Electrostatics & Gauss's law
The idea
Gauss's law repackages Coulomb's law as a statement about flux: the electric flux through any closed surface equals the enclosed charge divided by ε₀. Flux counts field lines crossing the surface, so the law says field lines begin and end only on charges. It is always true, but it becomes a calculation tool only when symmetry lets you argue the field's direction and constancy in advance — spheres, infinite lines, and infinite planes are the three workable geometries.
The method is to choose a Gaussian surface that matches the symmetry: one where E is perpendicular to the surface with constant magnitude on part of it, and parallel to the surface (zero flux) on the rest. Then the flux integral collapses to E times an area, and the unknown E pops out algebraically. The surface is purely imaginary — a mathematical bookkeeping boundary, not a physical object.
A persistent misconception: if the enclosed charge is zero, the field on the surface must be zero. False — only the net flux vanishes. A point charge sitting just outside your surface drives plenty of field through it; the inward and outward contributions merely cancel. Gauss's law constrains the total flux, never the field at individual points, unless symmetry adds the missing information.
Worked example
A very long, straight wire carries a uniform linear charge density of 4.0 μC/m. Use Gauss's law to find the electric field magnitude 0.50 m from the wire, taking ε₀ = 8.85 × 10⁻¹² C²/(N·m²).
- Exploit symmetry: by the cylindrical symmetry of an infinite line, E must point radially outward with the same magnitude at every point a distance r from the wire — so choose a coaxial cylindrical Gaussian surface of radius r and length L.
- Evaluate the flux: the two flat end caps contribute nothing (E lies parallel to them), and the curved side contributes E × 2πrL since E is perpendicular to it with constant magnitude.
- Apply Gauss's law with enclosed charge λL: E × 2πrL = λL/ε₀, and the arbitrary length L cancels, leaving E = λ/(2πε₀r).
- Substitute numbers: E = (4.0 × 10⁻⁶)/(2π × 8.85 × 10⁻¹² × 0.50) = (4.0 × 10⁻⁶)/(2.78 × 10⁻¹¹) ≈ 1.4 × 10⁵ N/C.
- Check the structure: the field falls as 1/r rather than 1/r² because the source is a line, not a point — doubling r halves E, and the units N/C confirm a field, not a force.
Answer. The field is about 1.4 × 10⁵ N/C, pointing radially away from the wire.
Check your understanding
- Why does the field of a line charge fall off as 1/r while a point charge falls off as 1/r²?
- What goes wrong if you try to use Gauss's law to find the field of an electric dipole or a finite rod?
- Why must the electrostatic field be zero inside conductor material, and what does Gauss's law then say about where excess charge sits?
- How can the flux through a closed surface be zero while the field on the surface is large — what would the field-line picture look like?
Build the foundations first
Electrostatics & Gauss's law builds on these concepts. If any feel shaky, start there.