1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Cathode ray tube, Voltage needed to move the beam impact spot

  1. May 9, 2017 #1

    Opo

    User Avatar

    1. The problem statement, all variables and given/known data
    Question: A cathode ray tube in a TV has an electron gun operating at 10 kV.

    a: What is the velocity (in m.s-1) at which the electrons are emitted? Answer:5.9x107

    b: How many volts are needed between the deflector plates to move the beam impact spot across a 70 cm screen if the screen is 120 cm from the deflector?

    2. Relevant equations
    Tan(θ)=vE/vI
    1/2mv2
    QE=qV

    3. The attempt at a solution
    First attempt:
    So first I worked out the angle the beam has to move which was 30.3 degrees. Then I used the equation Tan(angle) = velocity in the direction of the field/ initial velocity. Which was Tan(30.3) X 5.9x107 = 3.4x107

    Then I used 1/2 x 9.1x10-31 x (3.4x107 )2 to get 5.3x10-16 <I think I'm going wrong here. When I read through the lecture slides things become limited.

    then PE = qV, so PE/q = V, so 5.3x10-16 / 1.6x10-19 = 3287

    Then I was given this advice:
    "
    The problem is that velocity is a vector quantity, while energy is not.

    So it seems that you appropriately worked out the velocity in the direction of the field. (I'll call this ve, and the initial velocity vi)

    What you should do then is find the total velocity vtot=sqrt( ve2 + vi 2 ). You then use this number to find the total KE of the electron. Subtract the initial KE (find with vi) and this gives you the energy (PE) which must be added by the electric field."
    So I tried this, sqrt(5.9x1072+3.4x1072) = 6.81x107

    So Total KE = 0.5x9.1x10-31x6.81x1072 =2.11x10-15
    Initial KE = 0.5x9.1x10-31x5,9x1072 = 1.583855x10^-15

    2.11x10-15 - 1.583855x10-15 = 5.26145x10-16

    Then QE=qV, so QE/q = V

    So 5.26145x10-16/1.6x10-19 = 3288.40625 But this is also wrong.
     
  2. jcsd
  3. May 9, 2017 #2

    gneill

    User Avatar

    Staff: Mentor

    If the screen is 70 cm wide, doesn't the defection have to be only half of that in one direction, half in the other direction, in order to cover the full width of the screen?
     
  4. May 9, 2017 #3

    Opo

    User Avatar

    So how should I adjust my answer ? I went from 70cm to 35cm (if that's what you meant?) which changed the angle from 30.3ο to 16.3ο but the answer is still wrong.
     
  5. May 9, 2017 #4

    gneill

    User Avatar

    Staff: Mentor

    What transverse velocity did you find for this deflection angle? (That's the velocity component due to the deflection).

    I think you need to make a detailed sketch of the scenario. How much of the available potential difference can the electron "fall though" between the deflection plates if it enters the plates midway between them?
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Cathode ray tube, Voltage needed to move the beam impact spot
  1. Cathode Ray Tube (Replies: 1)

  2. Cathode Ray Tubes (Replies: 3)

Loading...