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UY1: Electric Field Of Ring Of Charge

Derive the on-axis electric field of a uniformly charged ring using symmetry and integration, including limiting-case checks.

  • 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 Of Ring Of Charge. 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 idea (Field of a Point Charge), symmetry + basic integration (Integration Techniques)
  • Outcomes: derive the on-axis field of a uniformly charged ring and check center/far-field behavior
  • Key result (on axis):
vector E(x) = (1/4πε₀)Qx/((x² + a²)^(3/2)) x hat
  • Common trap: forgetting that only the axial components add (transverse components cancel by symmetry)
  • At center (x = 0), field is zero by symmetry.

Motivation / intuition

The ring is the cleanest “distributed charge” example where symmetry kills two components automatically. It also becomes a building block for the next result: a uniformly charged disk is just a stack of rings.

2) Setup

  • Ring lies in the yz-plane, center at origin, radius a.
  • Point P is on x-axis at coordinate x.
  • Charge is uniformly distributed; use element dq.
  • Symmetry: transverse components cancel; only x-components survive.

3) Core derivation/explanation

For element dq at distance

r = square root of (x² + a²)

field magnitude is

dE = (1/4πε₀)dq/r²

Its x-component:

dEₓ = dE cos α = (1/4πε₀)(x dq)/((x² + a²)^(3/2))

Integrate around full ring:

Eₓ = (1/4πε₀)x/((x² + a²)^(3/2))∫ dq = (1/4πε₀)Qx/((x² + a²)^(3/2))

Direction:

  • For Q > 0 and x > 0, field is + x hat.
  • Reverse sign if Q < 0 or x < 0.

Checks:

  • x = 0 ⇒ E = 0 by symmetry.
  • x≫ a ⇒ E ≈ kQ/x² (behaves like a point charge).
Quick checks (units + limits/sign)
  • Units: [kQx/(x² + a²)^(3/2)] = (N m² C⁻²)C m/m³ = N/C.
  • Limits/signs: E(0) = 0; for x≫ a you recover the point-charge field; changing the sign of Q flips the direction of vector E.

4) Worked example(s)

A ring has Q = 6.0 nC and a = 0.080 m. Find field at x = 0.060 m.

E = ((8.99 × 10⁹)(6.0 × 10⁻⁹)(0.060))/([(0.060)² + (0.080)²]^(3/2))

Since (0.060)² + (0.080)² = 0.0100, denominator is (0.0100)^(3/2) = 0.0010. So:

E ≈ 3.24 × 10³ N C⁻¹

Direction is + x hat (for positive ring charge).

5) Practice set (with hints + answers)

  1. A ring has Q = +2.0 nC, a = 0.10 m. What is E at the center? Hint: symmetry. Answer: 0.

  2. Same ring, find far-field approximation at x = 1.0 m. Hint: use kQ/x². Answer: ≈ 18 N C⁻¹.

  3. If Q becomes negative, what changes in formula outcome? Hint: magnitude expression is same, sign sets direction. Answer: field direction reverses.

6) Summary + next steps

  • Symmetry removed two components and made the integral one-line.
  • The ring result smoothly connects center behavior and far-field point-charge behavior.
  • This method generalizes directly to disks (ring-stacking approach).

Next: Electric Field Of A Line Of Charge Previous: Electric Field Of An Electric Dipole Back To Electromagnetism