University Physics II · Electric Potential · 4.8
Conductors & Electric Potential
In equilibrium a conductor's interior field vanishes, so the whole body sits at a single potential. What is left is a surface charge, a perpendicular field of σ/ε₀ just outside, and boundary conditions that fix the rest.
Build the model
Connect the measurement to the mechanism.
One statement carries the topic: in electrostatic equilibrium no field survives inside conducting material, because any field would drive the free carriers until they cancelled it. Everything else is bookkeeping on that. Since V(b) − V(a) = −∫E·dl vanishes along any path lying in the metal, the whole body — interior, outer surface, cavity wall — is one equipotential, and the field just outside must meet it perpendicular.
Gauss's law converts the surface charge into that field, E = σ/ε₀, and the two boundary conditions — tangential field continuous, normal field jumping by σ/ε₀ — turn a conductor into a boundary you solve against rather than a charge you integrate. The consequences follow: cavities are screened, grounding fixes a potential rather than emptying a body of charge, and where a surface curves tightly one shared potential concentrates the field until air gives way.
- Simple definition
- In electrostatic equilibrium a conductor carries no interior field, so every point of it — inside the metal and on the surface — sits at one potential, any net charge lies on the surface, and the field just outside is perpendicular.
- Example
- Charge a 15 cm metal sphere to 20 nC and every point of the metal sits at kQ/R = 1.2 × 10³ V, while the field just outside the surface is 8.0 × 10³ V m⁻¹, everywhere pointing straight out of it.
Carriers keep moving until the field they build cancels whatever is applied.
Static case only; a wire carrying current has E = ρJ inside
A Gaussian pillbox straddling the surface: flux escapes outward only.
N C⁻¹ = V m⁻¹; ε₀ = 8.85 × 10⁻¹² C² N⁻¹ m⁻²; n̂ points out of the metal
Fix the potential of every conductor and the field between them is unique.
True at any charged surface; at a conductor E⊥, in = 0 and E∥ = 0
R = 0.15 m holding 20 nC: V = 1.2 × 10³ V and E = 8.0 × 10³ V m⁻¹.
V(∞) = 0; R in m; the isolated sphere has C = 4πε₀R
One shared potential gives the small sphere less charge but a stronger field.
Two far-apart spheres on a thin wire; a crude stand-in for curvature
What sits in the hole is mirrored on its wall and reappears outside.
From Gauss on a closed surface lying wholly in the metal, where E = 0
No field survives inside the metal
A metal carries roughly 10²⁸ mobile electrons per cubic metre. Put it in a field and they drift, pile up on the surfaces, and build a field of their own that opposes the applied one; the drift stops only when the total field inside the material is zero. That is what electrostatic equilibrium means, and it is fast: the ohmic estimate of the settling time is τ = ε₀ρ, which for copper with ρ = 1.7 × 10⁻⁸ Ω m gives 1.5 × 10⁻¹⁹ s. That figure sits well below the ~10⁻¹⁴ s between electron collisions, where the ohmic model stops being trustworthy, but either estimate finishes long before any measurement starts. Two consequences follow from Gauss's law. Any closed surface drawn entirely within the metal encloses zero flux, so it encloses zero net charge: excess charge can only sit on surfaces. And the assumption is doing real work — a copper wire carrying a current density of 1.0 × 10⁶ A m⁻² has E = ρJ = 0.017 V m⁻¹ inside it, small but not zero.
The whole body is one equipotential
Potential difference is a line integral of the field, V(b) − V(a) = −∫E·dl. Choose any path from a to b that stays in the metal and the integrand is zero everywhere along it, so V(b) = V(a). This is a statement about the volume, not just the skin: the centre of a solid conductor, its outer surface, and the wall of any cavity in it all share one value of V. Nothing about that value is special: it is set by the body's charge and by everything else in the room. Outside, the consequence is geometric. The field just beyond the surface must be perpendicular to it, because a tangential component would push the surface charge sideways and equilibrium would not have been reached; equivalently, moving along the surface would change V and the surface would not be an equipotential. So field lines meet a conductor at 90° and the equipotential surfaces of the surrounding space wrap it, with the conductor supplying one contour for free.
