In the figure, a uniform, upward-pointing electric field E of magnitude 4.00 In
ID: 2130452 • Letter: I
Question
In the figure, a uniform, upward-pointing electric field E of magnitude 4.00
In the figure, a uniform, upward-pointing electric field E of magnitude 4.00 times 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 = 4cm and separation d = 2.00 cm. An electron is then shot between the plates from the left edge of the lower plate. The initial velocity v0 of the electron makes an angle ?=45 degree with the lower plate and has a magnitude of 9.00 times 106m/s. The next electron has an initial velocity which has the same angle ?=45 degree with the lower plate and has a magnitude of 5.63 times 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 ( m).Explanation / Answer
a =F/m = 2.4 x 10-16/ 9.109 x 10-31 = 2.635 x 10^14 m/s2
vertical component of velocity = vsin45 = 2.72 x 10^6 m/s
0 - vsin45 ^2 = 2 x 2.635 x 10^14 x H
H = 0.01406 m or 1.41 cm
it is less den 2 cm.
in vertical direction to get time of flight
H = ut + at^2/2
0 = 2.72 x 10^6 x t - 2.635 x 10^14 x t^2 /2
t = 2.07 x 10-8 sec
d = vcos45 x t = 0.0562 m = 5.62 cm
but plate is only 4 cm
so it will NOT STRIKE ANY OF PLATES.
so time at which the particle leaves the space between the plates
t = 0.04 / vcos45 = 1.471 x 10-8 sec
H = vsin45 x t - at^2/2
H = 0.01152 m = 1.152 cm
This was my work for the problem withmy numbers. Try substituting yours in.