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Equipotential Surfaces

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Surfaces of constant V

An equipotential surface is the set of points sharing one value of V. Moving a charge anywhere along such a surface costs zero work, since W_field = −qΔV = 0. That immediately forces a geometric fact: the field must meet every equipotential at right angles. If E had a component tangent to the surface, it would do work on a charge sliding along it — contradicting ΔV = 0. Field lines and equipotentials therefore form an everywhere-perpendicular grid, with E pointing from high V to low V.

Spacing encodes field strength

Draw equipotentials at equal voltage steps ΔV. Where they crowd together, the potential changes rapidly with distance — a strong field; where they spread out, the field is weak. Quantitatively, over a small perpendicular step Δs, E ≈ ΔV/Δs (magnitude). Around a point charge the surfaces are concentric spheres that spread farther apart with distance, because the 1/r² field needs ever more room to drop each ΔV. It is exactly a topographic map: equipotentials are contour lines, and E is the steepness of the terrain.

Work and field from equipotentials
W_field = −qΔV = q(V_a − V_b), |E| ≈ |ΔV| / Δs
Δs is measured perpendicular to the surfaces. Along an equipotential, ΔV = 0 and the field does no work.
Worked example

Two parallel plates 2.0 mm apart hold a 12 V potential difference. Find the field between them, and the spacing of equipotential surfaces drawn every 3.0 V.

  1. 1.Between closely spaced plates the field is uniform, so ΔV = −∫E·dl reduces to V = E·d.
  2. 2.Solve symbolically: E = V/d.
  3. 3.Numerically: E = 12 V / (2.0 × 10⁻³ m) = 6.0 × 10³ V/m, directed from the + plate to the − plate.
  4. 4.Equipotentials in a uniform field are equally spaced planes parallel to the plates. Spacing per 3.0 V step: Δs = ΔV/E = 3.0/(6.0 × 10³) = 5.0 × 10⁻⁴ m.
  5. 5.Check: four 3.0 V steps × 0.50 mm = 2.0 mm, spanning the full gap. ✓
Answer: E = 6.0 × 10³ V/m; equipotential planes every 0.50 mm
Worked example

A −2.0 μC charge is moved slowly (starting and ending at rest) from the 100 V equipotential to the 300 V equipotential. How much work does the external agent do, and how much does the field do?

  1. 1.The path is irrelevant — only the endpoint potentials matter. ΔV = 300 − 100 = +200 V.
  2. 2.ΔU = qΔV = (−2.0 × 10⁻⁶ C)(+200 V) = −4.0 × 10⁻⁴ J.
  3. 3.With no change in kinetic energy, W_ext = ΔU = −4.0 × 10⁻⁴ J.
  4. 4.The field does W_field = −ΔU = +4.0 × 10⁻⁴ J.
  5. 5.Interpretation: a negative charge is attracted toward higher potential, so the field pulls it there and the agent must hold it back — hence the agent’s negative work.
Answer: W_ext = −4.0 × 10⁻⁴ J; W_field = +4.0 × 10⁻⁴ J

A conductor is one big equipotential

Inside a conductor in electrostatic equilibrium, E = 0 everywhere (Unit 1). Then for any two interior points, ΔV = −∫→E·d→l = 0 — the entire conductor, surface included, sits at a single potential. This is self-consistent with the perpendicularity rule: since the surface is an equipotential, the external field must meet it at 90°. On a non-spherical conductor the potential is uniform but the surface field is not: charge crowds onto regions of small radius of curvature, so E is strongest just outside sharp points.

Tip

When a problem hands you a map of equipotentials, read it like a hiker reads contours: field lines run perpendicular to the curves, from high V to low V, and the field is strongest where the curves are packed tightest. Estimate magnitudes with E ≈ ΔV/Δs between adjacent curves.

Checkpoint

Why must the electric field meet an equipotential surface at right angles?

Checkpoint

Equipotential surfaces around an isolated point charge, drawn at equal ΔV intervals, are:

Answer the 2 checkpoints as you read.

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