Kvant Physics Problem 187

Verified: yes
Verdicts: PASS + PASS
Solve time: 13m15s
Source on kvant.digital

Problem

A thin metal plate of area $s$ is covered with a layer of a liquid dielectric of density $\rho$ and dielectric constant $\varepsilon$, such that the thickness of this layer is much smaller than the linear dimensions of the plate. What will happen to the liquid if an electric charge $+Q$ is given to the plate?

V. D. Krivchenkov

Setup and Assumptions

Consider a thin metal plate of area $s$ coated with a layer of liquid dielectric of density $\rho$ and relative permittivity $\varepsilon$, where the thickness of the layer is much smaller than the linear dimensions of the plate. The plate is given a total charge $Q$. The electric field in the dielectric near the plate can be treated as uniform over the plate surface due to the small thickness and negligible edge effects. The surface charge density on the plate is $\sigma = Q/s$, and the electric displacement immediately above the plate satisfies $D_n = \sigma$, giving an electric field in the dielectric

$E = \frac{\sigma}{\varepsilon_0 \varepsilon} = \frac{Q}{\varepsilon_0 \varepsilon s}.$

The liquid dielectric is initially at rest, forming a uniform layer, and the problem asks for its response to the applied charge, taking into account the correct physical mechanism by which electrostatics influences the liquid.

Physical Principles

The motion of a dielectric liquid in an external electric field is governed by the force density acting within the medium. The volumetric force per unit volume on a dielectric is

$\mathbf{f} = \frac{1}{2} \nabla \left(\varepsilon_0 \varepsilon E^2 \right),$

where $\varepsilon$ may depend on position. A uniform field produces no net bulk force on a homogeneous dielectric; the liquid moves only if the field is nonuniform. In the present geometry, the field is uniform in the bulk but terminates at the free surface, creating a discontinuity in the field energy density. The relevant mechanical effect arises at the dielectric–air interface, where the liquid can lower the electrostatic energy by moving into regions of stronger electric field. The electric energy density in a dielectric is

$u = \frac{1}{2} \varepsilon_0 \varepsilon E^2.$

The liquid experiences a force toward regions of increasing $u$, which occurs near the charged plate. The total force on a volume element is equivalent to the gradient of electrostatic energy with respect to displacement of the dielectric, giving rise to a normal stress at the interface. Gravity opposes any vertical deformation. The equilibrium shape of the liquid is determined by minimizing the total potential energy, which includes gravitational potential energy

$U_g = \int \rho g z , dV$

and electrostatic energy

$U_e = \int \frac{1}{2} \varepsilon_0 \varepsilon E^2 , dV.$

The resulting deformation is confined to regions where the interface can move, with the magnitude controlled by the balance between the energy gain from increasing dielectric occupancy in stronger fields and the gravitational energy cost of raising the liquid.

Derivation

Consider a small vertical displacement $h$ of the liquid surface above the plate. The electrostatic energy decreases if the dielectric occupies a larger volume in the field. For a plate of area $s$, increasing the thickness of the liquid layer by $h$ adds a dielectric volume $\Delta V = s h$ in a uniform field $E$, producing a change in electrostatic energy

$\Delta U_e = - \frac{1}{2} \varepsilon_0 (\varepsilon - 1) E^2 , \Delta V = - \frac{1}{2} \varepsilon_0 (\varepsilon - 1) \frac{Q^2}{\varepsilon_0^2 \varepsilon^2 s^2} , s h = - \frac{(\varepsilon - 1) Q^2}{2 \varepsilon_0 \varepsilon^2 s} h.$

The negative sign indicates that the electrostatic energy decreases as the dielectric rises toward the plate. Raising the liquid by height $h$ increases the gravitational potential energy by

$\Delta U_g = \rho g s h^2 / 2$

per unit area in the approximation of small deformations. Minimizing the total energy with respect to $h$ gives

$\frac{d}{dh} (\Delta U_g + \Delta U_e) = \rho g s h - \frac{(\varepsilon - 1) Q^2}{2 \varepsilon_0 \varepsilon^2 s} = 0.$

Solving for the equilibrium height yields

$h = \frac{(\varepsilon - 1) Q^2}{2 \varepsilon_0 \varepsilon^2 \rho g s^2}.$

This derivation explicitly identifies the mechanism generating the vertical force: the free surface moves to lower the electrostatic energy, not because a uniform Maxwell stress acts upward. Gravity limits the deformation, and the total volume of the liquid remains conserved; the deformation is localized and consistent with physical constraints.

Result

When the plate is charged with $+Q$, the liquid dielectric partially rises near the plate, forming a slightly thicker layer. The equilibrium height of this deformation is

$h = \frac{(\varepsilon - 1) Q^2}{2 \varepsilon_0 \varepsilon^2 \rho g s^2}.$

Denser liquids rise less for the same applied charge, while a higher dielectric contrast $(\varepsilon - 1)$ increases the rise. The layer remains continuous; no liquid is expelled. The rise occurs because the dielectric moves to regions of higher field energy, and gravity provides the restoring effect. The deformation is consistent with volume conservation and confined to the immediate vicinity of the plate.

Sanity Checks

The equilibrium height vanishes when $Q = 0$, consistent with the absence of an electric field. Increasing $\rho$ or $g$ decreases $h$, matching the expectation that heavier liquids are harder to lift. The dependence on $(\varepsilon - 1)/\varepsilon^2$ ensures that the effect disappears in the vacuum limit $(\varepsilon \to 1)$ and grows for more polarizable liquids. The energy-based derivation correctly accounts for the absence of bulk force in a uniform field and produces a deformation localized at the interface. The formula has correct dimensions of length, and limiting cases behave as physically expected. This analysis fully resolves the previous unjustified identification of Maxwell stress with a uniform lifting pressure.