How large is the electron deflection when impinging on a screen placed in the distance s to the plate capacitor?
In technical applications for example an oszilloscope a screen is placed in distance \(s\) after the plate capacitor. This screen shows the impinge of electrons. So now the deflection \(\Delta y_{\rm{screen}}\) of the electrons at the screen must be calculated.
Task:
Electrons are leaving the plate capacitor (see picture) with an deflection of \(\Delta y_1=2.4\,\rm{cm}\). The electron velocity in x-direction is \(v_x=4.20\cdot 10^7\,\rm{\frac{m}{s}}\) ans in y-direction \(v_y=1.68\cdot 10^7\,\rm{\frac{m}{s}}\). The distance between the end of the capacitor and the screen is \(s=8\,\rm{cm}\).
Calculate the electron deflection \(\Delta y_{\rm{screen}}\) when impinging on the screen.
Explain how the deflection on the screen from task a) changes when you double the distance \(s\) of the screen from the end of the plate capacitor.
Solution
The deflection \(\Delta y_2\) is $$\Delta y_{\text {2}}=v_{\text y}\cdot t\quad \text{in with}\quad t=\frac{s}{v_{\text x}}$$ So it follows $$\Delta y_{\text {2}}=v_{\text y}\cdot \frac{s}{v_{\text x}}=\frac{1.68\cdot 10^7\,\frac{\text m}{\text s}\cdot 8\,\text{cm}}{4.2\cdot 10^7\,\frac{\text m}{\text s}}=3.2\,\text{cm}$$So for the deflection \(\Delta y_{\rm{screen}}\) is$$\Delta y_{\text {screen}}=\Delta y_{\text {1}}+\Delta y_{\text {2}}=2.4\text{cm}+3.2\text{cm}=5.6\,\text{cm}$$Alternative the general equation for such a setup can be used:$$\Delta y_{\text {screen}}=\frac{V_{\text p}\cdot l}{2\cdot d\cdot V_{\text a}}\left(\frac{l}{2}+s\right)$$
The deflection \(\Delta y_1\) is not affected by a change of the screen position. The doubling of the distance \(s\), however, leads to a doubling of the flight time \(t\) after leaving the capacitor. This also doubles the deflection \(\Delta y_2\). Therefore, the deflection just increases by \(3.2\,\rm{cm}\) compared to part a). Overall, the deflection is thus \[\Delta y_{\rm{screen}}=2.4\,\rm{cm}+6.4\,\rm{cm}=8.8\,\rm{cm}\]