
Physics II course sequence · Interactive lesson
Conductors, insulators, and polarization
Every object you bring near a charge answers it. A metal answers by moving electrons across the whole sample until the force on them dies; an insulator answers by leaning each molecule about a millionth of its own width. This lesson is about telling the two answers apart, and knowing how far either model can be pushed.
UPII-01.02 · Lesson 2 of 8 in this unit. Position is not a completion score.Build the model
Matter answers a field in two ways: charge that travels, and charge that only leans.
The key idea
A conductor carries charge that is free to travel through the whole sample, an insulator holds every charge bound to its own molecule, and polarization is the small separation of bound charge that a field produces inside neutral matter.
Example. Put copper and polystyrene in the same 1.0 × 10⁶ V m⁻¹ field: copper's electrons rearrange within about 10⁻¹⁴ s until the field inside the metal is zero, while polystyrene's molecules only lean, each electron cloud shifting some 10⁻¹⁵ m.
The complete physical model and assumptions
Bring a charged object near any piece of matter and something inside the matter moves; what moves depends on how tightly it is held. In a metal roughly one electron per atom belongs to no particular ion — some 10²⁹ carriers per cubic metre, free to cross the whole sample — so they flow until the force on them vanishes, which happens only when the field inside the conducting material has fallen to zero and every scrap of excess charge has been pushed onto the surface. In an insulator nothing travels at all: each electron cloud merely leans a fraction of a molecular width against its own nucleus, or each already-lopsided molecule turns slightly towards the field against the thermal jostling that randomizes it. Both responses leave the object exactly as neutral as it was, and both leave the near face carrying the sign opposite to the source, which is why a charged comb lifts scraps of metal foil and scraps of paper alike. The two ideal materials are then the two ends of a single continuum, separated by a number rather than by a mechanism — and neither ideal survives contact with a laboratory, because every real insulator leaks through a film of adsorbed water long before its bulk resistivity says it should.
1What mobile actually means
A cubic centimetre of copper holds about 8.5 × 10²² conduction electrons, one per atom, each belonging to the crystal rather than to any particular ion. That is 1.4 × 10⁴ C of mobile charge sitting in a piece of metal you can hold. Charging that same centimetre cube to 1 μC takes 6.2 × 10¹² electrons away, seven parts in a hundred thousand million of what is there, which is why a charged conductor is still, chemically and mechanically, ordinary copper. In an insulator every electron is bound to its own molecule and cannot go further than a fraction of a molecular width. The two names label the ends of one continuum: resistivity runs from 1.7 × 10⁻⁸ Ω m for copper to about 10¹⁶ Ω m for fused quartz, twenty-four decades, with semiconductors, salt water and damp air spread across the middle.
2A conductor in equilibrium is defined by what has stopped
A mobile carrier feels F = qE. If any macroscopic field survived inside the conducting material, those carriers would still be accelerating and the arrangement would not be static. So electrostatic equilibrium in a conductor means exactly one thing: E = 0 at every point inside the metal. Two results follow with no further machinery. Excess charge cannot rest in the interior, because charge in the interior would source a field there, so it collects on the outer surface — spray 25 nC onto a copper shell of outer radius 5.0 cm and it settles over 3.14 × 10⁻² m² as a surface density of 0.80 μC m⁻². And the field just outside must meet the surface at a right angle, since any component along the surface would drive the surface charge sideways and the arrangement would still be changing.
3How long instantly takes, and where the ideal model fails
Excess charge placed inside a conductor drains to the surface with a characteristic time τ = ε₀ρ, set by the material's resistivity alone. For copper, ρ = 1.68 × 10⁻⁸ Ω m gives τ = 1.5 × 10⁻¹⁹ s. That answer should make you suspicious: it is far shorter than the mean time between electron collisions, roughly 2.5 × 10⁻¹⁴ s, so the smooth continuum picture has outrun its own assumptions. The honest statement is that redistribution in a metal is over within about 10⁻¹⁴ s. Feed the same formula fused quartz, ρ ≈ 10¹⁶ Ω m, and τ = 8.9 × 10⁴ s, roughly a day — one expression covering both ideal materials, which differ only by a number. In practice a charged insulator leaks far sooner than that through a film of adsorbed water on its surface, which is why electrostatics demonstrations fail on a humid afternoon.
