
Physics II course sequence · Interactive lesson
Conductors, shielding and field mapping
Metals do something nothing else in this unit does: their charge moves. Follow that motion to its end and you can say where the charge ends up, what the field does at the surface, what a cavity feels, and how to read a field map taken with a probe.
UPII-02.08 · Lesson 8 of 8 in this unit. Position is not a completion score.Build the model
Free charge moves until the field that moved it is gone.
The key idea
A conductor is in electrostatic equilibrium when its free charges have stopped moving, which requires the electric field to be zero at every point inside the conducting material and all excess charge to sit on its surfaces.
Example. Drop an isolated copper slab into a 5.0 × 10⁴ N C⁻¹ field and about 10⁻¹⁹ s later its faces carry ±4.4 × 10⁻⁷ C m⁻², while a probe pushed into the metal reads nothing at all.
The complete physical model and assumptions
A conductor is a material whose charge is free to move, so it cannot hold still while a field pushes on it. Put a metal anywhere in a field and its electrons drift until the pushing stops, and that happens only when the total field — the external source plus the charge that has just rearranged — is zero at every point inside the material. That one condition does the rest of the work. No net charge can hide in the interior, so every excess charge is squeezed onto a surface. The field at the surface must be perpendicular, since any tangential part would keep sliding charge along it, and its magnitude reads σ/ε₀. An empty cavity inside the metal is dead whatever happens outside, which is what a Faraday cage sells. The model is exact only in equilibrium, and it is silent about the transient it assumes has already finished — about 10⁻¹⁹ s in copper, never in an insulator. It also fails for a wire carrying a steady current, where a small field inside the metal is precisely what keeps the charge moving.
1Equilibrium is a statement about motion
Copper offers roughly 8.5 × 10²⁸ free electrons in every cubic metre, none of them tied to one atom. Switch on a field inside that metal and every one of them is pushed, and the only way the pushing stops is for charge to pile up until the field it makes cancels the field that moved it. So E = 0 inside a conductor is not a property of the substance, it is the end state of a motion — and it arrives fast. The characteristic time is τ = ε₀ρ, which for copper's resistivity 1.7 × 10⁻⁸ Ω m is 1.5 × 10⁻¹⁹ s. Two consequences follow at once. Draw any closed surface inside the material: E = 0 everywhere on it, so no flux crosses it, and Gauss's law turns that into zero enclosed charge — which drives every excess charge out onto a surface. And a wire carrying a steady current is not in this state at all, because there a field of order 10⁻² V m⁻¹ inside the metal is exactly what keeps the electrons drifting.
2The interior zero is a cancellation, not a barrier
Nothing about a metal blocks a field. Superposition still holds term by term: the external source's field is present at every point inside the copper, as strong as it was before the metal arrived, and the induced surface charge contributes a second field that is equal and opposite there. Zero is a sum. Take a wide slab in a uniform external field of 5.0 × 10⁴ N C⁻¹. Its two faces acquire sheets of −σ and +σ, and a pair of sheets of density σ makes a field σ/ε₀ in the gap between them and nothing outside, so setting σ/ε₀ = E₀ gives σ = ε₀E₀ = 8.85 × 10⁻¹² × 5.0 × 10⁴ = 4.4 × 10⁻⁷ C m⁻². On a 10 cm × 10 cm face that is 4.4 nC — a countable amount of charge that really had to move. The same picture says what shielding cannot do: those induced charges push back on whatever made the external field, so the source still feels the metal.
3At the surface: perpendicular, and σ/ε₀ not σ/2ε₀
Just outside the metal the field has no component along the surface. If it had one, that component would drag surface charge sideways and the conductor would not be in equilibrium — so field lines always meet a conductor at 90°, and the surface is an equipotential. The magnitude comes from a thin pillbox straddling a patch of surface, one face just outside and one just inside. The inside face sits in dead metal and passes no flux, the sides pass none if the box is thin enough, so the whole flux EA leaves through the outer face and Gauss's law gives EA = σA/ε₀, that is E = σ/ε₀. The factor deserves a pause. An isolated charged sheet in free space gives σ/2ε₀ on each side; a conductor's surface gives twice that, because the patch's own σ/2ε₀ is joined by another σ/2ε₀ from all the rest of the conductor's charge — the same contribution that cancels the patch's field on the inside. So a meter reading 2.3 × 10⁴ N C⁻¹ just off the metal is reporting σ = 2.0 × 10⁻⁷ C m⁻² on the patch beneath it.
