In the figure, a uniform, upward-pointing electric field E of magnitude 5.00×10
ID: 1406986 • Letter: I
Question
In the figure, a uniform, upward-pointing electric field E of magnitude 5.00×103 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 = 5 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 8.79×106 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.
B)Another electron has an initial velocity which has the angle =45° with the lower plate and has a magnitude of 7.50×106 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)
electric field E=5*10^3 N/C
length of plate L=5cm
separation d=2cm
vo=8.79*10^6 m/sec
theta=45 degrees
acceletaion of particle is,
ay=q*E/m
ay=1.6*10^-19*5*10^3/(9.1*10^-31)
ay=8.8*10^14 m/sec^2
now,
y=vy*t-1/2*ay*t^2
0.02=(8.79*10^6)*sin(45)-1/2*8.8*10^14*t^2
(4.4*10^14)*t^2-(6.2*10^6)*t-(0.02)=0
====t=1.68*10^-8 sec
now,
horizontal distance x=vx*t
x=(8.79*10^6)*cos(45)*1.67*10^-8
x=0.1038 m
x=10.38 cm
here x is greater than L=5cm
hence, with these condition electron will hits the upper plate
b)
if vo=7.5*10^6 m/sec
y=vy*t-1/2*ay*t^2
0.02=(7.5*10^6)*sin(45)-1/2*8.8*10^14*t^2
(4.4*10^14)*t^2-(5.3*10^6)*t-(0.02)=0
====t=1.5*10^-8 sec
now,
horizontal distance x=vx*t
x=(7.5*10^6)*cos(45)*1.5*10^-8
x=0.0795 m
x=7.95 cm
here ,
x is greater than L=5cm
hence, electron will hits the upper plate