Fields from continuous charge

Three densities cover almost everything in this course. A line of charge carries λ = dq/dl in C m⁻¹, a surface carries σ = dq/dA in C m⁻², and a solid carries ρ = dq/dV in C m⁻³. When the distribution is uniform, each is just the total divided by the extent: a 30 cm rod holding 50 nC has λ = 1.67 × 10⁻⁷ C m⁻¹. When it is not uniform, the density is a function — λ(x) = λ₀x/L for a rod charged more heavily towards one end — and the total is recovered by integrating it back, Q = ∫ λ dl. The smoothing is safe because the numbers are enormous: a 1 cm² patch at σ = 40 nC m⁻² carries 4.0 × 10⁻¹² C, about 2.5 × 10⁷ electrons, so even a square-millimetre element still holds a quarter of a million of them and its graininess is invisible.

Each element contributes dE = (1/4πε₀)(dq/r²) r̂, where r̂ points from that element to the field point. Two elements in different places give dE vectors in different directions, so adding magnitudes counts a cancellation as a contribution and overstates the answer. Choose axes first, write the components dEₓ = dE cos α and dE_y = dE sin α with α read off the geometry, and integrate each component separately. Then express everything — r, cos α and dq — in terms of a single variable that sweeps the body exactly once: the position x along a rod, the polar angle φ around a ring, the radius r of an annulus inside a disc. The vector problem has become one or two ordinary definite integrals, and only the limits remain to be chosen.

Put a ring of radius R and total charge Q in a plane, and take the field point on the axis a distance z from the centre. Every element is the same distance r = √(z² + R²) away, so dE = k dq/(z² + R²) has the same size for all of them. Pair each element with the one diametrically opposite: the components perpendicular to the axis are equal and opposite and cancel exactly, and what survives is the axial share cos θ = z/√(z² + R²). Both factors are constants of the integration, so the integral collapses to ∫ dq = Q, giving E_z = kQz/(z² + R²)³⁄². Read both ends of that result. At z = 0 the field is zero, every element balanced by its partner. Differentiating gives a maximum where z² + R² = 3z², that is at z = R/√2; beyond it E falls off as kQ/z². For R = 4.0 cm and Q = 12 nC the peak is 2.59 × 10⁴ N C⁻¹ at z = 2.83 cm.

A uniformly charged disc is a nest of rings, so the ring result can be re-used instead of re-derived. The ring of radius r and thickness dr has area dA = 2πr dr, hence dq = σ 2πr dr, and its axial contribution is dE_z = k(σ 2πr dr) z/(r² + z²)³⁄². The integration variable is now r, running from 0 to R, with z held fixed as a parameter. The substitution u = r² + z², du = 2r dr, turns the integral into ½∫ u^(−3/2) du, so ∫₀^R r dr/(r² + z²)³⁄² = 1/z − 1/√(R² + z²). Multiplying back and using 2πk = 1/(2ε₀) gives E_z = (σ/2ε₀)[1 − z/√(z² + R²)]. The bracket is a pure number between 0 and 1: it reports what fraction of the sheet field survives at that height, 0.29 at z = R and 0.84 at z = R/6.

Every field integral should be checked at both ends. Far away, z ≫ R, expand the bracket binomially: z/√(z² + R²) = (1 + R²/z²)^(−1/2) ≈ 1 − R²/(2z²), so the bracket is about R²/(2z²) and E → σπR²/(4πε₀z²) = kQ/z². The disc has become a point charge, as it must. Close in, z ≪ R, the bracket tends to 1 and E → σ/(2ε₀), with no z in it at all — the uniform field of an infinite sheet. Numbers make the crossover concrete. For R = 6.0 cm and σ = 45 nC m⁻², the sheet value is 2.54 × 10³ N C⁻¹; at z = 1.0 cm the disc gives 2.12 × 10³ N C⁻¹, 16% below it, while the point-charge estimate gives 4.57 × 10⁴ N C⁻¹, twenty-one times too large. By z = 60 cm the point-charge estimate is right to 0.8%.

Symmetry runs out quickly. Move the field point off the axis of a ring and the integral becomes elliptic — no elementary answer exists — but the physics is unchanged, so add the elements directly. Take a ring of radius 5.0 cm carrying 10 nC, with the field point in the ring's plane 2.0 cm from the centre. Split the ring into N equal arcs, place each arc's charge Q/N at its midpoint, and sum the N vectors. With N = 4 the sum gives 3.27 × 10³ N C⁻¹; N = 8 gives 8.54 × 10³; N = 12 gives 8.74 × 10³; and from N = 24 onwards it has settled at 8.74 × 10³ N C⁻¹, pointing back towards the centre because the near arc, 3.0 cm away, beats the far arc at 7.0 cm. Convergence is the test: refine N until the answer stops moving, and keep the coarse values as the evidence that it has. Small N fails because an arc behaves like a point only when it is short compared with its distance to the field point.