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 v_{x} which is found from the kinetic energy which it gains as it accelerates through a difference of potential V_{a}. Therefore

[tex]K = \frac{1}{2}mv_x^2 = eV_a[/tex]

solve for [tex]mv_x^2[/tex] to get

[tex]v_x = \sqrt{\frac{2eV}{m}}[/tex]

The y-component of velocity is found through

[tex]F = ma_y = eE_y = eV_d/d[/tex]

[tex]a_y = \frac{eV_d}{md}[/tex]

[tex]y = a_y t^2 = \frac{eV_d}{md}t^2[/tex]

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

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.

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.

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

[tex]v_x = \sqrt{\frac{2eV_a}{m}}[/tex]

which can be written as

[tex]\frac{1}{v_x^2} = \frac{m}{2eV_a}[/tex]

The charge travels through the deflecting plates in a time [itex]T = L/v_x[/itex]

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

[tex]a_y = \frac{eV_d}{md}[/tex]

Therefore the particle's v_{y} increases from zero to v_{y} as

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