In the figure, a uniform, upward-pointing electric field E of magnitude 5.50×10
ID: 1655202 • Letter: I
Question
In the figure, a uniform, upward-pointing electric field E of magnitude 5.50×103 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 1.07×107 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 6.86×106 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
In the electric field
acceleration ay = Eq/m
ay = (5.5*10^3*1.6*10^-19)/(9.11*10^-31)
ay = 9.66*10^14 m/s^2
along horizantal
Xmax = vo^2*sin(2theta)/ay
Xmax = (1.07*10^7)^2 /(9.66*10^14)
Xmax = 0.118m = 11.8 cm > L
electron will not touch the lower plate
along vertical
Ymax = (vo*sin45)^2/2ay
Ymax = (1.07*10^7*sin45)^2/(2*9.66*10^14)
Y max = 2.9 cm
Ymax < d
the electron strikes the upper plate
from equation of projectile
y = x*tantheta - 0.5*ay*x^2/(vo^2*(costheta)^2)
0.02 = x*tan45 - 0.5*9.66*10^14*x^2/((1.07*10^7)^2*(cos45)^2)
x = 2.9 cm <<------answer
the electron will strike the upper plate at horizantal distance x = 2.9 cm
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along horizantal
Xmax = vo^2*sin(2theta)/ay
Xmax = (6.86*10^6)^2 /(9.66*10^14)
Xmax = 0.048m = 4.8 cm > L
electron will not touch the lower plate
along vertical
Ymax = (vo*sin45)^2/2ay
Ymax = (6.86*10^6*sin45)^2/(2*9.66*10^14)
Y max = 1.2 cm
Ymax < d
the electron will not strikes the upper plate
from equation of projectile
x = 0.04 m
y = x*tantheta - 0.5*ay*x^2/(vo^2*(costheta)^2)
y = 0.04*tan45 - 0.5*9.66*10^14*0.04^2/((6.86*10^6)^2*(cos45)^2)
y = 7.15 cm <<------answer
leave at the vertical position of7.15 cm