In the figure, a uniform, upward-pointing electric field E of magnitude 5.00 tim
ID: 251097 • Letter: I
Question
In the figure, a uniform, upward-pointing electric field E of magnitude 5.00 times 10^3 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 v_0, which makes an angle theta =45 degree with the lower plate and has a magnitude of 9.67 times 10^6 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
First of all we need to determine the time Ty-time required to travel a vertical distance of y = 0.02 m
and Tx-time required to travel a horizontal distance of x = 0.04 m.
The acceleration is found by equating F = qE = ma
a = qE/m
=(1.6 X 10-19)(5X103 )/(9.11X10-31)
= 8.79 X1014m/s^2.
We also need to isolate the x and y components of the velocity v0.
Vy = v0sin(45) = 9.67 X 106 m/s X sin45 = 6.8 X106 m/s
Vx = Vy =6.8 X106 m/s
Now we find Ty and Tx.
0.02 = 0+(6.8 X106 m/s)(Ty) --> Ty = 2.941 X10-9 s
0.04 = 0+(6.8 X106 m/s)(Tx) --> Tx = 5.88 X 10-9 s
Since Ty < Tx, the electron will in fact strike the plate at a horizontal distance
x = 0+(6.8 X106 m/s)(2.941 X10-9 s) = 0.019 m.