a) Will this electron strike one of the plates? If so, what is the horizontal di

Question

a) Will this electron strike one of the plates? If so, what is the horizontal distance from the left edge? If not enter the vertical position at which the particle leaves the space between the plates.

b) Another electron has an initial velocity which has the angle ?=45° with the lower plate and has a magnitude of 6.51×106 m/s. Will this electron strike one of the plates? If so, what is the horizontal distance from the left edge? If not enter the vertical position at which the particle leaves the space between the plates.

Explanation / Answer

The problem can be solved as follows:

(a) To determine whether or not the electron strikes one of the plates, we need to determine the time Ty required to travel a vertical distance of y = 0.02 m and the time Tx for a horizontal distance of x = 0.04 m.

If Ty < Tx, then the electron will strike the negative plate.
If Ty > Tx, the electron will not strike the plate and we will then determine the vertical distance at which the particle leaves the space between the plates.

For the most part, this is a kinematics problem, but we need to evaluate the vertical acceleration induced on the electron as it travels through the plates.

This acceleration is found by equating F = qE = ma --> a = qE/m = (1.6e-19)(4.0e3)/(9.11e-31) = 7.03e14 m/s^2.

We also need to isolate the x and y components of the velocity v0.

Vy = v0sin(45) = 8.42e6 *0.707 = 5.95e6 m/s
Vx = Vy = 5.95e6 m/s

Now we find Ty and Tx.

0.02 = 0+(5.95e6)(Ty)+(0.5)(7.03e14)(Ty)^2

=> Ty = 2.97e-9 s
0.04 = 0+(5.95e6)(Tx)

=> Tx = 6.72e-9 s

Since Ty < Tx,

The electron will in fact strike the plate at a horizontal distance of x = 0+(5.95e6)(2.97e-9) = 0.018 m.

(b)Repeat the process as explained above and again find the value of Tx and Ty, and Vx and Vy.

Then find the value of x, as above.

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