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Electric Dipoles

Key idea: Learn electric dipole moment, torque in a uniform electric field, and dipole potential energy, with worked examples and exam tips.

  • GCE A-Level H3 Physics 2027
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Learning objectives

  • define the magnitude of the electric dipole moment as the product of the charge and the separation
  • show an understanding of and use the torque on an electric dipole and the potential energy of an electric dipole to solve related problems
  • define the magnitude of the magnetic dipole moment for a current loop as the product of the current and the area of the loop
  • show an understanding of and use the torque on a magnetic dipole and the potential energy of a magnetic dipole to solve related problems

An electric dipole is a pair of equal and opposite charges separated by a small distance. In H3, the key skills are:

  • define the dipole moment, and
  • use torque and potential energy of a dipole in a (uniform) electric field.

1. Definitions (Must Know)

  • Electric dipole: charges + q and -q separated by distance d.
  • Dipole moment (magnitude): p = qd Unit: C m.
  • Dipole moment vector, vector p: points from -q to + q, with magnitude p = qd.
  • Uniform electric field: vector E has constant magnitude and direction in the region of interest.
  • Torque on a dipole in a uniform field (magnitude): τ = pE sin θ where θ is the angle between vector p and vector E.
  • Potential energy of a dipole in a uniform field: U = - vector p · vector E = -pE cos θ
  • Symbols used in this lesson: q (C), d (m), p (C m), vector E (N C⁻¹), τ (N m), U (J), W work done (J), θ (rad).

2. Key Ideas (What Earns Marks)

  • A uniform field exerts zero net force on an ideal dipole (forces cancel), but a non-zero torque (forces form a couple).
  • The torque tends to align vector p with vector E (stable at θ = 0).
  • Use energy to identify stable/unstable orientations:
    • minimum U at θ = 0 (aligned),
    • maximum U at θ = π (anti-aligned).

Quick comparison (uniform field):

QuantityResultWhat it means
Net force on ideal dipole0Translation doesn’t accelerate (in uniform vector E)
Torque on dipoleτ = pE sin θTends to rotate to align vector p with vector E
Potential energyU = -pE cos θStable at minimum (θ = 0)
Angled electric dipole in a uniform rightward electric field, with equal opposite forces and dipole moment pointing from negative to positive charge
The equal and opposite forces cancel translationally but act along different lines, producing a torque that aligns the dipole moment with the field.

3. Detailed Explanations

A. Why there is torque but no net force (uniform field)

Forces on the charges are: vector F₊ = +q vector E, vector F₋ = -q vector E So the net force is: vector Fₙₑₜ = vector F₊ + vector F₋ = vector 0

But the forces act at different points, producing a turning effect (a couple). The torque magnitude is: τ = pE sin θ

B. Potential energy and equilibrium angles

The potential energy is: U(θ) = -pE cos θ So:

  • θ = 0 gives Uₘᵢₙ = -pE (stable equilibrium),
  • θ = π gives Uₘₐₓ = +pE (unstable equilibrium).

4. Common Mistakes

  • Using p = q/d (wrong); it is p = qd.
  • Getting the direction of vector p wrong (it points from -q to + q).
  • Treating τ = pE regardless of angle (must include sin θ).
  • Forgetting the minus sign in U = -pE cos θ.

5. Exam Tips

  • Sketch the dipole and mark vector p and vector E; it prevents sign/direction errors.
  • Use torque for “rotating/turning” questions; use energy for “stable/unstable” or “work done to rotate”.
  • If the field is not uniform, the net force may not be zero (that is beyond this lesson unless stated).

6. Worked Examples

Modelled example 1

Dipole moment calculation

Core

Problem

A dipole consists of charges ±2.0 nC separated by 3.0 mm. Find its dipole-moment magnitude.
Study the worked solution

  1. Use the dipole definition

    Method

    p = qd.

    Reason

    Dipole-moment magnitude is charge magnitude times charge separation.

    Working

    p = qd

  2. Convert units

    Method

    q = 2.0 × 10⁻⁹ C and d = 3.0 × 10⁻³ m.

    Reason

    The SI unit of dipole moment is coulomb metre.

    Working

    nC = 10⁻⁹ C; mm = 10⁻³ m

  3. Calculate

    Method

    p = 6.0 × 10⁻¹² C m.

    Reason

    Multiply the converted charge and separation.

    Working

    p = (2.0 × 10⁻⁹)(3.0 × 10⁻³) = 6.0 × 10⁻¹² C m

Guided practice 2

Torque in a uniform field

About 4 min

Problem

A dipole with p = 4.0 × 10⁻⁸ C m is in a uniform field E = 2.0 × 10⁵ N C⁻¹ at 30°. Find the torque magnitude.

