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UY1: Using Gauss's Law For Common Charge Distributions

Apply Gauss's law to symmetric charge distributions: infinite line, infinite sheet, and oppositely charged parallel plates.

  • 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 Using Gauss’s Law For Common Charge Distributions. You reuse it when you build fields/potentials by symmetry or superposition, and when you connect fields to forces, energy, and circuits.

These are the standard symmetry cases where Gauss’s law gives compact field formulas.

1) At a glance

  • Prerequisites: Gauss’s law setup (Gauss’s Law (Simple Version)) and symmetry choices (Usage Of Gauss’s Law)
  • Outcomes: derive E for an infinite line, infinite sheet, and ideal opposite-sheet (parallel-plate) model
  • Core strategy: choose a Gaussian surface where E is constant on contributing areas
  • Key results: Eₗᵢₙₑ = λ/(2πε₀ r), Eₛₕₑₑₜ = σ/2ε₀, E_(between plates) = σ/ε₀
  • Common trap: trying to apply these exact formulas to finite objects without edge corrections

Motivation / intuition

These are the “canonical symmetries” because they turn the flux integral into simple algebra: you design the Gaussian surface so that vector E · d vector A is either constant or zero on each part.

2) Setup

Assume idealized infinite distributions:

  • line charge density λ (C/m)
  • sheet charge density σ (C/m²)
  • two infinite sheets with + σ and -σ

Use outward area convention on closed surfaces.

3) Core derivation/explanation

A) Infinite line charge

Use a cylindrical Gaussian surface (radius r, length L): E(2π rL) = (λ L)/ε₀ E = λ/(2πε₀ r) Direction is radial outward for λ > 0.

B) Infinite sheet charge

Use a pillbox of area A crossing the sheet. Flux exits through two faces: 2EA = (σ A)/ε₀ E = σ/2ε₀ Field magnitude is distance-independent for the ideal infinite sheet.

C) Two opposite infinite sheets (parallel-plate idealization)

Superposition:

  • between plates, fields add: E = σ/ε₀
  • outside plates, fields cancel (ideal infinite-sheet model): E = 0

This is the basis for uniform-field capacitor models.

Assumption: ideal infinite objects

These formulas are exact for the infinite models. Real objects are finite, so edge/fringing effects appear; the results are still good approximations far from edges when the relevant dimensions are large compared to your distance from the object.

Quick checks (units + limits/sign)
  • Units: λ/(ε₀ r) gives N/C and σ/ε₀ gives N/C.
  • Limits/signs: for λ > 0 or σ > 0, the field points away from the distribution; for opposite sheets, changing the sign of σ reverses the direction, while changing plate separation does not change E (ideal model).

4) Worked example(s)

For a line charge with λ = 4.0 × 10⁻⁹ C/m at r = 0.050 m: E = λ/(2πε₀ r) ≈ 1.44 × 10³ N/C

For a sheet with σ = 3.0 × 10⁻⁶ C/m²: E = σ/2ε₀ ≈ 1.69 × 10⁵ N/C

Between two opposite sheets of same |σ|: E = σ/ε₀ ≈ 3.39 × 10⁵ N/C

5) Practice set (with hints + answers)

  1. If distance from an infinite line doubles, what happens to E?
  2. For an infinite sheet, does E depend on distance?
  3. For opposite infinite sheets, what is the field outside (ideal model)?

Hints

  1. Inspect E ∝ 1/r.
  2. Use sheet result directly.
  3. Apply superposition of two equal magnitudes.

Answers

  1. It halves.
  2. No.
  3. Zero.

6) Summary + next steps

  • Symmetry turns Gauss’s law into algebra.
  • Line: E ∝ 1/r; sheet: constant E; opposite sheets: nearly uniform interior field.
  • These results feed directly into capacitor derivations.

Next: Capacitors And Capacitance Previous: Electric Field Of A Uniformly Charged Sphere Back To Electromagnetism (UY1)