An electron in an oscilloscope tube is traveling horizontally with a speed of 4.4×106m/s . It passes through a 2.1cm long region with an electric field pointing upward with a magnitude of 1500N/C.
what is the magnitude of the electron’s velocity when it emerges from the electric field region?
What is the direction of the electron’s velocity when it emerges from the electric field region?
I am supposed to just do all of these for you? I already took the course. Let's see your attempts so someone can tell where you are having trouble.
24m/s
To find the magnitude of the electron's velocity when it emerges from the electric field region, we need to use the basic principle of electric forces acting on a charged particle.
Step 1: Calculate the acceleration of the electron in the electric field.
From Newton's second law, the force experienced by a charged particle moving in an electric field is given by F = q * E, where F is the force, q is the charge of the particle, and E is the electric field strength.
Since the electron is negatively charged, its charge is -e (where e is the elementary charge). Therefore, the force on the electron is F = (-e) * E.
The electron experiences a constant force over the 2.1 cm region, so we can calculate the acceleration using Newton's second law: F = ma, where m is the mass of the electron and a is its acceleration.
Rearranging the equation, we have a = F / m.
Step 2: Calculate the change in velocity of the electron in the electric field.
Since the acceleration is constant, we can use the kinematic equation v^2 = u^2 + 2aΔs, where v is the final velocity, u is the initial velocity, a is the acceleration, and Δs is the displacement.
In this case, the electron starts from rest, so the initial velocity u = 0.
The displacement Δs is given as 2.1 cm = 0.021 m.
Rearranging the equation, we have v = sqrt(u^2 + 2aΔs).
Step 3: Plug in the given values and calculate.
Charge of an electron, q = -1.6 x 10^-19 C
Mass of an electron, m = 9.1 x 10^-31 kg
Electric field strength, E = 1500 N/C
Displacement, Δs = 0.021 m
Using F = q * E, we can find the force F on the electron:
F = (-1.6 x 10^-19 C) * (1500 N/C).
Next, using a = F / m, we can find the acceleration a:
a = F / m.
Finally, using v = sqrt(u^2 + 2aΔs), we can find the magnitude of the electron's velocity v when it emerges from the electric field region.
To determine the direction of the electron's velocity when it emerges from the electric field region, we can use the fact that the electric field points upward and apply the right-hand rule. Since the electron is negatively charged, it experiences a force in the opposite direction of the electric field. Thus, the direction of the electron's velocity will be opposite to the direction of the electric field.