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If a charge moves from one point on an equipotential surface to another, the electric field does no work on it, however long or curved the path. This one fact answers a whole family of board and JEE questions, and it explains why field lines always cross equipotentials at right angles.
Why is the work done on an equipotential surface zero?
The potential difference between two points is the change in potential energy of a charge moved between them, divided by the charge, so ΔU = qΔV, as OpenStax University Physics Volume 2 defines it.
An equipotential surface is a surface on which the potential has the same value at every point.
Put the two together. Start and end on the same equipotential and ΔV = 0, so the work is W = −qΔV = 0, exactly as OpenStax's section on equipotential surfaces states it. Because the electric force is conservative, only the start and end matter, not the route between them.
Why is the electric field perpendicular to an equipotential?
Take a tiny step dl along the surface. The work done by the field on the charge is qE·dl = qE dl cos θ, where θ is the angle between the field and the step. On the surface that work must be zero for every possible step. The charge is not zero and the field is not zero, so cos θ = 0 and θ = 90°.
OpenStax puts the result plainly: equipotential lines are always perpendicular to electric field lines (OpenStax 7.5). If the field had any component along the surface, moving along it would take work, and the potential would change. That is also why two equipotential surfaces can never intersect: the field at the crossing point would need two directions at once.
Worked example 1: a point charge
The potential of a point charge is V = kQ/r, with k = 8.99 × 10⁹ N·m²/C² (OpenStax 7.3). V depends only on r, so every sphere centred on the charge is an equipotential (OpenStax 7.5).
Let Q = +2 nC sit at the origin, and move a test charge q = +3 µC.
Along a circle of radius 0.30 m, from the top of the circle to its side: V stays at kQ/r = (8.99 × 10⁹)(2 × 10⁻⁹)/0.30 ≈ 59.9 V the whole way. ΔV = 0, so W = 0.
Radially outward from 0.30 m to 0.60 m: V falls to about 30.0 V, so ΔV ≈ −30.0 V. The field does W = −qΔV ≈ (3 × 10⁻⁶)(30.0) ≈ 9.0 × 10⁻⁵ J of work, positive because a positive charge is pushed away.
Out to 0.60 m, around that bigger circle, and back in to a different point on the 0.30 m circle: the out and in legs cancel, the arc costs nothing, and the total is zero again.
The third path is still zero, because the start and end points share one potential. For how the spheres space out as r grows, see our guide to the shape and spacing of equipotential surfaces.
Worked example 2: parallel plates
Between two charged parallel plates the field is uniform, and the equipotentials are evenly spaced flat planes parallel to the plates (OpenStax 7.5).
OpenStax's own example uses plates 6.50 mm apart with 500 V across them, so the 100 V equipotentials sit 1.3 mm apart (OpenStax 7.5). In a uniform field |V| = Ed (OpenStax 7.2), so E = 500/0.00650 ≈ 7.7 × 10⁴ V/m.
Slide a +2 µC charge 4 mm sideways, parallel to the plates: it stays on one plane, so W = 0.
Move it from the 100 V plane to the 300 V plane, 2.60 mm away: the work you must do is qΔV = (2 × 10⁻⁶)(200) = 4.0 × 10⁻⁴ J.
Move it diagonally between those same two planes: still 4.0 × 10⁻⁴ J. Only the sideways part of the path changed, and sideways is free.
What mistakes do students make with this?
Mixing up whose work it is. Work by the field is −qΔV; work by an external agent at constant speed is +qΔV.
Dropping the sign of the charge. An electron moving to a higher potential has its potential energy fall, not rise.
Thinking a long path means more work. Between two given equipotentials, every path gives the same answer.
Counting the sideways part of the path. In plate problems, split the displacement into along-the-field and across-the-field parts and drop the second.
If a question keeps going wrong at the sign or the setup, the Fermi Problem Library has a step-by-step solver that explains the reasoning rather than printing an answer. To see the right angles for yourself, the Fermi Simulation Lab has interactive simulations for ideas easier to see than read.
