved in Chapter 3 by separation of variables. Nothing new is needed; the previous deck did the work.
Both results are worth dwelling on. Inside, the field is perfectly uniform and points opposite to the polarisation — a depolarising field, which makes sense given negative charge at the top and positive at the bottom. Outside, it is exactly a perfect dipole field, with moment equal to the volume times P.
That there are no higher multipoles at all outside is genuinely surprising and worth noticing.
Worked example
When P varies with position, bound charge appears in the volume too. Take a sphere of radius R with a radially increasing polarisation.
\[ \mathbf{P} = k\,r\,\hat{\mathbf{r}} \]
Compute the volume bound charge with the spherical divergence, in coulombs per cubic metre
Why: Use the spherical form of the divergence, which carries the r-squared factor from Deck 1.
\[ \rho_b = -\nabla\cdot\mathbf{P} = -\frac{1}{r^2}\frac{\partial}{\partial r}(r^2\cdot kr) = -\frac{1}{r^2}(3kr^2) = -3k \]
Compute the surface bound charge at r = R, in coulombs per square metre
Why: On the outer surface the normal is r-hat, so the dot product just picks up the magnitude of P there.
\[ \sigma_b = \mathbf{P}\cdot\hat{\mathbf{n}}\big|_{r=R} = kR \]
Check that the object is neutral overall, in coulombs
Why: The volume charge is uniform at −3k throughout, and the surface charge is kR spread over the sphere's area.
\[ Q_{\text{vol}} = -3k\cdot\tfrac{4}{3}\pi R^3 = -4\pi kR^3, \qquad Q_{\text{surf}} = kR\cdot4\pi R^2 = 4\pi kR^3 \]
Verify: the two must cancel exactly, in coulombs
Why: They do — the total is zero, as it must be for a polarised but uncharged object. Polarisation only ever moves charge around inside an object; it never creates any. If your bound charges do not cancel, you have made an error.
This one exercises the volume bound charge, which the uniform case never touched.
The polarisation grows radially, so its divergence is non-zero and there is bound charge throughout the volume as well as on the surface. Use the spherical form of the divergence, remembering the r-squared factor from Deck 1.
The check at the end is the habit worth installing: compute both the total volume charge and the total surface charge, and confirm they cancel exactly.
They have to cancel, because polarisation only ever moves charge around inside an object — it never creates any. If bound charges do not sum to zero for an uncharged dielectric, something has gone wrong, and this is the fastest way to catch it.
Concept
At the atomic scale the field inside matter is wildly non-uniform: enormous near a nucleus, reversed between atoms, changing over distances of 10⁻¹⁰ m.
Figure (svg): Two panels: a jagged microscopic field varying violently at atomic scale, and the same data smoothed into a gentle macroscopic average.
The E in this chapter is the macroscopic field: the microscopic field averaged over a region containing thousands of molecules but still tiny by laboratory standards.
Griffiths shows that this average is exactly what you get by treating the material as a smooth distribution of bound charge — which is why the bound-charge picture is not an approximation but a theorem about averages.
This addresses a sharp question that most textbooks skate over.
At the atomic scale, the field inside matter is violently non-uniform: enormous near a nucleus, reversed between atoms, changing completely over distances of a tenth of a nanometre. Nothing in this chapter describes that field.
The E in these equations is the macroscopic field — the microscopic field averaged over a region containing thousands of molecules but still tiny by laboratory standards.
The reassuring part is that Griffiths proves this average is exactly what you get by treating the material as a smooth distribution of bound charge. So the bound-charge picture is not a convenient approximation; it is a theorem about averages, which is why it gives exact answers.
Trap
Asked whether the bound charge on a polarised block would show up on a measurement, a student answers:
Say no — bound charge is a mathematical device for replacing the dipole sum
Why: It was introduced as a way to avoid adding 10^29 dipole fields, so it sounds like an accounting trick.
But then a polarised block would exert no force on a nearby test charge, which is false — it does.
Bound charge is real, physical, measurable charge:
Keep the distinction that matters: bound charge cannot MOVE through the material
Why: Free charge conducts; bound charge is stuck with its molecule. That mobility difference — not reality — is what the word 'bound' names.
This is exactly why a charged balloon sticks to a neutral wall: the wall polarises, real bound charge appears on its surface, and it really does attract.
The misconception arises naturally, since bound charge was introduced as a device for avoiding a sum over dipoles.
The refutation is empirical: a polarised block really does exert a force on a nearby test charge. If bound charge were fictitious, it could not.
Three concrete points. It is made of real particles that really moved. It produces a measurable field. And you can even collect it — pull a polarised dielectric apart and the two halves carry net charge.
So what does 'bound' actually mean? Not unreal, but immobile: it cannot conduct through the material. That mobility difference, and nothing else, is what the word names.
The balloon on the wall closes it. The wall polarises, real bound charge appears on its surface, and the attraction is entirely real.
Picture it
Figure (svg): A polarised sphere with a uniform downward field inside opposing the polarisation, and a dipole field pattern outside.
The factor of one third is specific to a sphere. Other shapes have different depolarising factors — long thin needles polarised along their axis have almost none, which is why they are easy to polarise.
Use this to consolidate the worked example, because the two-region structure is the thing to hold on to.
Inside: perfectly uniform, magnitude P over three-epsilon-zero, pointing against the polarisation. Outside: exactly a point dipole field. At the surface: a jump, because there is bound surface charge sitting there.
The internal field opposing P is the feature to notice, and the reason is visible in the earlier figure — positive bound charge on top and negative underneath produce a field pointing downward inside, against the upward polarisation.
The factor of one third is specific to a sphere. Other shapes have different depolarising factors, and a long thin needle polarised along its axis has almost none — which is why needles are much easier to polarise than spheres.
Sorting
Sort into buckets
The two-question test: at a surface, does P have a component sticking out? In the volume, does P change from point to point? Those are the only two ways bound charge appears.
This drills the two-question test, which is the whole practical skill.
At a surface: does P have a component along the outward normal? If yes there is surface bound charge, if no there is none. In the volume: does P change from point to point? If yes there is volume bound charge, if no there is none.
Those two questions cover every case. The cylinder's curved side is the one worth thinking about — the normal is radial while P is axial, so the dot product vanishes and there is nothing there, even though the material is thoroughly polarised.
