A) In the figure, a uniform, upward-pointing electric field E of magnitude 3.00×

Question

A) In the figure, a uniform, upward-pointing electric field E of magnitude 3.00×103 N/C has been set up between two horizontal plates by charging the lower plate positively and the upper plate negatively. The plates have length L = 4 cm and separation d = 2.00 cm. Electrons are shot between the plates from the left edge of the lower plate. The first electron has the initial velocity v0, which makes an angle ?=45° with the lower plate and has a magnitude of 7.74×106 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.

B) Another electron has an initial velocity which has the angle ?=45° with the lower plate and has a magnitude of 5.26×106 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

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)(3e3)/(9.11e-31) = 5.27e14 m/s^2.

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

Vy = v0sin(45) = 5.47e6 m/s
Vx = Vy = 5.47e6 m/s

Now we find Ty and Tx.

0.02 = 0+(5.47e6)(Ty)+(0.5)(5.27e14)(Ty)^2 --> Ty = 3.17e-9 s
0.04 = 0+(5.47e6)(Tx) --> Tx = 7.3e-9 s

Since Ty < Tx, the electron will in fact strike the plate at a horizontal distance of x = 0+(5.47e6)(3.17e-9) = 0.01734 m.

Part 2

We repeat the process and find that that Ty < Tx. The electron strikes the plate at a horizontal distance of x = 0+(3.73e6)(4.14e-9) = 0.01544 m.

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