IPhO 1972 · Theory problem 3

Dielectric liquid rising between the plates of a charged capacitor

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A plane capacitor with rectangular plates is fixed in a vertical position having the lower part in contact with a dielectric liquid (fig. 3.1)
Determine the height, \(h\), of the liquid between the plates and explain the phenomenon.
The capillarity effects are neglected.
It is supposed that the distance between the plates is much smaller than the linear dimensions of the plates.

Figure 1

Fig. 3.1

It is known: the initial intensity of the electric field of the charged capacitor, \(E\), the density \(\rho\), the relative electric permittivity \(\varepsilon_r\) of the liquid, and the height \(H\) of the plates of the capacitor.
Discussion.

Official English text of the problem.

Answers

  • Problem 3 — Height: Assuming that the charged capacitor is isolated, put \(a=\varepsilon_r-1>0\). Let \(h_*\) be the unique nonnegative solution of \[ \boxed{h_*\left(1+a\frac{h_*}{H}\right)^2 =\frac{\varepsilon_0aE^2}{2\rho g}.} \] The equilibrium height is \(h=h_*\) when \(h_*<H\). The liquid fills the gap to the top, \(h=H\) in the ideal approximation, when \[ \boxed{E\ge E_c =\varepsilon_r\sqrt{\frac{2\rho gH}{\varepsilon_0(\varepsilon_r-1)}}.} \]

  • Problem 3 — Explanation and discussion: Polarization produces an upward electric force. At a partial-rise equilibrium, its pressure balances the hydrostatic pressure: \[ \boxed{\rho gh=\frac12\varepsilon_0(\varepsilon_r-1) \left[\frac{E}{1+(\varepsilon_r-1)h/H}\right]^2.} \] This equilibrium is unique and stable. There is no rise if \(E=0\) or \(\varepsilon_r=1\).

  • Problem 3 — Fixed-voltage alternative: If a source maintains the initial voltage instead, the result is \[ \boxed{h=\min\!\left\{H,\frac{\varepsilon_0(\varepsilon_r-1)E^2}{2\rho g}\right\}.} \]

Solution

Problem 3. The reference to the initial field, together with the absence of a specified voltage source, suggests an isolated capacitor, whose free charge remains constant. We first solve this case.

Let the plate width be \(b\), their separation be \(d\), and the external liquid surface be at \(z=0\). We treat the reservoir as sufficiently large that this level remains fixed. The liquid is a linear, homogeneous, incompressible dielectric; air has permittivity \(\varepsilon_0\). Fringing corrections to capacitance are neglected because \(d\) is much smaller than the plate dimensions.

Capacitance and field during the rise. When the liquid has risen to height \(h<H\), the liquid-filled and air-filled portions form two capacitors in parallel. Thus \[ C(h)=\frac{\varepsilon_0b(H-h)}{d} +\frac{\varepsilon_0\varepsilon_rbh}{d} =C_0\left(1+a\frac hH\right), \] where \[ C_0=\frac{\varepsilon_0bH}{d}, \qquad a=\varepsilon_r-1. \]

Initially \(V_0=Ed\), so the conserved charge is \[ Q=C_0Ed. \] Consequently, the field after the rise is \[ \mathcal E(h)=\frac{Q}{C(h)d} =\boxed{\frac{E}{1+ah/H}}. \] The field therefore decreases as liquid enters the isolated capacitor.

Equilibrium height. At fixed charge, the electric energy is \[ U_{\mathrm e}(h)=\frac{Q^2}{2C(h)} =\frac{\varepsilon_0bdHE^2}{2(1+ah/H)}. \]

The gravitational energy required to raise liquid from the reservoir into the gap is \[ U_{\mathrm g}(h) =\int_0^h \rho bd\,g z\,dz =\frac12\rho bdg h^2. \]

A static equilibrium minimizes the total energy. Hence \[ 0=\frac{d}{dh}(U_{\mathrm e}+U_{\mathrm g}) =bd\left[ \rho gh-\frac{\varepsilon_0aE^2}{2(1+ah/H)^2} \right]. \] Therefore \[ \boxed{\rho gh =\frac{\varepsilon_0aE^2}{2(1+ah/H)^2}}, \] or, equivalently, \[ \boxed{ h\left(1+a\frac hH\right)^2 =\frac{\varepsilon_0aE^2}{2\rho g}. } \]

This cubic equation determines the equilibrium height from the supplied quantities. Both \(b\) and \(d\) cancel, explaining why their values are unnecessary.

