In the figure, a uniform, upward-pointing electric field E of magnitude 5.00×10

Question

In the figure, a uniform, upward-pointing electric field E of magnitude 5.00×103N/Chas 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 9.10×106m/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.

Another electron has an initial velocity which has the angle ?=45° with the lower plate and has a magnitude of 6.91×106m/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.

Explanation / Answer

apply distance s = u cos theta * t

so time taken t = L/u cos theta

t = 4 e-2/(9.1 e 6 * cos 45)

t = 6.21 ns

now apply vertical distance travelled = Vt sin 45 - E eT^2/m

Y = 9.16e6 *6.21 e -9 * sin 45 - (4e3 * 1.6 e-19 * 6.47 e-9 * 6.47 e-9)/(9.11 e-31)

Y = 0.040 - 0.0029

Y = 3.711 cm-----------------<<<<<<<<<<<<<<<<<<Answer

so it reaches the plate as 3.71 is above 2cm, so it hits the upper plate

now Horizontal distance S = ut

H = 6.91 e 6 *cos 45 * 6.21 e-9 = 3.03 cm

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