At which point do the electrons leave the plate capacitor?
Often teachers want to know: how large is the electron deflection $\Delta y$ at the end of the plate capacitor? This question can be answered with the formula derived from the equations of motion:$${y(x)=\frac{V_{\text p}}{4\cdot \text{d}\cdot V_{\text a}}\cdot x^2}\qquad (1)$$
Task:
Calculate the electron deflection $\Delta y_1$ when leaving the electric field. Given: Acceleration Voltage Va = 5 kV Plate Voltage Vp = 2 kV Plate Distance d = 6 cm Length of Capacitor l = 12 cm
Solution
The deflection at the end of the capacitor must be calculated. So x = l = 12 cm. 12 cm must be plugged in (1) for x: $$\Delta y_1=\frac{2 \text{kV}}{4\cdot 6\text{cm}\cdot5\text{kV}}\cdot (12\text{cm})^2=2{,}4\,\text{cm}$$
Hypotheses
Experiment
