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UY1: Electric Field Of An Electric Dipole

Derive exact and far-field electric-field formulas for a dipole on axial and equatorial lines, with direction and sign clarity.

  • 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 An Electric Dipole. 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: dipole moment and torque/energy ideas (Electric Dipole), superposition from point charges (Coulomb’s Law)
  • Outcomes: compute vector E on axial/equatorial lines, choose exact vs far-field forms, and state direction correctly
  • Key result (far field): dipole fields fall as 1/r³ (faster than a point charge’s 1/r²)
  • Common trap: using the far-field forms when r is not much larger than a, or mixing axial vs equatorial directions
  • A dipole has charges + q and -q separated by 2a.
  • Dipole moment magnitude: p = 2aq (direction from -q to + q).
  • On the axial line (far field):
Eₐₓᵢₐₗ ≈ (1/4πε₀)2p/r³
  • On the equatorial line (far field):
E_eq ≈ (1/4πε₀)p/r³

with direction opposite to vector p.

Motivation / intuition

Most real objects are electrically neutral overall (Qₙₑₜ = 0). Far away, their leading electric effect is often a dipole, so these 1/r³ results are the standard “first approximation” for many neutral systems.

2) Setup

  • Place dipole on x-axis: -q at x = -a, + q at x = +a.
  • Use superposition of fields from each charge.
  • Keep vector directions explicit; many sign errors come from direction, not algebra.

3) Core derivation/explanation

Axial point

Let point P be at x > a on x-axis.

Eₐₓᵢₐₗ = (1/4πε₀)q[1/((x-a)²)-1/((x + a)²)]

Simplify:

Eₐₓᵢₐₗ = (1/4πε₀)4aqx/((x²-a²)²) = (1/4πε₀)2px/((x²-a²)²)

For x≫ a:

Eₐₓᵢₐₗ ≈ (1/4πε₀)2p/x³

Direction is along + x hat (same direction as vector p on the + side).

Equatorial point

Let point P be on y-axis at distance r from center. By symmetry, y-components cancel and x-components add:

vector E_eq = -(1/4πε₀)p/((r² + a²)^(3/2)) x hat

For r≫ a:

vector E_eq ≈ -(1/4πε₀)p/r³ x hat

The minus sign shows direction opposite to vector p.

Assumption: when can you use the far-field formulas?

The approximations Eₐₓᵢₐₗ ≈ k(2p/r³) and E_eq ≈ k(p/r³) require r≫ a (distance from the dipole much larger than half-separation).

Quick checks (units + limits/sign)
  • Units: [k p/r³] = (N m² C⁻²)(C m)/m³ = N/C.
  • Limits/signs: as r → ∞, E → 0 like 1/r³; on the +x axial side the field points with vector p, while on the equatorial line it points opposite vector p.

4) Worked example(s)

Given q = 2.0 μC, a = 0.020 m, find axial field at x = 0.30 m.

Exact:

Eₐₓᵢₐₗ = (1/4πε₀)q[1/((0.30-0.02)²)-1/((0.30 + 0.02)²)] ≈ 5.38 × 10⁴ N C⁻¹

Direction: + x hat.

Far-field estimate:

p = 2aq = 8.0 × 10⁻⁸ C m
E ≈ (1/4πε₀)2p/x³ = ((8.99 × 10⁹) 2(8.0 × 10⁻⁸))/(0.30)³ ≈ 5.33 × 10⁴ N C⁻¹

Close agreement confirms x≫ a approximation is reasonable.

5) Practice set (with hints + answers)

  1. A dipole has p = 3.0 × 10⁻⁸ C m. Find far axial field at r = 0.50 m. Hint: use Eₐₓᵢₐₗ ≈ k(2p/r³). Answer: E ≈ 4.31 × 10³ N C⁻¹.

  2. Same dipole, find far equatorial field magnitude at r = 0.50 m. Hint: E_eq ≈ k(p/r³). Answer: E ≈ 2.16 × 10³ N C⁻¹, opposite vector p.

  3. What is the far-field ratio Eₐₓᵢₐₗ/E_eq at equal r? Hint: compare coefficients. Answer: 2.

6) Summary + next steps

  • Dipole fields come from superposition of two point-charge fields.
  • Exact formulas are useful near the dipole; 1/r³ formulas are for far field.
  • Direction differs by location: axial follows vector p, equatorial opposes vector p.

Next: Electric Field Of A Ring Of Charge Previous: Electric Field Lines Back To Electromagnetism (UY1) Back to Electromagnetism (UY1)