8 min readConcept sheet2-mark diagrams and properties, 1-mark MCQ on work along an equipotential, and short questions using E = −dV/dr

Equipotential surfaces and the link between E and V

An equipotential surface has the same potential at every point. No work is done moving a charge along it, so the field is always perpendicular to it.

Work along an equipotentialW = q[V(B) − V(A)] = 0, since V(A) = V(B)

Field and potentialE = −dV/dr: E points in the direction in which V falls fastest

Size of the field|E| = |ΔV|/Δr, measured along the normal to the equipotentials; unit V m⁻¹

ShapesPoint charge: concentric spheres. Uniform field: parallel planes at right angles to E

What is an equipotential surface?

An equipotential surface is a surface on which the potential is the same at every point. For a point charge, V = kq/r depends only on r. So its equipotentials are concentric spheres centred on the charge. In a uniform field, they are parallel planes at right angles to the field. The field must be normal to an equipotential surface. If E had a part along the surface, moving a charge along it would need work, and the potential would change. That cannot happen on a surface of constant potential.

Properties of equipotential surfaces

No work is done in moving a charge along an equipotential surface.
The electric field is always perpendicular to the surface at every point.
Two equipotential surfaces never intersect. At the crossing, one point would have two potentials.
For equal steps of potential, the surfaces are closer together where the field is strong.
The surface of a conductor in electrostatics is an equipotential surface.

Derivation: relation between E and V

Take two close equipotential surfaces: A at potential V and B at V + δV, a small distance δl apart. The field E points from A to B, so δV is negative (V falls along E).

1
Move a unit positive charge from B to A along the normal, against the field.
2
Work done by the external force = force × distance = |E| δl.
3
This work equals the potential difference V(A) − V(B) = V − (V + δV) = −δV.
4
So |E| δl = −δV, giving |E| = −δV/δl = |δV|/δl.
5★
So, E points in the direction in which the potential decreases fastest, and its size is the change in potential per unit distance along the normal: E = −dV/dl.

★ key step

How to draw equipotential surfaces

1
Point charge: dashed concentric circles round +q. Draw solid radial field lines crossing them at right angles. For equal steps of V, the circles get farther apart going outward.
2
Uniform field: solid horizontal field lines with arrows; dashed vertical straight lines, equally spaced, crossing them at right angles.
3
Dipole: dashed closed curves round each charge, crowded near the charges. The perpendicular bisector is the V = 0 surface. Field lines cross every dashed curve at right angles.
4
Two equal positive charges: dashed closed curves round each charge. Farther out they merge into one curve shaped like a peanut, and far away they look like circles.

Worked example: field from potentials

Problem

(a) Two parallel equipotential planes at 30 V and 10 V are 5 mm apart. Find the field between them. (b) In a region, V = 5x² volt, where x is in metres. Find E at x = 2 m.

Working

(a) |E| = ΔV/Δr = (30 − 10) V ÷ 0.005 m = 4000 V m⁻¹.
It points from the 30 V plane towards the 10 V plane (towards lower potential).
(b) E = −dV/dx = −d(5x²)/dx = −10x.
At x = 2 m: E = −20 V m⁻¹, so the field is 20 V m⁻¹ in the −x direction.

Answer

(a) 4 × 10³ V m⁻¹, from the 30 V plane towards the 10 V plane; (b) 20 V m⁻¹ along −x

In real life

Contour lines on a map of the Western Ghats are like equipotentials: walk along one and your height does not change, and the steepest way down crosses them at right angles. Closely packed contours mean a steep slope, just as crowded equipotentials mean a strong field.

Exam angle

2 marks: “Draw equipotential surfaces for (a) a point charge and (b) a uniform electric field.” (a) Concentric spheres (dashed circles) centred on the charge, spreading apart outward. (b) Equally spaced parallel planes at right angles to the field lines.

2 marks: “Why can two equipotential surfaces never intersect? Why is the field normal to an equipotential surface?” At an intersection one point would have two potentials, which is impossible. If E had a part along the surface, moving a charge along it would need work, so V would change.

2 marks: “Show that the electric field points in the direction of steepest decrease of potential.” Between close equipotentials, |E| δl = −δV, so |E| = −dV/dl. E is largest along the normal and points towards lower V.

Remember it

Equipotentials and field lines always cross at 90°. The field runs downhill in potential: E = −dV/dr.