UY1: Electric Potential Of An Infinite Line Charge

Express potential for an infinite line charge using a reference radius and recover the standard radial electric field.

University Physics Year 1

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Learning objectives

Construct electric-field and potential models for discrete and continuous charge distributions.

This page gives the UY1 working model/result for Electric Potential Of An Infinite Line Charge. You reuse it when you build fields/potentials by symmetry or superposition, and when you connect fields to forces, energy, and circuits.

Module path: Electromagnetism (UY1)
Practice: UY1 Electromagnetism Quiz
Full routing: UY1 Assessment Map
Math toolkit: Mathematics for Undergraduate Physics

1) At a glance

Prerequisites: radial field of a line charge (Gauss symmetry) (Common Charge Distributions) and log integrals (Integration Techniques)
Outcomes: write potential as a difference relative to a reference radius, keep sign conventions consistent, and recover Eᵣ = -dV/dr
Key result (choose reference r₀): V(r)-V(r₀) = λ/2πε₀ ln(r₀/r)
Common trap: trying to use V(∞) = 0 (it diverges), or forgetting to state the chosen reference r₀
For an infinite line charge, absolute potential is not finite with V(∞) = 0.
You define potential relative to a reference radius r₀:

V(r)-V(r₀) = λ/2πε₀ ln(r₀/r)

The electric field is still well-defined:

Eᵣ = λ/(2πε₀ r).

Motivation / intuition

Infinite charge distributions have too much “stuff at infinity”, so absolute potential depends on how you choose the zero level. Physics only needs potential differences, and picking a reference radius r₀ makes the logarithm well-defined.

2) Setup

Assume:

Very long straight line charge with uniform linear charge density λ.
Cylindrical symmetry, so vector E = Eᵣ(r) r hat.
Choose outward radial direction as positive r hat.
Reference potential is set at finite radius r₀.

3) Core derivation/explanation

From Gauss’s law (or symmetry + Coulomb result):

Eᵣ(r) = λ/(2πε₀ r).

Potential difference between radii r₀ and r:

V(r)-V(r₀) = -∫_r₀^rEᵣ dr = -∫_r₀^rλ/(2πε₀ r) dr

V(r)-V(r₀) = λ/2πε₀ ln(r₀/r).

If we set V(r₀) = 0:

V(r) = λ/2πε₀ ln(r₀/r).

Sign convention:

For λ > 0, potential decreases as r increases.
For λ < 0, signs reverse.

Recover field from potential:

Eᵣ = -dV/dr = λ/(2πε₀ r).

Units: [λ/(2πε₀)] = V and ln(r₀/r) is dimensionless, so V has volts.
Limits/signs: r = r₀ ⇒ V(r)-V(r₀) = 0; for λ > 0, increasing r makes V decrease (log becomes more negative) and the field points outward (Eᵣ > 0).

4) Worked example(s)

Given λ = 3.0 × 10⁻⁶ C m⁻¹, r₀ = 0.10 m with V(r₀) = 0, find V at r = 0.25 m.

V(r) = λ/2πε₀ ln(0.10/0.25) = (5.39 × 10⁴) ln(0.4) ≈ -4.94 × 10⁴ V

Negative value is expected because r > r₀ for positive λ.

5) Practice set (with hints + answers)

Why can we not set V(∞) = 0 for an infinite line charge? Hint: examine ln(r₀/r) as r → ∞. Answer: potential diverges logarithmically, so the infinity reference is not finite.
If r doubles, how does V(r)-V(r₀) change for fixed r₀? Hint: use log properties. Answer: it changes by -(λ/2πε₀) ln 2.
Find Eᵣ at r = 0.050 m for λ = 1.0 × 10⁻⁶ C m⁻¹. Hint: substitute into Eᵣ = λ/(2πε₀ r). Answer: Eᵣ ≈ 3.60 × 10⁵ N C⁻¹ outward.

6) Summary + next steps

Infinite-line potential must be written as a potential difference relative to a finite reference radius.
The logarithmic potential and inverse-radius field are consistent through Eᵣ = -dV/dr.
Keep your reference radius explicit in exam solutions to avoid sign/reference mistakes.

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