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UY1: Coulomb's Law To Gauss's Law

Show how Coulomb's inverse-square field leads to Gauss's law for a point charge and clarifies flux invariance.

  • University Physics Year 1
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Learning objectives

  • Construct electric-field and potential models for discrete and continuous charge distributions.
Why this matters + quick links

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

1) At a glance

  • Prerequisites: point-charge field (Field of a Point Charge), flux integral (Gauss’s Law (Simple Version)), surface integrals (Vector Calculus)
  • Outcomes: compute flux through a sphere around a point charge, explain why it’s independent of radius, and connect it to Gauss’s law
  • Key result: for a point charge at the center of a sphere, Φ_E = ∮ vector E · d vector A = q/ε₀
  • Common trap: forgetting that d vector A is radial on a sphere (so vector E∥ d vector A) or dropping the sign for q < 0
  • For a point charge q, Coulomb’s law gives:
vector E = (1/4πε₀)q/r² r hat
  • On a spherical surface centered on the charge:
Φ_E = ∮ vector E · d vector A = q/ε₀
  • This is the key idea behind Gauss’s law.

Motivation / intuition

Gauss’s law looks like a new “integral rule”, but for a point charge it is really the same physics as Coulomb’s inverse-square law. The 1/r² falloff is special because the area of a sphere grows like r², so the total flux becomes radius-independent.

2) Setup

Choose a Gaussian sphere of radius r centered at charge q.

Conventions:

  • d vector A points outward.
  • For q > 0, vector E is outward so vector E · d vector A > 0.
  • For q < 0, flux is negative.

Because of spherical symmetry, E has constant magnitude everywhere on the sphere and is parallel to d vector A.

3) Core derivation/explanation

From Coulomb’s law:

E = (1/4πε₀)q/r²

Flux through the sphere:

Φ_E = ∮ vector E · d vector A = ∮ E dA = E∮ dA = E(4π r²)

Hence:

Φ_E = ((1/4πε₀)q/r²)(4π r²) = q/ε₀

So total flux does not depend on sphere radius.

This works because E ∝ 1/r² exactly; the area grows as r², so the factors cancel.

General Gauss’s law statement:

∮_S vector E · d vector A = q_enc/ε₀

for any closed surface S.

Quick checks (units + limits/sign)
  • Units: [Φ_E] = [vector E · vector A] = (N/C)m² = N m²/C and [q/ε₀] = N m²/C.
  • Limits/signs: doubling the sphere radius does not change Φ_E; if q < 0, the flux is negative (field lines enter the surface).

4) Worked example(s)

A charge q = 5.0 nC sits at the center of a sphere.

Total flux through the sphere:

Φ_E = q/ε₀ = (5.0 × 10⁻⁹)/(8.854 × 10⁻¹²) ≈ 5.65 × 10² N m²C⁻¹

If sphere radius doubles, flux is unchanged.

5) Practice set (with hints + answers)

  1. A centered point charge is enclosed by a cube. What is total electric flux through cube? Hint: Gauss’s law depends on enclosed charge, not shape. Answer: Φ_E = q/ε₀.

  2. Same setup as Q1. What is flux through one face of the cube? Hint: symmetry among six faces. Answer: Φ_face = q/(6ε₀).

  3. If enclosed charge is -2.0 nC, what is sign of total flux? Hint: outward normal convention. Answer: negative flux, Φ_E = -2.0 × 10⁻⁹/ε₀.

6) Summary + next steps

  • Coulomb’s law on a sphere gives flux q/ε₀ directly.
  • Flux independence from radius follows from inverse-square behavior.
  • Gauss’s law extends this result to any closed surface via enclosed charge.

Next: Usage Of Gauss’s Law Previous: Gauss’s Law (Simple Version) Back To Electromagnetism