Equipotentials and work

The force on a charge q is qE, so the work along a path is W = q ∫ E ⋅ dl from A to B. Take a single source: E is radial with magnitude kQ/r², and any small displacement splits into a radial part dr and a sideways part that the dot product kills, so the integral collapses to kQq ∫ dr/r² and depends only on the starting and finishing radii. Adding more sources adds more such integrals, so the property survives superposition, and for any static arrangement W = q(V_A − V_B). Put numbers on it: carrying 5.0 nC from a point at 240 V to a point at 60 V, the field does 5.0 × 10⁻⁹ × 180 = 9.0 × 10⁻⁷ J of work whether you go straight there, loop twice round the room, or take a week over it. Now set V_A = V_B. The work is zero, and the set of points that satisfies it is an equipotential.

Let the path end where it started. Then the two endpoint potentials are the same, so the work round any closed loop is zero, which is the statement ∮ E ⋅ dl = 0 for a static field. Two things follow. First, energy bookkeeping works: a charge that wanders off and returns to its starting point comes back with the speed it left with, and no arrangement of fixed charges can be run as an engine. Second, two different equipotentials can never intersect, because a point of intersection would have to carry two values of V at once. Contours can touch only where the field vanishes — at the saddle point midway between two equal positive charges a single contour crosses itself, and there E = 0, the one place on a map where perpendicular to the contour means nothing.

Write the work on a small step as dW = qE ⋅ dl = q|E| |dl| cos θ. Along an equipotential dW = 0 for every step dl that lies in the surface. The charge is not zero and the step is not zero, so wherever |E| ≠ 0 the only survivor is cos θ = 0, giving θ = 90°. Field lines therefore cross equipotentials at right angles everywhere the field exists, which is why a contour map and a field-line map are the same drawing read two ways. The consequence worth holding on to is dynamic, not merely geometric. A bead threaded on a wire bent to follow a contour arrives with the speed it set out with, because the electric force on it always acts sideways — exactly as a ball rolled along a level path on a hillside is steered by gravity without being sped up or slowed down.

Draw contours at equal steps ΔV and the map carries magnitudes as well as directions. Cross from one contour to the next along the shortest route, the perpendicular, and |E| ≈ ΔV/Δs⊥. On a map drawn every 20 V with the lines 4.0 mm apart, |E| ≈ 20 ÷ 0.0040 = 5.0 × 10³ V m⁻¹; where the same lines squeeze to 1.0 mm apart, |E| ≈ 2.0 × 10⁴ V m⁻¹. Two cautions. Measure the gap perpendicular to the contours: a ruler laid at a slant reports too large a spacing and so too small a field. And read the answer as an average across the gap rather than the value at its middle — for a point charge the estimate ΔV/Δr comes out exactly equal to the field at the geometric-mean radius √(r₁r₂), which sits a little inside the arithmetic midpoint.

A point charge: concentric spheres at r = kQ/V. For +2.0 nC the 100, 200, 300 and 400 V surfaces sit at 0.18, 0.090, 0.060 and 0.045 m, so equal steps in potential give radii that pile up towards the charge, matching a field that grows as 1/r². A uniform field between large parallel plates: flat planes parallel to the plates, evenly spaced, because equal steps in V need equal steps in distance when |E| is constant. And a pair of equal and opposite charges: the plane midway between them is the V = 0 equipotential, perfectly flat, with the field crossing it at right angles from the positive charge towards the negative one. Whatever the map, the field points from high potential towards low, straight down the steepest descent.

In electrostatic equilibrium nothing inside a conductor is moving. Suppose the field at its surface had a component along the surface: the free electrons would feel it and slide, which contradicts equilibrium. So the tangential field is zero, the whole surface sits at one potential, and two practical readings follow. Field lines meet any conductor at 90°, so a metal object placed in a field bends the nearby lines round to arrive square on. And charge can be moved anywhere over that surface without work — no energy is needed to redistribute charge on a charged sphere. Where a conductor is sharply curved the contours around it crowd together, so the field there is largest, which is why a point discharges into the air long before a flat face does.