Check
Reason it out first.
Check your understanding
A cylinder of material is uniformly polarised along its axis. Where is there bound charge?
Answer: A
Why: The volume bound charge is minus the divergence of P, and a uniform P has zero divergence, so the interior is neutral. On the curved side the normal is perpendicular to P, so σ_b = P·n̂ = 0 there. On the two end faces the normal is parallel (or antiparallel) to P, giving +P on one and -P on the other.
The two-question test applied to a cylinder.
Uniform polarisation means zero divergence, so no volume charge. On the curved side the normal is perpendicular to P, so nothing there either. Only the two flat end faces have a normal along P, and they carry plus and minus P.
Option D is half right, which is what makes it tempting. Uniform polarisation does kill the volume bound charge, but not the surface bound charge. Compressing the rule to 'uniform means no bound charge' is what goes wrong.
Section
Section 3
Concept
Inside a dielectric, Gauss's law still holds — but the charge in it must include the bound charge.
\[ \nabla\cdot\mathbf{E} = \frac{\rho}{\varepsilon_0} = \frac{\rho_f + \rho_b}{\varepsilon_0} \]
Figure (svg): A loop diagram: the field polarises the material, the polarisation creates bound charge, and the bound charge changes the field.
You cannot compute the bound charge until you know the field, and you cannot compute the field until you know the bound charge. The same circle that motivated Chapter 3 — and it has a much slicker escape this time.
Free charge is charge you put there deliberately: electrons on a capacitor plate, ions in a solution. It is the part you actually control.
This is the same shape of difficulty that motivated Chapter 3, and noticing the parallel helps.
Gauss's law still holds inside a dielectric, but the charge appearing in it has to include the bound charge. And the bound charge depends on the polarisation, which depends on the field, which is the thing you are trying to find.
The loop diagram makes it explicit: field polarises the material, polarisation creates bound charge, bound charge changes the field.
Free charge needs defining clearly here, because everything that follows depends on the distinction. Free charge is charge you deliberately put there — electrons pumped onto a capacitor plate, ions in solution. It is the part you control and the part you know.
Worked example
Substitute the expression for bound charge into Gauss's law
Why: Bound volume charge is minus the divergence of the polarisation, from Section 2.
\[ \varepsilon_0\nabla\cdot\mathbf{E} = \rho_f + \rho_b = \rho_f - \nabla\cdot\mathbf{P} \]
Collect both divergence terms on the left
Why: This is the whole trick: the awkward bound-charge term is itself a divergence, so it can be absorbed into the left-hand side.
\[ \nabla\cdot(\varepsilon_0\mathbf{E} + \mathbf{P}) = \rho_f \]
Name the combination in the bracket
Why: Give it a symbol and the equation becomes as simple as Gauss's law in vacuum, but with only the free charge on the right.
\[ \mathbf{D} \equiv \varepsilon_0\mathbf{E} + \mathbf{P} \quad\Rightarrow\quad \nabla\cdot\mathbf{D} = \rho_f \]
\[ \oint\mathbf{D}\cdot d\mathbf{a} = Q_{f,\text{enc}} \]
Verify what has been gained, in coulombs per square metre
Why: The bound charge has vanished from the equation entirely — not because it stopped existing, but because it has been folded into the definition of D. Now the symmetry tricks from Deck 2 work again, using only the free charge you already know.
Three lines, and the trick is reusable, so it is worth watching closely.
Substitute the bound charge into Gauss's law. The bound volume charge is minus the divergence of P — and that is the key observation, because it means the awkward term is itself a divergence.
So collect both divergences on the left. What emerges is a new vector field whose divergence is the free charge alone.
Name that combination D, and Gauss's law is restored to exactly its vacuum form with only the free charge on the right. The bound charge has not stopped existing; it has been folded into the definition.
What has been gained: the symmetry tricks from Deck 2 work again, using only the charge you already know about.
Explain it to yourself
The move that produced D is one of the slickest in the book, and it is worth being able to reconstruct it rather than memorise it.
Discussion prompt
Without looking, explain why absorbing the bound charge into a new field was possible at all. What special property did ρ_b have?
This is worth noticing as a general technique: when an unknown term in an equation is the derivative of something you can name, move it to the other side and rename the combination.
Worth doing as an active exercise, because the insight generalises well beyond this chapter.
The property that made it possible is that bound charge is itself a divergence. Had it been anything else, the two terms could not have been combined, and there would be no D at all.
The general technique is worth naming: when an unknown term in an equation turns out to be the derivative of something you can name, move it to the other side and rename the combination. Exactly the same manoeuvre produces the auxiliary field H in Chapter 6.
Notation
Annotate
On: \( \mathbf{D} = \varepsilon_0\mathbf{E} + \mathbf{P} \)
Two names to keep straight: E is the actual field, the thing that exerts force on a charge. D is an accounting device that makes Gauss's law usable — no force is ever proportional to D.
Griffiths is blunt about it: D has no deep physical meaning. It exists because the combination ε₀E + P happens to have only free charge as its source.
Take the definition apart, and be clear about what D is and is not.
It is epsilon-zero times the real field, plus the material's response. It bundles 'what you applied' and 'what the material did' into a single object whose only source is free charge.
The honest framing, which Griffiths gives himself, is that D has no deep physical meaning. It exists because that particular combination happens to have a convenient divergence.
One distinction to insist on: E is the actual field, the thing that exerts force on a charge. D never exerts a force on anything. Treating D as 'the field in a dielectric' leads to errors that are hard to diagnose later.
Worked example
Griffiths' Example 4.4. A long wire carrying line charge lambda, wrapped in rubber insulation out to radius a.
Figure (svg): A charged wire surrounded by a cylindrical layer of rubber insulation, with a cylindrical Gaussian surface drawn through the rubber.
Apply the D form of Gauss's law to a cylinder of radius s and length L, in coulombs
Why: The enclosed FREE charge is just what is on the wire — the bound charge in the rubber is invisible to this equation, which is exactly the point.
\[ D(2\pi s L) = \lambda L \quad\Rightarrow\quad \mathbf{D} = \frac{\lambda}{2\pi s}\hat{\mathbf{s}} \]
Notice where this formula is valid
Why: It holds both inside the rubber and outside it, because the free charge enclosed is the same in both regions. No knowledge of the rubber's properties was needed.
