Field Lines, Equipotentials & the Link Between Them

You’ll be able to

Interpret field-line density and direction
Explain why equipotential surfaces are perpendicular to field lines
Relate the potential difference between two points to the field and the separation

Reading field lines

Field lines point in the direction a positive test charge would be pushed, so they leave positive charges and enter negative ones. Their density represents field strength — closely spaced lines mean a strong field. Two rules follow: lines never cross, because the field has one direction at each point, and the number of lines leaving a charge is proportional to its magnitude, which is how relative charge is read off a diagram.

Equipotentials are perpendicular to field lines

An equipotential surface joins points at the same potential. Moving along one requires no work, since ΔV = 0. But the work done moving a charge is W = qE·d, which is zero only when the displacement is perpendicular to the field. So equipotentials must cross field lines at right angles — not as a convention but as a consequence. On a diagram, a family of curves crossing the field lines at 90° is the equipotential map.

Field from potential

for a uniform field: E = ΔV/d, in volts per meter · the field points from HIGH potential toward LOW

V/m and N/C are the same unit. Field points downhill on the potential landscape, which is why a positive charge released at rest moves toward lower potential.

The topographic-map analogy, and its limit

Equipotentials behave like contour lines on a map: closely spaced contours mean a steep slope, which here means a strong field. The analogy is genuinely useful — and it has one limit worth stating. A positive charge moves toward lower potential, like a ball rolling downhill, but a negative charge moves toward higher potential. So the landscape is inverted for negative charges, and describing potential as "electrical height" only works if you specify the sign of the charge.

Worked example

Two parallel plates 0.020 m apart are held at a potential difference of 120 V. Find the field between them and the work done moving a +2.0 μC charge from the negative to the positive plate.

1.Uniform field: E = ΔV/d = 120/0.020 = 6.0 × 10³ V/m, directed from the positive toward the negative plate.
2.Moving a positive charge toward the positive plate is moving against the field, so an external agent must do work.
3.W = qΔV = (2.0 × 10⁻⁶)(120).
4.W = 2.4 × 10⁻⁴ J of work done on the charge, raising its potential energy.

Answer: E = 6.0 × 10³ V/m and W = 2.4 × 10⁻⁴ J done by the external agent. The positive charge gains potential energy moving toward the positive plate, which is why it would accelerate back if released.

Watch out

Field lines are perpendicular to the surface of a conductor in electrostatic equilibrium, and the whole conductor is a single equipotential. A field line drawn meeting a conductor at an angle is wrong: any parallel component would drive charge along the surface until it vanished.

Checkpoint

Equipotential surfaces are always perpendicular to electric field lines because:

Checkpoint

The potential difference between two plates is doubled while their separation is unchanged. The field between them:

Checkpoint

A negative charge is released at rest in a uniform electric field. It moves toward:

On the exam

When asked whether potential energy rises or falls, use U = qV and keep the sign of q. Potential and potential energy move together for a positive charge and opposite for a negative one, and that sign is where most errors occur.

Answer the 3 checkpoints as you read.