Work done in moving a 1C positive charge from one point to another

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Homework Help Overview

The discussion revolves around calculating the work done by an electric field represented by the equation E = axX - ay2y when moving a unit positive charge from one point to another in three-dimensional space. The specific points of interest are p1 (-2,0,0) and p2 (5,-1,3).

Discussion Character

  • Conceptual clarification, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • The original poster attempts to determine the displacement vector and questions its correctness based on the positive and negative displacements in the x and y directions. Participants discuss the representation of infinitesimal displacement and the implications of negative signs in the work integral.

Discussion Status

Participants are actively engaging with the original poster's questions, providing clarifications about the definitions of displacement and work. There is an exploration of the correct application of signs in the equations, particularly regarding the work done by the electric field versus external agents.

Contextual Notes

There is an indication that the original poster may be unfamiliar with forum discussions, and the conversation includes references to textbook definitions and the importance of integration limits in the context of work calculations.

Abdulwahab Hajar
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Homework Statement


Determine the Work done by the electric field E = axX - ay2y in moving a unit positive charge from position p1 (-2,0,0) to position p2 (5,-1,3) the distances are in m

Homework Equations

The Attempt at a Solution


I'm not really experienced with forums therefore my attempted solution is an attached image.https://scontent-lht6-1.xx.fbcdn.net/v/t34.0-12/15683157_1201819119912120_1486498649_n.jpg?oh=0aad954d8ecb161d62c98aa8d1d685e7&oe=585F5529
is my displacement vector in my solution correct ? I did it as such because I have +ve displacement in the ax direction and -ve displacement in the ay direction.
 
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The general way to represent an infinitesimal displacement is ##\vec{dl} = dx \, \hat{a}_x + dy \, \hat{a}_y + dz \, \hat{a}_z ##. The quantities ##dx##, ##dy##, and ##dz## could be positive, negative, or zero depending on the direction of the displacement. When you integrate with respect to ##y## from 0 to -1, ##dy## will be negative. You should not write ##\vec{dl} = dx \, \hat{a}_x - dy \, \hat{a}_y + dz \, \hat{a}_z ## with a negative sign for the y component. This would imply that the y component of displacement would be in the ##+\hat{a}_y## direction when ##dy## is negative.
 
Also, can you explain why you have a negative sign in front of the integral in the expression ##W = -q\int{\vec{E} \cdot \vec{dl}}##? Keep in mind that you are asked to find the work done by the electric field, not the work done by an external agent.
 
Wow dude thanks for the reply..
W=−q∫⃗E⋅→dl well I got the minus from the book, it's just like −∫⃗E⋅→dl for electrical potential apparently..
Could you enlighten me perhaps and so as you said dl=dx^ax+dy^ay+dz^az always stands right the integration limits determine the rest .
Thank you sir
 
The definition of the work done by a force along a path from point ##a## to point ##b## is ##\int_a^b \vec{F} \cdot \vec{dl}##. There is no negative sign in the definition. Definitions of electric potential and electric potential energy will have a negative sign.
 
Awesome bro
Thanks a lot
 

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