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Charge on a square loop: Work

  1. Nov 11, 2013 #1
    1. The problem statement, all variables and given/known data
    A small bead is on square loop in the xz plane with dimensions (±R, 0, ±R). An electric field is turned that is ##\vec{E}(\vec{r})=-Cx\hat{z}##

    Calculate W
    2. Relevant equations

    ##W= \int q\vec{E}\bullet dl##

    3. The attempt at a solution
    Starting with z direction:

    [tex]q\int_{-R}^R -Cx\hat{z} (dx\hat{x} + dy\hat{y}+ dz\hat{z})[/tex]

    =##-Cq \frac {x^2}{2} |_{-R}^R=-CqR^2##

    I believe that the total work would just be 2 times this quantity because since the E field isn't in the correct direction in order to move the charge in the x direction. But this doesn't make sense to me within the problem context. Am I missing something?
     
  2. jcsd
  3. Nov 11, 2013 #2

    tiny-tim

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    hi scorpius1782! :smile:

    (type \cdot not \bullet :wink:)
    sorry, but that makes no sense :redface:

    you need to write a dot product

    try again (and use a specific arm of the loop) :smile:
     
  4. Nov 11, 2013 #3
    thank you for the \cdot! I couldn't remember what it was (kept thinking \vdot) so I gave up and used bullet.
    the vertical (z) part of the box from -R to R at x=-R
    ##\int_{-R}^R-Cqx\hat{z}\cdot dz \hat{z}##

    I didn't think of it before but since here x=-R I can put that in for x in this integral. I believe that since there is no change in x or y then dx and dy are zero.

    This means that ##CqR\int_{-R}^R\hat{z}\hat{z}dz=CqR(R+R)=2CqR^2##

    Then for x=R: ##\int_{R}^{-R}-CqR\hat{z}\cdot dz \hat{z}##
    ##=-CqR\int_{R}^{-R}\hat{z}\hat{z}dz=-CqR(-R-R)=-2CqR^2##

    Is this better?
     
  5. Nov 11, 2013 #4

    tiny-tim

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    yes :smile:
    erm :rolleyes:
     
  6. Nov 11, 2013 #5
    Hmm, seems I didn't distribute my negative sign there!
     
  7. Nov 11, 2013 #6

    tiny-tim

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    are you in the loop now? o:)
     
  8. Nov 12, 2013 #7
    I am, thank you. I assume since you made no comment that my understanding of the horizontal forcei is correct? That is where z=±R there is no work?
     
  9. Nov 12, 2013 #8

    tiny-tim

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    yes, for displacement of constant z, the component of the field along the wire is zero, so the work done is zero :smile:
     
  10. Nov 12, 2013 #9
    Thanks for the help!
     
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