4Polarization: a shift for some molecules, a turn for others
Two mechanisms, and they behave differently. Induced polarization stretches: the electron cloud of an atom slides bodily against its nucleus, giving p = αE. A hydrogen atom, with α/(4πε₀) = 0.667 × 10⁻³⁰ m³, in a field of 1.0 × 10⁶ V m⁻¹ takes on p = 7.4 × 10⁻³⁵ C m, a displacement of 4.6 × 10⁻¹⁶ m — about nine millionths of the atom's own radius. That tininess is why the response is linear. Orientational polarization turns: a molecule that is already lopsided, such as hydrogen chloride with p₀ = 3.6 × 10⁻³⁰ C m, is not stretched but swung round. In the same field the alignment energy p₀E = 3.6 × 10⁻²⁴ J is dwarfed at 300 K by kT = 4.1 × 10⁻²¹ J, so the mean alignment is only ⟨cos θ⟩ ≈ p₀E/(3kT) = 2.9 × 10⁻⁴ — yet even that leaves ⟨p⟩ = 1.0 × 10⁻³³ C m along the field, fourteen times the hydrogen atom's stretched moment. Heating weakens the turning and leaves the stretching alone, which is how the two are told apart in the laboratory.
5From molecular dipoles to a sheet of bound charge
Define the polarization P as the dipole moment per unit volume, P = np for n dipoles per cubic metre. Inside a uniformly polarized slab the positive end of each dipole sits beside the negative end of its neighbour and the pair contributes nothing; only the two faces are left uncancelled, carrying bound surface densities σb = P and −σb. Run that definition backwards to feel how small the underlying shift is. Suppose a 1.0 cm² face of such a slab is measured to carry 1.0 nC: σb = P = 1.0 × 10⁻⁹ C ÷ 1.0 × 10⁻⁴ m² = 1.0 × 10⁻⁵ C m⁻². With n = 1.0 × 10²⁸ molecules per cubic metre that is p = P/n = 1.0 × 10⁻³³ C m apiece, one electronic charge moved 6.2 × 10⁻¹⁵ m, about two parts in a hundred thousand of a molecular width. Two points to keep. The faces are equal and opposite, so the slab is still exactly neutral — polarization redistributes, it does not create. And bound charge cannot be drawn off down a wire, which is the operational difference between it and the free charge on a conductor.
6Grounding is a reservoir, not an eraser
Grounding means connecting a conductor to something so large that any charge your experiment handles leaves its state unmeasurably changed. The Earth carries about −4.6 × 10⁵ C spread over 5.1 × 10¹⁴ m² of surface; handing it 1 μC alters that by two parts in a million million. A grounded conductor can therefore take in or give up mobile charge freely, and its carriers settle wherever the force on them vanishes. Two consequences. Grounding works on conductors only: touch a wire to one corner of a charged plastic rod and you drain that corner, because nothing elsewhere in the rod can travel to the wire. And grounding does not automatically neutralize — with another charged body held nearby, the carriers settle into a distribution that carries net charge, and what is left is decided by the order in which the wire and the neighbour are removed.
Key equations
Read the units, assumptions and limits before substituting.
Equilibrium inside a conductor
E = 0 everywhere inside the conducting material
Any surviving field would still be pushing the mobile carriers, so nothing could be static.
E in V m⁻¹, equivalently N C⁻¹; true only once the redistributing current has stoppedMobile charge a conductor already holds
Qfree = neV
A cubic centimetre of copper carries 1.4 × 10⁴ C of it, so charging the sample disturbs almost none of it.
n = conduction electrons per m³, 8.5 × 10²⁸ m⁻³ in copper; e = 1.602 × 10⁻¹⁹ C; V = volume in m³Charge relaxation time
τ = ε₀ρ
Puts a number on how long instantly takes: 1.5 × 10⁻¹⁹ s for copper, 8.9 × 10⁴ s for fused quartz.
ρ = resistivity in Ω m, ε₀ = 8.85 × 10⁻¹² F m⁻¹, τ in sInduced dipole moment
p = αE
The linear response of bound charge — linear precisely because the displacement is so small.
p in C m, E in V m⁻¹, polarizability α in C m² V⁻¹, and α/(4πε₀) has the units of a volumeMean moment of a polar molecule
⟨p⟩ = p₀²E/(3kT)
Turning is the part thermal agitation fights, so this term alone fades as the sample is heated.
p₀ = permanent moment in C m, k = 1.381 × 10⁻²³ J K⁻¹, T in K; valid while p₀E ≪ kTBound surface charge on a polarized slab
σb = P = np
Interior ends cancel in pairs; only the two faces survive, equal and opposite, so the slab stays neutral.
P = dipole moment per volume in C m⁻², n = dipoles per m³, σb in C m⁻²Interactive lab
Predict first. Explore the model. Compare the result.
