Question

In the figure, a uniform, upward-pointing electric field E of magnitude 2.50×103 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 θ=45° with the lower plate and has a magnitude of 6.44×106 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.

Answer 1

We know,F = qE

Also, F = ma

ma = qE

a = qE/m

a = 1.6e-19 * 2.50e3 / 9.1e-31

a = 4.395e14 m/s²

Now, the vertical distance travelled by electron

from kinematics, we have

y = v²sin² / 2a

y = (6.44e6)² * sin²45 / 2* 4.395e14

y = 0.02359 m

Given d is 2 cm or 0.02 m

as y > d, the electron might hit the upper plate

Now, in the x direction, time taken when y = d = 2 cm = 0.02 m

t = vsin - sqrt ( v²sin² -2ad) / a

t = 6.44e6* sin 45 - sqrt (2.0736e13 - 1.758e13) /  4.395e14

t =  6.44e6* sin 45 - 1.776e6 / 4.395e14

t = 6.32e-9 sec

so,

x = v * cos*t

x = 0.02878

so, electron hits the upper plate at x = 2.878 cm