controls:

acceleration voltage Va

$v_0= \sqrt{2\cdot \frac{e}{m}\cdot V_{\text a}} =$ ${2.652\cdot 10^7}$ $\frac{\text m}{\text s}$

plate voltage Vp

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

schematic diagram:

Schematic layout of the experiment

Use the experiment to test and rate your hypotheses!
- the weaker E, the bigger the deflection
- the higher v0, the bigger the deflection
Real experiment about deflection of electrons in electric field of a plate capacitor