Reading fields from potential graphs and maps

Key idea: Electric field points down the potential gradient: E = −dV/dx. Equipotentials cross field lines at right angles, and moving along one changes no electric potential energy.

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  • GCE A-Level H2 Physics 2027
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Field strength and potential describe the same field in two ways, one as force per unit charge and one as energy per unit charge. They are linked: the field is strongest where the potential changes most steeply. That link lets you read a field from a graph of potential or from a map of equipotential lines.

Field strength is the negative potential gradient

Move a charge q a small distance Δ x along the direction of a field E. The field does work qEΔ x on it, and this equals the loss of electric potential energy, -qΔ V. So E = -(Δ V)/(Δ x), and in the limit

E = -dV/dx

The electric field strength at a point is equal to the negative of the potential gradient there. The minus sign says that the field points from high potential to low potential. This also shows why V m⁻¹ is the same unit as N C⁻¹.

Potential across a uniform field

Calculated straight potential profile decreasing from ten volts to zero across five centimetres.

Scroll across the figure to read all labels.

Calculated straight potential profile decreasing from ten volts to zero across five centimetres.Calculated straight potential profile decreasing from ten volts to zero across five centimetres.
Take V(0) = 10 V and V(0.050 m) = 0 V. The constant gradient is −200 V m⁻¹, so the signed field component Eₓ = −dV/dx is +200 V m⁻¹.
Open full-size graph
What to noticeThe minus sign gives field direction: towards increasing x and decreasing potential.
View figure data
Figure data: Potential across a uniform field
Distance from plate, x (m)V(x) = 10 − 200x
0.00010.0
0.0019.8
0.0029.6
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0.0490.2
0.0500.0

Here the potential falls from 10 V to zero over 0.050 m. The gradient is -200 V m⁻¹, so the field is + 200 V m⁻¹: 200 V m⁻¹ in the + x direction. When the graph is curved, the field at a point comes from the gradient of the tangent there.

For a point charge, V = Q/4πε₀r, and -dV/dr = Q/4πε₀r²: the same field you found from Coulomb’s law.

Worked example 1

Field from a falling potential

Problem

The potential falls steadily from 120 V at x = 0 to 40 V at x = 0.20 m. Find the electric field, and the direction of the force on a positive and on a negative charge in it.

Show full solution
  1. Potential gradient

    Method

    Divide the change in potential by the distance, final minus initial.

    Reason

    The potential changes steadily, so the gradient is the same everywhere.

    Working

    Answer

    (Δ V)/(Δ x) = (40-120)/0.20 = -400 V m⁻¹

  2. Field

    Reason

    E is the negative of the gradient.

    Working

    Answer

    E = +400 V m⁻¹, in the + x direction, towards lower potential

  3. Forces

    Working

    Answer

    The force on a positive charge is in the + x direction; the force on a negative charge is in the -x direction.

Try it yourself 2

A rising potential

Minimal support

Problem

The potential rises steadily from -60 V to + 20 V over a distance of 0.040 m in the + x direction. Find the field, taking + x as positive.

Find the field

Unit: V m⁻¹

Hints

Hint 1

sign

Which direction leads towards lower potential?

Show full solution
  1. Field

    Working

    Answer

    (Δ V)/(Δ x) = (20-(-60))/0.040 = +2000 V m⁻¹, so E = -2000 V m⁻¹: 2000 V m⁻¹ in the -x direction

Check your understanding 1 Zero field, non-zero potential

Explain why the field strength can be zero at a point where the potential is not zero.

Show answer

The field depends on the gradient of the potential, not on its value. Midway between two equal positive charges the fields cancel, so the potential graph is flat there, but the potential itself is positive.

Equipotentials

An equipotential is a line, or surface, joining points at the same potential. Moving a charge along an equipotential changes neither its potential nor its electric potential energy, so the field does no work on it. That is only possible if the force has no component along the equipotential: field lines always cross equipotentials at right angles.

When equipotentials are drawn at equal steps of potential, their spacing shows the field strength. Close together means a steep potential gradient and a strong field. Around a point charge, equipotentials are concentric circles that get further apart with distance. In the uniform field between parallel plates, they are equally spaced and parallel to the plates.

Reading an equipotential mapParallel dashed equipotentials are labelled 120, 90, 60 and 30 volts from left to right. Their successive spacings increase; equal 30-volt steps therefore represent a field that weakens to the right. Horizontal field arrows cross the lines at right angles towards lower potential.Equal potential steps; increasing spacing120 V90 V60 V30 VE points right: towards decreasing V
Reading an equipotential mapParallel dashed equipotentials are labelled 120, 90, 60 and 30 volts from left to right. Their successive spacings increase; equal 30-volt steps therefore represent a field that weakens to the right. Horizontal field arrows cross the lines at right angles towards lower potential.Equal potential steps; increasing spacing120 V90 V60 V30 VE points right: towards decreasing V
Schematic, not to scale. Equal 30 V steps make spacing comparable: the smallest perpendicular gap has the strongest field. The arrows show direction, not a quantitative field magnitude. Along each dashed line, electric field work is zero.

