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Equipotential Lines

Key idea: Explain equipotential lines/surfaces, relate them to electric field lines, and use E as the negative potential gradient (A Level Physics).

  • GCE A-Level H2 Physics 2027
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Learning objectives

  • Use the negative potential gradient and relate equipotentials to field lines.

1. Definitions (Must Know)

A. Equipotential line / surface

An equipotential line is a line joining points with the same electric potential (V).

In 3D, the equivalent is an equipotential surface.

B. Work done along an equipotential

If a charge moves along an equipotential, Δ V = 0, so:

Δ U = qΔ V = 0 W_field = -Δ U = 0

No work is done (and there is no change in potential energy).

C. Relationship to electric field lines

Equipotential lines/surfaces are perpendicular to electric field lines.

2. Key Ideas (What Earns Marks)

  • Moving along an equipotential gives Δ U = 0, so both field work and slow external work are zero.
  • Electric field strength is the negative potential gradient: E = -dV/dr (along the field direction).
  • In a uniform field between parallel plates: E = (Δ V)/d
  • Equal potential steps:
    • point charge: spacing between equipotentials increases as r increases,
    • uniform field: spacing between equipotentials is constant.
Direction and sign

vector E points in the direction of decreasing V (for a positive test charge). So on a V–distance graph, the gradient gives -E.

3. Detailed Explanations

A. Why equipotentials are perpendicular to field lines

Potential difference is energy per unit charge:

Δ V = W/q

Along an equipotential, Δ V = 0 ⇒ W = 0.

If the electric field had a component along the equipotential, it would do work on a charge moving along it. Therefore the field must have no tangential component, so it must be perpendicular.

B. Potential gradient and field strength

Along the direction of the field:

E = -dV/dr

So:

  • steep V change (large gradient) ⇒ large E,
  • gentle V change ⇒ small E.

C. Shapes you should recognise

  • Around a point charge: equipotentials are concentric circles (2D) / spheres (3D).
  • Between parallel plates (uniform field): equipotentials are straight, parallel lines, equally spaced.
Electric field lines and equipotentials around point chargesTwo panels compare a positive and negative point charge. Solid radial field lines point outwards from the positive charge and inwards towards the negative charge. Dashed concentric circles are equipotentials.Positive source chargeNegative source charge+Q−Qfield lineequipotentialE is tangent to field linesand points towards lower V
Field lines point away from a positive source charge and towards a negative source charge. Dashed equipotentials are perpendicular to the field lines; wider spacing farther away represents a smaller potential gradient and weaker field.

4. Common Mistakes

  • Saying “equipotential line = field line” (they are perpendicular).
  • Using E = Δ V/d for a point charge field (only for uniform fields).
  • Forgetting to distinguish Δ U = qΔ V from W_field = -qΔ V.

5. Exam Tips

  • If the diagram shows equal potential steps (e.g. 0 V, 10 V, 20 V, …), spacing tells you field strength:
    • closer spacing ⇒ stronger field.
  • State clearly whether you mean “work done by the field” (W_field = -qΔ V) or “work done by an external force” (Wₑₓₜ = qΔ V).

6. Worked Examples

Modelled example 1

Work done along an equipotential

Core

Problem

A + 3.0 nC charge moves along an equipotential line. Find the work done by the electric field and explain the geometry.
Study the worked solution

  1. Use the equipotential condition

    Method

    Set Δ V = 0.

    Reason

    Every point on the line has the same electric potential.

    Working

    Δ V = 0.

  2. Calculate field work

    Method

    Obtain zero work.

    Reason

    W_field = -qΔ V.

    Working

    W_field = -(3.0 × 10⁻⁹)(0) = 0 J

  3. Interpret the geometry

    Method

    State that the path is perpendicular to the field locally.

    Reason

    A field component along the path would do work and change potential.

    Working

    vector E⊥ equipotential.

Guided practice 2

Uniform field spacing

About 4 min

Problem

Between parallel plates, E = 2000 V m⁻¹. Find the separation between equipotentials differing by 50 V.

Try this before viewing the solution

Unit: m

Hints

Hint 1: uniform region
Use the magnitude relation E = Δ V/d.
View solution step by step

  1. Rearrange the gradient

    Method

    Use d = Δ V/E.

