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

  1. Dec 23, 2016 #1
    1. The problem statement, all variables and given/known data
    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

    2. Relevant equations


    3. 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.
     
  2. jcsd
  3. Dec 23, 2016 #2

    TSny

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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.
     
  4. Dec 23, 2016 #3

    TSny

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    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.
     
  5. Dec 23, 2016 #4
    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
     
  6. Dec 23, 2016 #5

    TSny

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    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.
     
  7. Dec 24, 2016 #6
    Awesome bro
    Thanks a lot
     
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