## 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.
$\frac{\text{deflection on x = 5 cm}}{\text{capacitor voltage}~V_\text{p}}~\text{in}~\frac{\text{cm}}{\text{kV}}$