At what angle does an electron leave the electric field?
Using the velocities in the x- and in the y-direction you can now also calculate the angle \(\theta\) at which the electrons leave the E-field of the plate capacitor.
Task:
When leaving the capacitor, an electron has a velocity in the x-direction of \(v_x=4.20\cdot 10^7\,\rm{\frac{m}{s}}\) ans in y-direction of \(v_y=1.68\cdot 10^7\,\rm{\frac{m}{s}}\).
Calculate the angle \(\theta\) at which an electron leaves the plate capacitor.
State what must hold for the velocity components \(v_x\) and \(v_y\) so that the exit angle is \(\theta=45°\).
Now the velocity \(v_x\) of the electrons is doubled. Explain if this halves the angle \(\theta\).
Solution
The tangent of the angle \(\theta\) ist $$\text{tan}(\theta)=\frac{v_{\text y}}{v_{\text x}}$$ ans therefore $$\theta=\tan^{-1}\left(\frac{v_{\text y}}{v_{\text x}}\right)=\tan^{-1}\left(\frac{1.68\cdot 10^7}{4.2\cdot 10^7}\right)=21.8°$$
The two velocity components \(v_x\) and \(v_y\) must be equal, only then the resulting velocity forms the angle bisector and thus provides an angle of \(\theta=45°\).
No, the angle \(\theta\) is not halved. Only the argument of \(\tan^{-1}\) is halved, but this does not lead to an angle half as large.
Hypotheses
Experiment