Surface charge sets the outside field
Take a short cylinder straddling the surface, one flat face of area A just outside, the other just inside the metal. The inner face contributes nothing because E = 0 there, the side wall contributes nothing because E is perpendicular to the surface, and Gauss's law leaves EA = σA/ε₀, so E = σ/ε₀ just outside. This is not the σ/2ε₀ of an isolated charged sheet: the local patch supplies σ/2ε₀, and all the other charge supplies a second σ/2ε₀ that adds outside and cancels inside. Written generally, the electrostatic boundary conditions say that across any charged surface the tangential field is continuous, the normal field jumps by σ/ε₀, and V is continuous. On a conductor they read E∥ = 0 and E⊥ = σ/ε₀. For the 15 cm sphere with 20 nC, σ = Q/(4πR²) = 2.0 × 10⁻⁸/0.283 = 7.1 × 10⁻⁸ C m⁻², and σ/ε₀ = 8.0 × 10³ V m⁻¹, matching kQ/R² exactly.
A cavity is screened from outside
Hollow the conductor out, leaving an empty cavity. Its entire wall sits at one potential V₀, and inside the empty region V obeys Laplace's equation, which admits no maximum or minimum in the interior of a region — so V is pinned to V₀ everywhere in the cavity and E = 0 there. The field-line version is quicker: a line would have to run from one part of the wall to another, and closing the loop through the metal would give a non-zero ∮E·dl. Two results drop out. The cavity wall carries no charge, and no arrangement of external charge, however large, produces a field in the hole. That is the Faraday cage, and it works one way only. Place q inside the cavity and a Gaussian surface in the metal, where E = 0, forces −q onto the cavity wall; conservation puts Qₙₑₜ + q on the outer surface, spread according to the outer shape alone. Sliding q around inside the cavity changes nothing measurable outside.
Grounding fixes the potential, not the charge
A ground connection is a conducting path to a reservoir large enough that its potential is unaffected by what flows, and that potential is defined as zero. Grounding therefore does not 'drain the charge' as a rule; it lets charge move until V = 0, and how much moves depends on the geometry. Ground the isolated 15 cm sphere sitting at 1.2 × 10³ V and all 20 nC leaves, giving Q = 0. Now take a neutral shell with q = +4.0 nC inside its cavity: the wall holds −4.0 nC, the outer surface +4.0 nC, and the shell is at some positive potential. Ground the outer surface and that +4.0 nC flows to earth. The shell and everything outside it are now at V = 0 with no external field at all — but the cavity is untouched, the wall still holds −4.0 nC, and the field inside the hole is exactly what it was. Grounding a shield kills its field outside; it cannot shield the inside from a charge already in there.
Sharp curvature concentrates the field
Join two distant spheres, radii R₁ and R₂, with a thin wire and they become one conductor at one potential: kQ₁/R₁ = kQ₂/R₂, so charge shares out as Q ∝ R, surface density as σ ∝ 1/R, and surface field as E = V/R ∝ 1/R. With R₁ = 6.0 cm, R₂ = 2.0 cm and 8.0 nC in total, Q₁ = 6.0 nC and Q₂ = 2.0 nC, both spheres sit at 8.99 × 10² V, and the surface fields are 1.5 × 10⁴ V m⁻¹ and 4.5 × 10⁴ V m⁻¹. The small sphere holds a quarter of the charge and three times the field. Two spheres on a wire are a caricature of one curved surface, but the trend survives: on a body at a single potential, σ and E peak where the radius of curvature is smallest. Dry air breaks down near 3 × 10⁶ V m⁻¹, and Vbreak = Ebreak R for an isolated sphere, so a 1.0 mm ball electrode ionises the air around it at only 3.0 × 10³ V while a 15 cm sphere holds 4.5 × 10⁵ V. Hence smooth, fat high-voltage terminals — and hence a lightning rod is a spike.
Change one variable at a time
Make the relationship visible.
Shrink the radius and watch the outside curve stay exactly where it is while the plateau climbs, then drag the probe inside the metal, where V stops changing and E reads zero.
BODY POTENTIAL kQ/R1199 V
SURFACE FIELD σ/ε₀7991 V m⁻¹
V AT PROBE599 V
E AT PROBE1998 V m⁻¹
Live interpretationBODY POTENTIAL kQ/R: 1199 V. SURFACE FIELD σ/ε₀: 7991 V m⁻¹. V AT PROBE: 599 V. E AT PROBE: 1998 V m⁻¹
Catch the common trap
Explain before calculating.
Two conducting spheres of radii 6.0 cm and 2.0 cm sit far apart and are joined by a long thin wire. A total of 8.0 nC is placed on the pair. In equilibrium, what charge and surface field does the 2.0 cm sphere have? Take k = 8.99 × 10⁹ N m² C⁻².
Choose an answer to test the model.
Practice & worked examples
Reason from the model, then test the result.
