Prerequisites

Students should be familiar with the concepts of electric field, force, acceleration, velocity, and displacement, and be familiar with Newton's second law and the properties of projectile motion in the earth's gravitational field.

Learning Outcomes

Students will develop an understanding of how an electric field affects the motion of charged particles and will learn that, if the field is uniform, the motion is analogous to projectile motion in a gravitational field. They will learn to make calculations on the motion for the case that the field is uniform and will gain additional experience in using the kinematical equations of motion with constant acceleration.

Instructions

Students should know how the applet functions, as described in Help and ShowMe.

The applet should be open. The step-by-step instructions in the following text are to be done in the applet. You may need to toggle back and forth between instructions and applet if your screen space is limited.

Motion of a Charged Particle in an Electric Field

Calculations

Appendix

Equations For Motion With Constant Acceleration

The applet and the lesson assumes that the particle is subject only to an electric force. The gravitational force is not included.

1. Reset the applet.

   Make the following settings. These settings are also displayed in Figure 1 below.

   • Move the particle to the bottom of the applet window near the left-hand corner.
   • Set the particle's initial velocity to magnitude 150 m/s and direction 60^o N of E.
   • Set the electric field to magnitude 75 V/m and direction 270^o N of E.
   • Set the particle's charge and mass to 2.0 C and 3.0 kg, respectively.

   Play the motion with the particle's path displayed. You should obtain a trajectory like that in Figure 1.

   

   Figure 1

2. The motion looks just like projectile motion in the earth's gravitational field! Indeed, the two kinds of motion are entirely analogous. Instead of a downward uniform gravitational field there, we have a downward uniform electric field here. Instead of a constant gravitational force equal to m there, we have a constant electric force equal to q here. In both cases, the result is a motion with constant downward acceleration, which implies projectile motion.

   Therefore, there is not much new for you to learn about the motion of a charged particle in a uniform electric field if you are familiar with projectile motion in a gravitational field.

   One difference is that electric charges can be either positive or negative, or zero, while mass is always positive. Let's see what effect a change in sign has on the motion of the particle.

3. Rewind the applet, and set the charge to -2.0 C. Don't change any of the other settings. Then click Play .

   You should see the particle's trajectory curving to the left instead of to the right and the particle accelerating upward, rather than downward as before. This is Trajectory 1 in Figure 2 below.

   

   Figure 2

4. The particle is now accelerating in the direction opposite to that of the electric field because its charge is negative. The particle is still performing projectile motion, but this time it is "falling" upward. There is no "peak" in the trajectory in this case because the particle is already "falling" at the start of the motion.

   If you want to see a "peak" in the trajectory, click Rewind and move the particle to the top left corner of the applet window. Then set the velocity direction to 60^o S of E. (Not N of E as before!) The initial configuration should look like that for Trajectory 2 in Figure 2.

   When you click Play, you should obtain Trajectory 2. The acceleration is again upward, as for Trajectory 1. Trajectory 2 is a mirror image of the trajectory in Figure 1.

   Note that, except for the direction of the initial velocity, all parameters are the same for the two trajectories in Figure 2: charge, mass, direction of the field. Therefore, the acceleration of the particle is the same also. See Equation (3) below for the acceleration.

5. The force exerted by an electric field on a particle of charge q is equal to

   = q . (1)

   Since this is the only force acting on the particle (we are neglecting the gravitational force throughout this Lesson), Newton's second law applied to the particle gives the equation

   q = m (2)

   whence the acceleration of the particle is equal to

   = (q/m) . (3)

   Equation (3) explains why reversing the sign of the charge changes the direction of the acceleration from downward to upward. When q is positive, the electric field vector in Equation (3) is multiplied by a positive scalar and therefore the acceleration is in the same direction as the electric field, downward in the situation considered so far. When q is negative, the electric field vector is multiplied by a negative scalar, and therefore the sign of (q/m) is reversed and the acceleration is upward.

   This theory also predicts that if we reverse the direction of the electric field while keeping the charge negative, the acceleration should reverse again to its original downward direction. Let's test this.

   Rewind the applet, and set the direction of to 90^o N of E, so that points upward. The particle is in the top left corner. Then click Play. You should obtain a trajectory that curves to the right which means the particle is accelerating downward.

   To reproduce the original trajectory from Figure 1, but this time with a negative charge and an upward directed electric field, Rewind the applet, move the particle to the bottom left corner, and set the particle's velocity to a direction of 60⁰ N of E.

