Irodov 3.202: A Voltage Lifts the Water – Amit & Deepali Sinha Foundation

Problem Irodov 3.202

A cylindrical capacitor stands upright with the lower ends of its electrodes touching the surface of water (permittivity $\varepsilon$ , density $\rho$ ), and is held at a constant voltage $V$ by a dc source. The gap $d$ between the electrodes is much smaller than their mean radius. Neglecting capillary effects, find the height $h$ to which the water rises in the gap.

The capacitor in vertical section standing on the water: the water rises by h in the narrow gap between the electrodes, and not inside the inner cylinder. — The capacitor in vertical section standing on the water: the water rises by $h$ in the narrow gap between the electrodes, and not inside the inner cylinder.

Given

constant voltage $V$
gap $d \ll$ mean radius
water: permittivity $\varepsilon$ , density $\rho$
capillarity neglected

Find

the rise

h

of the water in the gap

Solution

Step 1The gap is a plane capacitor

Let $R$ be the mean radius and $H$ the height of the electrodes, and let the water stand at a height $x$ in the gap. Because $d \ll R$ , the gap is a plane capacitor of width $2\pi R$ rolled into a tube; the water-filled part (permittivity $\varepsilon$ ) and the empty part are in parallel:

C(x) = \frac{2\pi R\,\varepsilon_0}{d}\left[\varepsilon x + (H - x)\right] \tag{1}

Step 2The upward force at constant voltage

Let the water rise by $dx$ , slowly. The source pushes the charge $V\,dC$ through itself and does the work $V^2dC$ ; the field energy $CV^2/2$ grows by $V^2dC/2$ ; the rest is the work $F\,dx$ of the electric force lifting the water:

\begin{aligned} V^2\,dC &= \frac{V^2}{2}\,dC + F\,dx\\ F &= \frac{V^2}{2}\,\frac{dC}{dx} \end{aligned}

With (1) the derivative is a constant, so the force is the same at every level:

F = \frac{V^2}{2}\cdot\frac{2\pi R\,\varepsilon_0(\varepsilon - 1)}{d} = \frac{\pi R\,\varepsilon_0(\varepsilon - 1)V^2}{d} \tag{2}

Step 3Balance against the weight of the raised water

The water stops rising when (2) holds up the weight of the column raised in the gap, of height $h$ , thickness $d$ and length $2\pi R$ round the circle:

\frac{\pi R\,\varepsilon_0(\varepsilon - 1)V^2}{d} = \rho g\,h\cdot 2\pi R\,d

The mean radius cancels:

h \approx \frac{\varepsilon_0(\varepsilon - 1)V^2}{2\rho g d^2}

(3)

The sign $\approx$ marks the plane-capacitor approximation $d \ll R$ .

Check

Dimensions: $\varepsilon_0V^2/d^2$ is (F/m)(V/m) $^2$ = J/m $^3$ = N/m $^2$ , a pressure, and dividing by $\rho g$ (N/m $^3$ ) leaves metres. Limit: with $\varepsilon = 1$ the liquid would feel no pull and $h = 0$ . The rise is just the field’s energy density difference $\varepsilon_0(\varepsilon - 1)E^2/2$ , with $E = V/d$ , balanced by the hydrostatic pressure $\rho g h$ .

Remark

The result is a pressure balance and owes nothing to the shape of the electrodes: any narrow gap of the same width at the same voltage draws a dielectric liquid up by the same height. For water, with $\varepsilon = 81$ , a gap of $1$ mm at $1$ kV gives $h \approx 3.6$ cm, a large effect, because $\varepsilon - 1$ is large. With ordinary water a current also flows through the column and heats it, but the rise is set by the permittivity alone.