Convert to E outside the insulation, in newtons per coulomb
Why: Outside there is no matter, so P is zero and D is simply epsilon-zero times E.
\[ \mathbf{E} = \frac{\mathbf{D}}{\varepsilon_0} = \frac{\lambda}{2\pi\varepsilon_0 s}\hat{\mathbf{s}} \qquad (s>a) \]
Verify — and note the honest limitation
Why: Outside the rubber the answer matches the bare-wire result from Deck 2, as it must. INSIDE the rubber, however, E cannot be found from D alone: you would need to know P, and nothing so far tells you what P is. That gap is what Section 4 fills.
A short example showing both the power and the limitation of D, which is why it is worth doing.
Apply Gauss's law for D with a cylindrical surface. The enclosed free charge is just what is on the wire — the bound charge in the rubber is invisible to this equation, which is the entire point. So D comes out immediately, without knowing anything at all about the rubber.
The formula holds both inside the insulation and outside it, because the enclosed free charge is the same in both regions.
Now the honest limitation, which is the real lesson. Outside the rubber, P is zero and E is simply D over epsilon-zero. Inside the rubber, E cannot be determined from D alone — you would need to know P, and nothing so far tells you what it is.
That gap is exactly what the linear-dielectric assumption in the next section fills, so this example is what motivates it.
Trap
Having seen that D obeys a Gauss's law, a student assumes it behaves like E in every other way:
Write a Coulomb's law for D, and assume its curl vanishes
Why: The two fields obey identical-looking divergence equations, so the rest ought to carry over.
\[ \nabla\times\mathbf{D} \overset{?}{=} 0, \qquad \mathbf{D} \overset{?}{=} \frac{1}{4\pi}\int\frac{\hat{\boldsymbol{\mathfrak{r}}}}{\mathfrak{r}^2}\rho_f\,d\tau' \]
Both are false in general. Griffiths titles this section 'A Deceptive Parallel' for a reason.
Take the curl of the definition and see what actually happens:
\[ \nabla\times\mathbf{D} = \varepsilon_0\underbrace{(\nabla\times\mathbf{E})}_{=\,0} + \nabla\times\mathbf{P} = \nabla\times\mathbf{P} \]
Conclude that D has a curl whenever P does
Why: There is no law forcing the curl of P to vanish, so in general D circulates — and a field with a curl has no potential and obeys no Coulomb-style superposition formula.
| property | E | D |
|---|---|---|
| divergence | total charge / ε₀ | free charge only |
| curl | always zero (electrostatics) | equals the curl of P — not generally zero |
| has a potential? | yes, V | not in general |
| Coulomb-style integral? | yes | no |
| exerts force on charge? | yes | never |
So use D for exactly one thing: applying Gauss's law when you know the free charge and have symmetry. For everything else, get back to E.
Griffiths titles his section 'A Deceptive Parallel', and the deception is worth demonstrating rather than just warning about.
Having seen that D obeys a Gauss's law identical in form to E's, it is natural to assume the rest carries over: a Coulomb-style integral, a vanishing curl, a potential.
Take the curl of the definition and see what happens. The curl of E is zero, so the curl of D equals the curl of P — and nothing forces that to vanish. So in general D circulates.
A field with a curl has no potential and obeys no Coulomb-style superposition formula. So D is useful for exactly one thing: applying Gauss's law when you know the free charge and have symmetry. The table lists what transfers and what does not, and the last row is the one to remember — D never exerts a force.
Discrimination
D only counts free charge, so getting this classification wrong makes every subsequent step wrong.
Sort into buckets
Item five is the useful edge case: charge sprayed onto an insulator cannot move, but it is still free charge — it was put there from outside rather than being produced by polarisation. 'Bound' means 'produced by polarisation', not merely 'immobile'.
This classification decides what goes on the right-hand side of Gauss's law for D, so getting it wrong invalidates everything downstream.
The rule is about origin, not mobility. Free charge is charge that was put there from outside, or that can move through the material. Bound charge is charge produced by polarisation.
Item five is the edge case worth thinking about. Charge sprayed onto an insulator cannot move at all, and yet it is free charge, because it was deposited from outside rather than produced by polarisation.
That example is the cleanest demonstration that 'bound' does not simply mean 'immobile'.
Concept
The same pillbox-and-loop arguments as Deck 3, run on D and E instead:
\[ D^{\perp}_{\text{above}} - D^{\perp}_{\text{below}} = \sigma_f \]
\[ \mathbf{E}^{\parallel}_{\text{above}} = \mathbf{E}^{\parallel}_{\text{below}} \]
Figure (svg): An interface between two dielectrics with the perpendicular component of D and the parallel component of E marked as the two continuous quantities.
Note the asymmetry: it is the perpendicular part of D and the parallel part of E that behave nicely. Mixing that up is a reliable way to get a wrong refraction angle for field lines at an interface.
With no free charge on the interface, the perpendicular part of D is fully continuous — which makes 'no free surface charge' one of the most useful phrases in a Chapter 4 problem.
The same pillbox-and-loop arguments as the last deck, now applied to D and E — and the asymmetry is the thing to notice.
It is the perpendicular component of D that behaves simply, jumping only by the free surface charge. And it is the parallel component of E that is continuous.
Those get mixed up frequently, and the result is refraction angles at interfaces coming out backwards. Worth writing both explicitly and saying which field goes with which component.
One very common special case: with no free charge on the interface, the perpendicular component of D is fully continuous. The phrase 'no free charge on the boundary' appears in most problems for exactly that reason.
Check
Take a moment.
Check your understanding
A parallel-plate capacitor has free charge ±Q on its plates and a slab of dielectric filling the gap. Which statement is true?
Answer: A
Why: Gauss's law for D involves only free charge, so with Q fixed the displacement is exactly what it would be with no dielectric: D = σ_f. But the dielectric polarises, and its bound surface charge is opposite in sign to the nearby plate, partially cancelling the field. So the actual field E is reduced by a factor of the dielectric constant.
This is the central conceptual question of the chapter, so it is worth thinking through all four options.