Check your understanding 2

Two equal and opposite charges are at equal distances on either side of the y-axis: positive on the left, negative on the right. Predict the potential on the y-axis, the field direction there, and the field’s work on a charge moved along that axis.

Potential on the y-axis
Field on the y-axis
Field’s work along the y-axis

Show answer

The signed potentials cancel on the y-axis. The field is non-zero and points towards the negative charge, perpendicular to the axis. Along that axis Δ V = 0, so the field’s work -qΔ V is zero. Zero potential is a value; zero field would require every directional potential gradient to vanish.

Test the equipotential prediction

Use the stated equal and opposite charges, leave Show equipotentials on, and keep Probe x at 0. Change Probe y or drag the test-charge probe along the y-axis. Compare V at the probe, E at the probe and the field arrow. The plotted charge-mode graphs follow a horizontal slice at the probe’s current height, so they show how potential changes across the axis rather than along your vertical path.

Point charges: A +2.00 nanocoulombs, B −2.00 nanocoulombs. Field lines leave positive charges and end on negative ones; dashed equipotentials cross them at right angles. The probe reads E = 5754 newtons per coulomb and V = 0 volts.

Source of the field
nC

0 removes the charge.

nC

0 removes the charge.

nC

0 removes the charge.

More settings
cm
cm
cm
cm
cm
cm
E at the probe
5754 N C⁻¹
Direction of E (from +x)
0 °
V at the probe
0 V
Distance to nearest charge
7.21 cm
Potential along a line through the probe
Field component along the same line

Try this

0 of 4 done
  1. Put two like charges on the stage and find the point where E = 0. (not done yet)

  2. With a single charge, record E at three or more distances. (not done yet)

  3. Make the field between the plates 2.0 × 10⁴ V m⁻¹. (not done yet)

  4. Make the beam leave the plates just clear of a plate. (not done yet)

Your readings

#r / cmE / N C−1V / VRemove
No readings yet. Set up a measurement, then record it.

Check your understanding 3

Use the model to explain why a zero-potential path can pass through a non-zero field. How does the horizontal potential graph account for the arrow at the axis?

Show answer

On the vertical symmetry axis, V = 0 but the field points horizontally from positive to negative. Along the axis potential is constant and no work is done; across the axis the potential falls, giving a non-zero horizontal field. The horizontal potential graph has a negative tangent gradient at the axis, consistent with positive Eₓ.

Guided practice 3

Spacing of equipotentials

About 2 min

Problem

In a uniform field of strength 2000 V m⁻¹, how far apart are two equipotentials that differ by 50 V?

Find the separation

Unit: m

Hints

Hint 1

rearrange

How does field magnitude relate potential difference to perpendicular distance?

Show full solution
  1. Separation

    Working

    Answer

    d = 50/2000 = 0.025 m, measured along the field, at right angles to the equipotentials

Check your understanding 4 Closest lines, strongest field?

A student says the two closest equipotential lines on a map always mark the strongest field. One pair differs by 10 V and is 5.0 mm apart; another differs by 100 V and is 10 mm apart. Is the student right?

Unit: V m⁻¹
Unit: V m⁻¹
Can the closest lines alone identify the strongest field?

Show answer

No. The first pair gives 10/0.0050 = 2.0 × 10³ V m⁻¹ and the second gives 100/0.010 = 1.0 × 10⁴ V m⁻¹, so the wider gap has the stronger field. Spacing compares fields only when the lines are drawn at equal steps of potential.

Check your understanding 5 Equipotentials round a point charge

The potential 0.10 m from a point charge is 200 V. How far from the charge is the 50 V equipotential? Find the field strength at both distances, and explain why (Δ V)/(Δ r) between them does not give the field at either.

Unit: m
Unit: V m⁻¹
Show the steps Optional working
Unit: m
Unit: V m⁻¹
Show the steps Optional working
Unit: m
Unit: V m⁻¹

Show answer

V ∝ 1/r, so Vr = 20 V m is constant and the 50 V equipotential is at r = 0.40 m. Since E = V/r for a point charge, E = 2000 V m⁻¹ at 0.10 m and 125 V m⁻¹ at 0.40 m. (150 V)/(0.30 m) = 500 V m⁻¹ is only an average, because the field is not uniform.

Check your understanding 6

Your secant estimate is 500 V m⁻¹, while the two local fields are 2000 and 125 V m⁻¹. Explain why these answers differ, using a sketch of V against r if helpful.

Show answer

The potential curve is not a straight line: its gradient changes with radius. The field at one point is minus the tangent gradient there. The secant across a finite interval gives the average field over that interval, so it need not equal either endpoint value. A much smaller interval around a chosen point approaches its local tangent gradient.

Common mistakes

  • Dropping the minus sign: the field points towards lower potential.
  • Thinking a large potential means a strong field; the field depends on how fast the potential changes.
  • Comparing line spacing on a map whose equipotentials are not at equal steps of potential.
  • Drawing field lines that cross equipotentials at an angle other than 90°.
  • Using (Δ V)/(Δ r) across a wide gap in a non-uniform field as if it were the field at a point.

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