= Solution
Consider a long uniform length $\ell$ of a <coaxial cylindrical capacitor>, with permittivity $\epsilon_0\epsilon_r$ and inner-cylinder charge per unit length $\lambda$. A cylindrical <Gaussian surface> gives
$$
D_r=\frac{\lambda}{2\pi r},\qquad
E_r=\frac{\lambda}{2\pi\epsilon_0\epsilon_r r},\qquad a<r<b.
$$
The <electric potential difference> is
$$
V=\int_a^bE_r\,dr=\frac{\lambda\ln(b/a)}{2\pi\epsilon_0\epsilon_r}.
$$
Thus its <capacitance> per unit length is $c(\epsilon_r)=2\pi\epsilon_0\epsilon_r/\ln(b/a)$.
In the <partially filled coaxial capacitor>, the vacuum section of length $L-x$ and the dielectric section of length $x$ share the same two conductors and hence the same <electric potential difference>. They are electrically in parallel, not in series. The radial <electric field> is tangent to the axial dielectric interface, consistent with the boundary condition in part (b). Adding their charges gives
$$
\boxed{C=c(1)(L-x)+c(\epsilon)x
=\frac{2\pi\epsilon_0}{\ln(b/a)}[L-x+\epsilon x].}
$$
This is the leading long-capacitor expression, neglecting end and interface-fringing corrections. The stated $x\gg b$ supports a long filled section; treating the other section as uniform also requires it to be well away from the end regions when those corrections matter.
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