With the free charge fixed, D is exactly what it would be with no dielectric present, because Gauss's law for D only counts free charge. But the dielectric polarises, and its bound surface charge partially cancels the plate charge, so the actual field E is reduced.
Option C is the one to unpack, because it reverses the roles. D sees only free charge; E sees everything, free and bound, which is precisely why E is the one that changes.
If option B looks right, that is a signal the dielectric is not yet doing anything in your mental picture, and it is worth going back a slide.
Section
Section 4
Real world
Chapter 4's structure is not accidental. Chapter 6 repeats it exactly for magnetism, and recognising the pattern now saves a great deal of work later.
| electric (Ch. 4) | magnetic (Ch. 6) | the shared idea |
|---|---|---|
| polarization P | magnetization M | dipole moment per unit volume |
| bound charge σ_b, ρ_b | bound current K_b, J_b | what the aligned dipoles leave uncancelled |
| displacement D = ε₀E + P | auxiliary field H = B/μ₀ − M | a field whose source is only the FREE stuff |
| ∇ · D = ρ_f | ∇ × H = J_f | Gauss / Ampere with the bound part hidden |
| P = ε₀χ_eE | M = χ_mH | linear response of the material |
| ε_r = 1 + χ_e | μ_r = 1 + χ_m | the factor by which matter changes things |
Discussion prompt
Given the table, guess what the magnetic analogue of 'bound surface charge σ_b = P · n̂' will look like.
One difference is worth flagging now: electric susceptibility is always positive (ε_r > 1), but magnetic susceptibility can be negative — diamagnets weaken the field they sit in. That asymmetry has no electrostatic counterpart.
Worth seeing even though Chapter 6 is beyond this course, because it reframes what you have just learned as a pattern rather than a list.
Every row is the same idea in two guises: dipole moment per unit volume, the leftover uncancelled sources, an auxiliary field whose source is only the free stuff, and a linear response coefficient.
Spotting the pattern makes magnetostatics dramatically easier, because a great deal of it is a relabelling of what is already here.
The prompt is a real test of structural understanding: predicting that bound surface charge becomes bound surface current, with the dot product turning into a cross product, is the sign that this is understood rather than memorised.
One asymmetry to flag. Electric susceptibility is always positive, but magnetic susceptibility can be negative — diamagnets weaken the field they sit in, and nothing electrostatic behaves that way.
Concept
Nothing so far said how much a material polarises. For most materials in ordinary fields, the answer is simple: proportionally.
\[ \mathbf{P} = \varepsilon_0\chi_e\mathbf{E} \]
electric susceptibility — The dimensionless constant chi-e measuring how readily a material polarises. Vacuum has zero; most solids and liquids run from about 1 to 80. Water is about 79.
Note carefully which field appears: E is the total field inside the material, including the contribution from the material's own bound charge. Not the field you applied from outside.
Linearity fails in strong fields, in ferroelectrics, and in crystals where the response depends on direction — but for the materials in this chapter it holds well.
This is an assumption about materials, not a law of nature — the first thing in this chapter that could be false.
For most materials in ordinary fields it holds well: the polarisation is proportional to the field, with a dimensionless constant called the susceptibility.
But note which field appears, because the next slide unpacks it. The E here is the total field inside the material, including the contribution from the material's own bound charge — not the applied field.
Linearity fails in strong fields, in ferroelectrics, and in crystals whose response depends on direction. Knowing that means recognising when a problem is quietly signalling one of those cases.
Socratic
This looks like a detail. It is the subtlest point in the chapter.
Discussion prompt
The material polarises because of a field — but its own bound charge contributes to that field. So which E belongs in the formula: the applied field, or the total field?
The practical payoff: never compute the applied field, then polarise, then correct, then re-polarise. Use D — which knows only the free charge — and divide by epsilon at the end. The self-consistency is already built in.
This is the subtlest point in the chapter and worth confronting rather than glossing over.
The material polarises because of a field, but its own bound charge contributes to that field. So the relation looks circular.
It is not circular; it is a self-consistency condition. You are looking for the field that reproduces itself once the resulting polarisation has been fed back in.
And the practical resolution is precisely why D was invented. Do not compute the applied field, polarise, correct, and re-polarise. Compute D from the free charge, then divide by epsilon at the end. The self-consistency is already built into that algebra, which makes the D route not just faster but conceptually cleaner.
Concept
Susceptibility is a bulk property, but it is built out of the single-atom polarizability from Section 1.
\[ \mathbf{P} = N\mathbf{p} = N\alpha\mathbf{E} \quad\Rightarrow\quad \chi_e \approx \frac{N\alpha}{\varepsilon_0} \]
N is the number of molecules per cubic metre. So a material polarises strongly if its molecules are individually floppy (large alpha), or if it simply has a lot of them per unit volume.
This is why gases have dielectric constants barely above 1 — air is 1.0006 — while the same molecules condensed into a liquid can reach tens. Nothing changed about the molecules; there are just a thousand times more of them in the same space.
The relation is approximate because each molecule sits in the field of its neighbours as well as the applied field. The refinement is the Clausius-Mossotti relation, which Griffiths sets as a problem.
This connects the bulk property back to the single-atom picture from the start of the deck, which is a loop that often gets left open.
Polarisation is the number density times the individual dipole moment, and the individual moment is the polarizability times the field. So susceptibility is roughly number density times polarizability, over epsilon-zero.
Which explains an observation worth quoting. Gases have dielectric constants barely above one — air is 1.0006 — while the same molecules condensed into a liquid can reach tens. Nothing about the molecules changed; there are simply a thousand times more of them per unit volume.
The relation is approximate, because each molecule also sits in its neighbours' fields. The refinement is the Clausius-Mossotti relation, which Griffiths sets as a problem.
