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UY1: Electric Field And Potential Of Charged Conducting Sphere

Derive electric field and potential inside and outside a charged conducting sphere using Gauss's law and equipotential behavior.

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

For an isolated conducting sphere, all excess charge sits on the outer surface in electrostatic equilibrium.

1) At a glance

  • Prerequisites: Gauss’s law, relation vector E = -∇ V
  • Outcomes: write piecewise formulas for E(r) and V(r)
  • Key results: E = 0 for r < R (inside conductor), and outside behaves like a point charge: E = kQ/r², V = kQ/r
  • Reference potential: V(∞) = 0
  • Common trap: assuming potential is zero inside; it is constant, not necessarily zero

Motivation / intuition

This is the simplest place where “conductor rules” and “Gauss rules” meet: the conductor forces E = 0 in the metal, while spherical symmetry outside makes the field look exactly like a point charge at the center.

2) Setup

A conducting sphere has radius R and total charge Q.

Because of spherical symmetry, for r ≥ R the field is radial and has constant magnitude over a spherical Gaussian surface.

3) Core derivation/explanation

For r ≥ R: E(4π r²) = Q/ε₀ ⇒ E(r) = (1/4πε₀)Q/r² Using V(∞) = 0, V(r) = (1/4πε₀)Q/r (r ≥ R)

For r < R (inside conducting material): E(r) = 0 Potential must be constant throughout the conductor and equal to surface value: V(r) = V(R) = (1/4πε₀)Q/R (r ≤ R)

So,

E(r) = { 0, r < R; (1/4πε₀)Q/r², r ≥ R
V(r) = { (1/4πε₀)Q/R, r ≤ R; (1/4πε₀)Q/r, r ≥ R
Quick checks (units + limits/sign)
  • Units: [E] = N/C from kQ/r² and [V] = V from kQ/r.
  • Limits/signs: as r → ∞, V → 0 (given V(∞) = 0); for Q > 0, E points outward and V > 0; V is continuous at r = R while E has a jump at the surface due to surface charge.

4) Worked example(s)

Given Q = 6.0 nC and R = 0.10 m.

At r = 0.20 m: E = (1/4πε₀)Q/r² ≈ 1.35 × 10³ N/C V = (1/4πε₀)Q/r ≈ 270 V

At r = 0.05 m (inside): E = 0, V = (1/4πε₀)Q/R ≈ 540 V

5) Practice set (with hints + answers)

  1. For a positively charged conducting sphere, direction of vector E outside?
  2. If Q doubles and R fixed, what happens to surface potential?
  3. True/false: E is continuous at r = R for a charged conductor surface.

Hints

  1. Radial symmetry plus charge sign.
  2. Use V(R) = kQ/R.
  3. Surface charge causes a jump in normal field.

Answers

  1. Radially outward.
  2. It doubles.
  3. False.

6) Summary + next steps

  • Outside a conducting sphere, field and potential are the same as a point charge at center.
  • Inside, field is zero but potential is constant and equal to surface potential.
  • This piecewise behavior is central for conductor electrostatics and capacitor models.

Next: Electric Field Of A Uniformly Charged Sphere Previous: Gauss’s Law For Conductors Back To Electromagnetism (UY1)