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

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Homework Help Overview

The discussion revolves around an electrostatics problem involving the motion of an electron in a cathode ray tube, specifically analyzing its trajectory under the influence of an electric field and gravity.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants describe the forces acting on the electron, including electrostatic force and gravitational force, and calculate net forces and accelerations. There are attempts to find vertical displacements during different stages of motion. Some participants question the correctness of their calculations and reasoning, particularly regarding the direction of forces and the resulting displacement.

Discussion Status

Participants are actively engaging with the problem, checking each other's reasoning and calculations. There is recognition of potential numerical errors and a focus on understanding the implications of the electron's negative charge in the context of the electric field direction. No consensus has been reached on the correct final displacement.

Contextual Notes

Participants note that the weight of the electron is negligible compared to the electrical force, and there is a mention of the correct answer being different from their calculations, indicating possible misunderstandings or errors in their approach.

lillybeans
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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?
 
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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.
 
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.
 
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.
 
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!
 

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