**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.

to find: work done by electric force

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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Question

**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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