Electrostatics problem (electron's motion in cathode ray tube)

In summary, the conversation is about a problem involving an electron moving between two plates with an electric field. The person is having trouble with their calculations, as they are getting a displacement above the mid-axis instead of below. They go through their steps and realize they forgot to account for the negative charge of the electron. After correcting their calculations, they still get the same answer. Another person suggests there may be a numerical error or that they are looking at the wrong answer. The original poster is relieved that the issue is not with their reasoning.
  • #1
lillybeans
68
1

Homework Statement



auuwzd.jpg


Homework Equations



Fq=E*q
kinematics equations...

The Attempt at a Solution



The problem is that in the end I get a value ABOVE the mid-axis instead of below. I don't know what I'm doing wrong. Please check my reasoning.

Stage 1: While it's in between the plates where there is an electric field...
1. Find Fq (electrostatic force). Fq=E*q=2.8*104*1.6*10-19=4.48*10-15N [up]
2. Find Fg (force due to gravity). Fg=9.1*10-31*9.8=8.92*10-30N [down]
3. Find Fnet acting on the electron and its vertical acceleration. Fnet=4.48*10-15N [up] - 8.92*10-30N = 4.48*10-15 [up]. Now find acceleration. Fnet=ma. a= 4.92*1015m/s2 [up]
4. Find its vertical displacement from beginning up to the end of the plates. d=vit+1/2at2. t=0.05m/2.7*10^7m/s = 1.9*10-9s. and sub the time found into the displacement equation, i get d=0.0089m up.

Stage 2: After it leaves the plates and enter the area where there is no electric field.
1. Find final velocity (near the end of the plates and at point where it is JUST about to enter the part without the plates). V2=V1+at=0+(4.92*1015m/s2)(1.9*10-9)=9,348,000m/s [up]
2. Find vertical displacement. This time only gravity is acting on it since there is no electric field to pull the electron "up". Time is found by dividing distance by horizontal velocity which is constant. so 0.015m/2.7*107=5.6*10-9s.
d=V1t+1/2at2
d=9.348*106(5.6*10-9)+1/2(-9.8)(5.6*10-9)
2
d=0.0523m up

total displacement = 0.0523m + 0.0089m = 0.06m above the horizontal axis. What's wrong?
 
Physics news on Phys.org
  • #2
lillybeans said:

Homework Statement



auuwzd.jpg


Homework Equations



Fq=E*q
kinematics equations...

The Attempt at a Solution



The problem is that in the end I get a value ABOVE the mid-axis instead of below. I don't know what I'm doing wrong. Please check my reasoning.

Stage 1: While it's in between the plates where there is an electric field...
1. Find Fq (electrostatic force). Fq=E*q=2.8*104*1.6*10-19=4.48*10-15N [up]
2. Find Fg (force due to gravity). Fg=9.1*10-31*9.8=8.92*10-30N [down]
3. Find Fnet acting on the electron and its vertical acceleration. Fnet=4.48*10-15N [up] - 8.92*10-30N = 4.48*10-15 [up]. Now find acceleration. Fnet=ma. a= 4.92*1015m/s2 [up]
4. Find its vertical displacement from beginning up to the end of the plates. d=vit+1/2at2. t=0.05m/2.7*10^7m/s = 1.9*10-9s. and sub the time found into the displacement equation, i get d=0.0089m up.

Stage 2: After it leaves the plates and enter the area where there is no electric field.
1. Find final velocity (near the end of the plates and at point where it is JUST about to enter the part without the plates). V2=V1+at=0+(4.92*1015m/s2)(1.9*10-9)=9,348,000m/s [up]
2. Find vertical displacement. This time only gravity is acting on it since there is no electric field to pull the electron "up". Time is found by dividing distance by horizontal velocity which is constant. so 0.015m/2.7*107=5.6*10-9s.
d=V1t+1/2at2
d=9.348*106(5.6*10-9)+1/2(-9.8)(5.6*10-9)
2
d=0.0523m up

total displacement = 0.0523m + 0.0089m = 0.06m above the horizontal axis. What's wrong?

Don't forget that an electron has a negative charge.
 
  • #3
Oops. Right, I forgot that the upward electric field is referring to a proton. thanks, I'm going to try again and see if it works out.
----

I just reversed the signs and in the end I got the same value, which is 6cm except now it's downwards. The correct answer is 2.5cm. I don't know why I'm not getting that.
 
  • #4
lillybeans said:
Oops. Right, I forgot that the upward electric field is referring to a proton. thanks, I'm going to try again and see if it works out.
----

I just reversed the signs and in the end I got the same value, which is 6cm except now it's downwards. The correct answer is 2.5cm. I don't know why I'm not getting that.

Good to see you got the same sized answer since the weight of the electron is insignificant compared to the electrical force involved.
There must be a numerical error somewhere [or you have looked at the wrong answer] It is not obvious to me where it is at the moment.
 
  • #5
Thank you, PeterO. Numerical errors I can deal with (well not really), I just can't stand an error in my reasoning. Thanks once again!
 

Related to Electrostatics problem (electron's motion in cathode ray tube)

1. What is a cathode ray tube (CRT)?

A cathode ray tube is a vacuum tube that uses an electron beam to create images on a screen. It was commonly used in older television sets and computer monitors.

2. What is the role of electrostatics in a CRT?

Electrostatics is the branch of physics that deals with the study of electric charges at rest. In a CRT, electrostatics is essential in controlling the motion of electrons to create an image on the screen.

3. How does an electron move inside a CRT?

Electrons are emitted from the cathode and accelerated towards the positively charged anode in a CRT. They are then focused and deflected using electrostatic and magnetic fields to create an image on the screen.

4. What is the purpose of the grid in a CRT?

The grid in a CRT is used to control the intensity of the electron beam. By varying the voltage on the grid, the number of electrons passing through it can be controlled, thus controlling the brightness of the image on the screen.

5. How does the deflection of electrons in a CRT work?

The deflection of electrons in a CRT is achieved by using magnetic fields. The electrons are deflected based on the strength and direction of the magnetic field, allowing them to scan across the screen and create an image.

Similar threads

Electrons in a cathode ray tube
    • Introductory Physics Homework Help
Replies
4
Views
1K
Cathode ray tube, Voltage needed to move the beam impact spot
    • Introductory Physics Homework Help
Replies
3
Views
1K
Cathode Ray Tube and magnetic field Question
    • Introductory Physics Homework Help
Replies
1
Views
2K
Field strength to deflect electron in a cathode-ray tube
    • Introductory Physics Homework Help
Replies
16
Views
7K
Calculate the deflection sensitivity in mm per volt
    • Introductory Physics Homework Help
2
Replies
44
Views
10K
Deflection of electron beam in cathode ray tube
    • Introductory Physics Homework Help
Replies
4
Views
5K
Archived Displacement and Momentum, particle in cathode ray tube with variables
    • Introductory Physics Homework Help
Replies
2
Views
4K
How do we know the mass of an electron?
    • Electromagnetism
Replies
28
Views
2K
Can a cathode ray tube be made from a light bulb?
    • Electrical Engineering
Replies
15
Views
4K
Kinematics Electrons In a Cathode Ray Tube
    • Introductory Physics Homework Help
Replies
2
Views
3K

Hot Threads

Recent Insights

Back
Top