In the figure, a uniform, upward-pointing electric field E of magnitude 4.50×10
ID: 1406634 • Letter: I
Question
In the figure, a uniform, upward-pointing electric field E of magnitude 4.50×103 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 9.10×106 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.
Another electron has an initial velocity which has the angle ?=45° with the lower plate and has a magnitude of 7.14×106 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.
PLEASE EXPLAIN YOUR STEPS
Explanation / Answer
use speed u = S/t
S = U cos thetas *t
or
time taken t = L/u cos theta
t = 4 e-2/(9.1 e 6 * cos 45)
t = 6.21 ns
vertical distance travelled = Vt sin 45 - E eT^2/m
Y = (9.1e 6 *6.21 e -9 * sin 45) - (4.5e3 * 1.6 e-19 * 6.47 e-9 * 6.21 e-9)/(9.11 e-31)
Y = 3.99 e-2 - 3.17 e -2
Y = 0.82 cm,--------------------<<<<<<<<<<<<<Answer
so it reaches the plate as 2.81 is below 2 cm, this reaches tghe bottomplate
-----------------------------------
so now Horizontal distance = u cos theta * t
H = 7.14 e6 * 6.21 e -9* cos 45
H = 3.13 cm----------------------<<<<<<<<<<<<<Answer