23.6.1 Motion in a Uniform Field

The motion of a charged particle in a uniform electric field is especially important for its basic simplicity and because of its many valuable applications. A uniform field is constant at all points—constant in both magnitude and direction—within the region of space where the charged particle is moving. It follows, from Equation 23.32, that a charged particle in a uniform electric field will move with constant acceleration. The magnitude of the acceleration is

a = \frac{q E}{m} = constant

(23.33)

where E is the electric field strength, and the direction of $\vec{a}$ is parallel or antiparallel to $\vec{E}$ , depending on the sign of q.

Identifying the motion of a charged particle in a uniform field as being one of constant acceleration brings into play all the kinematic machinery that we developed in Chapters 2 and 4 for constant-acceleration motion. The basic trajectory of a charged particle in a uniform field is a parabola, analogous to the projectile motion of a mass in the near-earth uniform gravitational field. In the special case of a charged particle moving parallel to the electric field vectors, the motion is one-dimensional, analogous to the one-dimensional vertical motion of a mass tossed straight up or falling straight down.

NOTE The gravitational acceleration $\vec{a}_{grav}$ always points straight down. The electric field acceleration $\vec{a}_{elec}$ can point in any direction. You must know the electric field $\vec{E}$ in order to learn the direction of $\vec{a}$ .

EXAMPLE 23.7 An electron moving across a capacitor

Two 6.0-cm-diameter electrodes are spaced 5.0 mm apart. They are charged by transferring $1.0 \times 10^{11}$ electrons from one electrode to the other. An electron is released from rest just above the surface of the negative electrode. How long does it take the electron to cross to the positive electrode? What is its speed as it collides with the positive electrode? Assume the space between the electrodes is a vacuum.

MODEL The electrodes form a parallel-plate capacitor. The charges on the electrodes cannot escape, but any additional charges between the capacitor plates will be accelerated by the electric field. The electric field inside a parallel-plate capacitor is a uniform field, so the electron will have constant acceleration.

VISUALIZE FIGURE 23.24 shows an edge view of the capacitor and the electron. The force on the negative electron is opposite the electric field, so the electron is repelled by the negative electrode as it accelerates across the gap of width d.

The figure illustrates an electron accelerated across a parallel plate capacitor.

FIGURE 23.24 Full Alternative Text

SOLVE The electrodes are not point charges, so we cannot use Coulomb’s law to find the force on the electron. Instead, we must analyze the electron’s motion in terms of the electric field inside the capacitor. The field is the agent that exerts the force on the electron, causing it to accelerate. The charge on the capacitor is $Q = N e$ , where N is the number of electrons transferred from one electrode to the other. The electric field strength inside a parallel-plate capacitor with charge Q is

E = \frac{η}{ϵ_{0}} = \frac{Q}{ϵ_{0} A} = \frac{N e}{ϵ_{0} π R^{2}} = 639, 000 N / C

The electron’s acceleration in this field is

a = \frac{e E}{m} = 1.1 \times 10^{17} m / s^{2}

where we used the electron mass $m = 9.11 \times 10^{- 31} kg$ from Table 22.1. This is an enormous acceleration compared to accelerations we’re familiar with for macroscopic objects. We can use one-dimensional kinematics, with $x_{i} = 0$ and $v_{i} = 0$ , to find the time required for the electron to cross the capacitor:

\begin{matrix} x_{f} & = & d = \frac{1}{2} a {(Δ t)}^{2} \\ Δ t & = & \sqrt{\frac{2 d}{a}} = 3.0 \times 10^{- 10} s = 0.30 n s \end{matrix}

The electron’s speed as it reaches the positive electrode is

v = a Δ t = 3.3 \times 10^{7} m / s

REVIEW We used e rather than $- e$ to find the acceleration because we already knew the direction; we needed only the magnitude. The electron’s speed, after traveling a mere 5 mm, is approximately 10% the speed of light.

