UY1: Gauss's Law (Simple Version)

Learn electric flux and the integral form of Gauss's law, including sign conventions and units for closed surfaces.

University Physics Year 1

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Learning objectives

Construct electric-field and potential models for discrete and continuous charge distributions.

This page gives the UY1 working model/result for Gauss’s Law (Simple Version). You reuse it when you build fields/potentials by symmetry or superposition, and when you connect fields to forces, energy, and circuits.

Module path: Electromagnetism (UY1)
Practice: UY1 Electromagnetism Quiz
Full routing: UY1 Assessment Map
Math toolkit: Mathematics for Undergraduate Physics

Gauss’s law links electric flux through a closed surface to the net charge inside that surface.

1) At a glance

Prerequisites: electric field (The Electric Field), dot product/flux (Vector Calculus)
Outcomes: compute electric flux and apply Gauss’s law in integral form (with correct sign)
Key result: ∮ vector E · d vector A = Q_encl/ε₀
Units check: [Φ_E] = N · m²/C, [Q/ε₀] = N · m²/C
Common trap: concluding Φ_E = 0 means vector E = 0 everywhere (it only implies net enclosed charge is zero)

Motivation / intuition

Flux is a “global count” of how much field leaves a closed surface. Gauss’s law says that global count is controlled only by the net enclosed charge, which is why it becomes powerful when symmetry makes the flux integral simple.

2) Setup

Take a closed surface (Gaussian surface). By convention, d vector A points outward.

For a small patch of surface, dΦ_E = vector E · d vector A = E cos φ dA where φ is the angle between vector E and outward d vector A.

Sign convention:

field leaving the surface gives positive flux
field entering the surface gives negative flux

3) Core derivation/explanation

For any surface (flat or curved), total flux is Φ_E = ∫ vector E · d vector A For a closed surface, Φ_E = ∮ vector E · d vector A Gauss’s law states ∮ vector E · d vector A = Q_encl/ε₀

Why this step matters

The closed-surface integral converts many local field details into one global balance statement. You can solve for flux (or E in symmetric cases) without tracking field direction at every point.

Important interpretation points:

only enclosed charge contributes to net flux
charges outside can change local vector E, but their total net contribution to closed-surface flux is zero
zero net flux does not imply vector E = 0 everywhere; it only implies net enclosed charge is zero

Using an open surface and still applying the closed-surface form ∮ vector E · d vector A.
Flipping flux signs by mixing inward/outward area-vector conventions.
Concluding Φ_E = 0 means no field at all, instead of no net enclosed charge.

Units: Φ_E must match Q_encl/ε₀ in N m²/C.
Limits/signs: if Q_encl < 0, net flux is negative (more field enters than leaves); changing the shape of the surface does not change Φ_E as long as Q_encl is unchanged.

4) Worked example(s)

A cube encloses a point charge q = +3.0 × 10⁻⁹ C.

By Gauss’s law, Φ_E = q/ε₀ = (3.0 × 10⁻⁹)/(8.85 × 10⁻¹²) ≈ 3.39 × 10² N m²/C

Positive result means net field lines leave the cube.
Shape of surface does not change this total flux as long as the enclosed charge is unchanged.

5) Practice set (with hints + answers)

A closed surface encloses -5.0 nC. Find net flux.
A closed surface encloses + 2.0 nC and -2.0 nC. Find net flux.
True or false: If net flux is zero, electric field must be zero everywhere on the surface.

Hints

Use Φ_E = Q_encl/ε₀ and keep the sign.
Use net enclosed charge first.
Think about equal in-and-out field lines.

Answers

Φ_E ≈ -5.65 × 10² N m²/C.
Φ_E = 0.
False.

6) Summary + next steps

Electric flux measures how much field passes through a surface (with sign).
For closed surfaces, total flux depends only on enclosed charge.
Gauss’s law is most powerful when symmetry makes ∮ vector E · d vector A easy.

Next: Coulomb’s Law To Gauss’s Law Previous: Equipotential Surfaces Back To Electromagnetism (UY1)