Irodov 3.373: Deflection by a Growing Field – Amit & Deepali Sinha Foundation

Problem Irodov 3.373

A proton, accelerated from rest through a potential difference $V$ , enters the uniform field of a parallel-plate capacitor, moving parallel to the plates, which are $l$ long in that direction. The field between the plates grows with time as $E = at$ , $a$ a constant, and the proton enters at $t = 0$ . Treating the proton as non-relativistic and neglecting edge effects, find the angle between its direction of motion before it enters the capacitor and after it leaves.

The capacitor with plates of length l and the proton p arriving along the midline with velocity v_0; the transverse field between the plates is E = at from the moment of entry. — The capacitor with plates of length $l$ and the proton $p$ arriving along the midline with velocity $\vect v_0$ ; the transverse field between the plates is $E = at$ from the moment of entry.

Given

$V$
$l$
$E = at$
entry at $t = 0$

Find

the angle

\alpha

between the directions of motion before and after the capacitor

Solution

Step 1The entry speed

The accelerating voltage gives the proton, of charge $e$ and mass $m$ , the kinetic energy $eV$ , so it enters with speed $v_0$ :

\frac{m v_0^2}{2} = eV \quad\Longrightarrow\quad v_0 = \sqrt{\frac{2eV}{m}} \tag{1}

Step 2The time spent between the plates

Along the plates there is no force, so the proton keeps $v_0$ in that direction and covers the length $l$ in the time

\tau = \frac{l}{v_0} \tag{2}

Step 3The transverse velocity gained

Across the plates the force is $eE = eat$ , growing from zero at entry. Integrating Newton’s second law over the transit, from $v_y = 0$ at $t = 0$ to $t = \tau$ :

\begin{aligned} m\frac{dv_y}{dt} &= eat, \\ v_y(\tau) &= \int_0^\tau \frac{ea}{m}\,t\,dt \end{aligned}

v_y = \frac{ea\tau^2}{2m}

(3)

Beyond the plates there is no field, so the proton leaves along the velocity it has at $t = \tau$ , resolved below.

The velocity on leaving the plates: the unchanged component v_0 along the entry direction x and the gained component v_y across the plates. The deflection is the angle α with x. — The velocity on leaving the plates: the unchanged component $v_0$ along the entry direction $x$ and the gained component $v_y$ across the plates. The deflection is the angle $\alpha$ with $x$ .

Step 4The deflection angle

The tangent of the angle is the ratio of the two components. Using (3), then (2) for $\tau$ and (1) for $v_0$ :

\begin{aligned} \tan\alpha &= \frac{v_y}{v_0} = \frac{ea\tau^2}{2m v_0} = \frac{eal^2}{2m v_0^3} \\ &= \frac{eal^2}{2m}\left(\frac{m}{2eV}\right)^{3/2} = \frac{eal^2}{2m}\cdot\frac{m}{2eV}\sqrt{\frac{m}{2eV}} \end{aligned}

\tan\alpha = \frac{al^2}{4V}\sqrt{\frac{m}{2eV}}

(4)

Check

Units: $a$ is a field per unit time, $\text{V}/(\text{m}\cdot\text{s})$ , so $al^2/V$ is in m/s, while $\sqrt{m/(eV)}$ is the reciprocal of a speed; the product is a pure number, as a tangent must be. Limit: $a \to 0$ gives $\alpha \to 0$ , no field and no deflection. Comparison: a constant field $E_c$ would give $v_y = eE_c\tau/m$ ; (3) is that with $E_c = a\tau/2$ , the field’s mean value over the transit.

Remark

The deflection depends only on the transverse impulse, the time integral of the force, and so only on the field averaged over the transit. A slower proton spends longer in a field that keeps growing, which is why $\tan\alpha$ goes as $v_0^{-3}$ , one power steeper than for a steady field ( $v_0^{-2}$ ): a time-varying deflector sorts particles by speed more sharply than a static one.