University Physics II · Gauss's Law · 3.3
Choosing a Gaussian Surface
Gauss's law is true for every closed surface and useful for almost none. The choice of surface is what turns a surface integral into one line of algebra, and only three symmetries will do it.
Build the model
Connect the measurement to the mechanism.
Gauss's law ∮E·dA = q(enc)/ε₀ holds for every closed surface you can imagine, which is why it usually hands you nothing. The unknown field sits inside the integral, and one scalar equation cannot determine a vector function of position. To get E out you must know in advance which way it points and where its magnitude is constant, and only symmetry supplies that.
Three families qualify: those unchanged by any rotation about a point, by rotation about and sliding along an infinite axis, or by sliding parallel to an infinite plane. Each forces the field into one direction and allows its magnitude one coordinate. You then assemble a closed surface entirely out of two kinds of piece: constant-field pieces, where E is normal and uniform so the flux is ±EA, and zero-flux pieces, where E lies in the surface or vanishes.
The integral collapses to one number times an area. Every other surface is still true and still useless.
- Simple definition
- A Gaussian surface is an imaginary closed surface chosen so that symmetry makes ∮E·dA trivial: on every piece, either E is constant in magnitude and perpendicular to the surface, or the flux through that piece is zero.
- Example
- For an infinite line charge, take a coaxial cylinder. The curved wall has E constant and outward-normal, giving Φ = E(2πsL); the two flat caps have their normals along the axis while E is radial, so they contribute nothing.
True for any closed surface. All the work is in making the left side factor.
ε₀ = 8.854×10⁻¹² C² N⁻¹ m⁻²; flux in N m² C⁻¹
This is the step that replaces an unknown function with one unknown number.
legal only where |E| is constant and E ∥ dA
Symmetry fixes the direction and the single coordinate the magnitude may use.
spherical · cylindrical (s = distance from axis) · planar
One piece, because every point on it is equivalent under rotation about the centre.
r in m; q(enc) = ∫ρ dV over the enclosed volume
Curved wall constant-field, flat caps parallel to a radial E.
λ in C m⁻¹; caps give zero flux and L cancels
Faces parallel to the sheet carry the flux, the side wall carries none.
σ in C m⁻²; second form when one face sits inside a conductor
Truth is cheap, solvability is not
∮E·dA = q(enc)/ε₀ holds for any closed surface, around any arrangement of charge. That universality is precisely why the law rarely yields a field: the unknown E sits inside the integral, and one scalar equation cannot pin down a vector function of position. Wrap a cube of side a around a point charge and everything the law says is correct. The flux is q/ε₀, and since the six faces are equivalent, each carries q/(6ε₀). You still cannot name E anywhere. On one face the outward-normal component is largest at the centre, directly out from the charge, and falls to 1/(3√3) ≈ 0.19 of that value at a corner: the distance grows by √3, cutting the magnitude by 3, and the field tilts away from the normal by a further factor 1/√3. The one step that converts flux into field is ∮E·dA = E∮dA = EA, and it is legal only where |E| is constant and E is perpendicular to the surface. Choosing a Gaussian surface means engineering that condition, not drawing a box around the charge.
The symmetry test: what the source cannot tell apart
The test is operational. Find a geometric operation — a rotation, a translation, a reflection — that leaves the entire charge distribution looking exactly as it did. The field must then look as it did too, since nothing producing it has changed, so any field component the operation would alter has to be zero. Run it three ways. If the distribution is unchanged by every rotation about one point, no tangential direction is distinguishable from any other, so E is radial and its magnitude can depend only on r: E = E(r)r̂. If it is unchanged by rotation about an infinite axis, by sliding along that axis, and by reflection in any plane perpendicular to it, then the circulating and axial components die and E = E(s)ŝ, with s the distance from the axis. If it is unchanged by every translation parallel to an infinite plane and by reflection in that plane, the in-plane components die and E = E(z)ẑ, reversing as you cross. Notice the price of the word entire: one stray external charge, one nearby plate, one finite end, and the argument is void.
Assemble the surface from two kinds of piece
A Gaussian surface is imaginary. It has no thickness, no charge and no physical existence; it need only be closed and let you evaluate the flux through it. So build it from pieces, each of which must be one of two kinds. A constant-field piece has E perpendicular to it with |E| the same everywhere on it, contributing ±EA. A zero-flux piece has E lying in the surface, or E = 0 there. Put the point whose field you want on a constant-field piece: the surface must pass through it, not merely contain it. Three standard builds follow. A sphere concentric with the symmetry centre is a single constant-field piece, Φ = E(4πr²). A coaxial cylinder of radius s and length L has a constant-field wall, Φ = E(2πsL), plus two zero-flux caps, and L cancels out. A pillbox straddling a sheet has two faces of area A parallel to the sheet and a zero-flux side wall, giving Φ = 2EA and E = σ/(2ε₀) — 56.5 N C⁻¹ for σ = 1.0 nC m⁻². Push one face inside a conductor, where E = 0, and that face becomes a zero-flux piece: Φ = EA and E = σ/ε₀, twice as large.
