Gauss's Law (Integral Form) from Coulomb

Derivation

Gauss's Law (Integral Form) from Coulomb

D-040 Home PU-102 Threads fields · symmetry Depends on Electric Field from Coulomb's Law, solid-angle-and-inverse-square-flux

Statement

For an electrostatic field built by superposing Coulomb fields, the outward flux of E through any closed surface S equals the net charge it encloses divided by ε₀: ∮_S E·dA = Q_enc/ε₀. The surface may have any shape and size; only the charge inside it survives, and each interior charge contributes the same amount independent of where inside it sits.

Why it matters

Coulomb's law is a statement about the field of one point charge; Gauss's law is a statement about the field of everything at once, encoded as a flux balance. Recasting Coulomb in this form is what makes electrostatics tractable: whenever a charge distribution has enough symmetry, a wisely chosen Gaussian surface reduces a hard vector integral to a one-line algebraic solve for E.

The derivation also exposes why the law is true. Nothing in it depends on the details of the surface — the whole result is geometry, driven by one fact: the Coulomb field falls off exactly as 1/r², so its flux through an area element is precisely the solid angle that element subtends. Gauss's law is inverse-square made global.

Assumptions

Electrostatics (static charges, superposition holds).If charges move or fields change in time, the field is no longer a pure sum of Coulomb terms; the flux identity still holds as a Maxwell equation, but this particular derivation from Coulomb no longer applies.

Exact inverse-square Coulomb field, E = (q/4πε₀) r̂/r².The step that turns (r̂·n̂)dA/r² into a solid angle uses the 1/r² exactly. Any other power leaves an explicit r-dependence and the flux then depends on surface shape, not just enclosed charge.

No charge lies exactly on S.A charge sitting on the surface subtends an ambiguous solid angle (2π for a locally smooth surface, less at an edge or corner). The flux is then discontinuous and Q_enc is ill-defined; the charge must be strictly inside or strictly outside.

The charge distribution is bounded and the field integrable on S.Charge escaping to infinity, or a field singularity landing on S, makes the flux integral diverge or converge conditionally; the clean equality can fail.

Derivation

E(r) = (q / 4πε₀) · r̂/r²

Coulomb field of a single point charge at the origin, the prior result electric-field-from-coulomb-law; r̂ points from the charge to the field point. A

Φ = ∮_S E·dA , dA = n̂ dA

Definition of the outward flux through the closed surface; n̂ is the outward unit normal and the origin (the charge) is taken inside S for now. A

Φ = (q / 4πε₀) ∮_S (r̂·n̂) / r² dA

Substitute step 1 into step 2 and pull the constants out of the integral; symbols only, no numbers yet. A

dΩ ≡ (r̂·n̂) dA / r² = cos θ dA / r²

Definition of the solid angle subtended at the origin by the element dA — the prior result solid-angle-and-inverse-square-flux. Here θ is the angle between r̂ and n̂; the factor cos θ projects dA onto the sphere of radius r, and the exact 1/r² is what makes this r-independent. B

Φ = (q / 4πε₀) ∮_S dΩ

Rewrite step 3 using the identification in step 4; the flux is now purely a total-solid-angle problem. B

∮_S dΩ = 4π (origin inside) , = 0 (origin outside)

The signed solid angle a closed surface subtends at a point is a topological invariant. From an interior point every direction is covered exactly once ⇒ 4π. From an exterior point a ray pierces S an even number of times; entry and exit crossings carry opposite signs of cos θ and equal |dΩ|, so they cancel in pairs ⇒ 0. C

Φ_{one charge inside} = (q / 4πε₀)(4π) = q / ε₀

Insert the interior value 4π from step 6; the two factors of 4π cancel exactly — the reason Coulomb's constant is written with a 4π in the first place. A charge outside contributes 0. A

Φ = ∮_S (∑_i E_i)·dA = ∑_i q_i,in/ε₀ = Q_enc/ε₀

By superposition the total field is the sum of single-charge Coulomb fields; flux is linear in E, so it distributes over the sum. Exterior charges each give 0, interior charges each give q_i/ε₀. For a continuous density, Q_enc = ∫_V ρ dV. C

Result

∮_S E·dA = Q_enc / ε₀ = (1/ε₀) ∫_V ρ dV

Reading. The net number of field lines threading out of any closed surface is fixed by the charge trapped inside it and by nothing else — not the shape of the surface, not where inside the charges sit, not any charge outside. Positive enclosed charge is a net source (outward flux); negative charge is a net sink.