    Reason

    The field is uniform between the plates.

    Working

    d = 50/2000

  2. Evaluate

    Method

    Obtain 0.025 m.

    Reason

    Volts divided by volts per metre gives metres.

    Working

    d = 0.025 m.

Common misconception 3

Equipotentials around a point charge

Find and correct the mistake

Learner claim

A point charge has V = 200 V at 0.10 m. A learner treats its field as uniform and subtracts voltages linearly to locate the 50 V equipotential. Diagnose the method and find the radius.

Try this before viewing the solution

Point-charge potential dependence

View solution step by step

  1. Use point-charge scaling

    Method

    Relate potential inversely to radius.

    Reason

    The source field is non-uniform.

    Working

    V₁/V₂ = r₂/r₁

  2. Find the new radius

    Method

    Obtain 0.40 m.

    Reason

    Quartering potential requires quadrupling radius.

    Working

    r₂ = (0.10)200/50 = 0.40 m

Examiner practice 4

Estimating E from a potential drop over distance

3 marks

Examination question

Along the field direction, potential falls by 180 V over 0.060 m in an approximately uniform region. Estimate field magnitude and state its relation to the signed gradient. [3 marks]

Try this before viewing the solution

View solution step by step

  1. Calculate gradient magnitude

    1 mark

    Method

    Divide potential change magnitude by distance.

    Reason

    The region is approximately uniform.

    Working

    |Δ V/Δ x| = 180/0.060

  2. State field magnitude

    1 mark

    Method

    Obtain 3.0 × 10³ V m⁻¹.

    Reason

    Field magnitude equals potential-drop magnitude per distance.

    Working

    |E| = 3.0 × 10³ V m⁻¹.

  3. Apply the negative gradient

    1 mark

    Method

    State that field points towards decreasing potential.

    Reason

    Eₓ = -Δ V/Δ x.

    Working

    vector E is opposite the direction of increasing V.

Challenge 5

Comparing field strengths from equipotential spacing

Minimal support

Independent transfer

Equal 20 V equipotential steps are spaced by 5.0 mm in region 1 and 10 mm in region 2. Compare the field magnitudes and explain what the diagram spacing communicates.

Try this before viewing the solution

Hints

Hint 1: same potential step
With equal Δ V, closer lines mean a larger Δ V/Δ x.
View solution step by step

  1. Calculate region 1

    Method

    Use the 5.0 mm spacing.

    Reason

    Convert millimetres to metres.

    Working

    |E₁| = 20/(5.0 × 10⁻³) = 4.0 × 10³ V m⁻¹

  2. Calculate region 2

    Method

    Use the 10 mm spacing.

    Reason

    The potential step remains 20 V.

    Working

    |E₂| = 20/(1.0 × 10⁻²) = 2.0 × 10³ V m⁻¹

  3. Compare

    Method

    State region 1 is twice as strong.

    Reason

    Its equal potential change occurs over half the distance.

    Working

    |E₁|/|E₂| = 2.

7. Mind Stretchers

Mind stretcher 1: Negative charge and potential directionExtension

Explain why a negative charge tends to move towards higher potential (even though vector E points towards lower potential).

Show Answer

vector E is defined using the force on a positive test charge, so it points towards lower potential.

For a negative charge, vector F = q vector E reverses direction (since q < 0), so it accelerates opposite to vector E, towards higher potential.

Mind stretcher 2: Gradient and field strength change with distanceExtension

On a V–r graph for a point charge, the gradient becomes less steep as r increases. What does this tell you about how E changes with r?

Show Answer

Since E = -dV/dr, a less steep gradient means smaller |E|. So field strength decreases as r increases (for a point charge, it follows E ∝ 1/r²).

Mind stretcher 3: Optional (Enrichment)Extension

A. Equipotential mapping (experiment context)

In labs, equipotential lines can be mapped using a conductive paper and a voltmeter by finding points with equal measured potential difference.

Continue with the next resource in this course.

Course and syllabus information
Course
GCE A-Level H2 Physics
Edition
GCE A-Level H2 Physics 2027