Count only the charge inside, but the field belongs to everyone
q(enc) is the net charge inside the surface. Get it from the density and the enclosed geometry: ∫ρ dV for a volume charge, λL for the length of line inside, σA for the patch of sheet inside, and signs included, so an enclosed −Q subtracts. Now keep the asymmetry straight. Only interior charge appears on the right, but the E on the left is the total field at the surface, produced by every charge there is. Exterior charges contribute exactly zero net flux, since their field lines enter and leave, yet they do change E point by point — which is why the symmetry test has to be applied to the whole arrangement, not just to the part you enclose. Crossing a boundary means redrawing. For a uniformly charged insulating sphere of radius R carrying Q, a surface at r < R encloses Q(r/R)³, so E = Qr/(4πε₀R³) and the field grows linearly; a surface at r > R encloses all of Q and E = Q/(4πε₀r²). Two regions, two surfaces, and the two expressions agree at r = R.
The choices that fail
A sphere that is not concentric with a spherically symmetric charge still encloses a definite charge and still obeys the law, but |E| varies across it, so nothing factors. A surface around a dipole, or around any two unequal charges, admits no operation that leaves the source invariant, so no direction is forced on the field. A shape chosen to echo the object rather than the symmetry of its field — a cube around a cubical charge — buys nothing. A cylinder around a finite rod fails quantitatively: at the midplane of a rod of length 2L, the infinite-rod result E = λ/(2πε₀s) is too large by a factor √(1 + (s/L)²), which is 0.5% at s = L/10 but 41% at s = L, and near the ends the field is not even radial. Drawing the surface through a point charge, where E is undefined, or exactly inside a charged surface layer, where the field is discontinuous, is undefined bookkeeping; move the face just outside instead. And a surface that has the field point in its interior rather than on it reports nothing about that point at all.
When no surface works, the law still does
Failing the symmetry test does not suspend Gauss's law; it removes the shortcut. The law goes on doing three jobs. It fixes the total flux through any closed surface from the enclosed charge alone, whatever the shape, which is a complete answer to any flux question — 1.0 nC enclosed gives 113 N m² C⁻¹ through a sphere, a cube or a crumpled bag alike. It runs backwards inside conductors: in electrostatic equilibrium E = 0 throughout the conducting material, so a surface drawn inside the metal has zero flux and therefore encloses zero net charge, which pins down induced charge on a cavity wall with no symmetry anywhere. And it constrains: zero enclosed charge forbids net flux, though it permits a field. What the law will not do is hand you E for an irregular distribution. There you fall back on superposition integrals or a numerical flux sum over a mesh, where watching the discrete sum converge to q(enc)/ε₀ tests the mesh, not the law.
Change one variable at a time
Make the relationship visible.
Set q₂ to zero: the two arrows become equal and E = q₁/(4πε₀R²) drops straight out. Then wind q₂ back up and slide it closer — the arrows split, the near one can reverse and point inward, and the flux readout never moves.
TOTAL FLUX Φ339 N m² C⁻¹
E⊥ FAR SIDE2896 N C⁻¹
E⊥ NEAR SIDE899 N C⁻¹
E⊥ GAP |FAR−NEAR|1997 N C⁻¹
Live interpretationTOTAL FLUX Φ: 339 N m² C⁻¹. E⊥ FAR SIDE: 2896 N C⁻¹. E⊥ NEAR SIDE: 899 N C⁻¹. E⊥ GAP |FAR−NEAR|: 1997 N C⁻¹
Catch the common trap
Explain before calculating.
Two equal point charges +q sit a distance d apart. You draw a spherical Gaussian surface of radius r < d centred on one of them, so it encloses that charge only. Which statement is correct?
Choose an answer to test the model.
Practice & worked examples
Reason from the model, then test the result.
EasyAn infinite straight wire carries a uniform λ = 4.0 nC m⁻¹. Choose a Gaussian surface, name which of its pieces carry flux, and find E at s = 2.0 cm from the wire.