Units check. [E·dA] = (N·C⁻¹)(m²) = N·m²·C⁻¹. On the right, [ε₀] = C²·N⁻¹·m⁻², so [q/ε₀] = C / (C²·N⁻¹·m⁻²) = N·m²·C⁻¹. Both sides carry units of N·m²·C⁻¹ (equivalently V·m). ✓

Limiting cases

Concentric sphere, single charge. With S a sphere of radius R about q, symmetry gives Φ = E·4πR²; setting this to q/ε₀ recovers E = q/4πε₀R² — the derivation runs backwards to Coulomb.
Charge fully outside. Q_enc = 0, so Φ = 0 for any surface, however strong the external field piercing it.
Charge on the axis of symmetry but off-centre. Flux is unchanged (still q/ε₀); only the distribution of E over S shifts.
Neutral enclosure. Equal positive and negative charge inside ⇒ Q_enc = 0 and zero net flux, even though E is nonzero everywhere on S.
Small surface shrinking onto a point of density ρ. Φ/V → ρ/ε₀, the seed of the differential form ∇·E = ρ/ε₀.

Breaks when

A charge lies on S. The subtended solid angle is neither 4π nor 0 but the fractional angle the surface presents there (2π for a locally flat patch → flux q/2ε₀, and still less at an edge or vertex). Q_enc becomes ambiguous and the equality fails as stated.
The force is not exactly inverse-square. A massive-photon (Proca/Yukawa) field E ∝ (1+μr)e^−μr r̂/r² no longer makes (r̂·n̂)dA/r² a pure solid angle; step 4 collapses and the flux acquires a volume term depending on the field throughout the enclosed region, not just the charge.
Time-varying / radiating sources. Once E is not a static Coulomb superposition, this Coulomb-based route is invalid (retardation matters). Gauss's law survives — but as an independent Maxwell equation, not as a consequence of this argument.
Unbounded charge or a singular field on S. If total enclosed charge diverges, or a point charge sits arbitrarily close to S, the flux integral fails to converge absolutely and the tidy result breaks.

Failure modes

Position-dependence error. Believing the flux depends on where inside the surface the charge sits. It does not — only on whether it is enclosed.
Counting external charges. Adding nearby outside charges into Q_enc. They contribute exactly zero net flux (step 6, outside case), no matter how close or how strong.
Dropping the sign of cos θ. Writing |cos θ| or |dΩ| in step 6, which destroys the entry/exit cancellation and wrongly gives nonzero flux for an exterior charge.
Confusing E with Φ. Concluding that if Q_enc = 0 then E = 0 on S. The field on the surface depends on all charges; only its net flux is set by the enclosed ones.
Open surfaces. Applying Φ = q/ε₀ to a surface that is not closed (a hemisphere alone, a disk). The 4π result needs a closed surface; an open one subtends a partial solid angle.
Using symmetry that isn't there. Pulling E out of ∮E dA = E·A when the field is not constant and normal over S. The flux law always holds; the algebraic shortcut needs a symmetry-matched surface.

Discussion

The physical content lives entirely in step 4. The Coulomb field spreads its 4π steradians of flux outward and thins as 1/r², exactly matching the way a fixed solid angle’s cross-section grows as r². The two powers cancel, so every closed surface around a charge catches the same total flux — the field lines are conserved between charges, terminating only on charge. Gauss's law is the accountant's statement of that conservation.

Logically, Gauss's law is stronger than Coulomb's law even though we derived it from Coulomb. Coulomb's law is the electrostatic special case; Gauss's law is one of the four Maxwell equations and holds even for wildly time-dependent fields, where no simple Coulomb expression exists. What this derivation establishes is the equivalence in electrostatics: given superposition, "inverse-square point field" and "flux = enclosed charge / ε₀" say the same thing.

Applying the divergence theorem, ∮_S E·dA = ∫_V (∇·E) dV, and matching integrands with Q_enc = ∫_V ρ dV gives the local form ∇·E = ρ/ε₀ — the integral law holding for every surface is exactly the content of the differential law at every point.

The exact 1/r² is not a convenience; it is a probe of deep physics. In distributional language ∇·(r̂/r²) = 4π δ³(r), so the point charge is a clean delta-function source and the 4π is the surface area of the unit sphere in three dimensions. Any deviation of the exponent from 2 would signal a nonzero photon mass; null-flux experiments (Cavendish's concentric spheres, refined by Williams, Faller and Hill) look for a would-be charge on the inner sphere and, finding none, bound the exponent to 2 to within parts in 10¹⁶ and the photon mass below ~10⁻¹⁴ eV/c².

Common misconceptions. Gauss's law does not say the field is easy to compute — it says the flux is. The field E at a point on S generally depends on all charges everywhere; it is only the surface-integrated normal component that forgets the outside world. Solving for E pointwise from Gauss's law is legitimate only when symmetry guarantees E is uniform and normal (or tangential, contributing nothing) across a matched surface.

