= Solution
Neglect end effects, so the <electrostatic potential> depends only on the cylindrical radius $r$. The radial <Laplace equation> is $(r\phi')'=0$, with general solution $A\log r+B$. Imposing the conductor <boundary conditions> gives
$$
\boxed{\phi(r)=\begin{cases}\displaystyle V\frac{\log(r/a)}{\log\lambda},&a<r<\lambda a,\\[6pt]\displaystyle V\frac{\log(2a/r)}{\log(2/\lambda)},&\lambda a<r<2a.\end{cases}}
$$
The radial <electric field> is $E_r=-\phi'$. Just inside the middle cylinder it is $-V/[\lambda a\log\lambda]$, directed towards the inner cylinder; just outside it is $V/[\lambda a\log(2/\lambda)]$, directed towards the outer cylinder. By <Gauss's law>, its <charge per unit length> is
$$
q'=2\pi\epsilon_0\lambda a\bigl(E_r(\lambda a+)-E_r(\lambda a-)\bigr)=2\pi\epsilon_0V\left(\frac1{\log\lambda}+\frac1{\log(2/\lambda)}\right).
$$
Hence the <capacitance per unit length> is
$$
\boxed{C(\lambda)=2\pi\epsilon_0\left(\frac1{\log\lambda}+\frac1{\log(2/\lambda)}\right).}
$$
The two <coaxial cylindrical capacitors> contribute in parallel because both connect the middle cylinder to the same grounded potential. This is the <coaxial capacitor with grounded inner and outer cylinders> construction.
For $\lambda=1+\delta$ with $\delta\to0^+$, $\log\lambda=\delta+O(\delta^2)$ while $\log(2/\lambda)\to\log2$. Therefore
$$
\boxed{C(1+\delta)\sim\frac{2\pi\epsilon_0}{\delta}.}
$$
The inner gap has width $a\delta$ and area per unit length approximately $2\pi a$. Its <parallel-plate capacitor> approximation gives exactly the same leading result; the outer gap contributes only a bounded correction.
For the <extremum>, set $u=\log\lambda$, $L=\log2$, so $0<u<L$. Then
$$
\frac{C}{2\pi\epsilon_0}=\frac1u+\frac1{L-u},\qquad
\frac{d}{du}\frac{C}{2\pi\epsilon_0}=-\frac1{u^2}+\frac1{(L-u)^2},\qquad
\frac{d^2}{du^2}\frac{C}{2\pi\epsilon_0}=\frac2{u^3}+\frac2{(L-u)^3}>0.
$$
Thus the unique stationary point is $u=L/2$. Strict <convexity> and divergence at both endpoints show it is a global minimum:
$$
\boxed{\lambda=\sqrt2,\qquad C_{\min}=\frac{8\pi\epsilon_0}{\log2}.}
$$
There is no maximum on $1<\lambda<2$.
Back to article page