4Cavities: shielded from outside, not from inside
Hollow the conductor out and the empty cavity is dead. No field line can cross it: such a line would leave the wall at one place and land at another, and closing the path back through the metal, where E = 0, would give a non-zero net push around a closed loop — something an electrostatic field never does. So the cavity wall stays bare and the cavity holds no field, whatever storm of charge rages outside. Now put +5.0 nC in the cavity of a shell whose own net charge is −2.0 nC. A closed surface drawn inside the metal must again enclose nothing, so the wall takes −5.0 nC, piled up to match wherever the charge sits. Conservation then leaves −2.0 + 5.0 = +3.0 nC on the outer surface, and that layer is plainly visible from outside: at 10 cm from the centre a test charge feels the field of 3.0 nC. Shielding is one-way for an isolated shell. Earth it and the outer layer drains away, the external field goes with it, and the screening finally works both ways.
5Sharp points hold the strongest field
Surface charge only spreads evenly when the surface is symmetric. Join two distant spheres of radii R₁ and R₂ with a wire and they become one conductor, so charge flows until the potentials match: Q₁/R₁ = Q₂/R₂. Since σ = Q/(4πR²), that means σ ∝ 1/R, or σ₁R₁ = σ₂R₂, and with E = σ/ε₀ the surface field is strongest where the radius is smallest. A 1.0 mm bump on a dome sitting at 9.0 kV sees roughly 9.0 kV ÷ 1.0 mm = 9.0 × 10⁶ V m⁻¹, three times the 3.0 × 10⁶ V m⁻¹ at which air breaks down, so it sprays charge into the air long before the smooth part of the dome is in any difficulty. That is why high-voltage hardware is built round and polished, why a lightning rod is a spike, and why the charge distribution on a real conductor has to be solved for rather than assumed.
6Mapping the field a conductor actually makes
Fields with no symmetry are measured, not derived. The bench method paints two electrodes onto resistive paper, drives a small steady current between them, and hunts with a probe for points at the same voltage; joining those points draws one equipotential contour. Three rules then read the map. Every electrode surface is itself one contour, because the metal is an equipotential. Field lines cross the contours at right angles and run from high V to low. And the spacing gives the magnitude, |E| ≈ ΔV/Δs measured along the shortest path between neighbours: contours drawn every 1.00 V and found 4.0 mm apart mean |E| ≈ 250 V m⁻¹, while the same contours crowded to 1.0 mm mean 1.0 × 10³ V m⁻¹. Inside a closed conducting electrode the probe finds one single voltage everywhere, with no contours at all to cross — the shielded interior showing up as a flat patch on a real measurement.
Key equations
Read the units, assumptions and limits before substituting.
Electrostatic equilibrium
E = 0 inside the conducting material
Any surviving interior field would keep pushing the free electrons, so equilibrium and a non-zero interior field cannot coexist.
E in N C⁻¹ (= V m⁻¹). True once the charges have stopped, not in a wire carrying a steady current.Field just outside the surface
E⊥ = σ/ε₀
A field meter held just off the metal reads the local σ, patch by patch, without opening the conductor up.
σ is the local surface density in C m⁻², ε₀ = 8.85 × 10⁻¹² F m⁻¹; the tangential component is zero.Charge induced on a cavity wall
qwall = −qᵢₙ
It is exactly the charge needed to kill the cavity charge's field throughout the metal, whatever shape the cavity has.
qᵢₙ is the charge held in the cavity and qwall the total on its wall, both in C; the wall layer is not uniform.Charge left on the outer surface
qₒᵤₜₑᵣ = Q + qᵢₙ
This layer is the only one the outside world sees, which is why an isolated shell still betrays the charge hidden in its cavity.
Q is the conductor's own net charge in C. Earthing the conductor drains qₒᵤₜₑᵣ to zero.Curvature concentrates charge
σ₁R₁ = σ₂R₂
The tighter radius carries the larger σ and so the stronger surface field — the whole physics of a lightning rod.
