Question

In the figure, a uniform, upward-pointing electric field E of magnitude 4.50

Answer 1

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)(4.5e3)/(9.11e-31) = 7.907e14 m/s^2.

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

Vy = v0sin(45) = 6.31e6 m/s
Vx = Vy = 6.31e6 m/s

Now we find Ty and Tx.

0.02 = 0+(6.31e6)(Ty)+(0.5)(7.907e14)(Ty)^2 --> Ty = 2.709e-9 s
0.04 = 0+(6.31e6)(Tx) --> Tx = 6.339e-9 s

Since Ty < Tx, the electron will in fact strike the plate at a horizontal distance of x = 0+(6.339e6)(2.709e-9) = 0.01717 m.