Concept
Substituting the linear relation into the definition of D collapses everything into one constant.
\[ \mathbf{D} = \varepsilon_0\mathbf{E} + \mathbf{P} = \varepsilon_0(1+\chi_e)\mathbf{E} \equiv \varepsilon\mathbf{E} \]
\[ \varepsilon = \varepsilon_0(1+\chi_e), \qquad \varepsilon_r = \frac{\varepsilon}{\varepsilon_0} = 1 + \chi_e \]
| material | dielectric constant ε_r | note |
|---|---|---|
| vacuum | 1 exactly | the baseline |
| air | 1.00059 | very nearly vacuum |
| paper | ≈ 3.7 | typical solid insulator |
| glass (Pyrex) | ≈ 4.7 | |
| water | ≈ 80 | strongly polar molecules |
| barium titanate | ≈ 1200 | used in high-value capacitors |
The dielectric constant is always greater than 1: matter always polarises in the direction that partially cancels the applied field, never in the direction that reinforces it.
Here the algebra collapses into something usable.
Substituting the linear relation into the definition of D gives D proportional to E, with the constant called the permittivity. Divide by epsilon-zero and you get the dielectric constant, which is the number quoted in tables.
Walk the table and make the values concrete. Air is essentially vacuum. Paper and glass are a few. Water is around eighty, because its molecules are strongly polar. Barium titanate is over a thousand, which is why it turns up in high-value capacitors.
Note that the dielectric constant is always greater than one — and the next slide is about why that is not an accident.
Intuition
Could a material exist whose polarisation strengthened the applied field?
Figure (svg): A dielectric slab in a field, with bound negative charge on the face nearest the positive plate, producing an internal field opposing the applied one.
No. The field pulls positive charge toward the negative plate and negative charge toward the positive one — so the bound charge that appears on each face always has the opposite sign to the plate it faces.
That configuration necessarily produces an internal field pointing against the applied one. The response fights its cause, so the net field is always weakened and epsilon-r is always at least 1.
Ask the question before looking at the answer: could a material exist that strengthens the field it sits in?
No, and the reason is visible in the picture. The field pulls positive charge toward the negative plate and negative charge toward the positive one, so the bound charge appearing on each face always has the opposite sign to the plate it faces.
That arrangement necessarily produces an internal field opposing the applied one. The response fights its own cause.
So the net field inside is always weakened, and the dielectric constant is always at least one. The magnetic case is genuinely different — diamagnets do the opposite — which makes this a real physical asymmetry rather than a mathematical necessity.
Worked example
Griffiths' Example 4.6. A parallel-plate capacitor is filled with material of dielectric constant epsilon-r. What happens to its capacitance?
Apply Gauss's law for D with a pillbox on one plate, in coulombs per square metre
Why: Only free charge enters, so D is unchanged by the presence of the dielectric.
\[ D = \sigma_f = \frac{Q}{A} \]
Convert D into the actual field E inside the material, in volts per metre
Why: Divide by the permittivity rather than by epsilon-zero. This is the step where the dielectric finally shows up.
\[ E = \frac{D}{\varepsilon} = \frac{Q}{\varepsilon_r\varepsilon_0 A} \]
Integrate across the gap and form the ratio
Why: Same geometry as the vacuum case, with an extra factor of epsilon-r in the denominator of E.
\[ V = Ed = \frac{Qd}{\varepsilon_r\varepsilon_0 A} \quad\Rightarrow\quad C = \frac{Q}{V} = \varepsilon_r\,\frac{\varepsilon_0 A}{d} \]
\[ C = \varepsilon_r\,C_{\text{vacuum}} \]
Verify with a physical reading, in farads
Why: Filling the gap multiplies the capacitance by the dielectric constant — a factor of 80 for water, over 1000 for barium titanate. The mechanism: for the same free charge the field is weaker, so the voltage is lower, so the ratio Q/V is bigger. This is why every real capacitor has a dielectric, and why the plate material matters less than the filling.
The most practically important calculation in the chapter, and every step reuses something already established.
Apply Gauss's law for D with a pillbox on one plate. Only free charge enters, so D is completely unchanged by the presence of the dielectric.
Then convert to the actual field by dividing by the permittivity rather than epsilon-zero. That single step is where the dielectric finally shows up, and it reduces E by a factor of the dielectric constant.
Integrate across the gap and form the ratio: the capacitance is multiplied by the dielectric constant.
The mechanism in words: for the same free charge, the field is weaker, so the voltage is lower, so the ratio of charge to voltage is larger. That is why every real capacitor has a dielectric, and why the filling matters more than the plate material.
Tweak it
Slide the dielectric constant and watch the capacitance and the internal field move in opposite directions.
Parameter explorer
Move ε_r from 1 (vacuum) up to 80 (water). What happens to C, and what happens to E inside for the same free charge?
\[ \varepsilon_r = {er}: \quad C = {er}\,C_0, \quad E = \frac{E_0}{{er}} \]
There is a practical ceiling: strong fields eventually tear electrons off the molecules and the material conducts. That dielectric strength — about 3 million volts per metre for air — is what sets a capacitor's voltage rating.
Drive this one and watch the two quantities move in opposite directions.
Raising the dielectric constant multiplies the capacitance and divides the internal field by the same factor, with the free charge held fixed. Ten times the epsilon-r means ten times the stored charge at the same voltage.
There is a practical ceiling, though, and it is what stops designers simply using the largest epsilon-r available. Strong fields eventually tear electrons off the molecules and the material starts to conduct.
That dielectric strength — about three million volts per metre for air — is what sets a capacitor's voltage rating, and it is a completely separate material property from the dielectric constant.
Worked example
Griffiths' Example 4.5. A metal sphere of radius a carries free charge Q, surrounded by linear dielectric out to radius b.
Figure (svg): A metal sphere with a thick dielectric shell around it, with three regions labelled: inside the metal, the dielectric shell, and vacuum beyond.
Get D everywhere from the free charge alone, in coulombs per square metre
Why: Spherical symmetry plus Gauss's law for D, and the dielectric never enters the calculation.
\[ \mathbf{D} = \frac{Q}{4\pi r^2}\hat{\mathbf{r}} \qquad (r > a) \]
Convert to E region by region, in newtons per coulomb
Why: Inside the shell divide by the permittivity of the material; outside it, by epsilon-zero.
\[ \mathbf{E} = \begin{cases} \dfrac{Q}{4\pi\varepsilon r^2}\hat{\mathbf{r}}, & a<r<b \\[10pt] \dfrac{Q}{4\pi\varepsilon_0 r^2}\hat{\mathbf{r}}, & r>b \end{cases} \]
Find the polarisation and the bound surface charges, in coulombs per square metre
Why: P is epsilon-zero chi-e times E, and the bound surface charge on each face of the shell is P dotted with that face's outward normal.
\[ \sigma_b = -\frac{\varepsilon_0\chi_e Q}{4\pi\varepsilon a^2} \;\text{at } r=a, \qquad \sigma_b = +\frac{\varepsilon_0\chi_e Q}{4\pi\varepsilon b^2} \;\text{at } r=b \]
Verify the total bound charge, in coulombs
Why: Multiplying each density by its sphere's area gives −ε₀χ_eQ/ε and +ε₀χ_eQ/ε, which sum to exactly zero — as they must, since the shell was electrically neutral to begin with. And outside the shell the field is unchanged from the bare sphere's, because the net bound charge enclosed is zero.