Explore the model
Try this. Raise E and the lean inside the slab and the bars on its faces grow together, in step, because p = αE. Then raise n instead: every molecule leans by exactly the same amount and only the bars grow, because σb = np counts how many dipoles a cubic metre holds.
Live results
- MOMENT PER MOLECULE p
- 5.01 10⁻³⁴ C m
- POLARIZATION P = σb
- 12.52 μC m⁻²
- BOUND CHARGE ON A 1.0 cm² FACE
- 1252 pC
Results follow this lesson’s model and assumptions. Displayed values are rounded.
Predict → change one variable → check
Plan a change before running it. Every other input stays at the saved baseline.
Saved baseline. Applied field E: 3.0 MV m⁻¹; Molecules per cubic metre n: 25 10²⁷ m⁻³
This experiment is kept only while this page is open. It does not change saved course progress.
Catch the common trap
Two slips travel together. The first is to read polarization as charging, so that the paper scraps a comb picks up are said to have taken some charge from it. They have not. Bound charge shifts by a fraction of a molecular width, the near face ends up with the sign opposite to the comb's and the far face with the same sign, and the scrap's net charge is what it always was, namely zero. Polarization redistributes; only a transfer of carriers charges anything. The second slip is to read E = 0 as a property metal has, rather than a conclusion about a state it has reached. The field inside a conductor vanishes because the mobile carriers keep moving while any field remains, so E = 0 is simply the statement that nothing is moving any more. Push a current through that same copper and there is a perfectly good field inside it, E = ρJ: at a current density of 1.0 × 10⁶ A m⁻², E = 1.7 × 10⁻² V m⁻¹, small but unmistakably not zero, and it has to be there to keep the carriers going. Electrostatic equilibrium is a hypothesis you check, not a label the material wears.
Practice & worked examples
3 authored examples. Make an attempt, then reveal the reasoning one step at a time.
EasyExample 1
An isolated copper sphere of radius 1.5 cm is given a charge of +4.0 nC. Copper has density 8960 kg m⁻³, molar mass 63.5 g mol⁻¹, and about one conduction electron per atom. How many electrons were removed, what fraction of the conduction electrons is that, and where on the sphere are they now missing from?
Sketch your answer or reasoning, then compare it with the authored solution. Your writing is not automatically graded or saved.
0 of 5 steps revealed.
- Charge leaves in units of e, so N = 4.0 × 10⁻⁹ C ÷ 1.602 × 10⁻¹⁹ C = 2.5 × 10¹⁰ electrons removed.
- Count what was available. V = (4/3)π(0.015 m)³ = 1.414 × 10⁻⁵ m³, so m = 8960 × 1.414 × 10⁻⁵ = 0.1267 kg = 126.7 g.
- That is 126.7 ÷ 63.5 = 1.995 mol, or 1.995 × 6.022 × 10²³ = 1.20 × 10²⁴ atoms, and one conduction electron each.
- Fraction removed: 2.5 × 10¹⁰ ÷ 1.20 × 10²⁴ = 2.1 × 10⁻¹⁴. Two conduction electrons in every hundred million million have gone.
- In equilibrium E = 0 inside the metal, so the shortfall sits on the outer surface: σ = 4.0 × 10⁻⁹ C ÷ 4π(0.015 m)² = 4.0 × 10⁻⁹ ÷ 2.83 × 10⁻³ = 1.4 × 10⁻⁶ C m⁻².
Answer. 2.5 × 10¹⁰ electrons, a fraction 2.1 × 10⁻¹⁴ of those available, leaving 1.4 μC m⁻² on the outer surface.
MediumExample 2
Helium has polarizability α/(4πε₀) = 0.205 × 10⁻³⁰ m³ and an atomic radius near 3.1 × 10⁻¹¹ m, and helium gas at 0 °C and 1 atm holds 2.69 × 10²⁵ atoms per m³. In a uniform field of 3.0 × 10⁶ V m⁻¹, find the induced moment per atom, the displacement of the two-electron cloud, and the bound charge on a 1.0 cm² face of the gas.
Sketch your answer or reasoning, then compare it with the authored solution. Your writing is not automatically graded or saved.
0 of 5 steps revealed.
- α = 4πε₀ × 0.205 × 10⁻³⁰ m³ = 0.205 × 10⁻³⁰ ÷ 8.988 × 10⁹ = 2.28 × 10⁻⁴¹ C m² V⁻¹.
- p = αE = 2.28 × 10⁻⁴¹ × 3.0 × 10⁶ = 6.84 × 10⁻³⁵ C m.
