Source: Chapter 6
Tags: Gauss's law, electric flux, Gaussian surface, spherical symmetry, cylindrical symmetry, planar symmetry, conductors in equilibrium, enclosed charge
Difficulty: Intermediate
Prerequisites: Chapter 5 notes (electric charge, electric field, Coulomb's law). Comfort with integration notation and surface area formulas.
Coulomb's law works for point charges, but becomes unwieldy for continuous charge distributions. Gauss's law is a more elegant tool: it relates the total electric flux through a closed surface to the charge enclosed inside. When the charge distribution has a nice symmetry (spherical, cylindrical, or planar), Gauss's law lets you find the electric field with far less effort than direct integration. This chapter is the bridge between "here is the field from a single charge" and "here is the field from any symmetric arrangement of charge."
Electric flux measures how much electric field passes through a surface. Gauss's law says the total flux through any closed surface equals the enclosed charge divided by ε₀. Combined with the right choice of Gaussian surface, it gives you the electric field for symmetric charge distributions in a few lines of algebra.
Area vector (A)
A vector whose magnitude equals the area of a surface and whose direction is perpendicular (normal) to the surface. Think of it as: the area vector tells you which way the surface "faces."
Electric flux (Φ)
The measure of how much electric field passes through a surface. For a uniform field and flat surface: Φ = E · A = EA cos θ. For a general surface: Φ = ∮ E · dA. In simple terms, flux counts field lines going through a surface.
Gaussian surface
An imaginary closed surface chosen to exploit the symmetry of a charge distribution. It is not a physical object. You pick its shape so that E is either constant across the surface or zero, making the flux integral easy to evaluate.
Gauss's law
The total electric flux through any closed surface equals the net charge enclosed divided by the permittivity of free space: Φ = Q_enclosed / ε₀. Think of it as: the number of field lines leaving a closed surface depends only on how much charge is trapped inside.
Spherical symmetry
A charge distribution that looks the same from any direction around a central point. The Gaussian surface is a sphere.
Cylindrical symmetry
A charge distribution that looks the same from any direction around a central axis and is uniform along that axis. The Gaussian surface is a cylinder.
Planar symmetry
A charge distribution that is uniform across an infinite flat plane. The Gaussian surface is a rectangular box (pillbox) straddling the plane.
Electrostatic equilibrium
The state of a conductor in which no charges are moving. In equilibrium, the electric field inside the conductor is zero, and any excess charge resides entirely on the surface.
The area vector A has magnitude equal to the surface area and direction perpendicular to the surface.
For a uniform field through a flat surface: Φ = EA cos θ, where θ is the angle between E and A.
When field lines enter a closed surface, the contribution to the flux is negative. When field lines leave, it is positive.
If no charge is enclosed, every field line that enters must also exit somewhere, so the net flux through the closed surface is zero.
For non-uniform fields or curved surfaces, the flux becomes an integral: Φ = ∮ E · dA.
Gauss's law: ∮ E · dA = Q_enclosed / ε₀
Q_enclosed is the total net charge inside the Gaussian surface. Charges outside contribute zero net flux.
The law is always true, but it is only useful for finding E when the problem has enough symmetry that E can be pulled out of the integral.
Spherical symmetry: Use a spherical Gaussian surface of radius r centred on the charge distribution.
The electric field is radial and has the same magnitude at every point on the sphere.
Flux simplifies to: E(4πr²) = Q_enclosed / ε₀.
Cylindrical symmetry: Use a cylindrical Gaussian surface of radius r and height h, coaxial with the charge distribution.
The field is radial (pointing away from or toward the axis) and has the same magnitude at every point on the curved surface.
Flux through the curved surface: E(2πrh) = Q_enclosed / ε₀.
The flat end caps contribute zero flux because E is parallel to them.
Planar symmetry: Use a rectangular pillbox that straddles the plane of charge.
The field is perpendicular to the plane and has the same magnitude on both sides.
Flux through the two end faces: E(2A) = Q_enclosed / ε₀, giving E = σ/(2ε₀) for a single infinite plane.
Inside a conductor in equilibrium, E = 0. If there were a field, it would push the free charges until they rearranged to cancel it.
Any excess charge on a conductor sits on its surface.
Just outside the surface of a conductor: E = σ/ε₀, directed perpendicular to the surface.
The charge density (and therefore the field) is greatest at regions with a smaller radius of curvature (sharper points).
Point charge: E(4πr²) = q/ε₀, so E = kq/r².
Point charge inside an uncharged metal shell:
Inside the cavity (r < inner radius a): E(4πr²) = q/ε₀. Field is as if the shell were absent.
In the metal itself: E = 0. The point charge draws -q to the inner surface and +q appears on the outer surface.
Outside the shell (r > outer radius b): E(4πr²) = q/ε₀. Looks like a point charge from the outside.
Point charge inside a metal shell carrying its own charge Q:
In the metal: E = 0.
Outside: E(4πr²) = (q + Q)/ε₀.
Charged metal shell with no point charge inside:
Inside the shell: E = 0.
In the metal: E = 0.
Outside: E(4πr²) = Q/ε₀.
Insulating sphere with uniform charge density ρ (total charge Q, radius R):
Inside (r < R): E(4πr²) = ρ(4/3 πr³)/ε₀, so E = ρr/(3ε₀). The field grows linearly with r.
Outside (r > R): E(4πr²) = Q/ε₀, so E = kQ/r². Looks like a point charge.