Physical explanation. Inserting dielectric increases the capacitance and lowers the electric energy at fixed charge. The corresponding upward force is \[ F_{\mathrm e} =-\left.\frac{dU_{\mathrm e}}{dh}\right|_Q =\frac12\varepsilon_0a\,bd\,\mathcal E(h)^2. \] Its electric pressure is therefore \[ p_{\mathrm e}=\frac{F_{\mathrm e}}{bd} =\frac12\varepsilon_0a\mathcal E(h)^2. \] The liquid rises until this pressure balances the hydrostatic pressure \(\rho gh\) associated with the elevated column.

Equating the initial and final total energies would generally describe a possible turning point during motion, rather than the static equilibrium. The equilibrium condition is the vanishing energy derivative used above.

Discussion.

For \(a>0\), define \[ f(h)=h\left(1+a\frac hH\right)^2. \] On \(h\ge0\), \[ f'(h)=\left(1+a\frac hH\right) \left(1+3a\frac hH\right)>0. \] Thus there is exactly one nonnegative solution. It is stable: increasing \(h\) increases the downward gravitational force while decreasing the upward electric force. Explicitly, \[ \frac{d^2(U_{\mathrm e}+U_{\mathrm g})}{dh^2} =bd\left[ \rho g+\frac{\varepsilon_0a^2E^2} {H(1+ah/H)^3} \right]>0. \]

The derivation applies to a surface lying between the plate ends. Substituting \(h=H\) into the equilibrium equation gives \[ H\varepsilon_r^2 =\frac{\varepsilon_0(\varepsilon_r-1)E_c^2}{2\rho g}, \] which yields the threshold \(E_c\) stated above. If \(E>E_c\), there is no partial-rise equilibrium below the top: the gap becomes completely filled. The detailed surface shape near or above the upper edges requires fringing fields, which the parallel-plate approximation does not determine.

When the capacitance changes only slightly, \[ a\frac hH\ll1, \] the field remains approximately equal to its initial value, and \[ \boxed{h\simeq\frac{\varepsilon_0(\varepsilon_r-1)E^2}{2\rho g}.} \] In particular, the rise tends to zero as \(E\to0\) or \(\varepsilon_r\to1\).

For an isolated capacitor, increasing the permittivity does not always increase the rise, because it also reduces the field. At fixed \(E,H\), the large-permittivity limit is \[ h\simeq \left(\frac{\varepsilon_0E^2H^2} {2\rho g(\varepsilon_r-1)}\right)^{1/3} \longrightarrow0. \] More quantitatively, put \[ B=\frac{\varepsilon_0E^2}{2\rho gH}, \qquad x=a\frac hH. \] The equilibrium equation gives \[ \frac hH=\sqrt B\,\frac{\sqrt x}{1+x} \le\frac{\sqrt B}{2}. \] Before imposing the geometrical limit \(h\le H\), the maximum over permittivity is therefore \[ h_{\max}=E\sqrt{\frac{\varepsilon_0H}{8\rho g}}, \] attained when \(x=1\), or \(a=2/\sqrt B\). If this maximum exceeds \(H\), complete filling is possible for a range of permittivities.

Finally, if a voltage source remains connected, \(V=Ed\) and \(\mathcal E=E\) stay constant. Accounting for the source, the effective electric potential energy is \(-CV^2/2\). Equilibrium then gives \[ \rho gh=\frac12\varepsilon_0(\varepsilon_r-1)E^2, \] leading to the fixed-voltage result in the Answers.

The solution was checked against the official solution. Where this solution differs from the official solution, the official solution contains an error or a misprint.