Lab: Positive Particle Moves Parallel to the Electric Field

This lab consists of four parts. In Part A, you will examine the motion of a positively charged particle when it initially moves parallel to the electric field.  In Part B you will examine the motion of a positively charged particle when it initially moves perpendicularly to the electric field. In Part C you will examine how a negatively charged particle will move when it starts perpendicular to the electric field.  Finally, in Part D you will try to get the particle to move with uniform motion.

 

Purpose - Part C

 

In this part of the lab activity you will use a computer simulation to collect data enabling you to answer the following question:

  • How does a negative particle move when the direction of its initial velocity is perpendicular to the direction of the electric field?
  • How can this motion be explained using physics principles?

Procedure

 

Step 1: If you still have the simulation open from Part B of this lab activity, continue with the next step.  Otherwise, re-open the Particle in an Electric Field simulation and enter the settings described in Step 1 of Part B.

 

Step 2: To return the particle to its initial position, press "rewind".

 

Step 3: Make the following adjustments to the initial settings:

  • Change the charge of the particle to -2.0 C.
  • Ensure that the magnitude of the Electric Field is 75 V/m and the direction of the Electric Field is 270o.
  • Set the direction of the initial velocity to 0o.  Enter a value of 150 m/s for the magnitude of the initial velocity.

If these settings have been properly applied, the screen should look like this:

 


 

Step 4: Given these initial conditions, predict how the particle will move if it is released.  Press the "play" button to confirm your prediction.

 

Step 5: Use "rewind" to return to the initial settings.  Repeat the previous step, but use "pause" to stop the motion at the instant the particle starts to leave the display.  Use "Simulation Data" to view information describing the particle at this location.

 

Step 6: Collect the following data:

 

elapsed time

 

charge

 

mass

Δ t  = ________s

 

q  = ____________C

 

m  = ___________kg


 

Express the remaining values in terms of x and y components:

 

 

electric field

 


acceleration

 

initial velocity

 

final velocity

 

displacement

 = ____________V/m

 

( a x ,  a y ) =_____________m/s 2

 

( v i x ,  v i y ) =   (150, 0) m/s

 

( Δd x ,  Δd y ) = __________m

 

( v f x ,  v f y ) = ____________m/s

 

Note that the acceleration due to gravity is ignored in this simulation.

 

Analysis

 

Self-Check

Answer the following self-check (SC) questions then click the "Check your work" bar to assess your response.

 

SC 14.

a. Use calculations to verify the x component of the displacement.

  1. Use calculations to verify the y component of the displacement.

SC 15.  

Use calculations to verify the y components of the final velocity.

 

SC 16.
Use calculations to verify the magnitude and direction of the final velocity as displayed in the velocity display panel.

 

SC 17.

In this part of the lab, a particle that had an initial velocity perpendicular to a field moved through a parabolic trajectory so that it accelerated in a direction opposite to the external field. Explain why this situation could never happen in a gravitational field.

 

Self-Check Answer

SC 14.

 a.

  Given

 

 

Required

The  x  component of the displacement.

 

Analysis and Solution

Since there is no  x  component of the electric field, there is no electrostatic force acting in the  x  direction. In other words the charge moves with uniform motion in the  x  direction.

 

 

Paraphrase

The  x  component of the displacement is calculated as 369 m, which matches the value provided by the simulation.

 

b.

Given

 

Required

The  y  component of the displacement.

 

Analysis and Solution

Since the  y  component of the electric field is in the negative  y  direction, the particle experiences an electrostatic force in the positive  y  direction. This is because the charge is  negative  and experiences a force  opposite  to the direction of the electric field. Since the electrostatic force is unbalanced, the negative particle accelerates in the positive  y  direction.

 

 

Paraphrase

The  y  component of the displacement is calculated as +151 m, which matches the value provided by the simulation.

 

SC 15.

 

Given

 

Required

The  y  component of the final velocity

 

Analysis and Solution

The  y  component of the electric field is in the negative  y  direction. However because the particle is negative, it experiences a force in the positive  y  direction. (See SC 14.b. for explanation) The electrostatic force in the positive  y  direction is unbalanced, so the negative particle accelerates in the positive  y  direction.

 

 

Paraphrase

The  y  component of the final velocity is calculated to be 123 m/s. This matches the value provided by the simulation.

 

SC 16.

 

Given


 

Required

The magnitude and direction of the final velocity.

 

Analysis and Solution

The magnitude of the velocity can be found by using the Pythagorean theorem. The direction can be found by using trigonometry.

 


 

Paraphrase

The final velocity is calculated to be 194 m/s [39 o  N of E], which matches the value provided by the simulation.

 

SC 17.  

There are two types of test bodies that can be accelerated in electric fields: positively charged test bodies and negatively charged test bodies. Positively charged particles accelerate in the same direction as the electric field, and negatively charged particles accelerate opposite to the direction of the field.

 

However, in gravitational fields there is only one kind of test body: an object with mass. Objects with mass always accelerate in the same direction as the  gravitational field. So it is impossible for a test mass to be injected at right angles to a gravitational field and then accelerate in a direction that is opposite to that of the field.