Insulating spherical shell (inner radius a, outer radius b, charge density ρ):
r < a: E = 0.
a < r < b: E(4πr²) = ρ × (4/3)π(r³ - a³)/ε₀.
r > b: E(4πr²) = Q_total/ε₀.
Non-uniform charge density ρ(r): Integrate the charge enclosed: Q_enclosed = ∫ ρ(r) 4πr² dr from 0 to r. Then apply Gauss's law as usual.
Insulating cylinder, uniform charge density ρ, radius R:
Inside (r < R): E(2πrh) = ρ(πr²h)/ε₀, so E = ρr/(2ε₀).
Outside (r > R): E(2πrh) = ρ(πR²h)/ε₀, so E = ρR²/(2ε₀r).
Non-uniform ρ(r): Integrate Q_enclosed = ∫₀ʳ ρ(r') 2πr'h dr'. Use dV = 2πr h dr.
Hollow insulating cylinder (inner a, outer b, charge density ρ):
r < a: E = 0.
a < r < b: E(2πrh) = ρπ(r² - a²)h/ε₀.
r > b: E(2πrh) = ρπ(b² - a²)h/ε₀.
Single infinite insulating plane with surface charge density σ: E = σ/(2ε₀), uniform on both sides.
Two infinite parallel planes with equal and opposite surface charge densities ±σ: fields add between the plates to E = σ/ε₀ and cancel to zero outside.
Quantity | Formula |
|---|---|
Electric flux (uniform) | Φ = EA cos θ |
Gauss's law | ∮ E · dA = Q_enc / ε₀ |
Spherical Gaussian surface area | 4πr² |
Cylindrical Gaussian curved area | 2πrh |
Field inside insulating sphere | E = ρr / (3ε₀) |
Field outside any spherically symmetric charge | E = kQ / r² |
Field outside infinite line/cylinder | E = λ / (2πε₀r) |
Field from infinite plane | E = σ / (2ε₀) |
Conductor surface field | E = σ / ε₀ |
The reason charge accumulates at sharp points on a conductor (and the field is strongest there) is the principle behind lightning rods: the rod's pointed tip creates a strong enough field to initiate a controlled discharge. Coaxial cables exploit cylindrical symmetry to contain electric fields entirely within the cable, shielding signals from external interference.
Students often think Gauss's law only works when there is symmetry. The law is always true. Symmetry is needed to make it useful for calculating E, but the flux-equals-enclosed-charge relationship holds for any closed surface and any charge distribution.
A common error is including charges outside the Gaussian surface when computing Q_enclosed. Only charges inside the surface matter for the flux.
Students sometimes forget that E = 0 inside a conductor in equilibrium, even if there is a charge nearby. The free charges in the conductor rearrange to make this happen.
When a point charge sits inside a metal shell, students sometimes think the shell "blocks" the field. It does not: outside the shell, the field is exactly as if the total enclosed charge were a point charge at the centre.
⚠️ Know which Gaussian surface to use for each symmetry type. If you pick the wrong shape, you cannot simplify the integral.
⚠️ Be able to state what happens inside a conductor in electrostatic equilibrium (E = 0, excess charge on the surface).
⚠️ The inside-outside pattern for insulating spheres is a very common exam question: field grows linearly with r inside, falls off as 1/r² outside.
⚠️ Problems with nested shells (a point charge inside a conducting shell, possibly carrying its own charge) appear frequently. Practise identifying the charge on each surface.
True or false: The electric flux through a closed surface that encloses no charge is always zero.
True.
Fill in the blank: Inside a solid conductor in electrostatic equilibrium, the electric field is ________.
Zero.
True or false: If you double the radius of a spherical Gaussian surface surrounding a point charge, the total flux through the surface doubles.
False. The flux depends only on the enclosed charge, which hasn't changed.
Fill in the blank: The electric field inside a uniformly charged insulating sphere increases ________ with distance from the centre.
Linearly.
Q: A non-conducting sphere of radius R has total charge Q spread uniformly throughout its volume. The electric field at the surface has magnitude E. What is the field at r = R/3?
A: Inside the sphere, E is proportional to r. At the surface (r = R), the field is E. At r = R/3, the field is E × (R/3)/R = E/3.
Q: A metal shell carries charge +Q on its outer surface. A point charge -q is placed at its centre. What is the charge on the inner surface of the shell?
A: The inner surface must carry +q (to cancel the field in the metal). The outer surface then carries Q - q.
Q: An infinite plane of surface charge density σ = 4 x 10⁻⁶ C/m² sits in free space. What is the electric field magnitude near the plane?
A: E = σ/(2ε₀) = (4 x 10⁻⁶) / (2 × 8.85 x 10⁻¹²) ≈ 2.26 x 10⁵ N/C.
Gauss's law is one of Maxwell's four equations and reappears in that context in the electromagnetic waves chapter (Ch. 16). The field results derived here feed directly into the electric potential chapter (Ch. 7), where you integrate E to find voltage. The conductor-in-equilibrium results are essential for understanding capacitors (Ch. 8).
Gauss's law, electric flux, Gaussian surface, closed surface, enclosed charge, permittivity of free space, ε₀, spherical symmetry, cylindrical symmetry, planar symmetry, conductor equilibrium, surface charge, shell theorem, insulating sphere, coaxial, pillbox, flux integral, PHYS 212