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The Timpson Lyceum · University Physics II (2026) – Electricity & Magnetism

Gauss’s Law – Lecture 3 (University Physics II)

Overview

This lecture introduces electric flux and Gauss's law, the first of Maxwell's four equations and one of the central results of the course. After briefly closing out the topic of electric dipoles, the focus shifts to thinking of an electric field as something that "flows" through a surface, and to counting how much field passes through a chosen surface. Gauss's law states that the net flux through any closed surface depends only on the charge enclosed — not on the surface's shape or the charge's position inside it. Because we are free to pick the surface, this turns hard field calculations into easy ones whenever the charge distribution has symmetry. We use it to recover Coulomb's law, to find the fields of charged spheres, infinite lines, and infinite planes, and to explain why excess charge on a conductor lives entirely on its surface.

Key Concepts

Electric dipole (chapter wrap-up). A dipole is two equal and opposite charges held a fixed distance apart (e.g. a water molecule). Placed in a uniform field, the two opposite forces produce a torque that tends to align the dipole with the field.

Electric flux ΦE\Phi_E — a measure of how much field passes through a surface.

Closed surfaces and sign of flux. For a closed surface the area vector always points outward.

Gauss's law. The net flux through any closed surface equals the enclosed charge over ε0\varepsilon_0: ∮E⃗⋅dA⃗=Qencε0\oint \vec E \cdot d\vec A = \frac{Q_{\text{enc}}}{\varepsilon_0}

Key results from Gauss's law.

Conductors in electrostatics.

Worked Example

Find the field a distance rr from an infinitely long line of charge with linear charge density λ\lambda.

Choose a coaxial cylinder of radius rr and length LL as the Gaussian surface. By symmetry E⃗\vec E points radially outward, so no flux passes through the end caps — all flux exits the curved side, whose area is 2πrL2\pi r L. The enclosed charge is λL\lambda L. Apply Gauss's law:

∮E⃗⋅dA⃗=Qencε0E (2πrL)=λLε0E=λ2πε0r\begin{aligned} \oint \vec E \cdot d\vec A &= \frac{Q_{\text{enc}}}{\varepsilon_0} \\ E\,(2\pi r L) &= \frac{\lambda L}{\varepsilon_0} \\ E &= \frac{\lambda}{2\pi\varepsilon_0 r} \end{aligned}

The length LL cancels, and the field falls off as 1/r1/r — the same answer the Coulomb's-law integral gave, but with far less work.

Summary