Let's consider two coaxial conducting cylinders separated by vacuum. The inner cylinder has linear charge density \(+ \lambda\) and radius \(r_a\) and and the outer cylinder has linear charge density \(-\lambda\) and inner radius \(r_b\).

The electric field due to the outer cylinder has no effect on the electric field between the cylinders and therefore the electric field between the cylinders is only due to the charge in the inner cylinder. And we know the electric field of an infinite line of charge. The same expression also applies to an infinite cylindrical charge distribution; if you enclose the charged conducting cylinder with a Gaussian surface the charge on the cylinder is the same as the charge concentrated along its axis. So the electric field at distance \(r\) between the cylinders is

\[E = k\frac{{2\lambda }}{r}\]

Now the potential difference of inner cylinder \(a\) with respect to the outer cylinder \(b\), \(V_{ab}\) is obtained by integrating the small potential difference \(\vec E \cdot d\vec r\) in a very small displacement \(d\vec r\) along the radial line from \(r_a\) to \(r_b\) (note that the electric field is radially outward so \(\vec E\) and \(d\vec r\) are parallel):

\[{V_{ab}} = \int\limits_{{r_a}}^{{r_b}} {E \cdot dr} = 2k\lambda \int\limits_{{r_a}}^{{r_b}} {\frac{{dr}}{r}} = 2k\lambda \ln \left( {\frac{{{r_b}}}{{{r_a}}}} \right)\]

If \(L\) is the length of the cylinders, \(\lambda = Q/L\) and \(C = Q/V_{ab}\) is

\[C = \frac{Q}{{{V_{ab}}}} = \frac{Q}{{2k(Q/L)\ln ({r_b}/{r_a})}} = \frac{L}{{2k\ln ({r_b}/{r_a})}}\]

As you can see from the above expression that the capacitance is proportional to the length of the cylinders and also depends on the radii \(r_a\) and \(r_b\).