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
Chapter1: Units, Trigonometry. And Vectors
Section: Chapter Questions
Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...
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*Figure 6.26*
First off, note that the acceleration \( a_x \) component is negative because it points to the left so \( a_x = -\frac{qE}{m} \). We choose the velocity squared equation above because we want to solve for the final velocity (and we are not given the time interval):
\[ v_{xf}^2 = v_{xi}^2 + 2a_x (x_f - x_i) = 0 - \frac{2Eq(x_f - x_i)}{m} = -\frac{2Eq x_f}{m} \]
Where we made \( x_i = 0 \) such that \( x_f = 0.0021m \) and that the proton starts at rest so \( v_{xi} = 0 \). We have to take the square root to get the final speed:
\[ v_{xf} = \pm \sqrt{-\frac{2Eq x_f}{m}} \]
And substituting the numbers we get:
\[ v_{xf} = \pm \sqrt{\frac{4.5 \times 10^4 \, N/C \, \times (1.602 \times 10^{-19} \, C) \, \times (-0.0021 \, m)}{-2 \times 1.67 \times 10^{-27} \, kg }} = \pm 1.35 \times 10^5 \, m/s \]
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 proton is moving extremely fast in this case because typical fields (like in an old TV set) are on this order (around \( 10^5 \, N/C \)).
**Question 7: Homework – Unanswered**
For the example above, how long (in nanoseconds) does it take the proton to travel the distance of 2](/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 = 4.5 × 10⁴ N/C.**
*Figure 6.26*
First off, note that the acceleration \( a_x \) component is negative because it points to the left so \( a_x = -\frac{qE}{m} \). We choose the velocity squared equation above because we want to solve for the final velocity (and we are not given the time interval):
\[ v_{xf}^2 = v_{xi}^2 + 2a_x (x_f - x_i) = 0 - \frac{2Eq(x_f - x_i)}{m} = -\frac{2Eq x_f}{m} \]
Where we made \( x_i = 0 \) such that \( x_f = 0.0021m \) and that the proton starts at rest so \( v_{xi} = 0 \). We have to take the square root to get the final speed:
\[ v_{xf} = \pm \sqrt{-\frac{2Eq x_f}{m}} \]
And substituting the numbers we get:
\[ v_{xf} = \pm \sqrt{\frac{4.5 \times 10^4 \, N/C \, \times (1.602 \times 10^{-19} \, C) \, \times (-0.0021 \, m)}{-2 \times 1.67 \times 10^{-27} \, kg }} = \pm 1.35 \times 10^5 \, m/s \]
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 proton is moving extremely fast in this case because typical fields (like in an old TV set) are on this order (around \( 10^5 \, N/C \)).
**Question 7: Homework – Unanswered**
For the example above, how long (in nanoseconds) does it take the proton to travel the distance of 2
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