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Math Help - work done in a vector field

  1. #1
    jut
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    work done in a vector field

    Determine the work done by the electric field E=i*x-j*2y in moving a unit positive charge from position P1(-2,0,0) to position P2(5,-1,3).

    The work is defined as the integral from p1 to p2 of E "dot" dl,

    \frac{W}{q}=-\int _{\text{P1}}^{\text{P2}}E\circ dl

    Answer from book: 9.5 J

    My attempted solution

    Break dl into constituent vectors: dl = i*dx+j*dy

    The dot product of E dot dl = x*dx - 2y*dy

    And the integration becomes,

    \frac{W}{q}=-\left[\int _{-2}^5x \text{dx}-\int _0^{-1}2y \text{dy}\right]= -19/2 = -9.5

    So the negative sign was removed b/c work cannot be negative?
    Last edited by jut; March 31st 2010 at 02:42 PM.
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  2. #2
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    Consider more simpler task:
    if E=i
    P1(0,0,0) to P2(1,0,0)
    According to your formular
    W/q=- \int^1_0 dx=-1.
    To me must be +1 (the field on the particle).
    Please check the definition of the work.
    The field on the particle
    or particle on the field.
    Last edited by zzzoak; March 31st 2010 at 03:43 PM.
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  3. #3
    jut
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    This is what my book says about work (see attachment)
    Attached Files Attached Files
    • File Type: pdf W.pdf (215.4 KB, 15 views)
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  4. #4
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    Quote Originally Posted by jut View Post
    Determine the work done by the electric field E=i*x-j*2y in moving a unit positive charge from position P1(-2,0,0) to position P2(5,-1,3).

    The work is defined as the integral from p1 to p2 of E "dot" dl,

    \frac{W}{q}=-\int _{\text{P1}}^{\text{P2}}E\circ dl

    Answer from book: 9.5 J

    My attempted solution

    Break dl into constituent vectors: dl = i*dx+j*dy

    The dot product of E dot dl = x*dx - 2y*dy

    And the integration becomes,

    \frac{W}{q}=-\left[\int _{-2}^5x \text{dx}-\int _0^{-1}2y \text{dy}\right]= -19/2 = -9.5

    So the negative sign was removed b/c work cannot be negative?
    Here is something for you to think about: What would be the work done if the charge was negative ....?
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