Try this before viewing the solution

Unit: N m

Hints

Hint 1: include the angular factor
The torque is not pE unless θ = 90°.
View solution step by step

  1. Choose the relation

    Method

    τ = pE sin θ.

    Reason

    Only the component perpendicular to the dipole moment produces torque.

    Working

    τ = pE sin θ

  2. Calculate

    Method

    τ = 4.0 × 10⁻³ N m.

    Reason

    pE = 8.0 × 10⁻³ N m and sin 30° = 0.5.

    Working

    τ = (4.0 × 10⁻⁸)(2.0 × 10⁵)(0.5) = 4.0 × 10⁻³ N m

Common misconception 3

Energy change on rotation

Find and correct the mistake

Learner claim

A dipole with pE = 0.20 J rotates from 0° to 60°. A learner says potential energy decreases because the final value is -0.10 J. Explain the comparison error.

Try this before viewing the solution

Unit: J

View solution step by step

  1. Find the initial energy

    Method

    Uᵢ = -0.20 J.

    Reason

    U = -pE cos θ and cos 0° = 1.

    Working

    Uᵢ = -0.20(1) = -0.20 J

  2. Find the final energy

    Method

    U_f = -0.10 J.

    Reason

    cos 60° = 0.5.

    Working

    U_f = -0.20(0.5) = -0.10 J

  3. Compare signed values

    Method

    Δ U = +0.10 J, so potential energy increases.

    Reason

    -0.10 J is greater than -0.20 J.

    Working

    Δ U = U_f-Uᵢ = (-0.10)-(-0.20) = +0.10 J

Examiner practice 4

Work done to rotate a dipole

4 marks

Examination question

A dipole has pE = 0.30 J. An external agent rotates it slowly from 30° to 120°. Find the work done, assuming no kinetic-energy change. [4 marks]

Try this before viewing the solution

Unit: J

View solution step by step

  1. Link work and energy

    1 mark

    Method

    Wₑₓₜ = Δ U.

    Reason

    Slow rotation with no kinetic-energy change makes external work equal the potential-energy increase.

    Working

    Wₑₓₜ = U₂-U₁

  2. Initial energy

    1 mark

    Method

    U₁ = -0.260 J.

    Reason

    Apply U = -pE cos θ at 30°.

    Working

    U₁ = -0.30 cos 30° = -0.260 J

  3. Final energy

    1 mark

    Method

    U₂ = +0.150 J.

    Reason

    cos 120° = -0.5 and the energy relation has a leading minus sign.

    Working

    U₂ = -0.30 cos 120° = +0.150 J

  4. Calculate work

    1 mark

    Method

    Wₑₓₜ = 0.41 J.

    Reason

    The energy rises from a negative to a positive value.

    Working

    Wₑₓₜ = 0.150-(-0.260) = 0.410 J

Challenge 5

Angle for maximum torque

Minimal support

Independent transfer

At what angle between vector p and vector E is an electric dipole’s torque magnitude maximum in a uniform field?

Try this before viewing the solution

Unit: °

Hints

Hint 1: maximise the angular factor
p and E are fixed, so ask when | sin θ| is largest.
View solution step by step

  1. Isolate the variable factor

    Method

    τ = pE sin θ.

    Reason

    Only sin θ changes with orientation for fixed p and E.

    Working

    |τ| = pE| sin θ|

  2. Maximise

    Method

    The torque magnitude is maximum at θ = 90°.

    Reason

    | sin θ| = 1 at a right angle in the 0° to 180° orientation range.

    Working

    sin 90° = 1 ⇒ τₘₐₓ = pE

7. Mind Stretchers

Mind stretcher 1: Which orientation is stable?Extension

Explain using energy why θ = 0 is stable but θ = π is unstable.

Answer

At θ = 0, U is a minimum. A small angular displacement increases U, so the system tends to return to the minimum (stable).

At θ = π, U is a maximum. A small displacement decreases U, so the dipole moves away from that orientation (unstable).

Mind stretcher 2: Why a non-uniform field can pull a dipoleExtension

In a uniform field, the net force on an ideal dipole is zero. Explain qualitatively why a dipole can experience a net force in a non-uniform electric field.

Answer

The forces on + q and -q have magnitudes qE evaluated at their positions. In a non-uniform field, the two charges sit in slightly different field strengths, so the forces no longer cancel perfectly and a net force can result (in addition to any torque).

8. Optional/Enrichment: Dipole Field in Space

The electric field of a dipole varies with position and falls roughly as 1/r³ far from the dipole. Detailed dipole-field expressions are not required for the H3 learning outcomes listed for this section unless the question provides the formula or explicitly guides you.

Next step

Continue to Magnetic Dipoles and reuse the same torque-and-energy structure with a current-loop moment.

Continue with the next resource in this course.

Course and syllabus information
Course
GCE A-Level H3 Physics
Edition
GCE A-Level H3 Physics 2027