Question

Consider an electron, of charge magnitude e = 1.602 10-19 C and mass me = 9.11 10-31 kg, moving in an electric field with an electric field magnitude E = 8 ✕ 102 N/C, similar to what Thana observed in the simulation. Let the length of the plates be L = 50 cm, and the distance between them be d = 20 cm. Find the maximum speed, v, the electron could be moving if it enters the space halfway between and parallel to the two plates to just barely strike one of the plates.>> If the field is pointing upward, which plate will Thana conclude the electron strikes at this speed?A] The lower plate, because the electron is attracted to the negative plate.B] The lower plate, because the electron is negatively charged. C] The upper plate, because we are only considering the magnitude of the electron charge, and magnitudes are always positive.D] The upper plate, because the electron charge magnitude is positive.

Answer 1

Answer: v= 7.509 x 10^6 m/s

B) the lower plate because the electron is negatively charged

Step-by-step 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