# Urgent CRT question

1. Feb 13, 2005

### Coldie

Our Physics review sheet includes two questions involving "accelerating voltage", "deflecting voltage", and "screen deflection", which are involved with Cathode Ray Tubes. The chapter on this in the book basically outlined what one was and how it worked, but it did not make use of any of these terms and I'm at a loss at how to solve the problems.

One of them is:
An accelerating voltage of 750 V produces a screen deflection of 4.2 cm on a CRT. If the deflecting voltage is kept constant but the accelerating voltage is increased to 1000 V, what will the deflection become?

There appear to be two voltages to keep track of, and so none of my equations seem to apply here. Would someone please explain to me what to do?

2. Feb 14, 2005

### dextercioby

What's the formula giving the deflection on the screen...?

Daniel.

3. Feb 14, 2005

### Coldie

The book gives none, and it's not on our equations sheet.

4. Feb 14, 2005

### dextercioby

That's weird.You could compute it,if one knew the geometry of the figure (how the (plane plates) capacitor is situated and how the particles enter the area with the deflecting field...

Sorry.I cannot help.

Daniel.

5. Feb 14, 2005

### Coldie

Is it possible that the Electric field remains constant, and therefore I could solve it via:
$$E = \frac{V}{D}$$

$$\frac{750}{.042} = \frac{1000}{D}$$

$$D = .056m$$?

6. Feb 14, 2005

### dextercioby

How could you prove that proportionality...?

Daniel.

7. Feb 14, 2005

### Coldie

If the electric field is constant in a Cathode Ray Tube, which I'm guessing it could mean by stating that "the deflecting voltage is kept constant", then would I need to?

8. Feb 14, 2005

### dextercioby

Okay,have it your way.Check the answer and if it is correct,then it's okay.

Daniel.

9. Feb 14, 2005

### Coldie

I'm sorry if I caused any offense... I was trying to explain my reasoning. Can you think of another way to do this with basic Electrostatics equations?

10. Feb 14, 2005

### dextercioby

On a second thought,your answer is wrong.I could give the justification,based on physical arguments.If the accleration potential is larger,then the initial velocity is bigger and the time the particle spends between the plates of the capacitor is smaller,therefore the deviation is smaller.
Ergo,it's the other way around...The deflection decreases...

Daniel.

11. Feb 14, 2005

### dextercioby

I just did.

Daniel.

12. Feb 14, 2005

### Coldie

Well, please note that we've done absolutely nothing on Cathode Ray Tubes in class, and he didn't even tell us to read the one section in our book that mentions them(which doesn't help to solve this question), so I really can't follow you at all. The only thing that I'm looking for is an equation to solve the problem.

13. Feb 14, 2005

### dextercioby

I've said to use the same equation,but with reversed outcomes:
$$D=\frac{0.042\times 750}{1000}$$

Daniel.

14. Feb 14, 2005

### pervect

Staff Emeritus
Much of the intent seems almost clear to me. The accelerating voltage determines the velocity of the electrons.

Most CRT's use magnetic deflection, but because the textook talks about "deflecting voltage", I'd assume that there is some constant E-fiield that deflects the electrons.

The deflection distance is just how far the electrons move on the scrreen due to the deflection voltage, i.e. you measure the distance the spot on the screen moves when you remove the deflection voltage.

Now, when you boost the accelerating voltage, you speed up the electrons, so they deflect less.

The problem is that we don't know the region of space in which the E-fields are located that cause the deflection. Two different approximations come to mind

1) The E-field is contant everywhere - this leads to a quadratic dependence of deflection on transit time. (Transit time is inversely proportional to velocity).

2) The E-field is only present during the first part of the flight. The lateral velocity imparted will be inversely proportional to the time it takes to traverse the electrodes. If we ignore the deflection that occurs during the acutal transit of the electrodes and assume that it's small, the deflection will be proportional to the lateral velocity the electron has when it leaves the electrode region (we multiply the lateral velocity times the distance the electron travels after it leaves the deflection area.)

I'd guess they problably want you to use case 2, and igonre the quadratic nature of the deflection that happens when th electron is actually moving between the charged deflecting plates. But this is just a guess.

15. Feb 14, 2005

### Coldie

Turns out we hadn't done it yet. The equation he gave us was $$y = k\frac{V_d}{V_a}$$, with y being the screen deflection distance and k being $$\frac{1}{4} x \frac{L^2}{d}$$ with L being the length of the deflecting plates and d being the distance between them, though this question didn't require an in-depth understanding of k. Had nobody seen this equation before?