The Electric Field of a Continuous Charge Distribution
A continuous model replaces many discrete charges with a charge density. Integrate the Coulomb field of small charge elements, using symmetry to cancel components when the geometry and density allow it.
What a learner can do afterwards
- Sets up the field integral for a line or ring of charge with a stated charge density
- Uses symmetry to argue that certain components cancel
- Checks that a finite distribution with nonzero total charge has a point-charge field as its leading far-field term
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The electric field vector E is the force F on a tiny positive test charge q divided by q: E = F/q. Its units are N/C. For one point charge Q, its magnitude is k|Q|/r², where k is Coulomb's constant and r is the source-to-point distance. It points away from positive Q and toward negative Q. Fields add as vectors: add each signed component, not the magnitudes when directions differ.
A continuous model smooths out discrete charges. Line density λ is charge per length (C/m): dq = λ dl. For constant λ, a segment's charge is λ times its length. Sheet density σ is charge per area (C/m²): dq = σ dA. Volume density ρ is charge per volume (C/m³): dq = ρ dV. Each element gives dE = k dq r̂/r². Here r is its distance to the observation point; r̂ points from the element to that point. Integrate each signed component over the object.
Take a thin ring of radius R with uniform positive line density λ. Its total charge is Q = 2πRλ. The z-axis passes through its centre, perpendicular to its plane; choose a point with z > 0. Each element is distance r = √(R²+z²) from this point. Diametrically opposite elements give equal and opposite sideways components. Their axial components add. Projection onto the axis multiplies k dq/r² by z/r, giving dE_z = k z dq/(R²+z²)^(3/2).
Let θ run around the ring in radians. A small arc has length R dθ, so dq = λR dθ. Cover the full ring with limits 0 and 2π. Then E_z = ∫₀²π kλRz/(R²+z²)^(3/2) dθ. The factor outside dθ is constant because R, z and λ do not change around the ring. Since ∫₀²π dθ = 2π, E_z = kQz/(R²+z²)^(3/2). The other components are zero. At the centre, z = 0 and the total field is zero.
For this ring, when z is much larger than R, (R²+z²)^(3/2) is approximately z³. Thus E_z is approximately kQ/z², the field of a point charge Q. Use this check for a finite distribution with nonzero total charge, at distances much larger than its size. A neutral object can still have a field, but the leading point-charge term is zero. An infinite line or sheet has no distance beyond its size, so this check does not apply.
Choose dq from the density, add field components, use symmetry where it applies, and check the far field of a finite charged object.
Lesson, questions and study cards written for this dataset. OpenStax University Physics Volume 2 was consulted for physics facts and problem structure; no textbook prose is reproduced.
Readings written for this dataset using OpenStax University Physics Volume 2 for factual study. Video references identify Khan Academy material; publisher reuse terms have not been verified. Study cards were drawn from the imported fact collection and checked against the revised readings.
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