I need help with part 2 of this question: 1) In the figure, a uniform, upward-po
ID: 2165967 • Letter: I
Question
I need help with part 2 of this question:
1) In the figure, a uniform, upward-pointing electric field E of magnitude 5.50?103N/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 1.05?107m/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.
Correct Answer: 2.59 * 10^(-2) m
2) Another electron has an initial velocity which has the angle ?=45? with the lower plate and has a magnitude of 6.60?106m/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.
0.006416 is the wrong answer. Any ideas?
Explanation / Answer
let find the time at which the velocity of particle is zero
v= u- at, time at which final velocity is zero
t= 4.67*10^6 / 9.67*10^14
t=.4829*10^(-8) = 4.8*10^-9
for this time it goes upwards thus distance covered
s = Ut -0.5* a* t^2
= 4.67*10^6 *4.8*10^-9 - 0.5*9.67*10^14*(4.8*10-9)2
=22.416*10^-3 -11.139*10^-3
s=11.277*10^-3
after that it goes downwards for the time (8*10^-9 - 4.8*10^-9) = 3.2*10^-9
s = ut +0.5* a*t2
s = 0 + 0.5 *9.67*10^14*(3.2*10^-9)2
downward distane travelled = 4.95 *10^-3
thus the distance of particle form lower plate is (11.277*10^-3 - 4.95 *10^-3) = 6.327*10^-3
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