(Figure included here) In the figure, a uniform, upward-pointing electric field
ID: 1655276 • Letter: #
Question
(Figure included here)
In the figure, a uniform, upward-pointing electric field E of magnitude 2.50×103 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 6.36×106 m/s.
1) 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) Another electron has an initial velocity which has the angle =45° with the lower plate and has a magnitude of 5.23×106 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
a = Fe / m = q E / m
a = (1.6 x 10^-19) (2.50 x 10^3) / (9.109 x 10^-31)
a = 4.40 x 10^14 m/s^2
Range = v0^2 sin(2 x 45) / a
= 0.092 m Or 9.2 cm
L is less then half of range.
H_max = (v0 sin45)^2 / 2 a
= 0.023 m or 2.30 cm
so it will strike the upper plate.
2) Rnage = 0.062 m = 6.2 cm
H_max = 0.0155 cm
Range > L
and H_max < d hence it will not strike any of plates.
t = (0.04) / (5.23 x 10^6 cos45) = 1.082 x 10^-8 s
y = (5.23 x 10^6 sin45)(1.082 x 10^-8) - (4.40 x 10^14)(1.082 x 10^-8)^2 / 2
y = 0.014 m Or 1.4 cm ......Ans
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