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Homework Help: Finding V from E

  1. Apr 23, 2013 #1
    1. The problem statement, all variables and given/known data
    Calculate V(ab) if E = (-5 y / x^3) ax + (5/x) ay + 4 az V/m.
    A(-8,3,2) and B(5,2,3).

    2. Relevant equations


    3. The attempt at a solution

    So I tried

    V=-[∫(-5y/x^3)dx {from 5 to -8}+ ∫(5x)dy {from 2 to 3} + ∫4dz {from 3 to 2}]

    The problem is that the limits are wrong. What should I do?
    Should it be the distance and not the literal point on the plane? like the first one, should it be 0 to 13, the second one 0 to 1, the third one 0 to 1?
    and what do I do with the remaining variables after the integration. For ex, in the first one I would have a y and on the second one a x
  2. jcsd
  3. Apr 23, 2013 #2

    Simon Bridge

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  4. Apr 23, 2013 #3


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    Check if the electric field is conservative. If it is not, the work between a and b depends on the path, and potential is not defined.

  5. Apr 23, 2013 #4
    It doesn't say anything about this.
  6. Apr 23, 2013 #5


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    Potential exist if the curl of the electric field is zero. Have you heard about that?
    If not, check if you copied the formulas for E correctly, or any path is given in the problem. If you were supposed to integral for the path you used, note that when determining the integral for a line segment, the initial and final values of all variables have to been substituted. See figure.


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  7. Apr 24, 2013 #6
    I ended up reading a lot from the book, and yes it is conservative.
    The book is just theory no example. So I don't really know the correct approach for this.
    I think I'm going to go with V=-[∫(-5y/x^3)dx {from 5 to -8}+ ∫(5x)dy {from 2 to 3} + ∫4dz {from 3 to 2}]

    and at the end, where I'm left with variables, substitute them for the coordinates from A
    Last edited by a moderator: Apr 24, 2013
  8. Apr 24, 2013 #7


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    The z component of the curl is not zero, so it is not conservative.

    The integral for the first line segment goes from (5,2,3) to (-8,2,3) (red line in the figure). The second integral goes from (-8,2,3) to (-8,3,3) (the blue line) and so on. You have to substitute the values of all variables.

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