here is the problem In the figure, a uniform, upward-pointing electric field E o

Question

here is the problem

In the figure, a uniform, upward-pointing electric field E of magnitude 4.5 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 V0, which makes an angle theta = 45 degree with the lower plate and has a magnitude of 9.20 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. Tries 1/10 Previous Tries Another electron has an initial velocity which has the angle theta = 45 degree with the lower plate and has a magnitude of 7.14 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. Tries 0/10

Explanation / Answer

Part A)

We need to first find the components of the velocity of the electron

Both vx and vy = 9.20 X 10^6(cos 45) = 6.51 X 10^6 m/s

The Electric Field provides a force pushing the electron down. The magnitude is

F = ma = qE

(9.11 X 10^-31)(a) = (1.6 X 10^-19)(4.5 X 10^3)

a = 7.90 X 10^14 m/s^2

The force is downward,so a is actually negative

Now we need to find out how much time would be required for the electron to get out of the plates using the x component, which does not change

Apply L = vt

.04 = (6.51 X 10^6)(t)

t = 6.15 X 10^-9 sec

Now we can figure out where it is in the y direction at that time.

Lets see how long it would take to get to the top plate

Find the velocity it would have if it could travel the 2 cm

vf^2 = vo^2 + 2ad

vf^2 = (6.15 X 10^6)^2 + (2)(-7.90 X 10^14)(.02)

vf = 2.49 X 10^6 m/s

Now we can find out how much time it takes to travel that distance

vf = vo + at

2.49 X 10^6 = 6.15 X 10^6 + (-7.90 X 10^14)(t)

t = 4.63 X 10^-9 s

Since that is less time then required to leave the plates, it will hit the top plate

To find out where, apply d = vt

d = (6.15 X 10^6)(4.63 X 10^-9)

d = .0285 m (2.85 cm)

So the electron hits the top plate at 2.85 cm from the left edge of the top plate.

Part B)

Same idea, new velocity

We need to first find the components of the velocity of the electron

Both vx and vy = 7.14 X 10^6(cos 45) = 5.05 X 10^6 m/s

The Electric Field provides a force pushing the electron down. The magnitude is

F = ma = qE

(9.11 X 10^-31)(a) = (1.6 X 10^-19)(4.5 X 10^3)

a = 7.90 X 10^14 m/s^2

The force is downward,so a is actually negative

Now we need to find out how much time would be required for the electron to get out of the plates using the x component, which does not change

Apply L = vt

.04 = (5.05 X 10^6)(t)

t = 7.92 X 10^-9 sec

Now we can figure out where it is in the y direction at that time.

Lets see how long it would take to get to the top plate

Find the velocity it would have if it could travel the 2 cm

vf^2 = vo^2 + 2ad

vf^2 = (5.05 X 10^6)^2 + (2)(-7.90 X 10^14)(.02)

This gives us a negative answer, so we have to try a different approach

Lets find out where the top of the path is when the y velocity is zero

0 = (5.05 X 10^6)^2 + (2)(-7.90 X 10^14)(d)

d = .0161 m (so it will not hit the top plate)

The time to go that distance is found using

vf = vo + at

0 = (5.05 X 10^6) + (-7.90 X 10^14)(t)

t = 6.39 X 10^-9 sec

If we double that time, (to come back down to where a plate would be) we would get 1.28 X 10^-8 sec

That means it would not hit the bottom plate either, it has left the area.

Now we can find out where

We need to know the distance the electron is in the y direction at the 7.92 X 10^-9 sec

d = vot + .5at^2

d = (5.05 X 10^6)(7.92 X 10^-9) + (.5)(-7.90 X 10^14)(7.92 X 10^-9)^2

d = .0152 m

Therefore it is .0152 m (1.52 cm) above the bottom plate when it leaves

This is .0048 cm below the top plate

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