In the figure, a uniform, upward-pointing electric field E of magnitude 3.50×103
ID: 1526752 • Letter: I
Question
In the figure, a uniform, upward-pointing electric field E of magnitude 3.50×103 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 =45° with the lower plate and has a magnitude of 8.39×106 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. 2.58×10-2 m You are correct.
Another electron has an initial velocity which has the angle =45° with the lower plate and has a magnitude of 5.16×106 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
vertical velocity = 5.16*10^6*cos(45 degrees)
As Edq = 3.5*10^3*0.02*1.6*10^-19 = 1.12*10^-17 J is greater than 0.5*(9.1*10^-31)*(5.27*10^6*cos(45 degrees))^2 =
0.63*10^-17
So, it will not hit the other plate
3I.5*10^3*d*1.6*10^-19 = 0.5*(9.1*10^-31)*(5.27*10^6*cos(45 degrees))^2
d = (0.63*10^-17)/(5.6*10^-16)
So, d = 0.0113 = 1.13 cm
So, t = 5.16*10^6*cos(45 degrees)/((3.5*10^3)*(1.6*10^-19)/(9.1*10^-31)) = 5.93*10^-9 s
So, x = 5.16*10^6*cos(45 degrees)*5.93*10^-9 = 2.16 cm
So, it will not hit the plate .... <------------answer
t' = 0.04/(5.16*10^6*cos(45 degrees)) = 1.10*10^-8 s
So, y = 5.16*10^6*cos(45 degrees)*(1.10*10^-8)-0.5*((3.5*10^3)*(1.6*10^-19)/(9.1*10^-31))*(1.10*10^-8)^2
= 0.29 cm <---------------answer(vertical position from the lower plate)