A horizontally projected beam of electrons travels at 55 km/s into the space between two horizontal, parallel plates. The plates are separated by 7.4 cm, and they are 12 cm long. The uniform electric field between the plates is 7.2 mV/m upward. If the beam enters the region at a height of 3.7 cm above the lower plate, at what height above the lower plate does the beam exit the region? The electron mass is 9.11 × 10^-31 kg. A) 0.9 cm B) 2.3 cm C) 3.4 cm D) 4.6 cm

Question

Quizplus · Accepted Answer

The electric force on each electron due to the electric field is given by:

F = Eq

where E is the electric field and q is the charge of the electron (-1.6 × 10^-19 C).

The acceleration of the electron due to this force is given by:

a = F/m

where m is the mass of the electron.

Using the acceleration and the initial velocity of the electron, we can find the time it takes for the electron to travel the distance between the plates:

d = 1/2at^2 + vt

where d is the distance between the plates, v is the initial velocity of the electron, and t is the time it takes to travel the distance.

Substituting the values we get:

7.4 × 10^-2 m = 1/2 ((1.6 × 10^-19 C)(7.2 × 10^3 V/m))/ (9.11 × 10^-31 kg) t^2 + (55 × 10^3 m/s) t

Solving for t we get:

t = 4.37 × 10^-7 s

Using this time, we can find the vertical displacement of the electron:

y = 1/2gt^2

where g is the acceleration due to gravity.

Since the electron is moving horizontally, its initial and final vertical velocities are zero. Therefore, the vertical displacement is given by:

y = 1/2gt^2 = 1/2(9.8 m/s^2)(4.37 × 10^-7 s)^2 = 9.0 × 10^-13 m = 0.9 nm

This displacement is very small compared to the distance between the plates, so the height of the electron above the lower plate when it exits is approximately the same as the height above the lower plate when it entered. Therefore, the answer is approximately 3.7 cm, or choice (C).

A Horizontally Projected Beam of Electrons Travels at 55 Km/s