In the figure, a uniform, upward-pointing electric field E of magnitude 4.00×103
ID: 2043269 • Letter: I
Question
In the figure, a uniform, upward-pointing electric field E of magnitude 4.00×103 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 7.86×106 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.
Another electron has an initial velocity which has the angle =45° with the lower plate and has a magnitude of 6.24×106 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
It looks a lot like were projecting an projectile... like when we kick a soccer ball. But, the force that acts on the negatively charged electron rather than gravity. Other than that, its just a projectile motion problem! Lets seem what we should use for the vertical acceleration in this case instead of “g”:'
Fy =|q|Ey = may
ay = |q|Ey/m = (1.6*10-19 )(4*103)/(9.11*10-31) = 7.02 *1014 m/s2
Now, lets explore the parabolic projectile motion. First, lets see where it is horizontally when itreaches a height if 2.00 cm (and if it even reaches that height!)
vx = 7.86*106 m/s
vy0 = 7.86*106 cos 45 = 4.96*106 m/s
y=2cm=0.02 m
y=vy0t+(1/2)ayt2
0.02 = (4.96*106)t + (1/2)(-7.02 *1014 )t2
t=7.06*10-9 sec
x=vxt =7.86*106 * 7.06*10-9 =0.055m
So, it reaches a height of 2.00 cm a horizontal distance of 5.5 cm from the launch point. Thus, ithits the top plate 5.5 cm from the left edge.