Mini Physics
Part of Mini EducationSwitch subject

UY1: Electric Field Of A Uniformly Charged Sphere

Derive the electric field inside and outside a uniformly charged insulating sphere, with clear dependence on radius.

  • University Physics Year 1
On this page

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 Electric Field Of A Uniformly Charged Sphere. You reuse it when you build fields/potentials by symmetry or superposition, and when you connect fields to forces, energy, and circuits.

Unlike a conductor, charge in an insulating sphere can be spread through the volume.

1) At a glance

  • Prerequisites: Gauss’s law, volume charge density
  • Outcomes: derive E(r) for r < R and r ≥ R
  • Key result: inside field grows linearly with r
  • Common trap: using full charge Q for an interior Gaussian surface

Motivation / intuition

Gauss’s law turns the “inside vs outside” problem into one idea: inside the sphere, the enclosed charge grows like r³, while the surface area grows like r², so E ∝ r. Outside, enclosed charge is constant, so E ∝ 1/r².

2) Setup

A sphere of radius R carries uniform volume charge density ρ with total charge Q.

ρ = Q/((4/3)π R³)

Use a concentric Gaussian sphere of radius r.

3) Core derivation/explanation

Outside region (r ≥ R)

Entire charge is enclosed: E(4π r²) = Q/ε₀ E(r) = (1/4πε₀)Q/r²

Inside region (r < R)

Only charge within radius r is enclosed: Q_encl = ρ((4/3)π r³) = Qr³/R³ Then E(4π r²) = Q_encl/ε₀ E(r) = (1/4πε₀)(Q/R³)r = (ρ r)/3ε₀

So,

E(r) = { (1/4πε₀)(Q/R³)r, r < R; (1/4πε₀)Q/r², r ≥ R

At r = R, both expressions match, so E is continuous.

Quick checks (units + limits/sign)
  • Units: inside form ρ r/(3ε₀) gives (C/m³)m/(C²/(N m²)) = N/C.
  • Limits/signs: E(0) = 0 (no preferred direction at the center); E(r) is continuous at r = R; for Q > 0 the field points radially outward.

4) Worked example(s)

Let Q = 12 nC and R = 0.06 m.

At r = 0.03 m (inside): E = (1/4πε₀)(Q/R³)r ≈ 1.50 × 10⁴ N/C

At r = 0.12 m (outside): E = (1/4πε₀)Q/r² ≈ 7.49 × 10³ N/C

5) Practice set (with hints + answers)

  1. For fixed Q,R, how does inside field depend on r?
  2. At r = R/2, what fraction of total charge is enclosed?
  3. Is field maximum at center or surface for this distribution?

Hints

  1. Look at interior formula.
  2. Use Q_encl = Q(r³/R³).
  3. Compare E(0) and E(R).

Answers

  1. E ∝ r.
  2. 1/8 of total charge.
  3. Maximum at the surface.

6) Summary + next steps

  • Exterior field behaves like a point charge Q at center.
  • Interior field is linear in r because enclosed charge scales as r³.
  • Correct enclosed-charge bookkeeping is the key Gauss-law skill here.

Next: Using Gauss’s Law For Common Charge Distributions Previous: Electric Field And Potential Of Charged Conducting Sphere Back To Electromagnetism (UY1)