How can I calculate the deflection of an electron in a cathode ray tube?

[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

(page 206 Problem 5-4-12)

Cathode-ray tube. Electric field deflection Show that thte deflection of an electron at the screen of a cathode-ray tube is given by

y = \frac{V_d l x}{2V_a d}

where

y = deflection distance at screen, m
V_d = deflecting potential, V
L = length of deflecting plate, m
x = distance from deflecting plates to screen, m
V_a = accelerating potential, V
d = spacing of deflecting plates, m

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 v_x which is found from the kinetic energy which it gains as it accelerates through a difference of potential V_a. 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/v_x. 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

 

[dextercioby]

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

Daniel.

 

[pmb_phy]

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

Daniel.

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

 

[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!

Pete

 

[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.

 

[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 v_y/v_x = y/x.

The charge is accelerated through a potential difference, V_a, in the x-direction and it follows from that and K = eV_a 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 v_y increases from zero to v_y as

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

Now divide each side by v_x to find

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

Substitute the expression for the square of v_x and substitute

v_y/v_x = y/x

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

Pete

 

[dextercioby]

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

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

Daniel.

 

[pmb_phy]

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

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

Daniel.

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

Pete

 

[dextercioby]

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


Pete

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

Daniel.

 

[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.