- Read that moment as the two electrons displaced bodily: p = (2e)d, so d = 6.84 × 10⁻³⁵ ÷ 3.204 × 10⁻¹⁹ = 2.1 × 10⁻¹⁶ m, which is 6.9 × 10⁻⁶ of the atomic radius. A shift that small is why p stays proportional to E.
- Polarization is moment per volume: P = np = 2.69 × 10²⁵ × 6.84 × 10⁻³⁵ = 1.84 × 10⁻⁹ C m⁻², and for a uniformly polarized slab σb = P.
- On a face of 1.0 cm² = 1.0 × 10⁻⁴ m²: Qb = 1.84 × 10⁻⁹ × 1.0 × 10⁻⁴ = 1.8 × 10⁻¹³ C, with −1.8 × 10⁻¹³ C on the opposite face, so the gas is still neutral.
Answer. p = 6.8 × 10⁻³⁵ C m, d = 2.1 × 10⁻¹⁶ m (about 7 × 10⁻⁶ of the atomic radius), and 0.18 pC of bound charge on each 1.0 cm² face.
HardExample 3
Liquid water at 300 K sits in a field of 1.0 × 10⁶ V m⁻¹. A water molecule carries a permanent moment p₀ = 6.2 × 10⁻³⁰ C m and an electronic polarizability volume α/(4πε₀) = 1.45 × 10⁻³⁰ m³, and the liquid holds 3.35 × 10²⁸ molecules per cubic metre. Find the stretched moment, the mean turned moment, the bound surface charge the two together predict, and the temperature at which stretching would catch up with turning.
Sketch your answer or reasoning, then compare it with the authored solution. Your writing is not automatically graded or saved.
0 of 5 steps revealed.
- Stretching: α = 4πε₀ × 1.45 × 10⁻³⁰ m³ = 1.45 × 10⁻³⁰ ÷ 8.988 × 10⁹ = 1.61 × 10⁻⁴⁰ C m² V⁻¹, so p = αE = 1.61 × 10⁻⁴⁰ × 1.0 × 10⁶ = 1.61 × 10⁻³⁴ C m.
- Turning: check the small-field condition first. p₀E = 6.2 × 10⁻³⁰ × 1.0 × 10⁶ = 6.2 × 10⁻²⁴ J against kT = 1.381 × 10⁻²³ × 300 = 4.14 × 10⁻²¹ J, a ratio of 1.5 × 10⁻³, so ⟨p⟩ = p₀²E/(3kT) = 3.84 × 10⁻⁵⁹ × 1.0 × 10⁶ ÷ 1.243 × 10⁻²⁰ = 3.09 × 10⁻³³ C m.
- Turning beats stretching by 3.09 × 10⁻³³ ÷ 1.61 × 10⁻³⁴ = 19, and the two add, so the mean moment per molecule is 3.09 × 10⁻³³ + 0.16 × 10⁻³³ = 3.25 × 10⁻³³ C m.
- P = n⟨p⟩ = 3.35 × 10²⁸ × 3.25 × 10⁻³³ = 1.09 × 10⁻⁴ C m⁻², and σb = P, so a face across the field should carry 109 μC m⁻² — some 6 × 10⁴ times the helium gas figure above, at a third of its field, because the liquid packs 10³ times more molecules in and each already carries a moment.
- The two mechanisms would match when p₀²/(3kT) = α, that is T = p₀²/(3kα) = 3.84 × 10⁻⁵⁹ ÷ (3 × 1.381 × 10⁻²³ × 1.61 × 10⁻⁴⁰) = 5.8 × 10³ K, about the temperature of the Sun's surface, so in water turning always wins. Measurement puts the real polarization about six times higher again, because each molecule sits in its neighbours' fields as well as the applied one — the independent-molecule count is a floor, not a prediction.
Answer. Stretched p = 1.6 × 10⁻³⁴ C m against a turned ⟨p⟩ = 3.1 × 10⁻³³ C m, nineteen times larger; together they give σb = 1.1 × 10⁻⁴ C m⁻², and the two would only balance near 5.8 × 10³ K.
Quick check
Select an answer and explain it before checking.
Review & continue
A conductor carries charge that is free to travel through the whole sample, an insulator holds every charge bound to its own molecule, and polarization is the small separation of bound charge that a field produces inside neutral matter.
- Explain the relationship without looking at the equation.
- Check a sign, a unit and a model assumption in your working.
- Return to the lab or worked example if a step is still unclear.
Optional supporting lessons
These chapter resources can help with earlier ideas; they are not additional required Physics II completion tasks.