controls:

acceleration voltage Va
plate voltage Vp

$E=\frac {V_{\text p}}{\text d}=$ ${0}$ $\frac {\text V}{\text m}$

schematic diagram:

For the deflection of the electrons holds: the higher $V_\text{p}$, the bigger the deflection y(x).

Check if we have also: $y(x) \sim V_\text{p}$.
For checking the acceleration voltage $V_\text{a}$ is set to 3,5 kV. Complete the table at bottom of the page, by modifying the capacitor voltage and getting the missing values out of the experiment.
capacitor voltage 0,8 kV 1,6 kV 2,4 kV 3,2 kV 4,0 kV
deflection on x = 5 cm cm cm cm cm cm
$\frac{\text{deflection on x = 5 cm}}{\text{capacitor voltage}~V_\text{p}}~\text{in}~\frac{\text{cm}}{\text{kV}}$