University Physics II · Capacitance and Dielectrics · 5.7

Cylindrical & Spherical Capacitors

Between nested conductors the field is not uniform, so ΔV is an integral rather than E d. Do it once for cylinders and once for spheres, and the capacitance falls out as pure geometry — a logarithm, and a difference of reciprocals.

Open the lab ↓Practice & worked examples ↓

Build the model

Connect the measurement to the mechanism.

Parallel plates hide the general method behind a uniform field. Curve the conductors and ΔV is no longer E d but ∫ₐᵇ E dr. The procedure never changes: put +Q on the inner conductor, use symmetry and Gauss's law to get E(r), integrate radially across to the other conductor, then divide, C = Q/ΔV.

Q cancels every time, which is the point — capacitance is geometry and permittivity, nothing else. For coaxial cylinders E falls as 1/r, the integral is a logarithm, and C/L = 2πε₀/ln(b/a): only the ratio of radii matters, and it matters weakly. For concentric spheres E falls as 1/r², the integral gives 1/a − 1/b, and C = 4πε₀ab/(b−a) depends on both radii rather than on the gap alone.

Squeeze either gap and both collapse to ε₀A/d; open the sphere's outer shell to infinity and C = 4πε₀a remains, the capacitance of a single conductor.

Simple definition: A cylindrical or spherical capacitor is two nested conductors separated by a gap. Its capacitance comes from integrating the non-uniform field between them: C = 2πε₀L/ln(b/a) for coaxial cylinders, C = 4πε₀ab/(b−a) for concentric spheres.
Example: A metre of RG-58 coaxial cable — inner radius 0.405 mm, shield radius 1.475 mm, polyethylene κ ≈ 2.26 — holds about 97 pF. With the same geometry empty, it would hold 43 pF.

Field integration recipeC = Q/ΔV, with ΔV = ∫ₐᵇ E·dr

Gauss's law supplies E(r), the radial integral supplies ΔV, and Q cancels to leave pure geometry.

ΔV in V, Q in C, C in F; integrate radially from the positive conductor outward

Coaxial cylindersE = λ/(2πε₀r), C/L = 2πε₀/ln(b/a)

A 1/r field integrates to a logarithm, so only the ratio b/a survives — and it enters weakly.

λ = Q/L in C m⁻¹; 2πε₀ = 55.6 pF m⁻¹; holds for a < r < b, far from the ends

Concentric spheresE = Q/(4πε₀r²), C = 4πε₀ab/(b−a)

A 1/r² field integrates to 1/a − 1/b, so both radii matter, not the gap b − a alone.

1/(4πε₀) = 8.99 × 10⁹ N m² C⁻²; a and b in m give C in F

Isolated conductor limitC = 4πε₀a as b → ∞

Take the outer shell away and a lone conductor still has capacitance, fixed by its radius.

a = 20 mm gives 2.2 pF; the Earth, a = 6371 km, gives 709 μF

Thin-gap limitC = ε₀A/d, with A = 4πab (spheres) or 2πr*L, r* = (b−a)/ln(b/a)

Both curved geometries collapse to the parallel-plate result when the gap is thin.

d = b − a; exact as written, and A → the conductor area once d ≪ a

Dielectric fillingC(κ) = κ C(vacuum)

Polyethylene, κ ≈ 2.26, lifts a 43.0 pF m⁻¹ coaxial line to 97 pF m⁻¹.

Uniform linear dielectric filling the gap; at fixed Q the field in the gap drops by κ

The recipe: Gauss's law, then a radial integral

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.

Coaxial cylinders: the integral is a logarithm

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.

Concentric spheres: a difference of reciprocals

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.

Two limits worth checking: thin gaps and lone conductors

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.

The field is strongest at the inner conductor

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.

What the derivation assumed

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.

Change one variable at a time

Make the relationship visible.

Interactive model

Inner conductor radius a1.2 mm

Shield inner radius b6.0 mm

Dielectric constant κ1.00

Hold a at 1.2 mm and drag the shield out from 5.0 to 9.0 mm. The gap more than doubles, from 3.8 mm to 7.8 mm, but ΔV climbs only from 25.7 V to 36.2 V and C/L falls from 39.0 to 27.6 pF m⁻¹ — b enters only inside the logarithm.

ln(b/a)1.609

CAPACITANCE C/L34.6 pF m⁻¹

PEAK FIELD ÷ MEAN FIELD2.49 ×

GAP VOLTAGE ΔV28.94 V

Live interpretationln(b/a): 1.609. CAPACITANCE C/L: 34.6 pF m⁻¹. PEAK FIELD ÷ MEAN FIELD: 2.49 ×. GAP VOLTAGE ΔV: 28.94 V

Catch the common trap

Explain before calculating.

An air-filled coaxial cable has inner radius a = 0.50 mm and outer radius b = 2.00 mm, giving 40.1 pF m⁻¹. The shield is rebuilt at b = 4.00 mm with a unchanged. What is the new capacitance per metre?

Choose an answer to test the model.

Practice & worked examples

Reason from the model, then test the result.

