Oscilloscope

The magnitude of the force on the electron is $F=Ee=6000\text{N/C}\cdot1.6\cdot10^{-19}\text{C}=9.6\cdot10^{-16}\text{N}$ .

The acceleration is $a=\frac{F}{m}=\frac{9.6\cdot10^{-16}\text{N}}{9.1\cdot10^{-31}\text{kg}}\approx1.05\cdot10^{15}\text{m/s}^2$

Since the motion is accelerated we must use instantaneous velocities to find the angle. We can use $2as=v_0^2+v_f^2$ to get the vertical component. We already have the horizontal component $v_0$ which equals $8\cdot10^6\text{m/s}$ . The initial vertical component is 0.

The distance travelled is $s=\frac{at^2}{2}$ and we can get the time by looking at the horizontal components: $t=\frac{x}{v_0}=\frac{0.04\text{m}}{8\cdot10^6\text{m/s}}=5\cdot10^{-9}\text{s}$

Solving, we get $s=1.3125\text{cm}$ and the acceleration is $a=1.05\cdot10^{15}\text{m/s}^2$ .

$v_f=\sqrt{2as}=\sqrt{2\cdot1.05\cdot10^{15}\text{m/s}^2\cdot0.013125\text{m}}=5.25\cdot10^6 \text{m/s}$ .

The angle is therefore $\tan\alpha=\frac{v_f}{v_0} \to \alpha=\arctan(0.65625) \to \alpha=33.27°\approx33°$

The electron will not be affected by the acceleration anymore since it left the oscilloscope so it will keep the instantaneous velocities we just calculated. Time needed to reach the screen from this point is $t_s=\frac{D}{v_0}=1.5\cdot10^{-8}\text{s}$ .

The additional vertical distance from the original course is $s_s=v_ft_s=7.875\text{cm}$

Total: $L=s+s_s=1.3125\text{cm}+7.875\text{cm}=9.1875\text{cm}\approx9\text{cm}$

So the answer is $\alpha+L=33+9=42$