Irodov 3.144: A Charged Capacitor Lifts the Liquid – Amit & Deepali Sinha Foundation

Problem Irodov 3.144

A parallel-plate capacitor is held with its plates horizontal: the lower plate is under the surface of a liquid and the upper plate is above it. The liquid has permittivity $\varepsilon$ and density $\rho$ . The plates are then given charges of surface density $\sigma$ (and $-\sigma$ ). By how much does the liquid level between the plates rise?

The horizontal capacitor before charging: the lower plate under the liquid, the upper plate above its free surface, a lead from the middle of each plate.

Given

liquid: permittivity $\varepsilon$ , density $\rho$
surface charge density $\sigma$ on the plates

Find

the rise

h

of the liquid level between the plates

Solution

Step 1The fields in the air and in the liquid

Neglecting edge effects, the field between the plates is vertical. The free charge $\sigma$ is fixed on the plates, and the normal component of $\vect D$ is continuous across the liquid surface, so $D = \sigma$ in both layers whatever the liquid level:

E_{\mathrm{air}} = \frac{\sigma}{\varepsilon_0}, \qquad E_{\mathrm{liq}} = \frac{\sigma}{\varepsilon_0\varepsilon} \tag{1}

The energy density $\varepsilon_0\varepsilon E^2/2$ is therefore $\sigma^2/2\varepsilon_0$ in the air and $\sigma^2/2\varepsilon_0\varepsilon$ in the liquid.

Step 2The field energy released as the liquid rises

In equilibrium the liquid has risen by h between the plates. The field in the liquid is ε times weaker than in the air, and the lines that stop at the raised surface end on its negative bound charge. — In equilibrium the liquid has risen by $h$ between the plates. The field in the liquid is $\varepsilon$ times weaker than in the air, and the lines that stop at the raised surface end on its negative bound charge.

Let the level between the plates, of area $S$ , stand a height $x$ above the outside level, and let it rise by $dx$ . A layer of air of volume $S\,dx$ is replaced by liquid, and with the charges fixed the field energy changes by

dW_{\mathrm{el}} = \left(\frac{\sigma^2}{2\varepsilon_0\varepsilon} - \frac{\sigma^2}{2\varepsilon_0}\right)S\,dx = -\frac{(\varepsilon - 1)\,\sigma^2}{2\varepsilon_0\varepsilon}\,S\,dx \tag{2}

The field energy falls, so the field pulls the liquid up.

Step 3The gravitational energy it costs

The added layer, of mass $\rho S\,dx$ , comes from the wide outer surface, whose level does not change, and is lifted through the height $x$ :

dU_{\mathrm{g}} = \rho g x\,S\,dx \tag{3}

Step 4Equilibrium

The liquid stops where a further rise no longer lowers the total energy, $dW_{\mathrm{el}} + dU_{\mathrm{g}} = 0$ at $x = h$ . With (2) and (3):

\rho g h\,S\,dx = \frac{(\varepsilon - 1)\,\sigma^2}{2\varepsilon_0\varepsilon}\,S\,dx

h = \dfrac{(\varepsilon - 1)\,\sigma^2}{2\varepsilon_0\varepsilon\,\rho g}

(4)

Check

With $\varepsilon = 1$ the liquid is electrically the same as air and $h = 0$ . In the limit $\varepsilon \to \infty$ (a conducting liquid) the rise tends to $\sigma^2/2\varepsilon_0\rho g$ : the liquid surface then carries $-\sigma$ itself and is pulled up by the pressure $\sigma^2/2\varepsilon_0$ of a charged conducting surface, which the column’s hydrostatic pressure $\rho g h$ balances. Dimensions: $\sigma^2/\varepsilon_0$ is a pressure and $\rho g$ a pressure per length, so $h$ is a length.

Remark

Read as a pressure balance, the field pulls up on the liquid surface with $(\sigma^2/2\varepsilon_0)(1 - 1/\varepsilon)$ , the difference between the energy densities above and below it: a field normal to a dielectric boundary pulls the boundary toward the side of lower permittivity. Because $\sigma$ is fixed, the pull does not depend on how high the liquid has climbed, which is why the rise is the simple ratio of that pressure to $\rho g$ .