EasyA coaxial line has an inner conductor of radius a = 0.50 mm inside a shield of inner radius b = 3.50 mm. Find the capacitance per metre with the gap empty, the capacitance of a 25 m drum of it, and the per-metre figure once the gap is filled with polyethylene, κ = 2.26.

C/L = 2πε₀/ln(b/a), with 2πε₀ = 55.63 pF m⁻¹. Only the ratio b/a enters, so the millimetres cancel without converting: b/a = 3.50/0.50 = 7.00.
ln 7.00 = 1.946, so C/L = (55.63 pF m⁻¹)/1.946 = 28.6 pF m⁻¹.
A 25 m drum is that per-metre value times the length: C = (28.59 pF m⁻¹)(25 m) = 715 pF = 0.715 nF.
Filling the gap with a uniform linear dielectric multiplies the vacuum result by κ: C/L = 2.26 × 28.59 pF m⁻¹ = 64.6 pF m⁻¹.

AnswerC/L = 28.6 pF m⁻¹ empty, giving 715 pF (0.715 nF) for the 25 m drum; 64.6 pF m⁻¹ once filled with polyethylene.

MediumConcentric spheres have inner radius a = 30.0 mm and outer radius b = 50.0 mm with vacuum between them. Find the capacitance. Then show that ε₀A/d reproduces it exactly when A = 4πab, find what a naive A = 4πa² would have given, and find the capacitance the inner sphere keeps once the shell is taken away.

C = 4πε₀ab/(b−a). Work out the geometry factor first: ab = (0.0300 m)(0.0500 m) = 1.50 × 10⁻³ m², and b − a = 0.0200 m, so ab/(b−a) = 0.0750 m.
Divide by 1/(4πε₀) = 8.99 × 10⁹ N m² C⁻²: C = (0.0750 m)/(8.99 × 10⁹) = 8.34 × 10⁻¹² F = 8.34 pF.
Now the parallel-plate form. A = 4πab = 4π(1.50 × 10⁻³ m²) = 1.885 × 10⁻² m² — the area of a sphere of radius √(ab) = 38.7 mm, the geometric mean of the two radii. Then ε₀A/d = (8.854 × 10⁻¹² F m⁻¹)(1.885 × 10⁻² m²)/(0.0200 m) = 8.34 × 10⁻¹² F. Identical, and exact — no thin-gap approximation was used.
Use the inner sphere's own area instead, A = 4πa² = 1.131 × 10⁻² m², and ε₀A/d = (8.854 × 10⁻¹²)(1.131 × 10⁻²)/(0.0200) = 5.01 × 10⁻¹² F = 5.01 pF — exactly 0.600 of the true value, 40.0% low. The gap here is two thirds of a, so which radius you feed in matters.
Take the shell to infinity: 1/b → 0 and C = 4πε₀a = (0.0300 m)/(8.99 × 10⁹) = 3.34 × 10⁻¹² F. A lone conductor still has capacitance.

AnswerC = 8.34 pF with the shell. The exact parallel-plate form with A = 4πab returns the same 8.34 pF, while A = 4πa² gives 5.01 pF, 40.0% low. Isolated, the inner sphere holds 3.34 pF.

HardA coaxial line has a fixed shield inner radius b = 3.00 mm and runs at ΔV = 5.00 kV. Find the peak field at the inner conductor when a = 1.00 mm; find the inner radius that minimises that peak field and the field it achieves; and compare the capacitance per metre of the two designs.

Eliminate the charge. E(a) = λ/(2πε₀a) and ΔV = (λ/2πε₀)ln(b/a), so dividing gives E(a) = ΔV/(a ln(b/a)) — the charge drops out and only the applied voltage and the geometry are left.
At a = 1.00 mm: b/a = 3.00 and ln 3.00 = 1.0986, so E(a) = (5.00 × 10³ V)/[(1.00 × 10⁻³ m)(1.0986)] = 4.55 × 10⁶ V m⁻¹.
Minimise over a at fixed b and fixed ΔV: E(a) is smallest where a ln(b/a) is largest, and d[a ln b − a ln a]/da = ln(b/a) − 1 = 0, so b/a = e and a = b/e = (3.00 mm)/2.7183 = 1.10 mm.
There ln(b/a) = 1 exactly, so E(a) = ΔV/a = e ΔV/b = (2.7183)(5.00 × 10³ V)/(3.00 × 10⁻³ m) = 4.53 × 10⁶ V m⁻¹ — only 0.5% below the a = 1.00 mm value. The optimum is genuinely flat.
Capacitance: C/L = 2πε₀/ln(b/a) = (55.63 pF m⁻¹)/1.0986 = 50.6 pF m⁻¹ at a = 1.00 mm, and (55.63 pF m⁻¹)/1 = 55.6 pF m⁻¹ at the optimum — 9.9% more capacitance for 0.5% less peak field.

AnswerE(a) = 4.55 × 10⁶ V m⁻¹ at a = 1.00 mm. The minimum sits at a = b/e = 1.10 mm with E(a) = 4.53 × 10⁶ V m⁻¹, just 0.5% lower, while C/L rises from 50.6 to 55.6 pF m⁻¹.