Gauss’s Law

You’ll be able to

Define electric flux and compute it through a surface
State Gauss’s law and explain why flux depends only on enclosed charge
Use Gaussian surfaces to find fields of symmetric distributions

Electric flux counts field lines through a surface

Electric flux Φ measures how much electric field passes through a surface. For a small patch of area dA (a vector normal to the patch), the flux is dΦ = E·dA = E·dA·cosθ, where θ is the angle between the field and the surface normal. Field parallel to the surface (θ = 90°) carries zero flux; field straight through (θ = 0°) carries the most. The total flux is the surface integral Φ = ∫ E·dA.

Gauss’s law

Φ = ∮ →E·d→A = Q_enc / ε₀

The circle on the integral means a closed surface. The net flux out of any closed surface equals the charge enclosed divided by ε₀ — nothing else matters.

Only the enclosed charge counts

Gauss’s law says the net flux through a closed surface depends only on the charge inside it, Q_enc, and not on where that charge sits or on any charges outside. A charge outside sends as many field lines into the surface as out, for zero net flux. This is why Gauss’s law is so powerful: choose a Gaussian surface matched to the symmetry so E is constant and parallel to dA, and the integral collapses to E·A = Q_enc/ε₀, letting you solve for E directly.

Worked example

Use Gauss’s law to find the field at distance r from a point charge q, recovering Coulomb’s law.

1.Exploit spherical symmetry: choose a Gaussian sphere of radius r centered on q. By symmetry E has the same magnitude everywhere on it and points radially, parallel to dA.
2.Then ∮ E·dA = E ∮ dA = E·(4πr²), the surface area of the sphere.
3.Set equal to Q_enc/ε₀: E·(4πr²) = q/ε₀.
4.Solve: E = q / (4πε₀ r²) = kq/r² — exactly Coulomb’s field.

Answer: E = q / (4πε₀ r²) = kq/r², radially outward

Checkpoint

A point charge +Q sits at the center of a cube. What is the electric flux through the entire closed surface of the cube?

Worked example

A solid spherical conductor carries net charge +Q in electrostatic equilibrium. Find the field (a) inside the conductor and (b) outside at radius r > R.

1.(a) Inside a conductor in equilibrium, charges rearrange until the field is zero — otherwise they would keep moving. Draw a Gaussian sphere of radius r < R: since E = 0 everywhere on it, ∮ E·dA = 0, so Q_enc = 0. All the excess charge must reside on the outer surface.
2.(b) For r > R, draw a Gaussian sphere enclosing all of Q. By symmetry E·(4πr²) = Q/ε₀.
3.Solve the outside case: E = Q / (4πε₀ r²) = kQ/r².
4.Interpretation: outside, the charged sphere acts exactly like a point charge Q at its center; inside, the field vanishes.

Answer: Inside (r < R): E = 0. Outside (r > R): E = kQ/r², as if all charge were a point at the center.

Checkpoint

In electrostatic equilibrium, why is the electric field zero everywhere inside a solid conductor?

Watch out

Gauss’s law is always true, but it only solves for E when symmetry (spherical, cylindrical, or planar) makes E constant and either parallel or perpendicular to the surface. Without that symmetry the flux integral cannot be simplified, even though Φ = Q_enc/ε₀ still holds.

On the exam

For a Gaussian-surface FRP: (1) name the symmetry and pick a matching surface, (2) argue E is constant over the part carrying flux, (3) write ∮ E·dA = E·A, (4) find Q_enc — using ρ, σ, or λ times the enclosed volume/area/length — and (5) solve E·A = Q_enc/ε₀. Missing the Q_enc step is the usual lost point.

Answer the 2 checkpoints as you read.