This exercises the full workflow — D first, then E, then P, then bound charges — so it is a good place to rehearse the order.
D comes from the free charge alone by spherical symmetry, and the dielectric never enters that step. Then convert to E region by region: divide by the permittivity inside the shell, by epsilon-zero outside it.
P follows from E, and the bound surface charges follow from P dotted with each face's outward normal. The two faces have opposite normals, which is why the two surface charges come out with opposite signs.
The verification is the neutrality check again: multiply each density by its sphere's area and confirm the total is zero. It has to be, because the shell was uncharged.
One consequence worth drawing out: outside the shell the field is unchanged from the bare sphere's, because the net bound charge enclosed is zero. A dielectric changes the field where it is, and nowhere else.
Concept
For a linear dielectric with free charge inside it, the volume bound charge has a tidy closed form.
\[ \rho_b = -\nabla\cdot\mathbf{P} = -\nabla\cdot\left(\varepsilon_0\chi_e\frac{\mathbf{D}}{\varepsilon}\right) = -\left(\frac{\chi_e}{1+\chi_e}\right)\rho_f \]
Read this carefully: bound volume charge appears only where there is free charge. In a homogeneous dielectric with no free charge inside it, all the bound charge lives on the surfaces.
And the bound charge is always opposite in sign and smaller in magnitude than the free charge that caused it — which is another way of saying the material partially screens whatever you put in it.
\[ \rho_{\text{total}} = \rho_f + \rho_b = \frac{\rho_f}{\varepsilon_r} \]
A tidy result with a memorable physical reading.
For a linear dielectric, the bound volume charge is proportional to the free volume charge, with a factor that is always less than one and always negative.
So bound volume charge appears only where free charge sits. In a homogeneous dielectric with no free charge inside it, all the bound charge is on the surfaces — which covers most problems you will meet.
And the bound charge always opposes the free charge that caused it, and is always smaller. The net effect is that the material partially screens whatever you put in it, reducing the total charge density by exactly the dielectric constant. That last equation is the cleanest one-line statement of what a dielectric does.
Concept
Griffiths' Example 4.7, the dielectric cousin of the metal sphere from Deck 4.
\[ \mathbf{E}_{\text{in}} = \frac{3}{\varepsilon_r + 2}\,\mathbf{E}_0 \]
Figure (svg): A dielectric sphere in a uniform field, with the lines slightly crowded inside and only mildly bent, alongside a note that a metal sphere excludes them entirely.
The interior field is uniform — a genuinely surprising result — and weaker than the applied field, but never zero.
Check the two limits. With epsilon-r = 1 (no material) the factor is 1 and nothing changes. As epsilon-r goes to infinity the factor goes to zero: a perfect dielectric behaves like a conductor, excluding the field completely. Everything real sits between.
Compare this directly against the metal sphere from the last deck, because the contrast is the lesson.
The interior field is uniform, which is surprising in itself, and it is weakened by a factor depending on the dielectric constant — but it is never zero.
Check the two limits, since they bracket the physics. With a dielectric constant of one there is no material and nothing changes. As the constant goes to infinity the interior field goes to zero, so a perfect dielectric behaves exactly like a conductor.
Everything real sits between those extremes, which is a good way to think of a dielectric: a partial conductor, with the dielectric constant measuring how far along that spectrum it sits.
Trap
Asked what the field is inside a dielectric-filled capacitor, a student reasons from the conductor case:
Say the field inside is zero, since the induced charge cancels the applied field
Why: In a conductor the induced charge cancels the interior field exactly, and the mechanism here looks identical.
\[ E_{\text{inside}} \overset{?}{=} 0 \]
But then the dielectric would have infinite capacitance and no capacitor would ever break down.
The cancellation is partial, because the charges cannot travel — they can only stretch:
\[ E_{\text{inside}} = \frac{E_0}{\varepsilon_r} \;\neq\; 0 \]
Compare the two mechanisms directly
Why: In a conductor, free charge flows until the interior field is exactly zero — it has as much charge as it needs and all the mobility in the world. In a dielectric, each molecule stretches by a fixed small amount set by its polarisability, and then stops.
| conductor | dielectric | |
|---|---|---|
| charge can | travel through the material | shift within a molecule only |
| interior field | exactly zero | reduced by ε_r |
| cancellation | complete | partial |
| ε_r equivalent | effectively infinite | typically 2 to 80 |
So a dielectric is a conductor's weaker cousin — and the dielectric constant measures exactly how much weaker.
The most consequential misconception in the chapter, and it comes from over-applying conductor intuition.
In a conductor, free charge flows until the interior field is exactly zero. It has unlimited mobility and as much charge as it needs, so the cancellation is complete.
In a dielectric, each molecule stretches by a fixed small amount set by its polarizability, and then stops. The cancellation is partial.
Notice what the wrong answer would imply: zero internal field would mean infinite capacitance and no capacitor would ever break down. Both are plainly false.
The comparison table is the summary, and the last row is the useful framing — a conductor is the infinite-epsilon limit of a dielectric.
Concept
The energy formula from Deck 3 needs modifying, because some of the work now goes into stretching molecules.
\[ W = \frac{1}{2}\int \mathbf{D}\cdot\mathbf{E}\,d\tau \]
For a linear dielectric this is (ε/2)∫E² — the vacuum formula with epsilon-zero replaced by epsilon.
