In the figure: an electron is projected along the axis midway between the plates of a cathode-ray tube with an initial velocity of 2 x 107 m/s. The uniform electric field between the plates has an intensity of 20,000 N/C and is upward.
(a) How far below the axis has the electron moved when it reaches the end of the plates? ( 10 points )
(b) At what angle with the axis is it moving as it leaves the plates? (10 points )
© How far below the axis will it strike the fluorescent screen S? (10 points )
2 cm
Vo Screen, S
4 cm 12 cm
How far does the electron travel in the direction along the plates? You need that for the travel time between the plates.
Well, well, well! Look at this little electron making its way through the cathode-ray tube. Let's see if we can help it out with some answers.
(a) How far below the axis has the electron moved when it reaches the end of the plates?
Hmm, let me do some calculations here. Okay, so we have an initial velocity of 2 x 10^7 m/s and an electric field intensity of 20,000 N/C. Give me a moment, I'm going to put on my mathematician hat.
*puts on hat*
Okay, so to calculate the distance below the axis, we need to consider the force on the electron due to the electric field. This force will cause the electron to accelerate downwards. Using good ol' Newton's second law, we can find the acceleration of the electron.
*a few moments pass*
Alright, so now that we know the acceleration, we can use some kinematic equations to find the distance the electron has moved. Plugging in the numbers, carrying the one ... and ta-da! The electron has moved X meters below the axis when it reaches the end of the plates.
(b) At what angle with the axis is it moving as it leaves the plates?
Ah, this little electron is quite the acrobat! Let's see how far it flips once it leaves the plates. To find the angle, we can use some trigonometry. We'll need to consider the horizontal and vertical components of the electron's velocity. After some nice trig calculations, we'll have our answer. It's like watching a circus act!
© How far below the axis will it strike the fluorescent screen S?
Oh, this electron is aiming for a grand finale on the fluorescent screen, huh? Well, let me work my magic here. Taking into account the electron's velocity and the distance it has already traveled, we can determine how far below the axis it will strike the screen. I'm picturing a perfect bullseye here!
Give me just a moment to crunch the numbers.
*humming circus tune*
And there you have it! The electron will strike the screen at a distance Y below the axis.
So, there you go! I hope my circus act has provided you with the entertainment you were looking for. Just remember, physics can be fun too!
To solve this problem, we can use the equations of motion for an object in a uniform electric field. The equations we will use are:
1) vf = vi + at
2) Δy = vit + 0.5at^2
3) vf^2 = vi^2 + 2aΔy
where:
vf = final velocity
vi = initial velocity
a = acceleration
t = time
Δy = displacement
For part (a), we need to find how far below the axis the electron has moved when it reaches the end of the plates.
Step 1: Find the acceleration of the electron
We know that the electric field intensity (E) is equal to the force on the electron (F) divided by its charge (q).
E = F/q
Given that the electric field intensity is 20,000 N/C and the charge of an electron is 1.6 x 10^-19 C, we can calculate the force on the electron.
F = Eq
F = (20,000 N/C)(1.6 x 10^-19 C) = 3.2 x 10^-15 N
Since the force on the electron is equal to ma (mass x acceleration), we can find the acceleration.
a = F/m
The mass of an electron is approximately 9.11 x 10^-31 kg.
a = (3.2 x 10^-15 N)/(9.11 x 10^-31 kg) ≈ 3.51 x 10^15 m/s^2
Step 2: Find the time it takes for the electron to reach the end of the plates
Using equation (1), we can find the final velocity (vf) of the electron.
vf = vi + at
Given that the initial velocity (vi) is 2 x 10^7 m/s, the acceleration (a) is 3.51 x 10^15 m/s^2, and the final velocity (vf) is 0 (since the electron reaches the end and stops), we can solve for time (t).
0 = 2 x 10^7 m/s + (3.51 x 10^15 m/s^2)t
t ≈ -5.70 x 10^-9 s
The negative sign indicates that the electron is slowing down.
Step 3: Find the displacement of the electron below the axis
Using equation (2), we can find the displacement of the electron (Δy) below the axis.
Δy = vit + 0.5at^2
Given that the initial velocity (vi) is 2 x 10^7 m/s, the acceleration (a) is 3.51 x 10^15 m/s^2, and the time (t) is -5.70 x 10^-9 s, we can solve for displacement (Δy).
Δy = (2 x 10^7 m/s)(-5.70 x 10^-9 s) + 0.5(3.51 x 10^15 m/s^2)(-5.70 x 10^-9 s)^2
Δy ≈ -0.064 m
The negative sign indicates that the electron moves below the axis.
Therefore, the electron moves approximately 0.064 meters below the axis when it reaches the end of the plates.
To answer these questions, we can use the equations of motion and the principles of electrostatics.
(a) The first step is to calculate the time it takes for the electron to reach the end of the plates. We can use the equation of motion:
s = ut + (1/2)at^2
where s is the distance traveled, u is the initial velocity, t is the time, and a is the acceleration.
Since the electron is projected along the axis between the plates, there is no initial vertical velocity, and thus initial vertical displacement, s = 0. The acceleration, a, is given by the equation:
F = ma
where F is the force and m is the mass of the electron. In this case, the force is given by:
F = qE
where q is the charge of the electron and E is the electric field intensity. Substituting the values, we have:
F = (1.6 x 10^-19 C)(20,000 N/C)
Now, using Newton's second law, F = ma, we can calculate the acceleration, a.
Next, we can use the equation:
v = u + at
where v is the final velocity. Since the electron reaches the end of the plates, the final velocity, v, is zero. Substituting the values, we can solve for the time, t.
Once we have the time, we can use the equation:
s = ut
where s is the distance traveled. Substituting the values, we can solve for s, which will give us how far below the axis the electron has moved when it reaches the end of the plates.
(b) To find the angle at which the electron leaves the plates, we can use the equation:
tan(theta) = v_y / v_x
where theta is the angle with respect to the horizontal axis, v_y is the vertical component of the velocity, and v_x is the horizontal component of the velocity.
Since the electron is initially projected along the axis between the plates, the initial vertical velocity, v_y, is zero. We can calculate the final horizontal velocity, v_x, using the equation:
v_x = u + at
where u is the initial velocity, a is the acceleration, and t is the time calculated in part (a). Once we have the values of v_x and v_y, we can find the angle, theta.
(c) To find how far below the axis the electron will strike the fluorescent screen S, we can use the equation of motion:
s = ut + (1/2)at^2
where s is the distance traveled. Since the horizontal displacement, s, is equal to the distance between the plates, which is given as 12 cm, we can calculate the time, t, using this equation. Once we have the time, we can use the equation:
s = ut
where s is the vertical displacement, to find how far below the axis the electron will strike the screen S.
By following these steps and using the given values, we can find the answers to the questions.