= Solution
In <electrostatics>, the <Maxwell equations in matter> are
$$
\nabla\cdot\mathbf D=\rho_f,\qquad \nabla\times\mathbf E=\mathbf0,
\qquad \mathbf D=\epsilon_0\mathbf E+\mathbf P.
$$
Here $\rho_f$ is free <charge density> and $\mathbf P$ is <polarization density>. For a homogeneous isotropic linear <dielectric>, the <constitutive equation> is $\mathbf D=\epsilon_0\epsilon_r\mathbf E$. Integrating the first equation over a thin pillbox straddling an interface, with unit normal $\mathbf n$ from medium 1 to medium 2, gives $\mathbf n\cdot(\mathbf D_2-\mathbf D_1)=\sigma_f$. Integrating the second around a vanishingly narrow rectangular loop gives $\mathbf n\times(\mathbf E_2-\mathbf E_1)=\mathbf0$. These are the <electrostatic interface boundary conditions>. When there is no free surface <electric charge>, the normal <electric displacement field> and tangential <electric field> are continuous; the normal <electric field> generally is not.
Neglect end <fringing field of a capacitor>. In either axial segment of the <coaxial cylindrical capacitor>, a cylindrical <Gaussian surface> gives $D_r=\lambda_f/(2\pi r)$, where $\lambda_f$ is free <electric charge> per length. Integrating the radial <electric field> between the conductors gives
$$
V=\frac{\lambda_f}{2\pi\epsilon_0\epsilon_r}\ln\frac ba,
\qquad C'=\frac{\lambda_f}{V}=\frac{2\pi\epsilon_0\epsilon_r}{\ln(b/a)}.
$$
The liquid and air segments share the same conductor <electric potential difference>, so their capacitances add in parallel. Thus the <partially filled coaxial capacitor> has
$$
\boxed{C(h)=\frac{2\pi\epsilon_0}{\ln(b/a)}[L-h+\epsilon h].}
$$
Here the question's $\epsilon$ is relative <permittivity>, and air's relative <permittivity> is approximated by one.
Write $x$ for the rise above the external level. The wide reservoir keeps that level approximately fixed. At constant applied <voltage>, one must include work done by the source: the appropriate electrical contribution to the mechanical potential is $-C V^2/2$. The upward <dielectric insertion force at fixed voltage> is consequently
$$
F_e=\frac12V^2\frac{dC}{dx}
=\frac{\pi\epsilon_0(\epsilon-1)V^2}{\ln(b/a)}.
$$
The excess <hydrostatic pressure> $\rho gx$ acts over the annular area $A=\pi(b^2-a^2)$. Alternatively its gravitational <potential energy> is $\rho gAx^2/2$, since liquid initially at the reservoir level must be raised through all heights from zero to $x$. Balancing the derivatives gives
$$
\boxed{x=\Delta h=\frac{\epsilon_0(\epsilon-1)V^2}{\rho g(b^2-a^2)\ln(b/a)}.}
$$
This assumes $h+x<L$, a communicating liquid reservoir, negligible capillarity, and a sufficiently long <capacitor> that end effects are small. Differentiating gives the <dielectric-rise voltmeter>'s sensitivity at a bias $V$:
$$
\boxed{\frac{dx}{dV}=\frac{2\epsilon_0(\epsilon-1)V}{\rho g(b^2-a^2)\ln(b/a)}.}
$$
To obtain its numerical value in $\mu\mathrm m\,\mathrm V^{-1}$, multiply the right-hand side in SI units by $10^6$. A convenient parametric scale, with $\beta=b/a$, is
$$
\frac{dx}{dV}=1.805\times10^{-3}\,
\frac{\epsilon-1}{(\beta^2-1)\ln\beta}
\left(\frac V{1\ \mathrm V}\right)
\left(\frac{10^3\ \mathrm{kg\,m^{-3}}}{\rho}\right)
\left(\frac{1\ \mathrm{mm}}a\right)^2
\left(\frac{9.81\ \mathrm{m\,s^{-2}}}g\right)
\ \mu\mathrm m\,\mathrm V^{-1}.
$$
For a concrete estimate, choose $a=1\,\mathrm{mm}$, $b=2\,\mathrm{mm}$, <mass density> $\rho=10^3\,\mathrm{kg\,m^{-3}}$, <relative permittivity> $\epsilon=2$ and operating <voltage> $V=100\,\mathrm V$. These are illustrative choices, not supplied apparatus data. They give
$$
\boxed{\frac{dx}{dV}\simeq0.087\,\mu\mathrm m\,\mathrm V^{-1},\qquad x\simeq4.3\,\mu\mathrm m.}
$$
Thus this millimetre-scale example requires fine optical <displacement> measurement for volt-level changes. The parametric formula states exactly how the estimate changes with the chosen geometry, liquid and bias.
The displacement is quadratic in <voltage>, cannot distinguish its sign, and has zero first-order sensitivity at zero bias. Smaller gaps and larger <relative permittivity> increase sensitivity, but capillarity, leakage, <dielectric breakdown>, temperature-dependent liquid properties, and finite available rise then limit performance. A nonconducting liquid avoids a sustained conduction load, although charging still draws <electric charge> and leakage is possible. A high <voltage> can give a readable displacement; a low-voltage instrument would require fine displacement measurement and careful calibration.
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