An electron with kinetic energy of 2x10-16J is moving to the right along the axis of a cathode-ray tube as shown in Figure 21-38. There is an electric field E= (2x 104N/C) j in the region between the deflection plates. Everything else, E=0. (a) How far away is the electron from the axis oft he tube when it reaches the end oft the plates? (b) At what angle is the electron moving with respect to the axis? (c) At what distance from the axis will the electron strike the fluorescent screen?
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Explanation:
The electron moves against the electric field because it experiences a force due to the field.
The acceleration experienced by the electron in y direction is a
y
=
m
qE
Thus, by kinematics relation Δy=
2
1
a
y
Δt
2
As the electron moves with a constant speed in x direction, Δt=
v
Δx
The velocity of the electron in terms of the kinetic energy is v=
m
2KE
Using the above relations for Δy, we have Δy=
2
1
m
qE
(
v
Δx
)
2
=
4×
q
KE
EΔx
2
Given
q
KE
=2000 V and E=1.2×10
4
N/C and Δx=1.5×10
−2
cm
We get, Δy=
4×2000
1.2×10
4
×(1.5×10
−2
)
2
=0.3375×10
−3
m≈0.34 mm
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