Difficulty: Introductory-Intermediate | Prerequisites: Electric fields, Coulomb's law, vector dot products, surface integrals (conceptual level).
This is your first real encounter with Gauss' law, one of the four Maxwell equations that govern all of electromagnetism. The core idea: instead of summing up contributions from every tiny charge (Coulomb's law, painful for distributed charges), you can exploit symmetry and a clever imaginary surface to find the electric field in one step. You need to be comfortable with electric field lines and the dot product before this will click. If you are not, revisit those topics first. Everything from capacitors to conductors later in the course leans on what you learn here.
Electric flux measures how much electric field passes through a surface. Gauss' law links the total flux through a closed surface to the charge inside it. When a charge distribution has spherical, cylindrical, or planar symmetry, Gauss' law gives you the electric field directly.
Electric flux (Φ)
The net amount of electric field flowing through a surface, defined as Φ = ∫ E · dA. In simple terms, think of it as counting how many field lines pierce through a surface, with direction taken into account.
Gaussian surface
An imaginary closed surface you choose to make a Gauss' law calculation tractable. It is not a physical object. You pick its shape to match the symmetry of the charge distribution so that E is constant over the surface and either parallel or perpendicular to dA everywhere.
dA** (area vector element)**
A tiny patch of surface area, represented as a vector pointing outward, perpendicular (normal) to the surface. Its magnitude is the area of the patch. Think of it as a small arrow sticking straight out of the surface at every point.
Gauss' law
The statement that the net electric flux through any closed surface equals the enclosed charge divided by ε₀: ∮ E · dA = Q_enc / ε₀. In plain terms, the total "flow" of field out of a closed surface tells you exactly how much charge is inside.
Permittivity of free space (ε₀)
A fundamental constant (8.85 × 10⁻¹² C²/N·m²) that appears in Coulomb's law and Gauss' law. It sets the scale for how strong electric fields are in vacuum.
Net flux
The difference between flux flowing out of and into a closed surface. If equal amounts of field enter and leave, the net flux is zero, regardless of how strong the field is inside.
The dot product E · dA = |E| dA cos θ, where θ is the angle between the field and the outward normal.
When E is parallel to dA (θ = 0°), flux through that patch is maximised.
When E is perpendicular to dA (θ = 90°), no flux passes through that patch. This is why a flat plane rotated 90° relative to a uniform field has zero net flux.
Flat plane aligned with field (dA** parallel to E):** There is a net flux. All of the field passes straight through.
Flat plane perpendicular to field (dA** at 90° to E):** Net flux is zero, because cos 90° = 0.
Closed sphere in a uniform field: Net flux is zero. Every field line that enters on one side exits on the other. The field inside is still E (non-zero), but Gauss' law only tells you about enclosed charge, which here is zero.
Positive net flux: more field lines leave the surface than enter. This means net positive charge is enclosed.
Negative net flux: more field lines enter than leave. Net negative charge is enclosed.
Zero net flux: no net charge enclosed (there could still be equal positive and negative charges inside).
Gauss' law is always true, but it is only a practical tool for finding the electric field when you can argue by symmetry that E has constant magnitude over your Gaussian surface and is everywhere parallel (or perpendicular) to dA. Three classic symmetries qualify:
Spherical symmetry: use a Gaussian sphere.
Cylindrical symmetry (infinite line or cylinder): use a Gaussian cylinder.
Planar symmetry (infinite sheet or slab): use a Gaussian "pillbox" (rectangular box straddling the surface).
If the problem lacks one of these symmetries, Gauss' law still tells you about enclosed charge but will not hand you E directly. You may need superposition instead.
For a Gaussian sphere of radius R enclosing charges Q₁ and Q₂ at its centre:
Surface area: ∫ dA = 4πR²
Gauss' law: |E| · 4πR² = (Q₁ + Q₂) / ε₀
Therefore: E = (Q₁ + Q₂) / (4πε₀R²) r̂
This recovers Coulomb's law for a point charge when Q₁ + Q₂ = Q.
When charges sit inside a sphere and a uniform external field E₀ is also present, the total field at any point is the vector sum of the field from the charges and E₀.
At a point on the sphere along the direction of E₀ (say point C, to the right): E = |E₀| x̂ + (Q₁ + Q₂)/(4πε₀R²) x̂. The two fields add or partially cancel depending on the sign of the enclosed charge.
At the point directly opposite (point D, to the left): the charge field reverses direction, so it adds to E₀ instead. Field at D > field at C when Q₁ + Q₂ < 0.
At points above and below (A and B): the charge field is vertical while E₀ is horizontal, so the magnitudes at A and B are equal (symmetric), though the vertical components point in opposite directions.
These are the building blocks for nearly every problem in the course. Memorise them.
E = ρR³ / (3ε₀r²), directed radially outward.
This is equivalent to a point charge Q = (4/3)πR³ρ at the centre.
E = λ / (2πε₀r), directed radially outward from the line.
