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.

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.