An electron passes location
h0.02m, 0.04m,−0.06mi, and 4 μs later is de-
tected at location h0.02 m, 1.84 m,−0.86 mi.
Find the average velocity of the elecron:
¯v = <vx, vy, vz>
If the electron continues to travel at this av-
erage velocity, find where it will be in another
9 μs :
~d = <dx, dy, dz>
Answer in units of m
dx=(xf-xi)
vx=dx/dt=dx/4E-6
similarlly, vy, vz.
then for the second part, knowing the velocities
xf=xi+vx*dt
To find the average velocity of the electron, we need to calculate its displacement over a given time interval. The displacement is given by the difference between the final and initial positions of the electron.
In this case, the initial position of the electron is (h0.02m, 0.04m, -0.06m), and the final position is (h0.02m, 1.84m, -0.86m). The time interval is 4 μs.
The average velocity of the electron is given by the displacement divided by the time interval:
¯v = (final position - initial position) / time interval
Let's calculate it step by step:
1. Calculate the displacement in the x-direction:
Δx = (final x-coordinate - initial x-coordinate) = (0.02m - 0.02m) = 0m
2. Calculate the displacement in the y-direction:
Δy = (final y-coordinate - initial y-coordinate) = (1.84m - 0.04m) = 1.80m
3. Calculate the displacement in the z-direction:
Δz = (final z-coordinate - initial z-coordinate) = (-0.86m - (-0.06m)) = -0.80m
Now we have the displacement vector: Δr = (0m, 1.80m, -0.80m)
4. Calculate the average velocity:
¯v = Δr / time interval = (0m, 1.80m, -0.80m) / 4 μs
To find where the electron will be in another 9 μs, we can use the average velocity vector to find the new displacement:
1. Multiply the average velocity vector by the additional time interval:
~d = ¯v * additional time interval = (0m, 1.80m, -0.80m) * 9 μs
Now we have the displacement vector after 9 μs: ~d = (0m, 16.20m, -7.20m)
Therefore, the electron will be at the position (h0.02m, 16.20m, -7.20m) after another 9 μs.