In the figure, a uniform, upward-pointing electric field E of magnitude 3.50 In

Question

In the figure, a uniform, upward-pointing electric field E of magnitude 3.50

In the figure, a uniform, upward-pointing electric field E of magnitude 3.50 times 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. An electron is then shot between the plates from the left edge of the lower plate. The initial velocity v0 of the electron makes an angle ?=45degree with the lower plate and has a magnitude of 8.15 times 106 m/s. Will the 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 ( m). The next electron has an initial velocity which has the same angle ?=45degree with the lower plate and has a magnitude of 5.68 times 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 ( m).

Explanation / Answer

downward acceleration on electron f/m=eq/m=6.15*10^14m/s^2

4*10^-2=5.76*10^6*t

t=.69*10^-8sec

vertical height=4*10^-2-1.46*10^-2=2.53cm

so it is going to hit the plate

2*10^-2=5.76*10^6t-3.075*10^14t^2

t=4.6*10^-9sec

horizontal distance=2.64cm from left plate

similarly for part B it wont hit the plate vertical position at which it leaves is .95cm

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