Cylindrical & Spherical Capacitors

Every capacitance calculation is the same four steps, and the parallel-plate case only makes them look easy. Put charge +Q on the inner conductor; the outer one carries −Q on its inner surface, so the field lives entirely in the gap and vanishes outside. Use the symmetry to choose a Gaussian surface on which E is constant and radial, and read off E(r). Integrate along a radial path from the inner conductor to the outer, ΔV = ∫ₐᵇ E dr — the path is free, so pick the radial one and the dot product becomes a plain product. Divide: C = Q/ΔV. Because E is proportional to Q, the charge cancels at that last step, which is why capacitance never depends on how much charge you put on. What survives is geometry and ε₀. For plates the field is uniform and the integral is E d; for nested cylinders and spheres it is not, and skipping the integral is the single most common way to get these two results wrong.

Take two long coaxial conductors of length L, inner radius a and outer radius b, with +Q spread along the inner one. A coaxial Gaussian cylinder of radius r and length ℓ encloses λℓ, with λ = Q/L, and E is radial and constant on its curved side, so E(2πrℓ) = λℓ/ε₀ and E = λ/(2πε₀r). Integrating outward, ΔV = (λ/2πε₀)∫ₐᵇ dr/r = (λ/2πε₀) ln(b/a). Then C = Q/ΔV = 2πε₀L/ln(b/a), or per unit length C/L = 2πε₀/ln(b/a), with 2πε₀ = 55.6 pF m⁻¹. Only the ratio b/a appears, so scaling both radii by the same factor leaves the capacitance per metre untouched. RG-58 cable has a ≈ 0.405 mm and b ≈ 1.475 mm, giving ln(b/a) = 1.293 and 43.0 pF m⁻¹ if the gap were empty; the polyethylene filling, κ ≈ 2.26, lifts that to 97 pF m⁻¹, near the 100 pF m⁻¹ on the datasheet.

Now nest two spheres, radii a and b, with +Q on the inner. A spherical Gaussian surface of radius r inside the gap encloses Q, so E(4πr²) = Q/ε₀ and E = Q/(4πε₀r²) — the field a point charge would make, because the shell outside contributes nothing within itself. Integrating, ΔV = (Q/4πε₀)∫ₐᵇ dr/r² = (Q/4πε₀)(1/a − 1/b), and dividing out Q leaves C = 4πε₀ab/(b−a). Unlike the coaxial result this is not a function of b/a alone: a length is built in through the product ab. Take a = 20.0 mm and b = 25.0 mm. Then ab/(b−a) = (5.00 × 10⁻⁴ m²)/(5.00 × 10⁻³ m) = 0.100 m, so C = 0.100/(8.99 × 10⁹) = 11.1 pF. Note how small that is: nested spheres of laboratory size store picofarads, which is why practical capacitors buy their capacitance with large areas at tiny separations instead.

Test every derived formula where you already know the answer. Let the gap d = b − a become small next to the radii. For the sphere, C = 4πε₀ab/d = ε₀(4πab)/d, and 4πab is exactly the surface area of a sphere of radius √(ab), the geometric mean of the two radii — so the spherical capacitor is a parallel-plate capacitor of that area, with no approximation at all. For the coax, write C = ε₀(2πr*L)/d with r* = (b−a)/ln(b/a), the logarithmic mean radius; the RG-58 numbers give r* = 0.828 mm, between a and b as it must be. In both cases the mean radius tends to the common radius as d shrinks, and C → ε₀A/d. Push the other way instead: send b → ∞ in the spherical result and C = 4πε₀a survives. A single isolated conductor has capacitance. A 20 mm sphere alone holds 2.2 pF against 11.1 pF with a shell at 25 mm, and the Earth, radius 6371 km, has 4πε₀a = 709 μF.

Because E falls as 1/r or 1/r², the largest field in either device sits on the inner conductor's surface, and that is where the insulation fails first. For the coax, E(a) = λ/(2πε₀a), and eliminating λ using ΔV gives E(a) = ΔV/(a ln(b/a)) — worth reading slowly, because a appears twice with opposite effects. Thinning the inner wire shrinks a but grows ln(b/a), so at fixed ΔV and fixed b the peak field is smallest where d[a ln(b/a)]/da = ln(b/a) − 1 = 0, that is b/a = e = 2.718, giving E(a) = e ΔV/b. The maximum is flat, though: RG-58's b/a = 3.64 sits well off that optimum, yet its peak field is only 3.7% above the best available, 1.91 kV m⁻¹ per applied volt against 1.84. Spheres behave the same way, with E(a) = ΔV b/(a(b − a)), minimised at a = b/2 where E(a) = 4ΔV/b. Converting those peak fields into a voltage rating is the next lesson's business.

Four assumptions carry these results. Symmetry: the inner conductor must be concentric, and a wire pushed off centre raises the field where the gap is narrowest while barely changing C, so eccentricity costs breakdown margin rather than capacitance. Infinite length: the coaxial answer ignores the ends, where field lines bulge outward, so quote it per metre and trust it for L ≫ b. A complete outer conductor: the induced −Q on the shell is what confines the field to the gap, and breaking or removing the shell drops the capacitance towards the isolated-conductor value. And a uniform linear dielectric filling the whole gap, so ε₀ → κε₀ everywhere; a partly filled gap needs series reasoning, and κ drifts with frequency and temperature. One dividend of the same logarithm: a coaxial line's characteristic impedance is (60 Ω/√κ) ln(b/a), which for RG-58 gives 52 Ω — the reason nominal 50 Ω cable has b/a near 3.5.