1.4 Electric Field Due to a Charge Distribution

From point charges to continuous distributions

For a continuous charge distribution, treat it as infinitely many infinitesimal charges dq, each producing a tiny field dE = k dq / r² in the direction from dq to the field point. The total field is found by integrating dE over the entire distribution, being careful to add components (usually only symmetric components survive).

Linear, surface, and volume charge density

Express dq using charge density: λ (C/m) for a line, σ (C/m²) for a surface, ρ (C/m³) for a volume. Then dq = λ dl, σ dA, or ρ dV respectively. Substituting these into dE = k dq/r² sets up the integral for E.

Using symmetry to simplify integrals

Exploit symmetry to cancel components before integrating — e.g., for a ring or rod on its perpendicular axis, horizontal components cancel by symmetry, leaving only the axial component. This turns a vector integral into a simpler scalar integral, often reducing to a standard result like E = kQx/(x² + R²)^(3/2) for a ring.

Key terms

dE for a point charge element

dE = k dq / r², directed from dq toward (or away from) the field point depending on sign.

Linear charge density λ

Charge per unit length (C/m); dq = λ dl for a line or rod.

Surface charge density σ

Charge per unit area (C/m²); dq = σ dA for a charged surface or disk.

Volume charge density ρ

Charge per unit volume (C/m³); dq = ρ dV for a charged 3D object.

Symmetry trick in field integrals

Identify canceling components (e.g., perpendicular components on a ring's axis) before integrating to simplify the vector integral into a scalar one.

Try an AP-style question

What is the field a distance r from an infinite line of charge with density λ?

1. λ/(ε₀r)
2. λr/(2ε₀)
3. λ/(4πε₀r²)
4. λ/(2πε₀r)