In the figure, a uniform, upward-pointing electric field E of magnitude 4.50×103 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 θ=45° with the lower plate and has a magnitude of 8.64×106 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 θ=45° with the lower plate and has a magnitude of 7.25×106 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.
To determine whether or not the electron strikes one of the
plates, we need to determine the time Ty required to travel a
vertical distance of y = 0.02 m and the time Tx for a horizontal
distance of x = 0.04 m.
If Ty < Tx, then the electron will strike the negative
plate.
If Ty > Tx, the electron will not strike the plate and we will
then determine the vertical distance at which the particle leaves
the space between the plates.
For the most part, this is a kinematics problem, but we need to
evaluate the vertical acceleration induced on the electron as it
travels through the plates.
This acceleration is found by equating F = qE = ma --> a = qE/m
= (1.6e-19)(4.5e3)/(9.11e-31) = 7.9e14 m/s^2.
We also need to isolate the x and y components of the velocity
v0.
Vy = v0sin(45) = 5.63e6 m/s
Vx = Vy = 5.63e6 m/s
Now we find Ty and Tx.
0.02 = 0+(5.63e6)(Ty)+(0.5)(6.15e14)(Ty)^2 --> Ty = 3.05e-9
s
0.04 = 0+(5.63e6)(Tx) --> Tx = 7104.78 s
Since Ty < Tx, the electron will in fact strike the plate at a
horizontal distance of x = 0+(5.63e6)(7104.78) = 5637104.79m.
Part 2
We repeat the process and find that that Ty < Tx. The electron
strikes the plate at a horizontal distance of x =
0+(3.05e6)(4.08e-9) = 0.012 m.
In the figure, a uniform, upward-pointing electric field E of magnitude 4.50×103 N/C has been set...
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[ Course Contents » ... » HW2 » An electron is then shot between the plates Notes Bookmark Evaluate Communicate Print Info FA POR In the figure, a uniform, upward-pointing electric field E of magnitude 5.00x103 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...
aAn electron is then shot between the plates Course Contents >> ... >> HW2 > An electron is then shot between the plates Notes Bookmark Evaluate Communicate Print Info V L- In the figure, a uniform, upward-pointing electric field E of magnitude 4.50X10' 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...
electron starts at the edge of a region with a uniform electric field. The region is between two An parallel met to the plane of the lower plate. from the lower plate towards the upper plate and has a value of 3.50 x 10'N/C. allic plates. The electron has an initial velocity of 5.00 x 10'm/s at an angle of 45° relative The plates are 2.00 cm apart and are very large. The electric field is On which plate and...
2. A uniform electric field, E, with magnitude 100 N/C exists between two charged plates. Assume the size of the plates is much larger than the distance between them, d - 1.0 cm. a) Indicate the signs of the charges on each plate. b) Assume an electron is placed at rest exactly between the two plates. How long does it take for the electron to strike one of the plates. Which direction does it move?