- Symmetry test: the wire is unchanged by rotation about its axis, by sliding along it, and by reflection in any plane perpendicular to it, so E = E(s)ŝ — radial, with a magnitude that may depend on s and on nothing else.
- Build the surface to match: a cylinder coaxial with the wire, radius s = 0.020 m and any length L. The curved wall is a constant-field piece — E is normal to it and |E| is the same at every point on it — with area 2πsL. The two flat caps are zero-flux pieces, since their normals lie along the axis while E is radial.
- Flux: Φ = E(2πsL) + 0 + 0.
- Enclosed charge: the surface contains a length L of wire, so q(enc) = λL. Gauss's law gives E(2πsL) = λL/ε₀, and L cancels — the answer cannot depend on how long a cylinder you chose to draw.
- E = λ/(2πε₀s) = 4.0 × 10⁻⁹ ÷ (2π × 8.854 × 10⁻¹² × 0.020) = 4.0 × 10⁻⁹ ÷ 1.113 × 10⁻¹² = 3.6 × 10³ N C⁻¹.
AnswerE = 3.6 × 10³ N C⁻¹, directed radially away from the wire; the curved wall carries all of the flux and the two caps carry none.
MediumA solid insulating sphere of radius R = 5.0 cm carries Q = 6.0 nC spread uniformly through its volume. Find E at r = 2.0 cm and at r = 10.0 cm from the centre.
- The charge is unchanged by every rotation about its centre, so E = E(r)r̂ and a concentric sphere is a single constant-field piece: Φ = E(4πr²).
- The surface must pass through the field point, and the two points lie on opposite sides of the boundary at r = R, so this is two surfaces enclosing two different charges.
- At r = 0.020 m the surface sits inside the charge and encloses the fraction (r/R)³: q(enc) = 6.0 nC × (2.0/5.0)³ = 6.0 × 0.064 = 0.384 nC.
- E = q(enc)/(4πε₀r²) = (8.988 × 10⁹ × 0.384 × 10⁻⁹) ÷ (0.020 m)² = 3.451 ÷ 4.0 × 10⁻⁴ = 8.6 × 10³ N C⁻¹.
- At r = 0.100 m the surface encloses all of Q: E = (8.988 × 10⁹ × 6.0 × 10⁻⁹) ÷ (0.100 m)² = 53.93 ÷ 1.00 × 10⁻² = 5.4 × 10³ N C⁻¹.
- The nearer point wins because inside E ∝ r while outside E ∝ 1/r², so the maximum sits on the sphere's own surface, at 53.93 ÷ 2.5 × 10⁻³ = 2.2 × 10⁴ N C⁻¹.
AnswerE = 8.6 × 10³ N C⁻¹ at r = 2.0 cm and E = 5.4 × 10³ N C⁻¹ at r = 10.0 cm, both radially outward.
HardA point charge q = +2.5 nC sits at the centre of a cube of side a = 12 cm. Find the flux through the whole cube and through one face, then decide whether Gauss's law hands you E on that face.
- Flux needs no symmetry at all: Φ = q/ε₀ = 2.5 × 10⁻⁹ ÷ 8.854 × 10⁻¹² = 2.8 × 10² N m² C⁻¹, whatever the shape or size of the box.
- The cube's own symmetry makes the six faces equivalent, so each takes Φ/6 = 282.4 ÷ 6 = 47 N m² C⁻¹.
- Now try to factor E out. At the centre of a face the charge is a/2 = 0.060 m away and E points straight along the normal: E⊥ = (8.988 × 10⁹ × 2.5 × 10⁻⁹) ÷ (0.060 m)² = 22.47 ÷ 3.6 × 10⁻³ = 6.2 × 10³ N C⁻¹.
- At a corner of the same face the distance is (a/2)√3 = 0.104 m, so |E| is smaller by 3 at 22.47 ÷ 1.08 × 10⁻² = 2.08 × 10³ N C⁻¹, and only cos θ = 1/√3 of it lies along the normal: E⊥ = 2.08 × 10³ × 0.577 = 1.2 × 10³ N C⁻¹.
- E⊥ runs 5.2 : 1 across a single face, so |E| is not constant on it and ∮E·dA ≠ EA. The two flux numbers are exact; the field number does not exist on this surface.
- Swap the cube for a concentric sphere and the same charge factors out in one line. The surface has to match the symmetry of the field, not the shape of the container.
AnswerΦ = 2.8 × 10² N m² C⁻¹ through the cube and 47 N m² C⁻¹ through each face; E cannot be extracted, because E⊥ falls from 6.2 × 10³ N C⁻¹ at the face centre to 1.2 × 10³ N C⁻¹ at a corner.