= Solution
For this <radially layered coaxial cable>, define
$$
L_1=\ln(c/a),\qquad L_2=\ln(b/c),\qquad
\Delta=\epsilon'L_1+\epsilon L_2,\qquad
C=\frac{V\epsilon\epsilon'}{\Delta}.
$$
Use the solution regular at the axis with bounded exterior <electric potential>, so that the empty interior and exterior have the constant <electric potentials> fixed by their adjacent conducting cylinders.
There is no <free charge> in either <electrical insulator> or on their interface. <Gauss's law> and continuity of normal <electric displacement field> give $D_\rho=C/\rho$ in both layers. The <electric potential difference> fixes the constant because
$$
V=\int_a^bE_\rho\,d\rho=C\left(\frac{L_1}{\epsilon}+\frac{L_2}{\epsilon'}\right).
$$
The <electric potential>, including the empty regions bounded by the conducting cylinders, is therefore
$$
\boxed{\phi(\rho)=\begin{cases}
V,&0\leq\rho\leq a,\\[2pt]
\displaystyle\frac V\Delta\left[\epsilon'\ln(c/\rho)+\epsilon\ln(b/c)\right],&a<\rho<c,\\[2pt]
\displaystyle\frac{V\epsilon}{\Delta}\ln(b/\rho),&c\leq\rho<b,\\[2pt]
0,&\rho\geq b.
\end{cases}}
$$
Both inner and outer <Dirichlet boundary conditions> hold, and the <electric potential> is continuous at the material interface, where
$$
\boxed{\phi(c)=\frac{V\epsilon\ln(b/c)}{\epsilon'\ln(c/a)+\epsilon\ln(b/c)}.}
$$
Taking the negative <gradient> of the <electric potential> yields
$$
\boxed{\mathbf E(\rho)=\begin{cases}
0,&\rho<a\ \text{or}\ \rho>b,\\[2pt]
\displaystyle\frac{V\epsilon'}{\Delta\rho}\mathbf e_\rho,&a<\rho<c,\\[2pt]
\displaystyle\frac{V\epsilon}{\Delta\rho}\mathbf e_\rho,&c<\rho<b.
\end{cases}}
$$
The radial <electric field> generally jumps at $\rho=c$, but $\epsilon E_\rho(c^-)=\epsilon'E_\rho(c^+)$, as required by the absence of free <surface charge density> there. The conducting cylinders carry opposite <linear charge densities> $\lambda=2\pi C$ and $-\lambda$.
For an <Amperian loop> of radius $\rho$, the enclosed <electric current> is zero, $I$, or zero according as $\rho<a$, $a<\rho<b$, or $\rho>b$. The <magnetostatic Ampère law> therefore gives $H_\varphi=I/(2\pi\rho)$ in the gap, and zero outside. Using the two <magnetic permeabilities>,
$$
\boxed{\mathbf B(\rho)=\begin{cases}
0,&\rho<a\ \text{or}\ \rho>b,\\[2pt]
\displaystyle\frac{\mu I}{2\pi\rho}\mathbf e_\varphi,&a<\rho<c,\\[2pt]
\displaystyle\frac{\mu'I}{2\pi\rho}\mathbf e_\varphi,&c<\rho<b.
\end{cases}}
$$
The tangential <magnetic field strength> is continuous at the material interface, while the tangential <magnetic field> need not be. At each ideal conducting surface and at the material interface, the displayed <electric field> and <magnetic field> formulas specify one-sided limits; a single interface value is unnecessary when the limiting values differ.
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