Potential from Field
A short lesson with diagrams, a worked example, a problem to try, and a quick check. Then quiz yourself on 4 practice questions.
Potential difference is minus the integral of the field along a path: ΔV = VB − VA = −∫AB E · dr. Moving along the field, the potential drops. In a uniform field, this is just ΔV = −Ed.
1. Integrating the field
2. Using it
Find E first (often with Gauss’s law), then integrate it. Pick a convenient reference: V = 0 at infinity for finite charges.
For an infinite line or sheet, V doesn’t go to zero at infinity, so only potential differences make sense.
E points toward lower V. A positive charge released from rest moves to lower V; a negative charge moves to higher V.
Moving perpendicular to E doesn’t change V: that’s an equipotential.
3. Example, then try it
A uniform 500 V/m field points in the +x direction. Find V(0.040 m) − V(0). Then, for the field graphed below (Ex = 6x), find V(3) − V(0).
Uniform: ΔV = −Ed = −(500)(0.040) = −20 V.
Figure: ΔV = −∫03 6x dx = −[3x²]03 = −27 V.
A long line of charge has λ = 4.0 nC/m. Point A is 0.10 m from it and point B is 0.30 m away. Find VA − VB, and the work needed to move a +2.0 nC charge slowly from B to A.
Field: from Gauss’s law, E = 2kλ/r, pointing away from the line.
Integrate from B to A: VA − VB = −∫0.300.10(2kλ/r) dr = 2kλ ln(0.30/0.10).
Numbers: 2(9.0 × 10⁹)(4.0 × 10⁻⁹) ln 3 = 72 × 1.10 ≈ 79 V. A is at higher potential (closer to the positive line).
Work: W = qΔV = (2.0 × 10⁻⁹)(79) ≈ 1.6 × 10⁻⁷ J.
Answer: VA − VB ≈ 79 V; about 1.6 × 10⁻⁷ J of work.
4. Common mistakes
You move in the same direction as the electric field. What happens to the electric potential?
ΔV = −∫E · dr. Moving along E makes the integral positive, so V decreases. (The potential doesn’t depend on what charge you carry.)
Ready to practice?
4 questions on Potential from Field, each with a hint and an explanation for every wrong answer.