Example: A proton is released from rest at the positive plate of a capacitor. What is the proton's speed when it reaches the negative plate? The capacitor is illustrated below with spacing d=2.1mm. The strength of the electric field is E=4.5x104N/C. F/₁ proton + d=2.1mm Q [6.26] First off, note that the acceleration x component is negative because it points to the left so ax=-qE/m. We choose the velocity squared equation above because we want to solve for the final velocity (and we are not given the time interval): 2Eqxj v²f = vi + 2ax (xƒ — xi) = 0 - 2Eq(x-x₁) m m Where we made x-0 such that xF-0.0021m and that the proton starts at rest so vx-0. We have to take the square root to get the final speed: 2Eqx Vrf = ± √ m And substituting the numbers we get 4.5 × 10¹N -2( C U₂f = ± √ -)(1.602x10-19C) (-0.0021m) 1.67x10-27kg = ±1.35 x 105 m/s Notice that the sign underneath the radical is positive (as required) and to get the rect velocity component we need to choose the negative sign (moving to the left). The electron is moving extremely fast in this case because typical fields (like in an old TV set) are on this order (around 105 N/C). Question 7 ✪ Homework Unanswered For the example above, how long (in nanoseconds) does it take the proton to travel the distance of 2.1mm between the plates? (Hint: use one of the kinematics equations.) Type your numeric answer and submit TERVE
Example: A proton is released from rest at the positive plate of a capacitor. What is the proton's speed when it reaches the negative plate? The capacitor is illustrated below with spacing d=2.1mm. The strength of the electric field is E=4.5x104N/C. F/₁ proton + d=2.1mm Q [6.26] First off, note that the acceleration x component is negative because it points to the left so ax=-qE/m. We choose the velocity squared equation above because we want to solve for the final velocity (and we are not given the time interval): 2Eqxj v²f = vi + 2ax (xƒ — xi) = 0 - 2Eq(x-x₁) m m Where we made x-0 such that xF-0.0021m and that the proton starts at rest so vx-0. We have to take the square root to get the final speed: 2Eqx Vrf = ± √ m And substituting the numbers we get 4.5 × 10¹N -2( C U₂f = ± √ -)(1.602x10-19C) (-0.0021m) 1.67x10-27kg = ±1.35 x 105 m/s Notice that the sign underneath the radical is positive (as required) and to get the rect velocity component we need to choose the negative sign (moving to the left). The electron is moving extremely fast in this case because typical fields (like in an old TV set) are on this order (around 105 N/C). Question 7 ✪ Homework Unanswered For the example above, how long (in nanoseconds) does it take the proton to travel the distance of 2.1mm between the plates? (Hint: use one of the kinematics equations.) Type your numeric answer and submit TERVE
College Physics
11th Edition
ISBN:9781305952300
Author:Raymond A. Serway, Chris Vuille
Publisher:Raymond A. Serway, Chris Vuille
Chapter16: Electrical Energy And Capacitance
Section: Chapter Questions
Problem 1CQ: A proton is released from rest in a uniform electric field. Determine whether the following...
Related questions
Question
![Example: A proton is released from rest at the positive plate of a capacitor. What is the proton's
speed when it reaches the negative plate? The capacitor is illustrated below with spacing
d=2.1mm. The strength of the electric field is E=4.5x104N/C.
E
proton +
d=2.1mm Q
[6.26]
First off, note that the acceleration x component is negative because it points to the left so ax=-qE/m. We
choose the velocity squared equation above because we want to solve for the final velocity (and we are
not given the time interval):
2Eq(x-x₁)
2Eqxf
: v²i + 2ax (xƒ − x₁) = 0 -
-
m
m
Where we made x-0 such that xF-0.0021m and that the proton starts at rest so vxF0. We have to take the
square root to get the final speed:
2Eqri
Vrf = ± √
m
And substituting the numbers we get
4.5 x 10¹N
-2(
C
-)(1.602x10-19C) (-0.0021m)
1.67x10-27kg
Vxf = ±
= ±1.35 x 105 m/s
Notice that the sign underneath the radical is positive (as required) and to get the correct velocity
component we need to choose the negative sign (moving to the left). The electron is moving extremely
fast in this case because typical fields (like in an old TV set) are on this order (around 105 N/C).
Question 7
Homework Unanswered
For the example above, how long (in nanoseconds) does it take the proton to travel the distance of 2.1mm between the plates?
(Hint: use one of the kinematics equations.)
Type your numeric answer and submit
F](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F0a5ac3c6-e4ab-423a-91e1-ff33bdae0d07%2F9324b5f9-416b-4407-a1b3-aa0d06ec110c%2Feo3o0v_processed.png&w=3840&q=75)
Transcribed Image Text:Example: A proton is released from rest at the positive plate of a capacitor. What is the proton's
speed when it reaches the negative plate? The capacitor is illustrated below with spacing
d=2.1mm. The strength of the electric field is E=4.5x104N/C.
E
proton +
d=2.1mm Q
[6.26]
First off, note that the acceleration x component is negative because it points to the left so ax=-qE/m. We
choose the velocity squared equation above because we want to solve for the final velocity (and we are
not given the time interval):
2Eq(x-x₁)
2Eqxf
: v²i + 2ax (xƒ − x₁) = 0 -
-
m
m
Where we made x-0 such that xF-0.0021m and that the proton starts at rest so vxF0. We have to take the
square root to get the final speed:
2Eqri
Vrf = ± √
m
And substituting the numbers we get
4.5 x 10¹N
-2(
C
-)(1.602x10-19C) (-0.0021m)
1.67x10-27kg
Vxf = ±
= ±1.35 x 105 m/s
Notice that the sign underneath the radical is positive (as required) and to get the correct velocity
component we need to choose the negative sign (moving to the left). The electron is moving extremely
fast in this case because typical fields (like in an old TV set) are on this order (around 105 N/C).
Question 7
Homework Unanswered
For the example above, how long (in nanoseconds) does it take the proton to travel the distance of 2.1mm between the plates?
(Hint: use one of the kinematics equations.)
Type your numeric answer and submit
F
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