# Cathode Ray Tube

1. Jan 22, 2005

### pmb_phy

Its funny how one can forget how to do very simple things.

Here is a problem I've been working on and have a mental block and can't get the darn solution. It's from Krauss's EM text
Note: Assume non-relativistic motion. I'm assuming beam enters between places with d/2 on each side

First thing to do is to find vx which is found from the kinetic energy which it gains as it accelerates through a difference of potential Va. Therefore

$$K = \frac{1}{2}mv_x^2 = eV_a$$

solve for $$mv_x^2$$ to get

$$v_x = \sqrt{\frac{2eV}{m}}$$

The y-component of velocity is found through

$$F = ma_y = eE_y = eV_d/d$$

$$a_y = \frac{eV_d}{md}$$

$$y = a_y t^2 = \frac{eV_d}{md}t^2$$

where t is the time it takes the particle to travel a distance l. So t = l/vx. Upon substitution this gives

$$y = a_y t^2 = \frac{eV_d}{md}\frac{l^2}{(ev_x)^2}$$

$$y = \frac{V_d}{d}\frac{l^2}{2V_a}$$

$$y = \frac{V_d l^2}{2V_ad}$$

That's about as far as I've gotten. Unless they were thinking that one should find v_y/v_x and note that v_y/v_x = y/x??

Pete

Last edited: Jan 22, 2005
2. Jan 22, 2005

### dextercioby

Okay,Pete,u're on the right track.But u didn't mention what your problem was... What's buggin'u??

Daniel.

3. Jan 22, 2005

### pmb_phy

Sorry Dan. I had to run off in the middle of doling that since my ride showed up early. When I got back I realize my error. So I'll work it out and paste it here anyway. I guess I just needed to describe it to someone else to figure what I did wrong.

Thanks

Pete

Last edited: Jan 22, 2005
4. Jan 23, 2005

### pmb_phy

I still need help here folks. I just can't see how to arrive at that answer. It must be all that snow outside! :rofl:

Pete

5. Jan 23, 2005

### dextercioby

The equations of motion (for connstant acceleration) are:
$$v_{x}(t)=v_{0,x}+a_{x}t$$

$$v_{y}(t)=v_{0,y}+a_{y}t$$

$$x(t)=x_{0}+v_{0,x}t+\frac{1}{2}a_{x}t^{2}$$

$$y(t)=y_{0}+v_{0,y}t+\frac{1}{2}a_{y}t^{2}$$

Chose axis of coordinates for your problem in a convenient way (par éxample,the Electric field should be along one of the 2 axis).You know the inital conditions and the expresion for the acceleration (electric force divided by mass).

You need to find the trajectory (namely that parabola arch).From the geometry of the problem,u should get the result pretty easily.

Good luck!!

Daniel.

6. Jan 23, 2005

### pmb_phy

It was much easier than I thought. I guess I just didn't make the connection that when the charge leaves the plates they travel in straight lines and therefore vy/vx = y/x.

The charge is accelerated through a potential difference, Va, in the x-direction and it follows from that and K = eVa that

$$v_x = \sqrt{\frac{2eV_a}{m}}$$

which can be written as

$$\frac{1}{v_x^2} = \frac{m}{2eV_a}$$

The charge travels through the deflecting plates in a time $T = L/v_x$

The charge will accelerate in the y-direction by the amount

$$a_y = \frac{eV_d}{md}$$

Therefore the particle's vy increases from zero to vy as

$$v_y = \frac{eV_d}{md}\frac{L}{v_x}$$

Now divide each side by vx to find

$$\frac{v_y}{v_y} = \frac{eV_d}{md}\frac{L}{v_x^2}$$

Substitute the expression for the square of vx and substitute

vy/vx = y/x

on the left and multiply by x and the solution yields. Yay!

Pete

Last edited: Jan 23, 2005
7. Jan 23, 2005

### dextercioby

$$y=\frac{1}{2}a_{y}t^{2}$$

Pay attention!!That's why it wasn't any 2 in your formula...

Daniel.

8. Jan 23, 2005

### pmb_phy

What is it that you didn't think I was paying attention to Dan?

Pete

9. Jan 23, 2005

### dextercioby

That's rotten.I cannot believe how u managed to pull it through... :uhh:

Daniel.

10. Jan 23, 2005

### krab

I know you've found the solution already, but here's a tidy way to do it.
$${eV_d\over d}=ma_y= m{d^2y\over dt^2} = mv^2\,y''=2eV_a\,y''$$
(which is actually the "paraxial ray approximation", prime is derivative wrt x)
So
$$y''={1\over 2d}{V_d\over V_a}$$
Integrate through the plates to get
$$\Delta y'={L\over 2d}{V_d\over V_a}$$
and integrate from end of plates to screen to get the desired answer.