example involves several concepts: the relationship between electric field and electric force, the relationship between force and acceleration, the definition of kinetic energy, and the kinematic relationships among acceleration, distance, velocity, and time.

SET UP: Figure 21.20 shows our coordinate system. We are given the electric field, so we use Eq. (21.4) to find the force on the electron and Newton’s second law to find its acceleration. Because the field is uniform between the plates, the force and acceleration are constant and we can use the constant-acceleration formulas from Chapter 3 to find the electron’s velocity and travel time. We find the kinetic energy using the definition .

EXECUTE: (a) Note that is upward (in the +y-direction) but is downward because the charge of the electron is negative. Thus Fy is negative. Because Fy is constant, the electron moves with constant acceleration ay given by

This is an enormous acceleration! To give a 1000-kg car this acceleration, we would need a force of about 2 × 1018 N (about 2 × 1014 tons). The gravitational force on the electron is completely negligible compared to the electric force.

(b) The electron starts from rest, so its motion is in the y-direction only (the direction of the acceleration). We can find the electron’s speed at any position using the constant-acceleration formula . We have v0y = 0 and y0 = 0, so the speed |vy| when y = −1.0 cm = −1.0 × 10−2 m is

The velocity is downward, so its y-component is vy = −5.9 × 106 m/s. The electron’s kinetic energy is

(c) From the constant-acceleration formula vy = v0y + ayt, we find that the time required is very brief:

(We could also have found the time by solving the equation for t.)

EVALUATE: This example shows that in problems about subatomic particles such as electrons, many quantities—including acceleration, speed, kinetic energy, and time—will have very different values from what we have seen for ordinary objects such as baseballs and automobiles.

Figure. 21.20 A uniform electric field between two parallel conducting plates connected to a 100-volt battery. (The separation of the plates is exaggerated in this figure relative to the dimensions of the plates.)

EXAMPLE 21.8 An electron trajectory

If we launch an electron into the electric field of Example 21.7 with an initial horizontal velocity v0 (Fig. 21.21), what is the equation of its trajectory?

SOLUTION

IDENTIFY: We found the electron’s acceleration in Example 21.7. Our goal is to find the trajectory that corresponds to that acceleration.

SET UP: The acceleration is constant and in the negative y-direction (there is no acceleration in the x-direction). Hence we can use the kinematic equations from Chapter 3 for two-dimensional motion with constant acceleration.

EXECUTE: We have ax = 0 and ay = (−e)E/m. At t = 0, x0 = y0 = 0, v0x = v0, and v0y. = 0; hence at time t,

Eliminating t between these equations, we get

EVALUATE: This is the equation of a parabola, just like the trajectory of a projectile launched horizontally in the earth’s gravitational field (discussed in Section 3.3). For a given initial velocity of the electron, the curvature of the trajectory depends on the field magnitude E. If we reverse the signs of the charges on the two plates in Fig. 21.21, the direction of reverses, and the electron trajectory will curve up, not down. Hence we can “steer” the electron by varying the charges on the plates. The electric field between charged conducting plates can be used in this way to control the trajectory of electron beams in oscilloscopes.

Figure. 21.21 The parabolic trajectory of an electron in a uniform electric field.

Step-by-step solution:

Step 1 of 3

(a)

Electric force on electron

Gravitational force

=

We see that

So, gravitational forces are negligible