Field of Continuous Charge Distributions
From point charges to integrals over lines, surfaces, and volumes
A charged wire does not have a single location where its charge lives — it is smeared along its entire length. By slicing the wire into infinitesimal pieces and summing their fields with an integral, we can compute the result exactly — and along the way we will discover why Gauss's law, coming next, is such a welcome shortcut.
Charge Density: Smearing Out the Points
In the previous lesson every charge was a point. Real charge distributions are rarely so tidy — a charged wire distributes its charge along a length, a charged plate spreads it across an area, and a charged cloud fills a volume. To handle these cases we introduce three charge densities: line charge density λ (C/m), surface charge density σ (C/m²), and volume charge density ρ (C/m³). Each density lets us write the infinitesimal charge dQ on a small piece of the distribution in terms of a geometric differential:
Building the Field Integral
Each infinitesimal element dQ behaves like a point charge, contributing dE = k(dQ/r²)r̂ to the total field. To find the total field we sum — integrate — all such contributions over the entire charge distribution:
Why This Is Harder Than It Looks
That integral looks innocuous, but it is a vector integral. The unit vector r̂ generally points in a different direction for each charge element dQ, which means you cannot simply pull it out of the integral. In practice you must choose a coordinate system, express r̂ in terms of its components, and integrate each component separately. Symmetry is your best ally — it often lets you argue that some components cancel, reducing the problem to a scalar integral.
This is exactly the vector calculus practiced in Module 1: setting up the geometry, parameterizing the source, writing the integrand, and evaluating. When the geometry is simple, the integral is tractable. When it is not, it can be fiendishly difficult — and that is precisely why Gauss's law, met in the next lesson, is so valuable.
A common mistake is to write E = ∫k(dQ/r²) and treat it as a scalar. This is only correct when symmetry guarantees that all contributions point in the same direction. In general, you must keep the vector form and resolve components.
The Infinite Line Charge
The classic example that illustrates both the power and the pain of the integral approach is the infinite line charge — a straight line of charge with uniform density λ extending to infinity in both directions. By symmetry, the field must point radially outward from the line and can depend only on the radial distance ρ from the line.
Setting up the integral in cylindrical coordinates, we take an element dQ = λ dz at position z along the line, at distance r = √(ρ² + z²) from the observation point. Substituting cos(θ) = ρ/r for the radial component and integrating from z = −∞ to z = +∞, the integral evaluates to:
Reading the Result
The field falls off as 1/ρ, not 1/r² as for a point charge — a slower decay, since as you move farther from the line, more of the line's charge contributes effectively to the field at your location. The direction is radially outward for positive λ.
Most importantly for the road ahead: this integral was tedious. We needed cylindrical coordinates, symmetry arguments, a trigonometric substitution, and a definite integral from −∞ to +∞. In the next lesson we will re-derive this exact same result using Gauss's law in about four lines of algebra. The reason we go through the integral first is so you can fully appreciate the shortcut when it arrives.
In this formula ρ is the radial distance from the line, in meters — not the volume charge density. This dual use of ρ is standard in electromagnetics textbooks. When ρ appears as a coordinate in cylindrical (ρ, φ, z), it is a distance; when it appears multiplying dV, it is a volume charge density.
- Identify the given values: λ = 4×10⁻⁹ C/m, ρ = 0.1 m, ε₀ = 8.854×10⁻¹² F/m.
- Use the infinite line charge formula: E = λ/(2πε₀ρ).
- Compute the denominator: 2πε₀ρ = 2π(8.854×10⁻¹²)(0.1) = 5.5636×10⁻¹².
- Divide: E = (4×10⁻⁹) / (5.5636×10⁻¹²) = 718.9 N/C.
- Use the infinite line charge formula: E = λ/(2πε₀ρ).
- Substitute: E = (10×10⁻⁹) / (2π × 8.854×10⁻¹² × 0.2).
- Compute the denominator: 2π(8.854×10⁻¹²)(0.2) = 1.11272×10⁻¹¹.
- Divide: E = (1×10⁻⁸) / (1.11272×10⁻¹¹) = 898.7 N/C.
Check your understanding
- Continuous charge distributions are described by line density λ (C/m), surface density σ (C/m²), or volume density ρ (C/m³).
- Each infinitesimal element dQ is treated as a point charge contributing dE = k(dQ/r²)r̂; the total field is the vector integral E = ∫k(dQ/r²)r̂.
- For an infinite line charge, the integral evaluates to E = λ/(2πε₀ρ), falling off as 1/ρ.
- The tediousness of the direct integral motivates Gauss's law, which can re-derive the same result far more efficiently.