Two spheres of radii R₁, R₂ in m joined by a wire and far apart; σ in C m⁻². Follows from Q₁/R₁ = Q₂/R₂.Reading a field off a measured map
|E| ≈ ΔV/Δs
Turns a voltmeter survey into a field map: crowded contours mean a strong field, and the direction runs from high V to low.
ΔV between neighbouring equipotentials in V, Δs their spacing in m along the normal; E in V m⁻¹.Interactive lab
Predict first. Explore the model. Compare the result.
Explore the model
Try this. Drag the probe in from x = 12 cm and watch the reading collapse the moment it crosses the face at {w|1} cm, then raise E₀ and see the outside trace and the induced σ climb together while the shielded stretch stays flat at zero.
Live results
- FIELD AT THE PROBE
- 40 kV m⁻¹
- INDUCED σ PER FACE
- 354 nC m⁻²
- CHARGE ON A 10 cm PATCH
- 3.54 nC
- PROBE DEPTH PAST THE FACE
- 0.0 cm
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. External field E₀: 40 kV m⁻¹; Conductor half-width: 5.0 cm; Probe position x: 8.0 cm
This experiment is kept only while this page is open. It does not change saved course progress.
Catch the common trap
The commonest wrong picture is that metal blocks the field, as though the external field stopped dead at the surface and the interior sat behind a shutter. Barrier and cancellation are not the same claim, and every way they differ can be tested. A barrier would be free, while cancellation is paid for in charge that has physically moved to the faces — cut the slab in two while it sits in the field, draw the halves apart, and each one comes out carrying net charge. A barrier would act instantly, while cancellation needs the rearrangement time, so a field that reverses faster than the free charge can follow is not screened at all; the electrostatic result is a statement about equilibrium, not about a field being switched. A barrier would be even-handed, while cancellation screens one way only: a cavity is protected from the outside, yet an isolated shell still advertises a cavity charge on its outer surface until somebody earths it. And a barrier would leave the source alone, whereas the charge that moved pulls back on whatever made the field, so the source feels the metal too.
Practice & worked examples
3 authored examples. Make an attempt, then reveal the reasoning one step at a time.
EasyExample 1
A hollow metal dome of radius 0.15 m stands on an insulating base, charged and then left alone. A field meter held just outside its surface reads 3.6 × 10⁵ N C⁻¹ pointing straight out of the metal. Find the surface charge density there, the total charge on the dome, and the field at a point 0.05 m from the centre.
Sketch your answer or reasoning, then compare it with the authored solution. Your writing is not automatically graded or saved.
0 of 4 steps revealed.
- Just outside a conductor the field is perpendicular to the surface with E = σ/ε₀, so σ = ε₀E = 8.854 × 10⁻¹² × 3.6 × 10⁵ = 3.19 × 10⁻⁶ C m⁻². It is positive, because the field points away from the metal.
- The dome is a sphere left to itself, so nothing breaks the symmetry and σ is the same on every patch. Its area is A = 4πR² = 4π(0.15 m)² = 0.2827 m².
- Total charge: Q = σA = 3.19 × 10⁻⁶ × 0.2827 = 9.0 × 10⁻⁷ C, that is 0.90 μC, and all of it sits on the surface.
- The point 0.05 m from the centre is in the empty cavity of a closed conductor, so the field there is zero — and it stays zero however much charge is brought up outside.
Answer. σ = 3.2 × 10⁻⁶ C m⁻², Q = 0.90 μC, and E = 0 at 0.05 m from the centre.
MediumExample 2
Two metal spheres of radii 12 cm and 3.0 cm stand far apart on insulating bases and are joined by a long thin wire. The pair carries 45 nC altogether. Find the charge, the surface charge density and the surface field of each sphere, then decide whether the air near either sphere breaks down at 3.0 × 10⁶ V m⁻¹.
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.
- The wire makes the two spheres one conductor, so charge flows until nothing is left to push it along the wire: the spheres finish at the same potential, which for widely separated spheres means Q₁/R₁ = Q₂/R₂, so Q₁/Q₂ = 12/3.0 = 4.
