Irodov 3.216: Cosmic Rays Drain a Capacitor – Amit & Deepali Sinha Foundation

Problem Irodov 3.216

A parallel-plate capacitor with air between its plates, a distance $d = 5.0$ mm apart, is charged to the voltage $V = 90$ V and then disconnected from the source. Natural ionization of air at standard conditions forms on average $\dot n_i = 5.0$ ion pairs per cm $^3$ per second, and at this voltage the current through the air is the saturation current. How long does it take the voltage across the capacitor to fall by $\eta = 1.0\%$ ?

The air capacitor, its plates a distance d apart. — The air capacitor, its plates a distance $d$ apart.

Given

$d = 5.0$ mm
$V = 90$ V
$\eta = 1.0\%$
$\dot n_i = 5.0$ cm $^{-3}$ s $^{-1}$
current saturated throughout the drop (assumed)
edge effects neglected (assumed)

Find

the time

t

Solution

Step 1The leakage current

At saturation every ion pair formed in the gap is collected before it can recombine, each pair carrying the charge $e$ across. With plate area $S$ , the gap holds the volume $Sd$ , so the current through the air is

I = e\,\dot n_i S d

It does not depend on the voltage, so it stays the same while the voltage falls by a small fraction and the current remains saturated (assumed).

Step 2The rate at which the voltage falls

The current drains the plates’ charge, $q = CV$ , with the capacitance $C = \varepsilon_0S/d$ :

\frac{\varepsilon_0 S}{d}\,\frac{dV}{dt} = -e\,\dot n_i S d

The area cancels:

\frac{dV}{dt} = -\frac{e\,\dot n_i d^2}{\varepsilon_0} \tag{1}

Step 3The time for a one per cent drop

The rate (1) is constant, so the drop $\eta V$ takes a time equal to the drop divided by the rate:

t = \frac{\eta V}{|dV/dt|} = \frac{\eta V}{e\,\dot n_i d^2/\varepsilon_0}

Numerically, with $\dot n_i = 5.0\times10^{6}$ m $^{-3}$ s $^{-1}$ and $1$ day $= 86\,400$ s:

t = \frac{8.85\times10^{-12}\times 0.010\times 90}{1.60\times10^{-19}\times 5.0\times10^{6}\times(5.0\times10^{-3})^2}\ \text{s}\times\frac{1\ \text{day}}{86\,400\ \text{s}}

t = \frac{\varepsilon_0\,\eta V}{e\,\dot n_i d^2} = 4.6\ \text{days}

(2)

Check

Units: $(\text{F/m})\cdot\text{V}$ is $\text{C/m}$ , and $e\dot n_i d^2$ is $\text{C}\,\text{m}^{-1}\text{s}^{-1}$ , so $t$ is in seconds. Is the current really saturated? With a typical air ion mobility of about $1.4$ cm $^2$ /(V $\cdot$ s), an ion crosses the gap in about $d^2/(uV)$ , which is $2$ ms. Without a field the ions would reach only $\sqrt{\dot n_i/r}$ , about $1.7\times10^{3}$ cm $^{-3}$ (with the air recombination coefficient $1.67\times10^{-6}$ cm $^3$ /s), and live about $1/(rn)$ , some $6$ minutes, before recombining. The ions are swept out some $10^{5}$ times faster than they recombine, so practically every pair is collected, as the saturation premise requires.

Remark

The leak grows as $d^2$ : one factor of $d$ because a wider gap holds more air to ionize, and one because the plates then carry less charge per unit area at the same voltage. Natural ionization, mostly from cosmic rays and ground radioactivity, is why a charged electroscope slowly loses its charge however well insulated it is; measuring that slow loss is how cosmic rays were discovered.