The field falls off as 1/r, not 1/r².
E = σ / (2ε₀), directed away from the surface.
The field is constant, independent of distance. This is why superposition with infinite planes is straightforward.
Result | Formula | Direction |
|---|---|---|
Flux definition | Φ = ∫ E · dA | — |
Gauss' law | ∮ E · dA = Q_enc / ε₀ | — |
Sphere of charge (outside) | E = ρR³ / (3ε₀r²) | Radially outward |
Infinite line charge | E = λ / (2πε₀r) | Radially outward |
Infinite plane of charge | E = σ / (2ε₀) | Normal to surface |
Sphere surface area | A = 4πR² | — |
Cylinder lateral area | A = 2πrL | — |
The infinite-plane result is why parallel-plate capacitors produce a nearly uniform field between their plates, which is the basis of most capacitor designs in electronics. Gauss' law for spheres is how physicists first confirmed that the gravitational and electric force laws share the same 1/r² form, because a uniform shell acts exactly like a point source from the outside.
Students often confuse net flux with the field strength inside a surface. A sphere in a uniform field has zero net flux, but the field inside is definitely not zero. Zero net flux means zero enclosed charge, full stop.
Students assume Gauss' law only works for spheres. It works for any closed surface, but it only lets you solve for E directly when there is enough symmetry.
Mixing up "the field is zero inside" with "the flux is zero through the surface." These are completely different statements. A conducting shell has zero field inside, while a sphere in a uniform field has zero flux but non-zero field everywhere.
Forgetting that dA always points outward for a closed surface. If you reverse the convention, you flip the sign of the flux.
⚠️ You will be asked to identify when Gauss' law is and is not useful for finding E. The answer hinges on symmetry, not on whether there is charge present.
⚠️ Expect problems that combine a symmetric charge distribution with a uniform external field. You must use superposition: find E from Gauss' law for the charges alone, then add E₀ as a separate vector.
⚠️ Know the three standard results (sphere, line, plane) cold. Many exam problems are layered applications of these.
⚠️ The sign of flux links directly to the sign of enclosed charge. If Q_enc < 0, the net flux is negative. This is tested both conceptually and in calculation.
True or false: If the net electric flux through a closed surface is zero, the electric field everywhere on that surface must also be zero.
Fill in the blank: The electric field due to an infinite line charge falls off as ______.
True or false: Gauss' law can be used to find the electric field of two point charges that are not at the centre of a sphere.
Fill in the blank: For a Gaussian surface enclosing net charge Q, the net flux is ______.
True or false: Rotating a flat Gaussian surface by 90° in a uniform field changes the net flux from non-zero to zero.
Answers: 1. False (a uniform field through a sphere gives zero flux but non-zero E). 2. 1/r. 3. False (insufficient symmetry to solve for E directly, though the law still holds). 4. Q/ε₀. 5. True.
Q: A Gaussian sphere of radius R encloses a total charge of -5 μC. What is the net electric flux through the sphere?
A: Φ = Q_enc / ε₀ = (-5 × 10⁻⁶) / (8.85 × 10⁻¹²) ≈ -5.65 × 10⁵ N·m²/C. The negative sign means field lines point inward on average.
Q: An imaginary sphere is placed in a uniform electric field with no charges inside. Is there a net flux through the sphere? Is the field inside zero?
A: Net flux is zero (no enclosed charge). The field inside is not zero; it equals the uniform external field E₀.
Q: Why does the electric field of an infinite plane not depend on distance from the plane?
A: Because a Gaussian pillbox argument shows that only the enclosed charge per unit area σ and the two flat faces of the pillbox matter. Moving the pillbox further away does not change the enclosed charge or the geometry, so E = σ/(2ε₀) everywhere.
Q: At point C on the surface of a sphere containing charges Q₁ + Q₂ < 0 in a uniform field E₀ (pointing in the +x direction), write the total field. C is on the +x side of the sphere at distance R from the centre.
A: E = |E₀| x̂ + (Q₁ + Q₂)/(4πε₀R²) x̂. Since Q₁ + Q₂ < 0, the second term is negative, so the total field at C is reduced compared to E₀ alone.
This material connects directly to conductors and capacitors (coming in the next few weeks), because the behaviour of charge on conductors is derived from the fact that E = 0 inside a conductor, which is itself enforced by Gauss' law. It also underpins the concept of electric potential, since knowing E from Gauss' law lets you integrate to find voltage differences. If you are comfortable with these three standard results (sphere, line, plane), the capacitor chapter will be significantly easier.
Electric flux, Gauss' law, Gaussian surface, ε₀, permittivity of free space, enclosed charge, Q_enc, flux integral, surface integral, area vector, dot product, net flux, positive flux, negative flux, spherical symmetry, cylindrical symmetry, planar symmetry, Coulomb's law from Gauss, superposition of fields, uniform electric field, P212, PHYS 212, University Physics, electromagnetism