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 \(\theta\).