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From V to E: the Gradient

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Differentiation undoes the integral

The definition V_b − V_a = −∫→E·d→l is an integral relationship, so the fundamental theorem of calculus runs it backward: each field component is the negative derivative of the potential along that direction, Eₓ = −∂V/∂x, and likewise for y and z. Compactly, E = −∇V, the negative gradient. The minus sign says the field points downhill — in the direction of steepest decrease of V — which is exactly why positive charges accelerate from high potential to low.

Field from potential (gradient)
Eₓ = −∂V/∂x, E_y = −∂V/∂y, E_z = −∂V/∂z; →E = −∇V
Units: 1 V/m = 1 N/C — the two field units are identical. For radial potentials, E_r = −dV/dr.

The strategy: easy scalar in, hard vector out

This relation completes the standard two-step method for continuous distributions: (1) compute V = ∫k·dq/r — a scalar integral with no components; (2) differentiate to get E. One derivative replaces an entire vector integral. Sanity check with the point charge: V = kq/r gives E_r = −d(kq/r)/dr = −kq·(−1/r²) = kq/r², Coulomb’s field recovered. The caveat: to take a derivative at a point you need V as a function in a neighborhood, so find V(x) symbolically before differentiating — one numerical value of V tells you nothing about E.

Worked example

Along the x-axis the potential is V(x) = 5x² − 2x (V in volts, x in meters). Find Eₓ(x), evaluate it at x = 0.50 m, and find where the field vanishes.

  1. 1.Differentiate: dV/dx = 10x − 2.
  2. 2.Apply the gradient relation: Eₓ = −dV/dx = 2 − 10x (in V/m).
  3. 3.At x = 0.50 m: Eₓ = 2 − 10(0.50) = −3.0 V/m — magnitude 3.0 V/m, pointing in the −x direction.
  4. 4.The field is zero where dV/dx = 0: x = 0.20 m, the minimum of the V(x) parabola.
  5. 5.Interpretation: for x > 0.20 m the potential climbs, so the field points backward (−x), pushing positive charges down the potential hill.
Answer: Eₓ = (2 − 10x) V/m; Eₓ(0.50 m) = −3.0 V/m; Eₓ = 0 at x = 0.20 m
Worked example

Use the ring potential V(x) = kQ/√(x² + R²) to find the on-axis field, and evaluate for Q = 5.0 nC, R = 0.40 m, x = 0.30 m.

  1. 1.Write V = kQ·(x² + R²)^(−1/2) and differentiate with the chain rule: dV/dx = kQ·(−½)(x² + R²)^(−3/2)·(2x) = −kQ·x/(x² + R²)^(3/2).
  2. 2.Then Eₓ = −dV/dx = kQ·x/(x² + R²)^(3/2) — exactly the result Unit 1 earned with a full vector integral and a symmetry argument. One derivative did the same job.
  3. 3.Limits check: Eₓ = 0 at x = 0 (V is maximal at the center, zero slope) and Eₓ ≈ kQ/x² for x ≫ R. ✓
  4. 4.Numerically: (x² + R²)^(3/2) = (0.25)^(3/2) = 0.125 m³, and kQ·x = (8.99 × 10⁹)(5.0 × 10⁻⁹)(0.30) = 13.5 V·m².
  5. 5.Eₓ = 13.5/0.125 ≈ 1.1 × 10² N/C, directed along +x away from the ring.
Answer: Eₓ = kQx/(x² + R²)^(3/2) ≈ 1.1 × 10² N/C at x = 0.30 m

Reading V(x) graphs

On a graph of V versus x, the field is the negative slope: steep descent means a strong field in +x, an uphill stretch means the field points in −x, and a flat region — however high the value of V — means E = 0. Extrema of V are field-free points. The complementary reading works too: on a graph of Eₓ versus x, the potential change between two points is minus the area under the curve. Slope and area are the two faces of the same fundamental theorem.

On the exam

Memorize the inverse pair and their graphical readings: V_b − V_a = −∫ₐᵇ →E·d→l (area under an Eₓ graph, negated) and Eₓ = −dV/dx (slope of a V graph, negated). AP free-response loves handing you one graph and demanding the other — check your minus sign at a point where you know which way the field must push a positive charge.

Checkpoint

Throughout some region of space the potential is 200 V at every point. The electric field in that region is:

Checkpoint

The potential along the x-axis is V(x) = 3x² (volts, x in meters). What is Eₓ at x = 2.0 m?

Answer the 2 checkpoints as you read.

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