Finding deflection of an electron through 2 charged plates when given initial velocity I've been trying to relate the initial velocity of an electron to the deflection created based on the electric field between 1 pair of plates. The 2nd half of page 3 of this pdf is what I'm concerned with. Now, I was trying to derive y (the deflection distance), where e is elementary charge, and let $\theta$ be the trajectory of the electron immediately after leaving the plates ($\sin{\theta}$ is equal to $\frac{\Delta y_1}{L}$ and probably equal to $\frac{v_y}{v}$ in the diagram, this could be part of the misunderstanding).
Explaining my approach for change in y-position through the plates only:
using $F=ma$ and $F=qE$ and $a=\frac{qE}{m}$ and $t=\frac{L}{v_x}$ where L is the length of a plate (total horizontal distance that the electric field between the plates acts on the electron) and $v_x$ is the initial velocity I get:
$$\Delta y_1=\frac{1}{2}\frac{qE}{m}\biggl(\frac{L}{v_x}\biggr)^2$$ by kinematics.
Accounting for the remaining distance (not between plates):
My thought was that $\Delta y_1=\frac{eEL^2}{2mv_x^2}$ is the change in the y-position while between the plates, so the remaining $D\sin{\theta}$ would be the extra change in y-position after leaving the plates:
$$y=D\sin{\theta} + \frac{eEL^2}{2mv_x^2}$$
vs what the pdf has:
$$y=\frac{eEL}{mv_x^2}\biggl(D+\frac{L}{2}\biggr)$$
This is almost the same as what I have, the only difference is I have $D\sin{\theta}$ instead of $D\frac{eEL^2}{2mv_x^2}$ and I'm trying to figure out my misunderstanding.
Can someone please explain where my misunderstanding is? I can't tell why my derivation is incorrect?
 A: What you did here was take the time that the charge moves from start of the plate to the end of it, which you had as $\frac{L}{v_x}$ But the  the time of flight of the electron to the screen is $\frac{L+D}{v_x}$ because $v_x$ stays the same from the point where the electron is released until it reaches the screen (not just the end of the plates). This is significant in finding the correct equation.
So
$$t=\frac{L+D}{v_x}$$ meaning $$\Delta y=\frac{1}{2}\frac{eE}{m}\biggl(\frac{L+D}{v_x}\biggr)^2$$ since $$y=\frac12 at^2$$ So
$$\Delta y=\frac12 \frac{eE}{m}\biggl(\frac{L^2+2LD+D^2}{v_x^2}\biggr) =\frac12 \frac{eE}{m}\biggl(\frac{2LD+L^2}{v_x^2}\biggr)$$ Note here that the author has assumed that since $D$ is very small (compared to the other dimensions) then $D^2\rightarrow 0$
Now if you factor out $\frac{2L}{v_x^2}$ gives $$\Delta y=\frac12 \frac{eE2L}{mv_x^2}({D+\frac{L}{2}})\\
\rightarrow \Delta y=\frac{eEL}{mv_x^2}({D+\frac{L}{2}})$$
A: I think that I need to write another answer to explain my original answer in more detail and to agree with the comments that @CottonHeadedNinnymuggins made with regard to the answer @josephh produced which I believe to be erroneous as it assumes a constant acceleration in the $y$ direction both inside and outside the electric field.
The deflection part of the experimental arrangement is as follows.

The motion of the electron consists of two parts.
Between $A$ and $B$ where there is a uniform electric field in the $-y$ direction an electron is accelerated in the $y$ direction with a constant acceleration of $\frac{eE}{m}$ and moves with a constant velocity $v_{\rm x}$ in the $x$ direction as there is no force on the electron in that direction.
Between $B$ and $C$ the electron moves with a constant velocity $v$ with components $v_{\rm x}$ and $v_{\rm y}$ on the assumption that there is no electric field in this region.
For region $AB$, applying $y = ut + \frac 12 at^2 \Rightarrow y_{\rm AB} = \frac 12 \frac {eE}{m} \left ( \frac {L}{v_{\rm x}}\right )^2$
The motion of an electron in the region $BC$ is linear and inclined at an angle $\theta$ to the straight through trajectory with $\tan \theta = \frac{v_{\rm y}}{v_{x}}$ the time taken to traverse this region is $t_{\rm BC}=\frac{D}{v_{\rm x}}$.
Now the $y$ component of velocity on entering the region $BC$ is $v_{\rm y} = \frac {eE}{m} t_{\rm AB}$ where $t_{\rm AB}= \frac{L}{v_{\rm x}}$.
So the extra deflection in region $BC$ is $y_{\rm BC} = \frac {eE}{m} \cdot \frac{L}{v_{\rm x}}\cdot \frac{D}{v_{\rm x}}$.
Hence $y_{\rm AC} = \dfrac{eEL}{mv_{\rm x}^2}\left (D+\dfrac{L}{2}\right)$
A: You have found yourself the cause of your misunderstanding,

($\sin{\theta}$ is equal to $\frac{\Delta y_1}{L}$ and probably equal to $\frac{v_y}{v}$ in the diagram, this could be part of the misunderstanding).

So let me clarify it for you so you really understand.
The only meaningful $\sin{\theta}$ is in fact $\frac{v_y}{v}$ , the local slope of the trajectory.
Integrating $v_y$ over the trajectory gives you $\Delta y$, but since $v_y$ is not constant over the entire trajectory, the quantity $\frac{\Delta y_1}{L}$ is not simply related to $\frac{v_y}{v}$. So $\frac{\Delta y_1}{L}$ is not a useful object during the calculation. It is the slope of the curve determined by $\frac{v_y}{v}$ at distance $L$ that gives you $\sin{\theta}$.
In fact, just by looking at the picture, you see that if the slope of the trajectory at $L$ is pushed back to the origin in $x$, you dont hit the initial point, but below that. So the slope is clearly larger than $\frac{\Delta y_1}{L}$ !
It turns out the slope is exactly twice this latter value, but this is in fact irrelevant, since $\frac{\Delta y_1}{L}$ is not a useful quantity.Of course ${\Delta y_1}$ is important, but dividing it by $L$ does not tell you anything useful.
