Result

Sphere and cylinder

(E4.2.2) E4.2 · Calculating Capacitance →

C_{sphere} = \frac{R}{k _{e}}, C_{cyl} = \frac{ℓ}{2 k _{e} ln ( b / a )}

Statement

An isolated sphere has capacitance set purely by its radius; a coaxial cable is a cylindrical capacitor.

Symbols and units

$R$	radius of an isolated sphere	m
$a, b$	inner and outer radii of the cylinders	m
$ℓ$	length of the cable	m
$k_{e}$	Coulomb constant	8.99×10⁹ N·m²/C² (=1/4πε₀)
$C_{sphere}, C_{cyl}$	capacitance of the sphere, of the cylinder	F

When it applies

Isolated sphere, or a coaxial pair long enough to ignore its ends.

Don't use when

The isolated-sphere and coaxial-pair formulas assume no other nearby conductors — a nearby grounded object or second conductor changes the capacitance from this ideal-isolation value.

Derivation

Sphere: the potential of an isolated conductor carrying Q is

V = k_{e} \frac{Q}{R}

Divide out Q:

C = \frac{Q}{V} = \frac{R}{k _{e}} = 4 π ε_{0} R

Cylinder: from the long-line result, the field between inner radius a and outer b is

E = \frac{2 k _{e} λ}{r}

Integrate across the gap — this is where the logarithm appears:

Δ V = \int_{a}^{b} \frac{2 k _{e} λ}{r} d r = 2 k_{e} λ ln \frac{b}{a}

With Q = λℓ, divide:

C = \frac{λ ℓ}{2 k _{e} λ ln ( b / a )} = \frac{ℓ}{2 k _{e} ln ( b / a )}

Worked examples

Coaxial cable, per metre

a = 0.5 mm, b = 1.75 mm:

C = \frac{1}{2 ( 8.99 \times 1 0 ^{9} ) ln 3.5} \approx 44 pF/m

A typical datasheet figure, and the reason long cables load high-frequency signals.

Watch out for

The Earth, radius 6.4 × 10⁶ m, has a capacitance of only about 700 µF. Farads really are huge units.

Connects to

Long line of charge Charged sphere Parallel plates