7. Motion of charged particles in an electric fieldSavchenko Formulas, chapter 7 of 14, 20 formulas
Sections follow the book, and within a section the most used formulas come first. An italic problem number means the formula appears in its statement. Rest the cursor on a number to see the statement.
$$\vec a = \frac{q\vec E}{m}, \qquad y = \frac{qE\,t^2}{2m}, \qquad y = \frac{qE\,L^2}{2mv^2}$$
$\displaystyle t = \frac{2mv\sin\alpha}{qE}\ \text{(время полёта против поля)}$$\displaystyle x = \frac{mv^2\sin2\alpha}{qE}\ \text{(дальность)}$
q, m
charge and mass of the particle
E
field strength
L
length of the field region
v
entry speed
In a uniform field a charge moves with constant acceleration $qE/m$, like a body under gravity, and every ballistics formula carries over with $g$ replaced by $qE/m$. The transverse shift during the transit time $L/v$ is $qEL^2/2mv^2$, so for one accelerating voltage the deflection depends on neither mass nor charge. A charge in gravity and an electric field together hangs where $qE = mg$.
$\displaystyle S = \frac{y}{U} = \frac{Ll}{2dU_0}\ \text{(чувствительность трубки)}$
U
deflecting voltage
d
gap between the plates
l
length of the plates
L
distance from the plates to the screen
U0
accelerating voltage, $mv^2 = 2eU_0$
Between the plates the electron gains a transverse speed $eUl/mdv$, then flies straight, and the spot on the screen adds the shift inside to the slope times $L$. Through the accelerating voltage $eU_0 = mv^2/2$ mass and charge drop out, and the sensitivity $Ll/2dU_0$ depends only on the tube's geometry. This is the oscilloscope.
A field normal to the layer's boundaries changes only the normal velocity component, the tangential one is kept, and the field's work $qEd$ is the gain in kinetic energy. This gives a refraction law for the trajectory at the boundary, much like light. The same holds for entering a region at a different potential.
$$\omega = \sqrt{\frac{2qE}{ml}}, \qquad T = 2\pi\sqrt{\frac{ml}{2qE}}$$
q
charges at the rod's ends
l
length of the rod
E
the field
m
mass at each end
A dipole or a charged rod in a field tends to align with it, the torque $qEl\sin\theta$ is proportional to the angle for small angles, and harmonic oscillations follow with the frequency from $I\ddot\theta = -qEl\theta$. A charge between two fixed charges or a pendulum above a charge oscillates the same way, the frequency built from the restoring forces expanded to first order in the displacement.
$$p_\perp = \int F_\perp\,dt = \frac{qe}{2\pi\varepsilon_0 v r}, \qquad \alpha \approx \frac{p_\perp}{p} = \frac{qe}{2\pi\varepsilon_0 m v^2 r}$$
$p_\perp$
transverse momentum picked up during the passage
r
impact parameter
v
speed of the particle, nearly unchanged
$\alpha$
small deflection angle
A fast particle passes a charge almost in a straight line, and its deflection is found by integrating the transverse force over time along the unperturbed path. The integral $\int E_\perp\,dt = \frac{1}{v}\int E_\perp\,dz$ for a point charge is $q/2\pi\varepsilon_0 v r$, and conveniently $\int E_\perp\,dz$ follows from Gauss's law for a cylinder of radius $r$. The deflection angle is the ratio of transverse to longitudinal momentum.
$$\frac{mv^2}{r} = eE = \frac{e\,U_0}{r\ln(R_2/R_1)}, \qquad U = \frac{U_0}{2\ln(R_2/R_1)}$$
U0
voltage between the cylinders
R1, R2
their radii
U
the electron's accelerating voltage, $eU = mv^2/2$
The field between coaxial cylinders falls as $1/r$, and an electron entering tangentially moves on a circle when the centripetal force $mv^2/r$ equals $eE$. Since $E r$ is constant, the condition does not depend on the orbit's radius and reduces to a relation between the capacitor's voltage and the electron's energy.
A thin electron lens turns every paraxial ray by an angle proportional to its distance from the axis, $\Delta\theta = -y/f$, and that is enough for rays from one point to meet again. Adding the angles before and after the lens gives the same formula as for an optical lens, with all its consequences for images. A negative $f$ means a diverging lens.
Near the axis $\operatorname{div}\vec E = 0$ gives a radial field $-\frac{r}{2}\partial E_z/\partial z$, so wherever the longitudinal field changes a particle feels a force proportional to its distance from the axis. Integrating it over the transit time gives a transverse momentum $\propto r$, that is a lens of focal length $4U/(E_2 - E_1)$. A hole with the stronger field beyond it converges the beam, with the weaker one diverges it.
A parallel ray is bent by the first lens through $y/f_1$, travels a distance $d$ closer to the axis, and the second lens adds its own turn. Adding the slopes gives the system's formula, in which for $d \to 0$ the powers simply add. A converging and a diverging lens of equal $f$ together converge.
$$f = R\,\frac{U_0}{V}\ \text{(сплошной шар)}, \qquad f = 2R\left(\frac{U_0}{V}\right)^2\ \text{(сфера)}$$
R
radius of the ball
V
potential of the ball
U0
accelerating voltage of the beam
x
distance of the ray from the centre
A thin beam passing through a charged ball at small potential picks up a transverse momentum which, by Gauss's law for a cylinder along the ray, is proportional to the distance from the centre. So the ball acts as a lens, for a solid ball only the crossed charges contribute and the focus is $R U_0/V$, for a thin shell the two punctures contribute with opposite signs and only a second-order effect remains.