Parallel electrodes such as those in Example 23.7 are often used to accelerate charged particles. If the positive plate has a small hole in the center, a beam of electrons will pass through the hole and emerge with a speed of $3.3 \times 10^{7} m / s$ . This is the basic idea of the electron gun used until quite recently in cathode-ray tube (CRT) devices such as televisions and computer display terminals. (A negatively charged electrode is called a cathode, so the physicists who first learned to produce electron beams in the late 19th century called them cathode rays.)

EXAMPLE 23.8 Deflecting an electron beam

An electron gun creates a beam of electrons moving horizontally with a speed of $3.3 \times 10^{7} m / s$ . The electrons enter a 2.0-cm-long gap between two parallel electrodes where the electric field is $\vec{E} = (5.0 \times 10^{4} N / C, d o w n)$ . In which direction, and by what angle, is the electron beam deflected by these electrodes?

MODEL The electric field between the electrodes is uniform. Assume that the electric field outside the electrodes is zero.

VISUALIZE FIGURE 23.25 shows an electron moving through the electric field. The electric field points down, so the force on the (negative) electrons is upward. The electrons will follow a parabolic trajectory, analogous to that of a ball thrown horizontally, except that the electrons “fall up” rather than down.

The figure illustrates an electron moving through the electric field inside a parallel-plate capacitor.

FIGURE 23.25 Full Alternative Text

SOLVE This is a two-dimensional motion problem. The electron enters the capacitor with velocity vector $\vec{v}_{0} = v_{0 x} î = 3.3 \times 10^{7} î m / s$ and leaves with velocity $\vec{v}_{1} = v_{1 x} î + v_{1 y} ĵ$ . The electron’s angle of travel upon leaving the electric field is

θ = {tan}^{- 1} (\frac{v_{1 y}}{v_{1 x}})

This is the deflection angle. To find $θ$ we must compute the final velocity vector $\vec{v}_{1}$ .

There is no horizontal force on the electron, so $v_{1 x} = v_{0 x} = 3.3 \times 10^{7} m / s$ . The electron’s upward acceleration is

\begin{matrix} a_{y} & = & \frac{e E}{m} = \frac{(1.60 \times 10^{- 19} C) (5.0 \times 10^{4} N / C)}{9.11 \times 10^{- 31} k g} \\ = & 8.78 \times 10^{15} m / s^{2} \end{matrix}

We can use the fact that the horizontal velocity is constant to determine the time interval $Δ t$ needed to travel length 2.0 cm:

Δ t = \frac{L}{v_{0 x}} = \frac{0.020 m}{3.3 \times 10^{7} m / s} = 6.06 \times 10^{- 10} s

Vertical acceleration will occur during this time interval, resulting in a final vertical velocity

v_{1 y} = v_{0 y} + a_{y} Δ t = 5.3 \times 10^{6} m / s

The electron’s velocity as it leaves the capacitor is thus

\vec{v}_{1} = (3.3 \times 10^{7} î + 5.3 \times 10^{6} ĵ) m / s

and the deflection angle $θ$ is

θ = {tan}^{- 1} (\frac{v_{1 y}}{v_{1 x}}) = 9.1 °

REVIEW We know that the electron beam in a cathode-ray tube can be deflected enough to cover the screen, so a deflection angle of $9 °$ seems reasonable. Our neglect of the gravitational force is seen to be justified because the acceleration of the electrons is enormous in comparison to the free-fall acceleration g.

By using two sets of deflection plates—one for vertical deflection and one for horizontal—a cathode-ray tube could steer the electrons to any point on the screen. Electrons striking a phosphor coating on the inside of the screen would then make a dot of light.

23.6.1 Motion in a Uniform Field

EXAMPLE 23.7 An electron moving across a capacitor

FIGURE 23.24 An electron accelerates across a capacitor (plate separation exaggerated).

EXAMPLE 23.8 Deflecting an electron beam

FIGURE 23.25 The deflection of an electron beam in a uniform electric field.