In the figure, a uniform, upward electric field E of magnitude 2.50×10 3 N/C has
ID: 2001889 • Letter: I
Question
In the figure, a uniform, upward electric field E of magnitude 2.50×103N/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. 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 ?=45° with the lower plate and has a magnitude of 6.55×106m/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.
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The next electron has an initial velocity which has the same angle ?=45° with the lower plate and has a magnitude of 5.15×106m/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
E = electric field = 2500 N/C
m = mass of electron
q = charge on electron
a = acceleration of electron = qE/m = (1.6 x 10-19) (2500) /(9.1 x 10-31) = 4.4 x 1014 m/s2 downward
Along the vertical direction :
Voy = initial velocity = Vo Cos45 = (6.55 x 106) Sin45 = 4.63 x 106 m/s
ay = acceleration along Y-direction = - 4.4 x 1014 m/s2
Vfy = final velocity = 0
Y = distance at which the electron comes to stop
using the equation
Vfy2 = Voy2 + 2 ay Y
02 = (4.63 x 106 )2 + 2 (- 4.4 x 1014 ) Y
Y = 0.024 m
so the electron hits the upper plate
using the equation
Vfy= Voy + at
0 = 4.63 x 106 + (- 4.4 x 1014 ) t
t = 1.05 x 10-8 sec
Distance travelled along X-direction is given as
X = Vox t = 4.63 x 106 (1.05 x 10-8 )= 0.0486 m