Irodov 3.148: Water Rising Through a Capacitor – Amit & Deepali Sinha Foundation

Problem Irodov 3.148

A cylindrical air capacitor is kept at a constant voltage $V = 200$ V and is lowered, with its axis vertical, into a vessel of water (permittivity $\varepsilon = 81$ ) at the speed $v = 5.0$ mm/s. The gap between its electrodes is $d = 2.0$ mm and their mean radius is $r = 50$ mm, with $d \ll r$ . Find the current in the lead wires while the capacitor goes in.

The cylindrical capacitor in axial section, lowered with velocity v into water: electrode gap d, mean radius r. The gap is drawn far wider relative to r than in the problem. — The cylindrical capacitor in axial section, lowered with velocity $\vect{v}$ into water: electrode gap $d$ , mean radius $r$ . The gap is drawn far wider relative to $r$ than in the problem.

The voltage is held fixed, so the charge on the electrodes is $q = CV$ and changes only as the capacitance changes. Water entering the gap raises the capacitance, and the extra charge has to come through the leads.

Given

$V = 200$ V, constant
$v = 5.0$ mm/s
$d = 2.0$ mm
$r = 50$ mm, $d \ll r$
water, $\varepsilon = 81$ (table value)
water non-conducting (assumed)

Find

the current

I

in the leads

Solution

Step 1The capacitance with the gap partly filled

With $d \ll r$ the gap is a thin layer, and a band of it of height $h$ acts as a parallel-plate capacitor of plate area $2\pi r h$ and separation $d$ . Let $x$ be the height of the gap already under water and $L$ the electrodes’ length. The field in the gap is $V/d$ both above and below the water surface, so the two parts are capacitors in parallel, the lower one filled with water of permittivity $\varepsilon = 81$ (table value):

\begin{aligned} C(x) &= \frac{\varepsilon_0\varepsilon\cdot 2\pi r x}{d} + \frac{\varepsilon_0\cdot 2\pi r\,(L - x)}{d}\\ &= \frac{2\pi\varepsilon_0 r}{d}\bigl(L + (\varepsilon - 1)\,x\bigr) \end{aligned} \tag{1}

Step 2The current from the rate of change of charge

The water surface climbs up the gap at the rate the capacitor sinks, $dx/dt = v$ . Differentiating $q = CV$ at constant $V$ with (1):

\begin{aligned} I &= \frac{dq}{dt} = V\,\frac{dC}{dx}\,\frac{dx}{dt}\\ &= V\,\frac{2\pi\varepsilon_0 r\,(\varepsilon - 1)}{d}\,v \end{aligned}

Numerically, with $\varepsilon - 1 = 80$ , $r = 0.050$ m, $v = 5.0 \times 10^{-3}$ m/s, $V = 200$ V and $d = 2.0 \times 10^{-3}$ m:

I = \frac{2\pi \times 8.85 \times 10^{-12} \times 80 \times 0.050 \times 5.0 \times 10^{-3} \times 200}{2.0 \times 10^{-3}}\ \text{A} = 1.1 \times 10^{-7}\ \text{A}

I \approx \frac{2\pi\varepsilon_0(\varepsilon - 1)\,rvV}{d} \approx 0.11\ \mu\text{A}

(2)

The current charges the capacitor further: it flows from the source onto the electrode at the higher potential.

Check

Units: $\varepsilon_0 r/d$ is a capacitance per unit length, F/m; times a speed it is F/s, and times a voltage C/s, an ampere. Without the condition $d \ll r$ the capacitance per unit length of a coaxial pair is $2\pi\varepsilon_0/\ln(b/a)$ , with electrode radii $a = r - d/2$ and $b = r + d/2$ ; for $d \ll r$ the logarithm is $\ln\bigl((1 + d/2r)/(1 - d/2r)\bigr) \approx d/r$ , which reproduces (2), the relative error being of order $(d/r)^2/12$ , here about $10^{-4}$ .

Remark

The measured current is a direct reading of the permittivity of the liquid, $\varepsilon - 1 = Id/(2\pi\varepsilon_0 rvV)$ , and the same principle runs capacitive level gauges: two coaxial tubes standing in a tank, whose capacitance grows linearly with the liquid level by (1). Real water conducts a little, which adds a steady leakage current through the submerged part; the charging current found here is the part that is present only while the level changes.