6. Exercise 1. Rewind the applet, and set the charge equal to 0. According to Equation (3), the particle's acceleration should now be 0. Click Play. Do you observe the kind of motion you would expect?
7. Exercise 2. Rewind the applet, and set the direction of the electric field to horizontal (so far it has been vertical). Observe the particle's motion for various initial velocities. Do you observe what you would expect?

Question 1. Suppose a charged particle starting from rest at time t = 0 is moving in a uniform electric field. The parameters describing the situation are as follows.

• : 75.0 V/m in the positive x-direction
• q: 3.0 C
• m: 1.5 kg
• (0): 100 m/s with direction pointing downward.

These conditions are illustrated in Figure 3 below.

Figure 3

What is the particle's velocity at t = 1.80 s?

Answer. We are dealing with a two-dimensional problem with constant acceleration.

By far the easiest way to work out the answer is by working with x and y components.

The x and y components of the initial velocity can be written down without calculation:

(v_x, v_y)(0) = (0, -100) m/s. (4)

The acceleration is given by Equation (3) above. It is in the positive x-direction since is in this direction and since q is positive. Thus,

= (a_x, a_y) = (a, 0) (5)

where the magnitude a of the acceleration is equal to

a = (q/m) E = (3/1.5) x 75 = 150 m/s² . (6)

Equation (A1) from the Appendix, combined with the values in Equations (4)-(6), gives for the x and y components of the velocity at t = 1.80 s

v_x(1.80) = v_x(0) + a_x t = 0 + 150x1.80 = 270 m/s, (7)

v_y(1.80) = v_y(0) + a_y t = -100 + 0x1.80 = -100 m/s . (8)

Results (7) and (8) give us the velocity of the particle at t = 1.80 s.

Question 2. Given the same conditions as in Question 1, what is the particle's displacement during the 1.80-s time interval from the start of the motion?

Answer. Equation (A4) from the Appendix, combined with the values in Equations (4)-(6), gives for the x and y components of the displacement at t = 1.80 s

d_x(1.80) = v_x(0) t + (a_x/2) t² = 0 + (150/2)x1.80² = 243 m, (9)

d_y(1.80) = v_y(0) t + (a_y/2) t² = -100x1.80 + 0 = -180 m . (10)

Results (7) to (10) for the velocity and the displacement of the particle at t = 1.80 s are illustrated in Figure 4 below.

Figure 4

>>>>> Appendix <<<<<

Figure A1

Axes and Components. In the following equations for the motion of a particle with constant acceleration it is assumed that an x,y coordinate system has been chosen and that the particle has the same acceleration everywhere, with constant x and y components a_x and a_y.

Sign Changes. In the diagram in Figure A1 above, the direction of the acceleration vector is such that both the x and y components of the vector are positive. In another situation, the direction of the vector may be different, so that one or both of the x and y components may be negative. (Or the direction of the vector may be as shown, but the directions of the x and y axes may have been chosen differently, again causing one or both components of the vector to be negative.) The equations below apply to all such situations. No signs in these equations need to be changed.

2D vs. 1D. The following equations are formulated for the 2D case, motion in two dimensions. However, they also apply to 1D motion, motion along a straight line. In this case, it will often be convenient to choose either the x or y axis to be along the line of motion. Assuming the motion is along the x axis, you will need only the equations for the x components. All y components in any of the equations will be zero. Similarly, if the motion is restricted to the y-axis, just use the y equations and ignore the x equations.

Acceleration, Velocity, and Position vs. Time:

Table 1

In vector notation, the equations for the x and y components of velocity can be combined into a single equation,

(t) = (0) + t. (A1)

Similarly, the equations for the x and y position coordinates can be combined into one vector equation,

(t) = (0) + (0)t + (/2) t². (A2)

Here, denotes the position vector whose components are (x, y).

Introducing the displacement vector (t) from the start of the motion at time t = 0 to time t which is equal to the change in the position vector,

(t) = (t) - (0), (A3)

Equation (A2) can be rewritten as

(t) = (0)t + (/2) t². (A4)

Speed vs. Displacement:

Dv² = v²(t) - v²(0) = 2a_xDx + 2a_yDy

= 2a_x[x(t) - x(0)] + 2a_x[y(t) - y(0)] (A5)

v² = v_x² + v_y². (A6)

For 1D motion, say, along the x axis, the speed vs. displacement relation reduces to

Dv² = v²(t) - v²(0) = 2a_xDx = 2a_x[x(t) - x(0)] . (A7)