Gauss's Law
Source lecture(s): SC134 Lec 4
Intuition
Count the field lines leaving a closed surface. The answer depends only on how much charge is inside — not on where it sits, not on its shape, not on what else is nearby.
It is Coulomb's law restated, and the restatement is worth a great deal: the inverse-square law is equivalent to "flux is conserved in empty space", because the area of a sphere grows as \(r^2\) exactly as fast as the field falls as \(1/r^2\). The two effects cancel and the flux through any enclosing surface is the same.
Why the surface shape does not matter
Field lines from a point charge are continuous and end only on charge. Any closed surface enclosing the charge is pierced by all of them, however crumpled it is. Any surface not enclosing it is pierced twice by each line — in and out — contributing zero net flux.
That is the whole proof, and it explains the apparently magical feature: a charge just outside your Gaussian surface contributes exactly nothing to the total flux, even though it certainly contributes to \(\mathbf{E}\) at every point on it. Flux is blind to outside charge; the field is not.
Using it: symmetry is everything
Gauss's law is always true and only sometimes useful. To extract \(E\) you must pull it outside the integral, which requires a surface on which \(E\) is constant and either parallel or perpendicular to \(d\mathbf{A}\). In practice that means three cases:
| Symmetry | Gaussian surface | Result |
|---|---|---|
| Spherical (point, sphere, shell) | sphere | \(E = \dfrac{Q}{4\pi\varepsilon_0r^2}\) |
| Cylindrical (line, cylinder) | coaxial cylinder | \(E = \dfrac{\lambda}{2\pi\varepsilon_0r}\) |
| Planar (sheet, slab) | pillbox | \(E = \dfrac{\sigma}{2\varepsilon_0}\) |
Note the striking dependence: point charge \(\propto 1/r^2\), line \(\propto 1/r\), infinite sheet independent of distance. The sheet's field does not weaken as you walk away — because as you recede, more of the sheet comes into view at a compensating rate.
Two classic consequences
A uniform shell exerts no force inside. For \(r < R\) the enclosed charge is zero, so \(E = 0\) everywhere inside a spherical shell. Newton needed a page of geometry for the gravitational version; Gauss's law gives it in a line.
Charge on a conductor lives on the surface. In electrostatic equilibrium \(\mathbf{E} = 0\) inside a conductor (otherwise charges would move). Take a Gaussian surface just inside the boundary: zero field means zero flux means zero enclosed charge. All excess charge is on the surface — which is why a Faraday cage works, and why the field just outside a conductor is \(\sigma/\varepsilon_0\) (not \(\sigma/2\varepsilon_0\) — the conductor suppresses the field on one side, so the whole flux goes outward).
Worked example
A solid insulating sphere of radius \(R\) carries total charge \(Q\) spread uniformly. Find \(E\) both inside and outside.
Outside (\(r > R\)): enclosed charge is all of \(Q\), so \(E = Q/4\pi\varepsilon_0r^2\) — identical to a point charge at the centre.
Inside (\(r < R\)): the enclosed charge scales with volume, \(Q_{\rm enc} = Q(r/R)^3\). Then
The field rises linearly from zero at the centre, peaks at the surface, then falls as \(1/r^2\). The two expressions agree at \(r = R\), as they must.
Common mistakes
- Thinking outside charges do not affect \(\mathbf{E}\) on the surface. They do — they just contribute zero net flux. Gauss's law constrains the integral, not the pointwise field.
- Using it without symmetry. For a charged cube, Gauss's law is true and useless: \(E\) is not constant on any convenient surface, so it cannot be factored out.
- Confusing \(\sigma/\varepsilon_0\) with \(\sigma/2\varepsilon_0\). The isolated sheet gives \(\sigma/2\varepsilon_0\) (flux both ways); a conductor's surface gives \(\sigma/\varepsilon_0\) (flux one way only).
- Forgetting flux is a signed quantity. Inward flux is negative; a surface with equal charges of both signs inside has zero net flux.
Related concepts
- Electric flux — the quantity being counted
- Coulomb's law — the equivalent statement
- Electric field · Conductors & insulators
- Gauss's law for magnetism — the same idea with zero on the right
- Maxwell's equations — this is the first one
- Divergence theorem (PHY621) — the differential form
Knowledge graph position
Prerequisites: electric field, electric flux, Coulomb's law. Leads to: conductors, capacitance, Maxwell's equations.
Quiz
Q1 (conceptual). A charge sits just outside a closed surface. What is the flux through it, and what is the field on it?
Answer
The net flux is zero — every field line that enters also leaves. But the field at each point on the surface is definitely non-zero. Gauss's law constrains only the integral, which is exactly why it is useless for finding \(\mathbf{E}\) without symmetry.
Q2 (computational). A uniformly charged sphere has \(R = 0.1\) m and \(Q = 5\) nC. Find \(E\) at \(r = 0.05\) m and \(r = 0.2\) m.
Answer
Inside: \(E = Qr/4\pi\varepsilon_0R^3 = (8.99\times10^9)(5\times10^{-9})(0.05)/(0.001) = 2.25\times10^{3}\) N/C. Outside: \(E = Q/4\pi\varepsilon_0r^2 = (8.99\times10^9)(5\times10^{-9})/0.04 = 1.12\times10^{3}\) N/C. The field is larger inside at \(r=R/2\) than outside at \(r=2R\) — it peaks at the surface.
Q3 (MCQ). The field of an infinite charged sheet is independent of distance because:
- (a) the sheet has infinite charge
- (b) as you recede, more of the sheet subtends your position, compensating the \(1/r^2\) falloff
- (c) Gauss's law only applies to sheets
- (d) the field is actually zero
Answer
(b). Each element still falls as \(1/r^2\), but the area contributing within a given solid angle grows as \(r^2\). The two cancel exactly, leaving \(\sigma/2\varepsilon_0\) at any distance.