 Solution

Neglect end effects, so the electrostatic potential depends only on the cylindrical radius $r$. The radial Laplace equation is $(r{\varphi }^{\prime }{)}^{\prime }=0$, with general solution $A\log r+B$. Imposing the conductor boundary conditions givesThe radial electric field is $E_r=-{\varphi }^{\prime }$. 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 isHence the capacitance per unit length isThe 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 $\delta \to 0^{+}$, $\log \lambda =\delta +O({\delta }^2)$ while $\log (2/\lambda )\to \log 2$. ThereforeThe 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 $u=\log \lambda$, $L=\log 2$, so $0<u<L$. ThenThus the unique stationary point is $u=L/2$. Strict convexity and divergence at both endpoints show it is a global minimum:There is no maximum on $1<\lambda <2$.