Gauss's Law: Electric Flux and Gaussian Surfaces (PHY 102)

A Shortcut Through Symmetry

Coulomb's law gives the field of a single point charge, but what about a long charged wire, a charged sheet or a charged ball? Adding up the fields of millions of tiny charges is hard work. Gauss's law offers a beautiful shortcut: when a charge distribution is symmetric, the field can be found in a few lines. It is one of Maxwell's four equations of electromagnetism and a core topic in first-year university physics (PHY 102 and similar courses). This tutorial explains electric flux, states Gauss's law, shows how to choose a Gaussian surface, and applies the law to a point charge, a charged sphere, a line of charge, a plane sheet and a conductor, with worked examples. It assumes you have read the tutorial on electric fields and Coulomb's law.

Electric Flux

Electric flux Φ measures how much electric field passes through a surface. Picture the field lines as a flow: the flux counts how many lines cross the surface.

For a uniform field E through a flat area A:

Φ = EA cos θ

where θ is the angle between the field and the normal (the perpendicular) to the surface. The unit is N m²/C (equivalently V m).

When the field is perpendicular to the surface (θ = 0°), the flux is maximum: Φ = EA.
When the field runs along the surface (θ = 90°), no lines cross it, so Φ = 0.
For a closed surface, the normal points outwards, so flux leaving the surface is positive and flux entering is negative.

For a curved surface or a varying field, divide the surface into tiny pieces dA and add up: Φ = ∮ E · dA. The circle on the integral sign means the surface is closed.

Worked example 1

A uniform field of 500 N/C passes through a flat surface of area 0.2 m². The field makes an angle of 60° with the normal to the surface. Find the flux.

Φ = EA cos θ = 500 × 0.2 × cos 60° = 50 N m²/C.

Statement of Gauss's Law

Gauss's law states that the total electric flux through any closed surface equals the total charge enclosed by the surface divided by ε₀:

∮ E · dA = Q_enc ÷ ε₀

where ε₀ = 8.85 × 10⁻¹² C² N⁻¹ m⁻² is the permittivity of free space. Some key consequences:

Only the charge inside the surface matters. Charges outside contribute field lines that enter and leave again, giving zero net flux.
The shape and size of the closed surface do not matter. Any closed surface around the same charge has the same flux.
If a closed surface encloses no net charge, the net flux through it is zero, even if the field is not zero on the surface.

Worked example 2

A charge of 8.85 nC sits at the centre of a cube. Find the flux through the whole cube and through one face.

Φ = Q/ε₀ = 8.85 × 10⁻⁹ ÷ 8.85 × 10⁻¹² = 1000 N m²/C. By symmetry each of the six faces receives one sixth: about 167 N m²/C.

How to Use Gauss's Law

Gauss's law is always true, but it only gives the field easily when you choose a clever imaginary closed surface, called a Gaussian surface, where:

The field has the same magnitude everywhere on the surface (or on parts of it), so E can be taken outside the integral.
The field is either perpendicular to the surface (flux = EA) or parallel to it (flux = 0) on each part.

The method has four steps: identify the symmetry, choose a Gaussian surface that matches it, write the flux in terms of E, and set it equal to Q_enc/ε₀. The three symmetries that work are spherical (use a sphere), cylindrical (use a cylinder) and planar (use a "pillbox" cylinder through the plane).

Application 1: Point Charge

Surround a point charge Q with a sphere of radius r centred on it. By symmetry E is the same everywhere on the sphere and points radially outwards, perpendicular to the surface. So:

E × 4πr² = Q/ε₀, giving E = Q ÷ (4πε₀r²)

This is exactly Coulomb's law, derived in one line. Gauss's law and Coulomb's law are two ways of expressing the same physics.

Application 2: Charged Sphere

Outside the sphere (r ≥ R)

For any spherically symmetric charge Q of radius R, a Gaussian sphere of radius r encloses all of Q, so E = kQ/r², where k = 1/(4πε₀). A charged sphere behaves, from outside, as if all its charge were concentrated at its centre.

Inside a conducting sphere (r < R)

On a conductor, all the charge sits on the outer surface. A Gaussian sphere inside encloses no charge, so E = 0 inside.

Inside a uniformly charged insulating sphere (r < R)

If the charge is spread evenly through the volume, a Gaussian sphere of radius r encloses only the fraction r³/R³ of the charge. Then E × 4πr² = (Qr³/R³)/ε₀, so:

E = kQr ÷ R³

The field grows linearly from zero at the centre to a maximum at the surface, then falls off as 1/r² outside.

