Gauss's law, explained with examples

By Ashraf Bhura, Co-Founder, STEM-Tutor · Updated October 2026

Gauss's law links the electric field on a closed surface to the charge inside it. With the right choice of surface, it turns hard electric field calculations into a few lines, and it is a core topic in AP Physics C: Electricity and Magnetism.

Electric flux

Electric flux Φ_E measures how much electric field passes through a surface. For a uniform field through a flat area,

Φ_E = E A cosθ

where θ is the angle between the field and the normal to the surface. For curved surfaces or non-uniform fields, add up the contributions from small pieces: Φ_E = ∫ E · dA. Field lines leaving a closed surface count as positive flux; lines entering count as negative.

The statement of Gauss's law

∮ E · dA = Q_enc / ε₀

The total electric flux out of any closed surface equals the charge enclosed by that surface divided by ε₀ = 8.85 × 10⁻¹² C²/(N m²). Charges outside the surface contribute zero net flux, because every field line from them that enters the surface also leaves it.

Gauss's law is always true, but it is only useful for calculating E when the charge distribution has strong symmetry: spherical, cylindrical or planar. Then you can choose a Gaussian surface on which E is constant in size and either perpendicular or parallel to the surface, so the integral becomes simple multiplication.

Choosing a Gaussian surface

• Spherical symmetry (point charges, charged spheres): use a concentric sphere.
• Cylindrical symmetry (long lines or cylinders of charge): use a coaxial cylinder.
• Planar symmetry (large sheets of charge): use a small cylinder, a “pillbox”, crossing the sheet.

A four-step method

1. Identify the symmetry of the charge distribution.
2. Choose a Gaussian surface through the point where you want E, matching that symmetry.
3. Write the flux as E times the area where the field crosses the surface, ignoring parts where the field runs parallel to it.
4. Find the charge enclosed, set flux equal to Q_enc/ε₀, and solve for E.

Worked examples

Example 1: a point charge

Use Gauss's law to find the field at distance r from a point charge Q.

• Choose a sphere of radius r centred on Q. By symmetry, E is the same everywhere on it and points straight outwards.
• Flux = E × 4πr² = Q/ε₀, so E = Q/(4πε₀r²). This is Coulomb's law, recovered from Gauss's law.

Example 2: a uniformly charged solid sphere

A solid insulating sphere of radius R carries charge Q spread uniformly through its volume. Find E inside and outside.

• Outside (r > R): the Gaussian sphere encloses all of Q, so E = Q/(4πε₀r²), as if all the charge were at the centre.
• Inside (r < R): the enclosed charge is Q(r³/R³). So E × 4πr² = Qr³/(ε₀R³), giving E = Qr/(4πε₀R³).
• Inside, E grows linearly from zero at the centre; outside, it falls as 1/r². The two formulas agree at r = R.

Example 3: a long line of charge

Find the field at distance r from a long straight wire with charge per unit length λ.

• Choose a cylinder of radius r and length L around the wire. E points radially outwards, so there is no flux through the flat ends.
• Flux through the curved side = E × 2πrL = λL/ε₀.
• E = λ/(2πε₀r), falling as 1/r, more slowly than a point charge.

Example 4: an infinite sheet

Find the field near a large flat sheet with surface charge density σ.

• Use a pillbox of end area A straddling the sheet. The field is perpendicular to the sheet, so flux passes only through the two ends.
• 2EA = σA/ε₀, so E = σ/(2ε₀).
• The field does not depend on distance from the sheet. Two oppositely charged parallel sheets, like a capacitor, give E = σ/ε₀ between them and zero outside.

Example 5: a numerical flux

A closed surface surrounds charges of +5.0 nC and −2.0 nC, and a +3.0 nC charge sits outside it. Find the total flux through the surface.

• Only enclosed charge counts: Q_enc = 3.0 nC.
• Φ = 3.0 × 10⁻⁹/8.85 × 10⁻¹² ≈ 339 N m²/C.

Conductors in electrostatic equilibrium

Gauss's law explains several facts about conductors. Inside a conductor in equilibrium, E = 0; otherwise free charges would keep moving. Taking a Gaussian surface just inside the conductor's surface encloses no net charge, so any excess charge sits on the outer surface. Just outside the surface, the field is perpendicular to it with size E = σ/ε₀. A hollow conductor shields its interior from external fields, which is the principle of a Faraday cage.

Gauss's law and Coulomb's law

For static charges, Gauss's law and Coulomb's law contain the same physics; each can be derived from the other. Gauss's law is more general, though: it is one of Maxwell's four equations, which describe all of electromagnetism, including light. A similar law for gravity explains why a uniform planet's field outside it acts as if all its mass were at the centre.

Common mistakes

• Using Gauss's law to find E for an irregular charge distribution; it only simplifies with symmetry.
• Including charges outside the Gaussian surface in Q_enc.
• Forgetting that the flat ends of a cylinder carry no flux for a line of charge.
• Using the total charge instead of the enclosed fraction inside a solid sphere.
• Thinking zero flux means zero field everywhere on the surface.

Practice questions

Try each one before opening the answer.

Frequently asked questions

Is Gauss's law on AP Physics C: E&M?

Yes. Gauss's law is a central topic of AP Physics C: Electricity and Magnetism, usually with spherical, cylindrical and planar symmetry.

Is Gauss's law in A level Physics?

Not by name in most A level Physics specifications, although the inverse square fields it predicts are covered. It is standard in first-year university physics.

Key takeaways

• Flux Φ_E = ∫ E · dA measures field through a surface.
• Gauss: total flux out of a closed surface = Q_enc/ε₀.
• Use symmetry to choose a surface where E is constant.
• Point charge: E ∝ 1/r²; line: E ∝ 1/r; sheet: E constant.
• Inside a conductor, E = 0 and excess charge sits on the surface.

Related guides

• Coulomb's law
• Faraday's law
• AP Physics C: E&M review