The difference between this and the vacuum expression is the energy stored in the polarised molecules themselves, held like the energy in a compressed spring. It is returned when the field is removed.
\[ W = \frac{1}{2}CV^2 = \frac{1}{2}\varepsilon_r C_0 V^2 \]
At the same voltage, a dielectric-filled capacitor stores epsilon-r times the energy — which is precisely why capacitors are built with them.
The energy formula changes inside a dielectric, and the reason is physical rather than formal.
Some of the work now goes into stretching molecules against their internal restoring forces, not just into establishing the field. That stored molecular energy is real, and it is returned when the field is removed.
The formula becomes one half the integral of D dotted into E, which for a linear dielectric is the vacuum expression with epsilon in place of epsilon-zero.
The practical consequence: at the same voltage, a dielectric-filled capacitor stores epsilon-r times the energy. That is the entire commercial reason capacitors are built with dielectrics rather than air gaps.
Concept
The two boundary conditions together make field lines bend when they cross from one dielectric into another.
Figure (svg): Field lines crossing an interface between two dielectrics, bending toward the interface normal on the low-permittivity side.
\[ \frac{\tan\theta_2}{\tan\theta_1} = \frac{\varepsilon_2}{\varepsilon_1} \]
Combine the two conditions to get the bending law
Why: The parallel component of E is continuous and, with no free surface charge, the perpendicular component of D is too. Taking the ratio of the two gives the tangent relation directly.
Lines bend away from the normal in the material with the larger permittivity — the exact analogue of Snell's law in optics, and for closely related reasons.
A nice payoff from the two boundary conditions, and the optical analogy makes it memorable.
Combine them: the parallel component of E is continuous, and with no free surface charge the perpendicular component of D is continuous too. Taking the ratio gives the tangent relation directly.
The result is that lines bend away from the normal in the material with the larger permittivity — structurally the same as Snell's law, and for related reasons.
Not a coincidence, either. Light is an electromagnetic field, and its refraction at an interface comes from these same boundary conditions applied to a time-varying field. Chapter 9 makes that precise.
Concept
Figure (svg): An animation of a dielectric slab sliding in and out between two capacitor plates, pulled inward by the curved fringing field at the plate edge.
The interior field is uniform and horizontal, so it cannot pull the slab sideways. The force comes entirely from the fringing field at the plate edges, which is not uniform.
That non-uniform field polarises the slab's leading edge and then pulls the induced dipoles toward the stronger field — inward, by the Section 1 rule.
\[ F = -\frac{dW}{dx} = \frac{\varepsilon_0\chi_e w V^2}{2d} \]
The energy method gets the answer without ever computing the messy fringing field — you only need to know how the total energy depends on how far the slab is in.
The mechanism here surprises most people, so watch the animation before reading the explanation.
The field in the middle of the capacitor is uniform and horizontal, so it cannot pull the slab sideways at all. If that were the whole field, there would be no force whatsoever.
The force comes entirely from the fringing field at the plate edges, which is non-uniform. It polarises the slab's leading edge and then pulls the induced dipoles toward the stronger field — inward, by the rule from Section 1.
And here is the elegance of the energy method: you can get the force from how the total energy depends on how far the slab is inserted, without ever calculating the messy fringing field. That is a general and very useful technique.
Counterexample
P = ε₀χ_eE is an assumption, not a law, and three important classes of material violate it.
| material class | what breaks | example |
|---|---|---|
| ferroelectrics | polarisation persists with NO field, and shows hysteresis | barium titanate |
| anisotropic crystals | P is not parallel to E — susceptibility becomes a tensor | calcite, quartz |
| strong-field regime | the response saturates, then the material breaks down | any dielectric near its limit |
Figure (svg): A hysteresis loop: polarisation plotted against applied field, tracing a loop that does not pass through the origin on the return path.
A ferroelectric remembers. Its polarisation depends on its history, not just on the current field — which makes it useless for the algebra in this section, and extremely useful for memory devices.
Discussion prompt
Which of the three assumptions — linear, homogeneous, isotropic — does each failure above break?
Everything in Section 4 assumes the material is linear, homogeneous and isotropic. When a problem says 'linear dielectric', it is granting you all three.
Honest limits on the theory, and all three failure modes turn up in practice.
Ferroelectrics keep a polarisation with no field at all, and show hysteresis — their state depends on history rather than only on the present field. That is what the loop in the figure means, and it is why they are used for memory devices.
Anisotropic crystals respond differently in different directions, so the susceptibility becomes a tensor and P need not even be parallel to E.
And every material saturates and then breaks down in a strong enough field.
The prompt maps each failure onto which of the three assumptions — linear, homogeneous, isotropic — it violates. When a problem says 'linear dielectric', it is granting all three, and knowing that tells you exactly what you are allowed to assume.
Estimation
Air breaks down at about 3 × 10⁶ V/m. Above that, the field rips electrons off molecules and the air conducts — a spark.
Predict first
Two capacitor plates are held 1 mm apart in air. Roughly what is the highest voltage you can put across them before it arcs over?
Correct: About 3000 V
Why: The field between parallel plates is V/d, so V_max = E_max × d = (3 × 10⁶ V/m)(10⁻³ m) = 3000 V. This is why a capacitor's datasheet always quotes a voltage rating alongside its capacitance: exceed it and the dielectric conducts, usually destroying the component.
Designers therefore face a genuine trade-off. A thinner gap raises capacitance (C goes as 1/d) but lowers the voltage rating (V_max goes as d). Solid dielectrics win on both counts, which is why nobody builds air-gap capacitors except for tuning.
This connects an abstract material property to a number an engineer would actually use.
Air breaks down at about three million volts per metre. For parallel plates the field is voltage over separation, so the maximum voltage is that field times the gap.
A one-millimetre air gap therefore holds off about three thousand volts, which is why a capacitor's datasheet always quotes a voltage rating alongside its capacitance.
And it explains why real components look the way they do. A thinner gap raises capacitance but lowers the voltage rating, since one goes as one over d and the other as d. Solid dielectrics win on both counts, which is why air-gap capacitors are used only for tuning.
Matching
Match the pairs
Why: This is the whole logic of the chapter in three lines. D is computed first because it needs only the free charge you already know. P follows from the material's response to whatever field results. And E, the only one that exerts a force, responds to every charge in the problem — which is exactly why it was hard to compute directly and why D was invented to get at it.
