Capacitance of a cylindrical capacitor

Problem

Challenge: capacitance of a cylindrical capacitor. A cylindrical capacitor has two coaxial plates of radii $r_a = 1\;\text{mm}$ (inner) and $r_b = 5\;\text{mm}$ (outer), length $L = 0{,}3\;\text{m}$. Calculate the capacitance by integrating $\vec E$ obtained from Gauss’s theorem on a coaxial cylinder. This is the typical geometry of a coaxial cable.

Solution

Field between the plates (Gauss’s theorem). Let us place charge $+Q$ on the inner plate, with linear density $\lambda = Q/L$. As the Gaussian surface we choose a coaxial cylinder of radius $r$ (with $r_a < r < r_b$) and length $L$. By symmetry the field is radial and uniform over the lateral surface; the end caps do not contribute to the flux. Hence: $E\cdot 2\pi r L = \frac{Q}{\varepsilon_0} \quad\Longrightarrow\quad E(r) = \frac{Q}{2\pi\varepsilon_0 L}\,\frac{1}{r}.$

Potential difference (integration). We integrate the field from the inner plate to the outer one: $V = \int_{r_a}^{r_b} E\,\dd r = \frac{Q}{2\pi\varepsilon_0 L}\int_{r_a}^{r_b}\frac{\dd r}{r} = \frac{Q}{2\pi\varepsilon_0 L}\ln\frac{r_b}{r_a}.$

Capacitance. By definition $C = Q/V$, and the charge $Q$ cancels out: $C = \frac{2\pi\varepsilon_0 L}{\ln(r_b/r_a)}.$

Numerical calculation. With $r_b/r_a = 5$ and $\ln 5 \approx 1{,}609$: $C = \frac{2\pi\cdot 8{,}85\cdot 10^{-12}\cdot 0{,}3}{1{,}609} = \frac{1{,}668\cdot 10^{-11}}{1{,}609} \approx \ev{C \approx 10{,}4\;\text{pF}}.$ A small value (of order picofarads), typical of a length of coaxial cable: the capacitance grows with the length $L$ and decreases logarithmically as the ratio between the radii increases.

Links

Topics: Campo elettrico e potenziale Concepts: Capacità e condensatore · Teorema di Gauss Skills: Applicazione del teorema di Gauss Methods: Teorema di Gauss Objects: Filo rettilineo infinito