= Solution
Let $R_1<R_2$ and put charge per unit length $+\lambda_q$ on the inner shell and $-\lambda_q$ on the facing outer conductor. For a long coaxial <Gaussian surface> of radius $r$ in the gap, <Gauss's law> gives
$$
E(r)\,2\pi rL=\frac{\lambda_qL}{\epsilon_0},\qquad E(r)=\frac{\lambda_q}{2\pi\epsilon_0r}.
$$
The <electric potential difference> between the shells is
$$
V_1-V_2=\int_{R_1}^{R_2}E(r)dr=\frac{\lambda_q}{2\pi\epsilon_0}\log\frac{R_2}{R_1}.
$$
With total inner charge $Q=\lambda_qL$, the <capacitance> is
$$
\boxed{C=\frac{2\pi\epsilon_0L}{\log(R_2/R_1)}}.
$$
This assumes a vacuum or approximately air-filled gap and negligible fringing at the ends. A uniform linear dielectric replaces $\epsilon_0$ by its permittivity. The expression depends on the shell-to-shell potential difference, not on the arbitrary common potential reference.
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