= Solution
Introduce cylinder length $\ell$ and line charge density $\lambda=Q/\ell$; only charge per length matters in the long-cylinder approximation. A coaxial cylindrical <Gaussian surface> gives $E_r2\pi r\ell=Q/\epsilon_0$. Choosing the outer cylinder's <electric potential> to be zero,
$$
\boxed{\mathbf E(r)=\frac{\lambda}{2\pi\epsilon_0r}\hat{\mathbf r},\quad V(r)=\frac{\lambda}{2\pi\epsilon_0}\ln\frac br,\quad a<r<b.}
$$
The <capacitance per unit length> follows from $C=Q/[V(a)-V(b)]$:
$$
\boxed{\frac C\ell=\frac{2\pi\epsilon_0}{\ln(b/a)}.}
$$
If $Q$ is intended as a line charge rather than a finite total charge, replace $\lambda$ by that $Q$. Neglecting end effects, equal opposite cylinder charges make the exterior <electric field> vanish; the potential difference between the conductors remains the quantity defining their <capacitance>.
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