Notice the order that implies for solving problems: D first, then E, then P if you need the bound charges. Never the other way round.
The logic of the chapter compressed into three lines, and it also gives the order to solve problems in.
D is determined by the free charge and the geometry alone, which is why it gets computed first. P follows from whatever field results, together with the material's susceptibility. And E responds to every charge present, free and bound alike.
That last point is why E was hard to compute directly, and why D was invented as a way in.
So the workflow is fixed: D first, then E, then P if you need bound charges. Never the other way round.
Comparison
Comparison matrix
| E | D | P | |
|---|---|---|---|
| what it is | the real field | an accounting field | the material's response |
| units | N/C = V/m | C/m² | C/m² |
| its divergence gives | total charge/ε₀ | free charge | minus the bound charge |
| curl | zero | equals curl of P | not generally zero |
| exerts force? | yes | no | no |
| in vacuum | unchanged | equals ε₀E | zero |
The one-line summary: use D to get through Gauss's law, then convert to E to do physics.
The consolidation grid for the whole chapter, and filling the blanks unaided is a good check.
The divergence rows are the definitional core: E sees all charge, D sees only free charge, and P's divergence is minus the bound charge.
The curl row is the corrective to the deceptive parallel — only E is guaranteed curl-free.
And the force row is the one to leave with. Only E exerts a force. D and P are descriptive devices, however useful they are.
The one-line summary at the bottom is the takeaway: use D to get through Gauss's law, then convert to E to do physics.
Real world
Discussion prompt
Why is water such a good solvent for salts, in terms of this chapter?
These four make the case that this chapter is applied physics rather than an abstract extension.
The cell membrane is worth a moment, because it is a striking application of a formula from two slides ago. A lipid bilayer is a dielectric only a few nanometres thick, and capacitance goes as one over the thickness — so the capacitance per unit area is enormous, and that is what makes nerve signalling possible.
The microwave oven is the polar-molecule mechanism at work: water's permanent dipole gets torqued back and forth by an oscillating field, and the resulting molecular friction is the heat.
The prompt about water as a solvent ties the chapter together. A dielectric constant of eighty reduces the attraction between dissolved ions by a factor of eighty, which is enough for thermal motion to keep them apart.
Pattern
If a problem gives you no symmetry, D will not help and you are back to solving Laplace's equation — now with the dielectric boundary conditions from Section 3 rather than conductor ones.
The workflow to follow every time, and the order is not negotiable.
Identify the free charge, get D from Gauss's law using the symmetry, convert to E by dividing by the right constant in each region, get P if you need it, and find the bound charges from P.
The last step is the check: for a neutral dielectric, the total bound charge has to come to zero. It is quick, and it catches most errors.
One honest limitation, mirroring Deck 2. If the problem has no symmetry, D will not help, and you are back to solving Laplace's equation — now with the dielectric boundary conditions from Section 3 rather than conductor ones.
Check
Final question of the course.
Check your understanding
A capacitor is charged, then DISCONNECTED from its battery, and a dielectric slab is slid into the gap. What happens to the voltage across it?
Answer: A
Why: Disconnected means the free charge Q is fixed. The capacitance rises to ε_rC₀, so V = Q/C falls by a factor of ε_r. Physically, the slab's bound charge partially cancels the plate charge, weakening the field, and less field across the same gap means less voltage. This is also why the slab is pulled in: the stored energy Q²/2C falls, and systems move toward lower energy.
The critical word is 'disconnected', exactly as in the capacitor question in Deck 3 — so this rewards having learned to ask what is being held fixed.
Disconnected means the free charge is fixed. The capacitance rises by the dielectric constant, so the voltage falls by the same factor.
The physical story is worth having alongside the algebra: the slab's bound charge partially cancels the plate charge, weakening the field, and less field across the same gap means less voltage.
And the energy observation explains the force. The stored energy falls, and systems move toward lower energy, which is precisely why the slab gets pulled in.
Exit ticket
Discussion prompt
Three sentences, no formulas: what is polarisation, what is bound charge, and what is D for?
If those three hold up, Chapters 1 through 4 are complete: the mathematics, the field, the potential, the techniques, and matter's response.
A final check on the chapter's three central ideas.
Polarisation is the average dipole moment per unit volume — how far, and in which direction, the material's charges have been stretched. Bound charge is the real uncancelled charge left at surfaces, and wherever the polarisation is non-uniform, once all the interior dipoles have cancelled against each other. And D is a repackaging of epsilon-zero E plus P whose only source is free charge.
If those three come out cleanly, Chapters 1 through 4 are genuinely complete: the mathematics, the field, the potential, the solution techniques, and matter's response to all of it.
Recap
| the result | the formula |
|---|---|
| induced dipole | p = αE, with α = 4πε₀a³ for the model atom |
| torque on a dipole | N = p × E |
| bound charges | σ_b = P·n̂ ; ρ_b = −∇·P |
| uniformly polarised sphere | E_in = −P/3ε₀ ; outside, a perfect dipole |
| displacement | D = ε₀E + P ; ∇·D = ρ_f |
| linear dielectric | P = ε₀χ_eE ; D = εE ; ε_r = 1 + χ_e |
| filled capacitor | C = ε_rC₀ |
| dielectric sphere in a field | E_in = 3E₀/(ε_r + 2) |
Chapters 5 and 6 do all of this again for magnetism: currents instead of charges, the Biot-Savart law instead of Coulomb's, the vector potential instead of V, and magnetisation instead of polarisation. The structure is deliberately parallel — which means the work you have just done gets used twice.
Worth taking stock across all five decks rather than just this one.
You began with vector calculus that looked like pure mathematics, and every piece of it has now been spent: divergence on Gauss's law, curl on the existence of the potential, gradient on recovering the field, the delta function on point charges, and separation of variables on boundary-value problems.
Go down the results table and check each entry has a picture attached rather than being a memorised formula.
Then look ahead honestly. Chapters 5 and 6 repeat this entire structure for magnetism — currents instead of charges, Biot-Savart instead of Coulomb, a vector potential instead of a scalar one, and magnetisation instead of polarisation. The parallel is deliberate on Griffiths' part, which means the work you have just done gets used twice.
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