Irodov 3.372: A Ramp Voltage Launches an Electron – Amit & Deepali Sinha Foundation

Problem Irodov 3.372

The plates of a parallel-plate capacitor are a distance $l = 5.0$ cm apart. At $t = 0$ an electron leaves one plate with negligible velocity, and from that moment a voltage that accelerates it, rising as $V = at$ with $a = 100$ V/s, is applied between the plates. With what velocity does the electron arrive at the other plate?

The capacitor with gap l; the electron starts at rest on the lower plate, and the voltage V = at between the plates accelerates it toward the upper one. — The capacitor with gap $l$ ; the electron starts at rest on the lower plate, and the voltage $V = at$ between the plates accelerates it toward the upper one.

Given

$V = at$
$a = 100$ V/s
$l = 5.0$ cm
$v(0) = 0$

Find

the velocity

v

on reaching the opposite plate

Solution

Step 1The force grows with time

Between the plates the field is uniform and equals the voltage over the gap, so it rises in proportion to $t$ . The force on the electron, $eE$ , points toward the far plate, and Newton’s second law along the gap reads

\begin{align*} E &= \frac{V}{l} = \frac{at}{l}, \\ m\frac{dv}{dt} &= eE = \frac{ea}{l}\,t \tag{1}\end{align*}

Step 2Velocity and distance as functions of time

Integrating (1) from rest at $t = 0$ gives the velocity:

v(t) = \int_0^t \frac{ea}{ml}\,t'\,dt' = \frac{ea}{2ml}\,t^2 \tag{2}

Integrating the velocity once more, from $x = 0$ , gives the distance from the starting plate:

x(t) = \int_0^t \frac{ea}{2ml}\,t'^2\,dt'

x(t) = \frac{ea}{6ml}\,t^3

(3)

Step 3The time of flight across the gap

The electron reaches the far plate at the moment $\tau$ when $x(\tau) = l$ . Setting (3) equal to $l$ and solving for $\tau$ :

\frac{ea}{6ml}\,\tau^3 = l \quad\Longrightarrow\quad \tau = \left(\frac{6ml^2}{ea}\right)^{1/3}

Step 4The speed on arrival

Putting $\tau$ into (2), all the powers combine under one cube root:

\begin{aligned} v(\tau) &= \frac{ea}{2ml}\left(\frac{6ml^2}{ea}\right)^{2/3} \\ &= \left[\frac{e^3a^3}{8m^3l^3}\cdot\frac{36\,m^2l^4}{e^2a^2}\right]^{1/3} \end{aligned}

Numerically, with $e/m = 1.76\times10^{11}$ C/kg,

v = \left[\frac{9}{2}\,(1.76\times10^{11})(100)(0.050)\right]^{1/3} \text{ m/s} = \left(3.96\times10^{12}\right)^{1/3} \text{ m/s}

v = \left(\frac{9eal}{2m}\right)^{1/3} \approx 16 \text{ km/s}

(4)

Check

Units: $ea l/m$ is $\text{C}\cdot(\text{V/s})\cdot\text{m}/\text{kg} = (\text{J}/\text{kg})\cdot(\text{m/s})$ , that is $(\text{m/s})^2\cdot\text{m/s}$ , so its cube root is a speed. Independent route through $\tau$ : $\tau = (6ml^2/(ea))^{1/3} = (8.5\times10^{-16})^{1/3} \text{ s} = 9.5\ \mu\text{s}$ , and $3l/\tau = 0.15/(9.5\times10^{-6})$ m/s $= 1.6\times10^4$ m/s, the same. The speed is $5\times10^{-5}$ of $c$ , so the non-relativistic law is safe.

Remark

In a static field the electron would arrive with kinetic energy $eV$ . Here the voltage on arrival is $a\tau \approx 0.95$ mV, yet the electron carries only three quarters of $e\,a\tau$ , because it crossed most of the gap early, while the field was still weak:

\frac{mv^2/2}{e a\tau} = \frac{m}{2ea\tau}\cdot\frac{9l^2}{\tau^2} = \frac{9ml^2}{2ea\tau^3} = \frac{3}{4}

A ramped voltage is a poor accelerator for exactly this reason: the field is strongest when the particle has almost nowhere left to go.