= Solution
Define $V=\phi(r_2)-\phi(r_1)$ and $\eta=\ln(r_2/r_1)>0$. Cylindrical <symmetry> and <Gauss's law> imply a radial <electric field> proportional to $1/r$. Integrating $\mathbf E=-\nabla\phi$ fixes the sign:
$$
\phi(r)-\phi(r_1)=\frac{V\ln(r/r_1)}\eta,\qquad
\boxed{\mathbf E(r)=-\frac{V}{\eta r}\hat{\mathbf r}.}
$$
If the inner cylinder has <linear charge density> $\lambda_q$, a coaxial <Gaussian surface> gives $2\pi r E_r=\lambda_q/\epsilon_0$, so $\lambda_q=-2\pi\epsilon_0V/\eta$. <Capacitance> is positive regardless of the chosen terminal polarity:
$$
\boxed{C'=\frac{|\lambda_q|}{|V|}=\frac{2\pi\epsilon_0}{\ln(r_2/r_1)}.}
$$
End effects are neglected for the long cylinders, as required by the cylindrical approximation.
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