At which point do the electrons leave the plate capacitor?
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.
Solution
The deflection at the end of the capacitor must be calculated. So \(x=l=12\,\rm{cm}\) must be plugged in (1) for x: $$\Delta y_1=\frac{2\,\text{kV}}{4\cdot 6\,\text{cm}\cdot 5\,\text{kV}}\cdot (12\,\text{cm})^2=2.4\,\text{cm}$$
If the entry point shifts downward by \(1\,\rm{cm}\), the exit point also shifts downward by \(1\,\rm{cm}\), because the deflection itself does not depend on the entry point. The force on the electrons is the same everywhere in the homogeneous electric field.
The plate voltage \(V_{\rm{p, max}}\) must be calculated, so equation (1) must be solved for \(V_{\rm{p}}\). The maximum possible deflection with centered injection is \(\Delta y=3\,\rm{cm}\). Thus follows:\[V_{\rm{p, max}}=\frac{4\cdot d\cdot \Delta y\cdot V_{\rm{a}}}{x^2}\]\[\Rightarrow V_{\rm{p, max}}=\frac{4\cdot 6\,\rm{cm}\cdot 3\,\rm{cm}\cdot 5\,\rm{kV}}{\left(12\,\rm{cm}\right)^2}=2.5\,\rm{kV}\]
Hypotheses
Experiment
