UY1: Electric Potential Of A Line Of Charge

Derive the potential of a uniformly charged finite line segment and connect it to the electric field by differentiation.

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 A Line Of 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: scalar potential idea (Electric Potential) and the integral ∫ dy/square root of (x² + y²) (Integration Techniques)
Outcomes: derive the finite-line potential, keep sign/reference consistent, and differentiate to recover Eₓ = -dV/dx
Key result (finite line, V(∞) = 0):

V(x) = λ/4πε₀ ln((square root of (x² + a²) + a)/(square root of (x² + a²) -a))

Common trap: forgetting the log argument must be dimensionless/positive, or using the field formula in place of the potential formula
A uniformly charged line segment of length 2a lies on the y-axis.
At point P(x,0), the potential is:

V(x) = λ/4πε₀ ln((square root of (x² + a²) + a)/(square root of (x² + a²) -a))

with λ = Q/(2a).

Sign convention: if λ > 0, then V > 0 for the standard reference.

Motivation / intuition

For continuous charges, potential is often easier than field because you don’t have to resolve vector components: dV = k dq/r is a scalar. The price you pay here is a logarithm, which is the natural “integral of 1/r” behavior for line-like geometry.

2) Setup

Assume:

Charge is uniformly distributed from y = -a to y = +a.
Observation point is P(x,0) with x > 0.
Use Coulomb potential element:

dV = (1/4πε₀)dq/r

where

dq = λ dy, r = square root of (x² + y²) .

3) Core derivation/explanation

Integrate contributions from all elements:

V = λ/4πε₀∫₋ₐ^ady/(square root of (x² + y²))

Use

∫dy/(square root of (x² + y²)) = ln(y + square root of (x² + y²)),

then evaluate limits:

V(x) = λ/4πε₀ ln((square root of (x² + a²) + a)/(square root of (x² + a²) -a)).

Field check for x > 0:

Eₓ = -dV/dx = (1/4πε₀)(2λ a)/(x square root of (x² + a²)) = (1/4πε₀)Q/(x square root of (x² + a²)).

Direction is + x hat for positive line charge when x > 0.

Units: [λ/(4πε₀)] = (C/m)(N m² C⁻²) = N m/C = V; the ln(…) term is dimensionless.
Limits/signs: for x≫ a, the line looks like a point charge Q = 2aλ so V ≈ kQ/x; the sign of λ sets the sign of V everywhere.

4) Worked example(s)

Given Q = 4.0 μC, 2a = 0.60 m, and x = 0.20 m:

a = 0.30 m, λ = Q/2a = 6.67 × 10⁻⁶ C m⁻¹

square root of (x² + a²) = square root of (0.20² + 0.30²) = 0.3606

V = λ/4πε₀ ln((0.3606 + 0.30)/(0.3606-0.30)) ≈ 1.60 × 10⁵ V

5) Practice set (with hints + answers)

If x increases while Q and a stay fixed, does V increase or decrease? Hint: inspect the log ratio. Answer: decreases.
Write V(x) in terms of Q and a only (replace λ). Hint: use λ = Q/(2a). Answer:

V(x) = Q/(8πε₀ a) ln((square root of (x² + a²) + a)/(square root of (x² + a²) -a)).

Why can this derivation use scalar addition directly? Hint: think about potential vs field. Answer: electric potential is scalar, so contributions add algebraically.

6) Summary + next steps

The finite-line potential comes from a single integral of 1/square root of (x² + y²).
Differentiating the result gives the correct on-axis electric field and sign.
This expression leads naturally to the infinite-line case with a reference radius.

Next: Electric Potential Of An Infinite Line Charge Previous: Electric Potential Of A Ring Of Charge Back To Electromagnetism (UY1)