Fields from Continuous Charge Distributions

You’ll be able to

Set up field integrals using charge density (λ, σ, ρ) and dq
Exploit symmetry to cancel components before integrating
Derive the field of a charged ring and an infinite line of charge

From a sum to an integral

A real charged object holds countless charges, so superposition becomes an integral. We slice the object into infinitesimal pieces dq, each acting as a point charge that contributes a field dE = k·dq / r², then integrate over the whole object: E = ∫ dE = ∫ k·dq / r² (r̂). The art is expressing dq through a charge density: λ = charge per length (dq = λ·dx), σ = charge per area (dq = σ·dA), or ρ = charge per volume (dq = ρ·dV).

Field of a continuous distribution

→E = ∫ k·dq / r² (r̂), dq = λ dx = σ dA = ρ dV

A vector integral: resolve dE into components and integrate each. Symmetry often makes one component vanish.

Let symmetry kill the hard components

The vector integral looks daunting until symmetry helps. On the axis of a symmetric object, for every dq there is a partner whose field cancels the perpendicular (off-axis) component of the first. Only the components along the symmetry axis survive, so the vector integral collapses to a single scalar integral of dEₓ = dE·cosθ. Recognizing this cancellation before you integrate is the whole game.

Worked example

A ring of radius R carries total charge Q uniformly. Find the electric field at a point on its axis a distance x from the center.

1.Each element dq is the same distance r = √(x² + R²) from the field point, so dE = k·dq / (x² + R²).
2.By symmetry, radial components from opposite elements cancel; only the axial (x) component survives: dEₓ = dE·cosθ, where cosθ = x / √(x² + R²).
3.So dEₓ = k·dq·x / (x² + R²)^(3/2). The factor x/(x² + R²)^(3/2) is the same for every element.
4.Integrate: Eₓ = [k·x / (x² + R²)^(3/2)] ∫ dq = k·x·Q / (x² + R²)^(3/2), since ∫ dq = Q.
5.Check limits: at x = 0, E = 0 (center of ring, by symmetry); for x ≫ R, E ≈ kQ/x² (the ring looks like a point charge).

Answer: E = k·Q·x / (x² + R²)^(3/2), directed along the axis

Tip

Always test a derived field in limits you already know. Far away (x ≫ R) a finite object must look like a point charge, giving kQ/r². A field expression that fails this check has an error.

Checkpoint

On the axis of a uniformly charged ring, at what point is the electric field zero?

Worked example

Find the field a perpendicular distance d from an infinitely long line of charge with linear density λ. (Set up the integral; the standard result is E = λ / (2πε₀ d).)

1.Place the line on the x-axis and the field point P at perpendicular distance d. An element dq = λ dx sits at position x, a distance r = √(x² + d²) from P.
2.By symmetry the components along the line cancel in pairs (x and −x), leaving only the perpendicular component: dE⊥ = (k·λ dx / (x² + d²))·cosθ, with cosθ = d / √(x² + d²).
3.So E⊥ = ∫₋∞⁺∞ k·λ·d dx / (x² + d²)^(3/2).
4.The standard integral ∫ dx/(x² + d²)^(3/2) = x / [d²√(x² + d²)], evaluated from −∞ to +∞ gives 2/d².
5.Therefore E⊥ = k·λ·d·(2/d²) = 2kλ/d = λ / (2πε₀ d), using k = 1/(4πε₀).

Answer: E = λ / (2πε₀ d) = 2kλ / d, directed radially away from the line (for λ > 0). Note the 1/d dependence, not 1/d².

Checkpoint

The field of an infinite line of charge falls off as 1/d, while a point charge falls off as 1/r². Why is the line’s field weaker-decaying?

On the exam

A continuous-distribution FRP earns its points in the setup: state dq in terms of a density, draw dE and identify which component survives by symmetry, write correct integration limits, and only then integrate. Show the symmetry argument explicitly — graders reward it.

Answer the 2 checkpoints as you read.