Kvant Physics Problem 187

A thin horizontal metallic plate of area $s$ carries a charge $+Q$, so the free surface charge density is $\sigma = \dfrac{Q}{s}$ in $\mathrm{C,m^{-2}}$.

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Problem

A thin metallic plate of area $s$ is covered with a layer of a liquid dielectric with 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 the plate is given an electric charge $+Q$?

V. D. Krivchenkov

Setup and Assumptions

A thin horizontal metallic plate of area $s$ carries a charge $+Q$, so the free surface charge density is $\sigma = \dfrac{Q}{s}$ in $\mathrm{C,m^{-2}}$. The plate is covered by a liquid dielectric of density $\rho$ in $\mathrm{kg,m^{-3}}$ and relative permittivity $\varepsilon$. The liquid forms a layer of thickness $h$ that is much smaller than the characteristic linear size of the plate, so edge effects are neglected and all fields are treated as locally one-dimensional and normal to the surface.

The region above the liquid is air, treated as vacuum with permittivity $\varepsilon_0$. The system is considered in electrostatic equilibrium. The liquid is assumed incompressible, nonconducting, and able to move without viscous constraints so that a mechanical equilibrium profile can be established. Gravity acts downward with acceleration $g$.

The unknown is the change in the liquid configuration, specifically the equilibrium thickness $h$ of the dielectric layer on the charged plate and the direction of motion of the liquid under the influence of the electric field.

Physical Principles

Gauss’s law for the electric displacement field in matter is used in the form

$\nabla \cdot \mathbf{D} = \rho_{\mathrm{free}},$

with the boundary condition that the normal component of $\mathbf{D}$ has a discontinuity equal to the free surface charge density.

The constitutive relation in a linear isotropic dielectric is

$\mathbf{D} = \varepsilon_0 \varepsilon \mathbf{E}$

inside the liquid and $\mathbf{D} = \varepsilon_0 \mathbf{E}$ in air.

The electrostatic energy density expressed in terms of $\mathbf{D}$ is

$u = \frac{1}{2}\mathbf{E}\cdot\mathbf{D} = \frac{D^2}{2\varepsilon_0 \varepsilon}$

in the dielectric and

$u = \frac{D^2}{2\varepsilon_0}$

in air.

The mechanical effect of the field on the dielectric is obtained from the change of electrostatic energy with respect to the volume occupied by the dielectric. The resulting effective pressure is

$p = -\frac{\partial U}{\partial V},$

evaluated at fixed free charge.

Hydrostatic equilibrium in the liquid requires

$p_{\mathrm{electric}} = \rho g h.$

Derivation

The charged plate fixes the normal component of the displacement field in the surrounding space. Since the plate is a conductor carrying total charge $Q$, the free surface charge density is uniform and equals $\sigma = Q/s$. Gauss’s law applied to a pillbox straddling the conductor surface gives

$D_n = \sigma.$

Inside the dielectric layer the magnitude of the electric field is therefore

$E_{\mathrm{d}} = \frac{D}{\varepsilon_0 \varepsilon} = \frac{\sigma}{\varepsilon_0 \varepsilon},$

while in air above the liquid

$E_{\mathrm{a}} = \frac{D}{\varepsilon_0} = \frac{\sigma}{\varepsilon_0}.$

The electrostatic energy density in the dielectric is

$u_{\mathrm{d}} = \frac{D^2}{2\varepsilon_0 \varepsilon},$

and in air it is

$u_{\mathrm{a}} = \frac{D^2}{2\varepsilon_0}.$

Replacing a layer of air by dielectric of the same volume changes the energy density by

$\Delta u = u_{\mathrm{d}} - u_{\mathrm{a}} = \frac{D^2}{2\varepsilon_0}\left(\frac{1}{\varepsilon} - 1\right).$

This decrease in energy corresponds to a mechanical pressure pulling the dielectric into the region of stronger field. The magnitude of the effective pressure acting on the liquid surface follows from the energy change per unit volume taken with opposite sign,

$p = -\Delta u = \frac{D^2}{2\varepsilon_0}\left(1 - \frac{1}{\varepsilon}\right).$

Substituting $D = \sigma$ gives

$p = \frac{\sigma^2}{2\varepsilon_0}\left(1 - \frac{1}{\varepsilon}\right).$

With $\sigma = Q/s$, this becomes

$p = \frac{Q^2}{2\varepsilon_0 s^2}\left(1 - \frac{1}{\varepsilon}\right).$

The liquid rises along the plate until the electric pressure is balanced by the hydrostatic pressure of the elevated column,

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

Solving for $h$ gives

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

Result

The dielectric liquid is drawn toward and spreads over the charged plate, forming a thicker layer whose equilibrium thickness is

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

All quantities carry SI units: $Q$ in $\mathrm{C}$, $s$ in $\mathrm{m^2}$, $\varepsilon_0$ in $\mathrm{F,m^{-1}}$, $\rho$ in $\mathrm{kg,m^{-3}}$, $g$ in $\mathrm{m,s^{-2}}$. The resulting $h$ is in $\mathrm{m}$.

Sanity Checks

Dimensional consistency follows from $\sigma^2/\varepsilon_0$ having units $\mathrm{C^2,m^{-4}} / \mathrm{F,m^{-1}} = \mathrm{N,m^{-2}}$, which is pressure, and division by $\rho g$ yields length in meters.

In the limit $\varepsilon \to 1$, the pressure vanishes and $h \to 0$, corresponding to the disappearance of any dielectric preference relative to air. In the limit $\varepsilon \to \infty$, the pressure approaches $\sigma^2/(2\varepsilon_0)$, giving the maximal attraction expected for a perfect dielectric.

The expression shows quadratic growth with charge $Q$, consistent with electrostatic pressure scaling as field squared, since $E \sim \sigma$. The most sensitive point of failure in the derivation lies in treating the displacement field as uniform and equal to $\sigma$; any neglect of boundary conditions at the dielectric-air interface would incorrectly modify the factor $\left(1 - \frac{1}{\varepsilon}\right)$ and change the predicted pressure.