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:

A plate capacitor is \(l=12\,\rm{cm}\) long, the distance between the plates is \(d=6.0\,\rm{cm}\) and the plate voltage is \(V_{\rm{p}}=2.0\,\rm{kV}\).

- Calculate the electorn deflection $\Delta y_1$ at the end of the plate capacitor if the acceleration voltage is \(V_{\rm{a}}=5\,\rm{kV}\) and electrons are injected in the middle between the two plates.
- Explain how the exit location changes when the electrons are no longer fired into the plate capacitor exactly in the center, but \(1\,\rm{cm}\) further down.
- Calculate how large the plate voltage \(V_{\rm{p}}\) may be at maximum during the central injection with \(V_{\rm{a}}=5\,\rm{kV}\), so that the electrons just leave the capacitor at the end and do not hit the capacitor plate.