Irodov 3.374: Turning Back in a Sloping Field – Amit & Deepali Sinha Foundation

Problem Irodov 3.374

A particle of specific charge $q/m$ , initially at rest, moves along a straight line under an electric field directed along that line, of strength $E = E_0 - ax$ , where $a$ is a positive constant and $x$ is the distance from the starting point. Find:

the distance the particle covers before it comes to rest;
its acceleration at that moment.

Given

$q/m$
$E = E_0 - ax$
$a > 0$
$v(0) = 0$

Find

(a) the distance covered before the particle stops · (b) its acceleration then

Solution

Step 1Kinetic energy as a function of position

Since $x$ is the distance covered from the start, the particle moves toward positive $x$ , which needs $qE_0 > 0$ . For it ever to stop the force must reverse along the way, which with $a > 0$ needs $q > 0$ , and therefore $E_0 > 0$ , as drawn. Writing the acceleration as $v\,dv/dx$ and integrating from the start, where $v = 0$ :

\begin{aligned} m v\frac{dv}{dx} &= q(E_0 - ax), \\ \int_0^v m v'\,dv' &= \int_0^x q(E_0 - ax')\,dx' \end{aligned}

\frac{mv^2}{2} = q\left(E_0x - \frac{ax^2}{2}\right)

(1)

Step 2(a) Where the particle stops

The particle gains speed while $E > 0$ , up to $x = E_0/a$ , and loses it beyond. It is at rest again where the kinetic energy (1) returns to zero at some $x \neq 0$ :

E_0x - \frac{ax^2}{2} = x\left(E_0 - \frac{ax}{2}\right) = 0

x_s = \frac{2E_0}{a}

(2)

In terms of the potential, $\varphi(x) = -E_0x + ax^2/2$ , the particle stops where $\varphi$ climbs back to its starting value, as the graph shows.

The potential along the line, with φ(0) = 0: lowest at x = E_0/a, where E = 0, and back to its starting value at x = 2E_0/a, where the particle stops. Drawn for E_0 > 0. — The potential along the line, with $\varphi(0) = 0$ : lowest at $x = E_0/a$ , where $E = 0$ , and back to its starting value at $x = 2E_0/a$ , where the particle stops. Drawn for $E_0 > 0$ .

Step 3(b) The acceleration at the stopping point

The acceleration is the force per unit mass. Putting $x_s$ from (2) into the field:

w = \frac{q}{m}\left(E_0 - a\cdot\frac{2E_0}{a}\right)

w = -\frac{qE_0}{m}

(3)

Its magnitude is $qE_0/m$ , the same as at the start, and it points back toward the starting point.

Check

Units: $E_0/a$ is a field over a field per length, a length. A second route: the equation of motion is $\ddot x = -(qa/m)\,(x - E_0/a)$ , simple harmonic motion about $x = E_0/a$ with $\omega^2 = qa/m$ . Starting at rest from $x = 0$ , the amplitude is $E_0/a$ , so the far turning point is $2E_0/a$ and the acceleration there is $-\omega^2(E_0/a) = -qE_0/m$ , both as found.

Remark

The particle oscillates for ever between $x = 0$ and $x = 2E_0/a$ , with period $2\pi\sqrt{m/(qa)}$ , which depends on the field’s slope and not on $E_0$ . A field that varies linearly with position is an electrostatic spring: its potential is a quadratic well, the kind a Penning trap uses to hold an ion along its axis.