A partici charge and mass M ers an electric field with speed Vo and is slowed down to zero speed over a distance d. Find the work done on the particle by the electric force. QEd -MV₂²/2 MV ²/2 O Fed
A partici charge and mass M ers an electric field with speed Vo and is slowed down to zero speed over a distance d. Find the work done on the particle by the electric force. QEd -MV₂²/2 MV ²/2 O Fed
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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![**Problem Description:**
A particle of charge \( Q \) and mass \( M \) enters an electric field with speed \( V_0 \) and is slowed down to zero speed over a distance \( d \). Find the work done on the particle by the electric force.
**Options:**
- \( \) \( QEd \)
- \( \) \(-MV_0^2/2\)
- \( \) \( MV_0^2/2 \)
- \( \) \( Fd \)
- \( \) \( 0 \)
**Explanation:**
To solve for the work done on the particle by the electric force, we can use the work-energy principle. This principle states that the work done on an object is equal to the change in its kinetic energy.
Initially, the particle has kinetic energy given by:
\[ KE_{\text{initial}} = \frac{1}{2}MV_0^2 \]
Since the particle comes to rest, its final kinetic energy is:
\[ KE_{\text{final}} = 0 \]
The work done by the electric force (\( W \)) is the change in kinetic energy:
\[ W = KE_{\text{final}} - KE_{\text{initial}} = 0 - \frac{1}{2}MV_0^2 = -\frac{1}{2}MV_0^2 \]
So, the correct option is:
\( \highlight{-\frac{1}{2}MV_0^2} \)
This choice is consistent with the principle that the work done by the electric force is negative since it slows down the particle.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F7a85083f-c711-4fe4-a0ea-d798364bd78d%2Ff44b22bb-dcb9-405f-9f49-63c92af3592e%2Fj0goc2_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Problem Description:**
A particle of charge \( Q \) and mass \( M \) enters an electric field with speed \( V_0 \) and is slowed down to zero speed over a distance \( d \). Find the work done on the particle by the electric force.
**Options:**
- \( \) \( QEd \)
- \( \) \(-MV_0^2/2\)
- \( \) \( MV_0^2/2 \)
- \( \) \( Fd \)
- \( \) \( 0 \)
**Explanation:**
To solve for the work done on the particle by the electric force, we can use the work-energy principle. This principle states that the work done on an object is equal to the change in its kinetic energy.
Initially, the particle has kinetic energy given by:
\[ KE_{\text{initial}} = \frac{1}{2}MV_0^2 \]
Since the particle comes to rest, its final kinetic energy is:
\[ KE_{\text{final}} = 0 \]
The work done by the electric force (\( W \)) is the change in kinetic energy:
\[ W = KE_{\text{final}} - KE_{\text{initial}} = 0 - \frac{1}{2}MV_0^2 = -\frac{1}{2}MV_0^2 \]
So, the correct option is:
\( \highlight{-\frac{1}{2}MV_0^2} \)
This choice is consistent with the principle that the work done by the electric force is negative since it slows down the particle.
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