Answer:
xmax = 9.5cm
Explanation:
In this case, the trajectory described by the electron, when it enters in the region between the parallel plates, is a semi parabolic trajectory.
In order to find the horizontal distance traveled by the electron you first calculate the vertical acceleration of the electron.
You use the Newton second law and the electric force on the electron:
[tex]F_e=qE=ma[/tex] (1)
q: charge of the electron = 1.6*10^-19 C
m: mass of the electron = 9.1*10-31 kg
E: magnitude of the electric field = 4.0*10^2N/C
You solve the equation (1) for a:
[tex]a=\frac{qE}{m}=\frac{(1.6*10^{-19}C)(4.0*10^2N/C)}{9.1*10^{-31}kg}=7.03*10^{13}\frac{m}{s^2}[/tex]
Next, you use the following formula for the maximum horizontal distance reached by an object, with semi parabolic motion at a height of d:
[tex]x_{max}=v_o\sqrt{\frac{2d}{a}}[/tex] (2)
Here, the height d is the distance between the plates d = 2.0cm = 0.02m
vo: initial velocity of the electron = 4.0*10^6m/s
You replace the values of the parameters in the equation (2):
[tex]x_{max}=(4.0*10^6m/s)\sqrt{\frac{2(0.02m)}{7.03*10^{13}m/s^2}}\\\\x_{max}=0.095m=9.5cm[/tex]
The horizontal distance traveled by the electron is 9.5cm