EasyA solid metal sphere of radius 12 cm carries 45 nC and sits far from anything else. Find the potential of the metal, the potential at its centre, the field just outside the surface, and the surface charge density. Take k = 8.99 × 10⁹ N m² C⁻² and ε₀ = 8.85 × 10⁻¹² C² N⁻¹ m⁻².
- In equilibrium E = 0 inside the metal, so no path through the metal changes V: centre, bulk and surface share one value, V = kQ/R.
- V = (8.99 × 10⁹)(45 × 10⁻⁹)/(0.12) = 404.55/0.12 = 3371 V = 3.4 × 10³ V — and the centre reads exactly the same 3.4 × 10³ V.
- Just outside, E = V/R = 3371/0.12 = 2.8 × 10⁴ V m⁻¹, perpendicular to the surface and pointing out of it.
- Cross-check against the surface charge: σ = ε₀E = (8.85 × 10⁻¹²)(2.81 × 10⁴) = 2.5 × 10⁻⁷ C m⁻², which is 45 nC spread over 4πR² = 0.181 m² — the same number.
AnswerV = 3.4 × 10³ V everywhere in the metal, centre included; E = 2.8 × 10⁴ V m⁻¹ just outside, radially outward; σ = 2.5 × 10⁻⁷ C m⁻².
MediumA thick conducting spherical shell has inner radius 4.0 cm and outer radius 10.0 cm and carries a net charge of −3.0 nC. A point charge of +8.0 nC is fixed at the centre of the cavity. Find the charge on the cavity wall, the charge on the outer surface, the field just outside the shell, and the potential of the metal.
- Draw a Gaussian sphere lying entirely in the metal. E = 0 there, so the flux is zero and the enclosed charge is zero: the +8.0 nC in the cavity forces qᵢₙₙₑᵣ = −8.0 nC onto the wall.
- Charge is conserved on the shell: qₒᵤₜₑᵣ = Qₙₑₜ − qᵢₙₙₑᵣ = −3.0 − (−8.0) = +5.0 nC, which is the qₒᵤₜₑᵣ = Qₙₑₜ + qcavity rule.
- Outside r = 10.0 cm the enclosed total is +8.0 − 8.0 + 5.0 = +5.0 nC, so E = kQ/r² = (8.99 × 10⁹)(5.0 × 10⁻⁹)/(0.100)² = 44.95/0.0100 = 4.5 × 10³ V m⁻¹, radially outward.
- The whole shell is one equipotential, so evaluate V at the outer surface: V = kQ/r = 44.95/0.100 = 4.5 × 10² V — the same on the outer surface, in the bulk metal, and on the cavity wall.
AnswerCavity wall −8.0 nC, outer surface +5.0 nC; E = 4.5 × 10³ V m⁻¹ just outside; the entire shell sits at +4.5 × 10² V.
HardA 9.0 cm sphere and a 1.5 cm sphere sit far apart, joined by a long thin wire, and charge is added until the air breaks down. Dry air fails at 3.0 × 10⁶ V m⁻¹. Find which sphere sparks first, the common potential at which it happens, and the total charge the pair is holding at that moment.
- The wire makes one conductor at one potential V, and for each sphere Esurface = V/R. The fields therefore rank as 1/R: the small sphere runs R₁/R₂ = 9.0/1.5 = 6 times the field of the large one, so it sparks first.
- Set its surface field to the breakdown value: V = Ebreak R₂ = (3.0 × 10⁶ V m⁻¹)(0.015 m) = 4.5 × 10⁴ V. The 9.0 cm sphere is then at only E = 4.5 × 10⁴/0.090 = 5.0 × 10⁵ V m⁻¹, a sixth of the way to breakdown.
- One shared potential shares the charge as Q = VR/k: Q₁ = (4.5 × 10⁴)(0.090)/(8.99 × 10⁹) = 4.51 × 10⁻⁷ C and Q₂ = (4.5 × 10⁴)(0.015)/(8.99 × 10⁹) = 7.51 × 10⁻⁸ C.
- Total Q = 4.51 × 10⁻⁷ + 7.51 × 10⁻⁸ = 5.3 × 10⁻⁷ C, and Q₁/Q = 0.090/0.105 = 6/7 of it sits on the big sphere — most of the charge, least of the field.
AnswerThe 1.5 cm sphere sparks first, at V = 4.5 × 10⁴ V (45 kV); the pair is then holding 5.3 × 10⁻⁷ C, of which 4.5 × 10⁻⁷ C is on the 9.0 cm sphere.