$$\ddot r + \frac{e\rho}{2m\varepsilon_0}\,r = 0, \qquad \omega = \sqrt{\frac{e\rho}{2m\varepsilon_0}}$$
$\rho$
charge density of the ions the beam crosses
r
distance of the electron from the axis
e, m
electron charge and mass
Inside a uniformly charged cylinder the field grows linearly with radius, $\rho r/2\varepsilon_0$, and an electron of opposite sign is pulled to the axis by a force proportional to its displacement. Its radial motion is harmonic, the beam focuses within a quarter period, and past the column it continues along the tangent to the cosine curve.
$$m\dot v = qE_0\sin\omega t, \qquad v = \frac{qE_0}{m\omega}(1 - \cos\omega t), \qquad x_{\max} = \frac{2qE_0}{m\omega^2}$$
$\displaystyle W_{\max} = \frac{q^2E_0^2}{2m\omega^2}\ \text{(наибольшая энергия, влёт в нужной фазе)}$$\displaystyle \langle v\rangle = \frac{qE_0}{m\omega}\cos\varphi\ \text{(дрейф зависит от фазы влёта)}$
E0
field amplitude
$\omega$
its frequency
$\varphi$
phase of the field when the particle enters
A charge's velocity in an alternating field is the integral of the acceleration, and to the oscillating part $qE_0/m\omega$ a constant is added that depends on the phase at which the particle found itself in the field. An electron entering at a zero of the field drifts at the oscillation's amplitude speed, and the largest energy it can carry away is $2q^2E_0^2/m\omega^2$. The displacement amplitude $qE_0/m\omega^2$ falls with frequency.
If the field reverses during the transit, the accumulated transverse momentum is the integral of a sine over $\tau$ and carries a factor $\sin(\omega\tau/2)$. At $\omega\tau = 2\pi n$ the deflection vanishes for any phase, and the tube's sensitivity to a fast signal falls as $\sin x/x$. Hence an oscilloscope's limiting frequency of order $v/l$.
A free electron in the field $E_0\sin\omega t$ oscillates in antiphase with amplitude $eE_0/m\omega^2$, and the polarization $nex$ points against the field, so a plasma's $\varepsilon$ is below one and negative below the plasma frequency, where waves are reflected. For electrons bound in atoms $\omega_0^2$ is added to $-\omega^2$, giving the dispersion of dielectrics.
Charged bodies interact at a distance, so the system's momentum is always conserved while kinetic energy trades with Coulomb energy. The bodies are closest when their velocities have become equal, and that state is found as in an inelastic collision, only with the potential energy added. This decides whether a particle reaches a ball, how fast released charges fly apart and how much goes into friction and a spring.
For a system of freely moving charges the sum of kinetic energies and pairwise Coulomb energies is conserved, so the speeds after release come from the initial energy, and momentum conservation and symmetry split it between the particles. Heavy particles pick up almost no speed, electrons carry nearly all the energy. When a metal or a dielectric is present, the energy of the induced charges enters the balance too, and a change of a body's self-energy is counted through the field energy.
$\displaystyle \mu = \frac{m_1 m_2}{m_1 + m_2}\ \text{(приведённая масса)}$$\displaystyle m v\rho = m v' r_{\min}\ \text{(момент импульса при нецентральном сближении)}$
$r_{\min}$
smallest distance between the charges
$\mu$
reduced mass
$v_{\text{отн}}$
relative speed far away
$\rho$
impact parameter
At closest approach the relative velocity along the line of centres is zero, and the kinetic energy of relative motion $\mu v^2/2$ has gone entirely into the Coulomb energy $q_1q_2/4\pi\varepsilon_0 r_{\min}$. The centre-of-mass motion takes no part, so for two identical particles approaching at equal speeds $\mu = m/2$, while for one hitting a particle at rest half the energy stays in the centre-of-mass motion. Off-centre, angular momentum conservation is added.
$$\frac{mv^2}{r} = \frac{Ze^2}{4\pi\varepsilon_0 r^2}, \qquad K = -\frac{U}{2}, \qquad E = -\frac{Ze^2}{8\pi\varepsilon_0 r}$$
$\displaystyle K = \frac{Ze^2}{8\pi\varepsilon_0 r}, \qquad U = -\frac{Ze^2}{4\pi\varepsilon_0 r}$$\displaystyle E_{\text{связи}} = -E = \frac{Ze^2}{8\pi\varepsilon_0 r}$
$Z e$
charge of the nucleus
r
orbit radius
K, U
kinetic and potential energies
E
total energy, negative for a bound state
On a circular orbit the Coulomb attraction is the centripetal force, hence $mv^2 = Ze^2/4\pi\varepsilon_0 r$, that is the kinetic energy is half the size of the potential energy. The total energy is negative and equals $-K$, and to strip the electron one must give it $K$. Two electrons circling a common centre, and positronium, are treated the same way, only with a separation $2r$ between the particles.
On a ring's axis the transverse forces from its elements cancel and the longitudinal ones add with the factor $z/\sqrt{z^2+R^2}$. Near the centre the force is linear in the displacement, so an opposite charge oscillates harmonically along the axis, far away the ring acts as a point charge. Across the axis a like charge is, on the contrary, unstable at the centre.
displacement of the electron layer relative to the ions
$\omega_p$
plasma frequency
Shifting all electrons of a layer by $x$ relative to the ions exposes a charge $\pm nex$ per unit area at the faces and sets up inside a field $nex/\varepsilon_0$ that pulls the electrons back. The equation of motion is harmonic at the plasma frequency, independent of the layer's thickness and of the amplitude. The same frequency limits the passage of radio waves through the ionosphere.