- With Q₁ + Q₂ = 45 nC that gives Q₂ = 45/5 = 9.0 nC and Q₁ = 36 nC: the big sphere holds four times the charge.
- Densities: σ₁ = 36 × 10⁻⁹ ÷ [4π(0.12)²] = 1.99 × 10⁻⁷ C m⁻², and σ₂ = 9.0 × 10⁻⁹ ÷ [4π(0.030)²] = 7.96 × 10⁻⁷ C m⁻². A quarter of the charge on the small sphere, four times the density — σ ∝ 1/R.
- Surface fields from E = σ/ε₀: E₁ = (1.99 × 10⁻⁷) ÷ (8.854 × 10⁻¹²) = 2.2 × 10⁴ V m⁻¹, and E₂ = 9.0 × 10⁴ V m⁻¹.
- Both are far under 3.0 × 10⁶ V m⁻¹, so the air holds. The small sphere gets there first: scaling every charge by (3.0 × 10⁶) ÷ (9.0 × 10⁴) = 33 puts about 1.5 μC on the pair, at which point it sprays charge while the large sphere is still at 7.5 × 10⁵ V m⁻¹.
Answer. Q₁ = 36 nC with E₁ = 2.2 × 10⁴ V m⁻¹; Q₂ = 9.0 nC with E₂ = 9.0 × 10⁴ V m⁻¹. Neither breaks down: about 1.5 μC on the pair would be needed to start a corona at the small sphere.
HardExample 3
A spherical metal shell of inner radius 4.0 cm and outer radius 6.0 cm carries a net charge of −2.0 nC, and a +5.0 nC point charge is fixed at the centre of its cavity. (a) Find the charge on each surface. (b) Find the field magnitude at r = 2.0 cm, 5.0 cm and 10.0 cm. (c) The shell is then earthed; recompute the outer charge and the fields at r = 10.0 cm and r = 2.0 cm.
Sketch your answer or reasoning, then compare it with the authored solution. Your writing is not automatically graded or saved.
0 of 6 steps revealed.
- (a) Put a closed spherical surface inside the metal, at r = 5.0 cm. It lies in dead metal, so E = 0 on it, no flux crosses it, and it encloses nothing: qwall + 5.0 nC = 0, giving qwall = −5.0 nC on the cavity wall.
- The shell's own charge is fixed at −2.0 nC, so what is left for the outer surface is qₒᵤₜₑᵣ = −2.0 − (−5.0) = +3.0 nC.
- (b) At r = 2.0 cm only the point charge lies inside, and the shell's layers are spherical shells outside that radius, which contribute nothing there: E = kq/r² = 8.99 × 10⁹ × 5.0 × 10⁻⁹ ÷ (0.020)² = 1.1 × 10⁵ N C⁻¹, radially outward.
- At r = 5.0 cm the point is inside the conducting material, so E = 0. At r = 10.0 cm everything within acts as a single point charge of 5.0 − 2.0 = 3.0 nC: E = 8.99 × 10⁹ × 3.0 × 10⁻⁹ ÷ (0.10)² = 2.7 × 10³ N C⁻¹, outward.
- (c) Earthing lets charge flow between shell and ground, and it flows until no field is left outside the shell. That empties the outer surface: qₒᵤₜₑᵣ = 0, and the shell now holds −5.0 nC in total, all of it on the cavity wall.
- So at r = 10.0 cm the field is now zero — the outside world can no longer tell the cavity holds anything. At r = 2.0 cm nothing has changed: E is still 1.1 × 10⁵ N C⁻¹, because the cavity is screened from everything the earthing did.
Answer. (a) −5.0 nC on the wall, +3.0 nC outside. (b) 1.1 × 10⁵ N C⁻¹, 0, and 2.7 × 10³ N C⁻¹. (c) Outer surface 0, field at 10.0 cm now 0, field at 2.0 cm unchanged at 1.1 × 10⁵ N C⁻¹.
Quick check
Select an answer and explain it before checking.
Review & continue
A conductor is in electrostatic equilibrium when its free charges have stopped moving, which requires the electric field to be zero at every point inside the conducting material and all excess charge to sit on its surfaces.
- 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.