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.

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 $$\theta$$.