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

AI Thread Summary
The discussion focuses on calculating the work done by an electric field in moving a unit positive charge between two points. A user shares their displacement vector and seeks confirmation on its correctness, particularly regarding the signs of the components. Clarifications are provided on the proper representation of the infinitesimal displacement vector and the integration limits, emphasizing that the direction of displacement should align with the signs of the components. The conversation also addresses the presence of a negative sign in the work integral, explaining that it pertains to definitions of electric potential rather than the work done by the electric field itself. Overall, the thread highlights key concepts in calculating work in electric fields and the importance of accurately representing vectors.
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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