# Homework Help: Cathode ray tube, Voltage needed to move the beam impact spot

1. May 9, 2017

### Opo

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. May 9, 2017

### 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?

3. May 9, 2017

### Opo

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.

4. May 9, 2017

### 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?

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