UY1: Electric Field Of Line Of Charge

Derive the electric field of a uniformly charged finite line using component integration and symmetry with clear sign conventions.

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 Field Of 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: Coulomb field element + superposition (Coulomb’s Law), symmetry and definite integrals (Integration Techniques)
Outcomes: set up dq = λ dy, use symmetry to cancel components, and recover the far-field kQ/x² limit
Key result (perpendicular bisector):

vector E(x) = (1/4πε₀)Q/(x square root of (x² + a²)) x hat

Common trap: losing a factor of 2 by integrating only half the rod without doubling, or mixing Q with λ
Symmetry: transverse components cancel, so the field points along ±x hat on the perpendicular bisector.

Motivation / intuition

This is the first “real” continuous-distribution skill: you replace a sum of many tiny Coulomb contributions by an integral, and you use symmetry to avoid doing vector algebra for every element.

2) Setup

Rod lies on y-axis from -a to + a.
Field point is P(x,0) with x > 0.
Linear charge density:

λ = Q/2a

Differential charge element: dq = λ dy.

3) Core derivation/explanation

Distance from element at y to point P:

r = square root of (x² + y²)

Differential field magnitude:

dE = (1/4πε₀)dq/r²

Only x-component survives after symmetry:

dEₓ = (1/4πε₀)(x dq)/((x² + y²)^(3/2))

Integrate from -a to a:

Eₓ = (1/4πε₀)xλ∫₋ₐ^ady/((x² + y²)^(3/2))

Using

∫dy/((x² + y²)^(3/2)) = y/(x² square root of (x² + y²))

we get

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

Direction is away from rod if Q > 0, toward rod if Q < 0.

Units: [kQ/(x square root of (x² + a²))] = (N m² C⁻²)C/m² = N/C.
Limits/signs: if x≫ a, then square root of (x² + a²) ≈ x and E ≈ kQ/x² (point-charge limit); sign of Q sets the direction along ±x hat.

4) Worked example(s)

A rod has total charge Q = 4.0 μC and length 2a = 0.60 m, so a = 0.30 m. Find field at x = 0.20 m.

E = ((8.99 × 10⁹)(4.0 × 10⁻⁶))/((0.20) square root of ((0.20)² + (0.30)²))

square root of 0.13 = 0.3606 ⇒ x square root of (x² + a²) = 0.0721

E ≈ 4.99 × 10⁵ N C⁻¹

Direction: + x hat for positive rod.

5) Practice set (with hints + answers)

If the same rod is observed at very large x, what should the field approach? Hint: far away it looks like a point charge. Answer: E ≈ kQ/x².
Keep Q fixed and double a while evaluating at same x. Does E increase or decrease? Hint: denominator has square root of (x² + a²). Answer: decreases.
A negative rod has the same magnitude of Q. What changes? Hint: sign affects direction only. Answer: same magnitude, direction flips toward the rod.

6) Summary + next steps

Finite-line fields are solved by component integration with symmetry.
The result transitions correctly to point-charge behavior at large distance.
Direction and sign come from charge sign and your chosen axis orientation.

Next: Electric Field Of Uniformly Charged Disk Previous: Electric Field Of A Ring Of Charge Back To Electromagnetism (UY1)