Why can the trajectory of the electrons be described with $\bbox[5px,border:2px solid red]{y(x)=\frac{V_{\text{p}}}{4\cdot d\cdot V_{\text{a}}}\cdot x^2}$ ?
First we inspect the motions in x- and y-direction separately:
x-Richtung:
After leaving the Electron Gun no force affects the motion in horizontal direction. So there is a linear motion, which can be described with following equations:$$\begin{equation}x(t)= v_0\cdot t\end{equation}$$
$$\begin{equation}v(t)= v_0\end{equation}$$
$$\begin{equation}a(t)=0\end{equation}$$
$v_0$ is the velocity of the electrons when they leave the Electron Gun:$$\begin{equation} v_0=\sqrt{2 \cdot \frac {e}{m_e}\cdot V_{\text a}} \end{equation}$$
y-direction:
In y-direction the electric field of the plate capacitor affects the motion of the electrons:$$\begin{equation} {F_{el}=E\cdot e =\frac{V_{\text p}\cdot e}{\text{d}}} \end{equation}$$
So there is an acceleration described by following equations:$$\begin{equation}y(t)=\frac{1}{2}a_y\cdot t^2\end{equation}$$
$$\begin{equation}v_y(t)=a_y\cdot t\end{equation}$$
$$\begin{equation}a_y(t)=\frac{F_{el}}{m_e}=\frac{V_{\text p}\cdot e}{m_e\cdot \text{d}}\end{equation}$$
Elimination of t:
Now we combine the motions. Therefor we eliminate the variable t out of equation (6)
To do so, we transform and insert equation (1). This yield:$$\begin{equation}y(x)=\frac{1}{2} \cdot \frac{a_y}{{v_0}^2}\cdot x^2\end{equation}$$
Plugging in (8) instead of $a_y$ yields the equation:$$\begin{equation}y(x)=\frac{1}{2} \cdot \frac{V_{\text p}\cdot e}{m_e\cdot \text d\cdot {v_0}^2}\cdot x^2\end{equation}$$
Squaring and plugging in equation (4) in place of ${v_0}^2$ and cancel $m_e$ and $e$ yield:$$\begin{equation}\bbox[5px,border:2px solid red]{y(x)=\frac{V_{\text p}}{4\cdot \text{d}\cdot V_{\text a}}\cdot x^2} \end{equation}$$
Hypotheses
Experiment