Worked example 3

An insulating sphere of radius 0.10 m carries a charge of 5 nC spread uniformly through its volume. Find the field at r = 0.05 m, at the surface, and at r = 0.20 m. (Take k = 9 × 10⁹ N m² C⁻².)

Inside: E = kQr/R³ = 9 × 10⁹ × 5 × 10⁻⁹ × 0.05 ÷ 0.001 = 2250 N/C.

At the surface: E = kQ/R² = 45 ÷ 0.01 = 4500 N/C.

Outside: E = kQ/r² = 45 ÷ 0.04 = 1125 N/C.

Application 3: Infinite Line of Charge

A long straight wire carries charge λ per metre. By symmetry the field points radially outwards from the wire. Choose a cylinder of radius r and length L around the wire.

The curved surface: E is perpendicular and constant, so flux = E × 2πrL.
The two flat ends: E is parallel to them, so flux = 0.

The enclosed charge is λL, so E × 2πrL = λL/ε₀, giving:

E = λ ÷ (2πε₀r) = 2kλ ÷ r

The field of a line falls off as 1/r, more slowly than the 1/r² of a point charge.

Worked example 4

A long wire carries 2 μC/m. Find the field 10 cm from it.

E = 2kλ/r = 2 × 9 × 10⁹ × 2 × 10⁻⁶ ÷ 0.1 = 3.6 × 10⁵ N/C.

Application 4: Infinite Plane Sheet

A large flat sheet carries charge σ per square metre. The field is perpendicular to the sheet on both sides. Choose a pillbox: a short cylinder with flat ends of area A, one on each side of the sheet.

The curved side is parallel to the field, so its flux is zero.
Each end has flux EA, giving a total of 2EA.

The enclosed charge is σA, so 2EA = σA/ε₀, giving:

E = σ ÷ (2ε₀)

Remarkably, the field does not depend on the distance from the sheet. This is why the field between two large parallel plates carrying +σ and −σ is uniform. There, the two sheets' fields add between the plates, giving E = σ/ε₀, and cancel outside.

Worked example 5

A large sheet carries a surface charge density of 1.77 μC/m². Find the field near it. What is the field between two parallel plates carrying ±1.77 μC/m²?

Single sheet: E = σ/2ε₀ = 1.77 × 10⁻⁶ ÷ (2 × 8.85 × 10⁻¹²) = 1.0 × 10⁵ N/C. Between the plates: E = σ/ε₀ = 2.0 × 10⁵ N/C.

Gauss's Law and Conductors

Gauss's law explains several facts about conductors in electrostatic equilibrium:

The field inside the material of a conductor is zero. Otherwise, free electrons would move.
So a Gaussian surface just inside a conductor encloses no net charge, which means all excess charge lies on the surface.
Just outside the surface, the field is perpendicular to it, with magnitude E = σ/ε₀, found using a pillbox with one end inside the conductor.
A hollow conductor shields its interior from outside fields. This is the principle of the Faraday cage, which protects sensitive equipment and explains why a car is fairly safe in a lightning storm.

Gauss's Law: Practice Questions

A closed surface encloses charges of +6 nC, −2 nC and +1 nC. Find the net flux. (ε₀ = 8.85 × 10⁻¹² C² N⁻¹ m⁻²)
A uniform field of 300 N/C passes through a 0.5 m² flat surface parallel to the surface. What is the flux?
Find the field 0.3 m from the centre of a conducting sphere of radius 0.1 m carrying 2 nC. What is the field 0.05 m from the centre?
Find the field 5 cm from a long wire carrying 1 μC/m.
Explain why the net flux through a closed surface in a uniform field is zero.

Answers: 1. Q_enc = +5 nC, so Φ = 5 × 10⁻⁹ ÷ 8.85 × 10⁻¹² ≈ 565 N m²/C. 2. Zero, because θ = 90° to the normal. 3. Outside: E = 9 × 10⁹ × 2 × 10⁻⁹ ÷ 0.09 = 200 N/C. Inside a conductor: 0. 4. E = 2kλ/r = 2 × 9 × 10⁹ × 10⁻⁶ ÷ 0.05 = 3.6 × 10⁵ N/C. 5. No charge is enclosed, so every field line that enters the surface also leaves it.

Conclusion

Gauss's law states that the total flux through a closed surface equals the enclosed charge divided by ε₀. It is always true, but it becomes a powerful tool only when symmetry lets you choose a Gaussian surface on which the field is constant and perpendicular. Spherical symmetry gives E = kQ/r² outside a sphere, cylindrical symmetry gives E = λ/2πε₀r for a line, and planar symmetry gives E = σ/2ε₀ for a sheet. The same law explains why charge lives on the surface of a conductor and why the field inside is zero.