Answer: v= 7.509 x 10^6 m/s
B) the lower plate because the electron is negatively charged
Explanation:
From the question
Electronic charge (q) =1.602 x 10^-19c
Electric field intensity (E) = 8 x 10² = 800N/C
Mass of electron (m) = 9.11 x 10^-31 kg
Length of plate (L) = 50cm=0.5m
Distance between plates (D) = 20cm = 0.2m
Since the electron is entering a uniform electric field, the resulting motion will be of a constant acceleration and can be defined by the equations of motion with a constant acceleration.
From newton's law of motion.
F= ma
The force (F) is coming from the electric field which is
F=Eq.
Thus F = 800 x 1.602 x 10^-19
F = 1.2816 x 10^-16 N
Acceleration of electron (a) = F/m where m is the mass of electron (given above)
Hence
a = 1.2816 x 10^-16 / 9.11 x 10^-31
a= 1.41 x 10¹⁴ m/s².
Using Newton laws of motion to get velocity, we recall that v²= u² + 2ad
Where v is final velocity, u is initial velocity (zero in this case because the electron starts it motion from rest), a= acceleration, d= distance traveled (which in this case is distance between plates)
v² = 0² + 2(1.41 x 10¹⁴) x 0.2
v² = 5.64 x 10¹³
Thus v = √ 5.64 x 10¹³
v = 7.509 x 10^6 m/s
Since the electric field is upwards it denotes that the positive plate is downward and the negative is upward ( this is because electric flux from a positive charge has an outward flow and that from a negative charge has an inward flow. So from our questions, if the electric field is upward, it means it is starting from the bottom plate which will be positive) the electron (which is negatively charged) will be attracted to the positive plate which is downward for this question of ours.
So therefore, the electron is attracted to the downward plate because it (electron) is negative
Option B
