Field from Potential
A short lesson with diagrams, a worked example, a problem to try, and a quick check. Then quiz yourself on 4 practice questions.
Going the other way, the field is minus the slope of the potential: Ex = −dV/dx. The field points downhill in potential, it’s strongest where V changes fastest, and it’s always perpendicular to equipotentials.
1. Slope of V
2. Reading graphs and maps
On a V(x) graph, read the slope, not the value. E is zero wherever the graph is flat, and largest where it’s steepest.
Force on a charge: F = qE. A positive charge accelerates downhill on the V graph; a negative charge, uphill.
On an equipotential map drawn at equal voltage steps, the field is strongest where the lines crowd together, and field lines cross them at right angles.
3. Example, then try it
The potential along the x-axis is V(x) = 3x³ − 6x (V, with x in m). Find the electric field at x = 2 m.
dV/dx = 9x² − 6, so at x = 2: 36 − 6 = 30 V/m.
Ex = −30 V/m: 30 V/m in the −x direction.
Use the V(x) graph below. Find E in each region. What force acts on a +2.0 µC charge at x = 6 m, and on one at x = 3 m?
0 to 2 m: slope = 20/2 = +10 V/m, so E = −10 V/m (toward −x).
2 to 4 m: slope 0, so E = 0, even though V = 20 V.
4 to 8 m: slope = −40/4 = −10 V/m, so E = +10 V/m.
Forces: at x = 6 m, F = qE = (2.0 × 10⁻⁶)(10) = 2.0 × 10⁻⁵ N toward +x. At x = 3 m, no force.
Answer: −10, 0 and +10 V/m; 2.0 × 10⁻⁵ N at 6 m, none at 3 m.
4. Common mistakes
In a region where the V(x) graph is flat at 50 V, what is the electric field?
Ex = −dV/dx = 0 where V doesn’t change, whatever value V has.
Ready to practice?
4 questions on Field from Potential, each with a hint and an explanation for every wrong answer.