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Potential Difference in a varying field

  1. Sep 12, 2009 #1
    1. The problem statement, all variables and given/known data

    problem.jpg

    2. Relevant equations

    [tex]V_f - V_i = - \int_i^f{\vec E \cdot d\vec s}[/tex]

    3. The attempt at a solution

    Can I assume that the [itex]V_i[/itex] is 0?
    [tex]V= - \int{\vec (4x) \cdot d\vec s}[/tex]
    [tex]V= - \int{\vec (4x) \cdot d\vec s}[/tex]
    [tex]V= -4x(s)[/tex]
    [tex]V=-4x(2.51)[/tex]
    [tex]V=-10.02x[/tex]

    At this point, what is done?
     
  2. jcsd
  3. Sep 12, 2009 #2

    kuruman

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    You need limits of integration. What are they? Also you need a path from A to B. You can pick any path, but some are more convenient than others. Finally, it would pay to write the electric field and element ds in unit vector notation first, then take the dot product between them.
     
  4. Sep 12, 2009 #3
    So we call point C the origin (0,0), and take the path from A-->C and then from C-->B, finding the potential difference between the points on the first segment, then the difference on the second segment.

    [tex]
    V_c - V_a = - \int_a^c{\vec E \cdot d\vec s}
    [/tex]
    [tex]
    V_c - V_a = - \int_a^c{(4x)(\hat x) \cdot (ds) (\hat y)}
    [/tex]
    [tex]
    V_c - V_a = {0}
    [/tex]

    And then from C-->B

    [tex]
    V_c - V_b = - \int_b^c{(4x)(\hat x) \cdot (ds) (\hat x)}
    [/tex]
    [tex]
    V_c - V_b = - \int_b^c{(4x)(ds)}
    [/tex]
    [tex]
    V_c - V_b = - (4x)\int_b^c{(ds)}
    [/tex]
    [tex]
    V_c - V_b = - (4x)(s)}
    [/tex]

    And now what?
     
  5. Sep 12, 2009 #4

    kuruman

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    The integral from A to C is correctly calculated to be zero. For the second integral, you have element ds. Look at your path, you are moving along the x-axis, what is a better name for ds? What are your limits of integration?
     
  6. Sep 12, 2009 #5
    dS is the path from c-->b.

    Obviously, i don't know how to set up the integral.
     
  7. Sep 12, 2009 #6

    kuruman

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    No, ds is a small increment along the path from C to B. This path is along the x-axis, so if you chop it up into many small increments, you are chopping up the x-axis into small increments and a small increment along the x-axis is traditionally called ...
     
  8. Sep 12, 2009 #7
    (I meant the integral of dS, the sum of all the dS's was the path. sorry.)

    A small differential on the x-axis would be dx.
     
  9. Sep 12, 2009 #8

    kuruman

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    Correct, it is dx. So when you go from C to B x varies from what to what value? These are your limits of integration.
     
  10. Sep 12, 2009 #9
    So..what its..

    [tex]V=\int_0^{1.2}{E} dx[/tex]
    [tex]V=\int_0^{1.2}{4x} dx[/tex]
    [tex]V=4(\frac{x^2}{2}|_0^{1.2})[/tex]
     
  11. Sep 12, 2009 #10

    kuruman

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    Looks good except for the minus sign that you dropped. Now finish the evaluation and you are done.
     
  12. Sep 12, 2009 #11
    So I'm confused. Is the answer 2.88 or -2.88? It's asking for the potential difference Vb-Va
     
  13. Sep 12, 2009 #12

    kuruman

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    When you go from A to C there is no change in the potential because you are moving in a direction perpendicular to the electric field. When you go from C to B you are moving with the electric field lines. Now, do electric field lines point from high to low potential or the other way around? Answer that and you have the sign of the difference and should confirm the result of your integration.
     
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