Example 23.11 An Accelerated Electron SOLVE IT (A) Find the acceleration of the
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Question
Example 23.11 An Accelerated Electron
SOLVE IT
(A) Find the acceleration of the electron while it is in the electric field.
Conceptualize This example differs from a previous one because the velocity of the charged particle is initially perpendicular to the electric field lines. (In a previous example, the velocity of the charged particle is always parallel to the electric field lines.) As a result, the electron in this example follows a curved path as shown in the figure. The motion of the electron is the same as that of a massive particle projected horizontally in a gravitational field near the surface of the Earth.
Categorize The electron is a particle in a field (electric). Because the electric field is uniform, a constant electric force is exerted on the electron. To find the acceleration of the electron, we can model it as a particle under a net force.
Analyze From the particle in a field model, we know that the direction of the electric force on the electron is downward in the figure, opposite the direction of the electric field lines. From the particle under a net force model, therefore, the acceleration of the electron is downward.
The particle under a net force model was used to develop the accelerating particle equation in the case in which the electric force on a particle is the only force. Use this equation to evaluate the y component of the acceleration of the electron:
ay =
Substitute numerical values:
ay =
ay = m/s2
(B) Assuming the electron enters the field at time t = 0, find the time at which it leaves the field.
Categorize Because the electric force acts only in the vertical direction in the figure, the motion of the particle in the horizontal direction can be analyzed by modeling it as a particle under constant velocity.
Analyze Solve the equation for position as a function of time for the time at which the electron arrives at the right edges of the plates:
xf = xi + vxt t =
Substitute numerical values:
t =
=
t = s
(C) Assuming the vertical position of the electron as it enters the field is yi = 0 what is its vertical position when it leaves the field?
Categorize Because the electric force is constant in the figure, the motion of the particle in the vertical direction can be analyzed by modeling it as a particle under constant acceleration.
Analyze Describe the position of the particle at any time t:
yf = yi + vyit +
ayt2
Substitute numerical values:
yf = 0 + 0 +
at2 = cm
Finalize If the electron enters just below the negative plate in the figure and the separation between the plates is less than the value just calculated, the electron will strike the positive plate.
Notice that we have used four analysis models to describe the electron in various parts of this problem. We have neglected the gravitational force acting on the electron, which represents a good approximation when dealing with atomic particles. For an electric field of 180 N/C, the ratio of the magnitude of the electric force eE to the magnitude of the gravitational force mg is on the order of 1012 for an electron and on the order of 109 for a proton.
MASTER ITHINTS: GETTING STARTED | I'M STUCK!
Find the speed of the electron as it emerges from the field.
Determine the extra component of velocity caused by the electric field force over the time the particle passes between the plates. Use that to find the speed. m/s
Explanation / Answer
a )
vi = 2.20 X 106 m/s
and E = 180 N/C
The horizontal length of the plates is = 0.100 m
using equation F = eE and F = m ay
m ay = eE
ay = eE/m
ay = 1.6 X 10-19 X 180 / 9.1 X 10-31
ay = 3.164 X 1013 m/sec2
b )
x = Vox t + 1/2 aot2
t = 0.1 / 2.20 X 106
t = 4.5454 X 10-8 sec
c )
yf = - y
= - (Voy t + 1/2 ayt2 )
= - ( 0 X 4.5454 X 10-8 + 0.5 X 3.164 X 1013 X (4.5454 X 10-8)2 )
= - 0.0326 m
= - 3.26 cm
d )
vy = Voy + ay t
= 0 + 3.164 X 1013 X (4.5454 X 10-8)
= 1.438 X 106 m/s
v = ( (2.20 X 106)2 + (1.438 X 106)2 )1/2
v = 2.628 X 106 m/sec
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