Worked examples

Example 1 — Flux is independent of the surface radius. A point charge q = 5.0 nC sits at the centre of a sphere of radius R = 0.10 m. Find the outward flux, and confirm it does not change if R is doubled.

Φ = ∮_S E·dA = E(R) · 4πR² , E(R) = q/4πε₀R²

Spherical symmetry: E is radial and constant in magnitude over S, so the dot product is E dA and pulls out. A

Φ = (q/4πε₀R²)(4πR²) = q/ε₀

The R² and 4π cancel — radius has left the expression, matching the general result. A

Φ = (5.0×10⁻⁹ C) / (8.854×10⁻¹² C²N⁻¹m⁻²)

Insert numbers only now. A

Φ = 5.6×10² N·m²·C⁻¹ ≈ 565 N·m²·C⁻¹

Reading. Doubling R to 0.20 m quarters E but quadruples the area, leaving Φ = 565 N·m²·C⁻¹ unchanged — the flux belongs to the charge, not the sphere.

Example 2 — Symmetry and solid angle: flux through one face of a cube. A charge q = 12 nC sits at the centre of a cube. Find the flux through a single face.

Φ_total = q/ε₀

The cube is a closed surface enclosing q, so the general result applies whatever the side length. A

Φ_face = Φ_total/6 = q/6ε₀

By the cube’s symmetry the six faces are equivalent; each subtends the same solid angle 4π/6 = 2π/3 sr at the centre, so each carries one-sixth of the flux. B

Φ_face = (12×10⁻⁹) / [6(8.854×10⁻¹²)]

Numbers substituted after the symbolic reduction. A

Φ_face ≈ 226 N·m²·C⁻¹

Reading. No integration over the awkward square face was needed: the solid-angle bookkeeping of the derivation does the work, giving q/6ε₀ directly. Total over six faces returns 12×10⁻⁹/8.854×10⁻¹² = 1355 N·m²·C⁻¹ = q/ε₀.

Problems

A point charge q = 2.0 nC sits at the centre of a sphere of radius r. What is the flux, and how does it change when r is doubled?
Solution
Flux depends only on enclosed charge: Φ = q/ε₀ = (2.0×10⁻⁹)/(8.854×10⁻¹²) = 226 N·m²·C⁻¹. Doubling r leaves it unchanged: 226 N·m²·C⁻¹.
A closed surface encloses two charges, +3.0 nC and −8.0 nC; a third charge +5.0 nC lies just outside it. Find the net flux.
Solution
Only enclosed charge counts: Q_enc = +3.0 − 8.0 = −5.0 nC; the external +5.0 nC contributes zero net flux. Φ = Q_enc/ε₀ = (−5.0×10⁻⁹)/(8.854×10⁻¹²) = −565 N·m²·C⁻¹. The negative sign means net inward flux.
A charge q = 6.0 nC sits at the centre of the flat circular base of a hemispherical surface. Find the flux through the curved (hemispherical) part alone.
Solution
From a point on the base plane, the curved surface caps the entire upper half-space and subtends solid angle 2π sr. Using step 5, Φ = (q/4πε₀)(2π) = q/2ε₀. Numerically Φ = (6.0×10⁻⁹)/[2(8.854×10⁻¹²)] = 339 N·m²·C⁻¹. (This is a charge on the boundary case — note the flat base carries no flux since E is parallel to it.)
A charge q = 8.0 nC sits at one corner of a cube. Find the flux through the whole cube.
Solution
At a corner the cube occupies one octant of the surrounding space, subtending 4π/8 = π/2 sr. Hence Φ = (q/4πε₀)(π/2) = q/8ε₀. Numerically Φ = (8.0×10⁻⁹)/[8(8.854×10⁻¹²)] = 113 N·m²·C⁻¹. (Seven other cubes would be needed to surround the charge fully, together giving q/ε₀.)
A uniform charge density ρ = 1.0 μC·m⁻³ fills a ball of radius a = 0.20 m. Find the flux through a concentric spherical Gaussian surface of radius r = 0.10 m (inside the ball).
Solution
The enclosed charge is only that within r: Q_enc = ρ·(4/3)πr³ = (1.0×10⁻⁶)·(4/3)π(0.10)³ = 4.19×10⁻⁹ C. Then Φ = Q_enc/ε₀ = (4.19×10⁻⁹)/(8.854×10⁻¹²) = 473 N·m²·C⁻¹. (For any r > a the flux would saturate at the total charge ρ(4/3)πa³/ε₀.)