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Line Integral + Potential Difference Problem Need Help Please

  1. Feb 7, 2009 #1
    1. The problem statement, all variables and given/known data
    The cyclotron is a device for accelerating charged particles. It requires changing electric fields and a magnetic field, but it can be modeled using (non-physical) static electric fields and potentials as we will do in this problem. Keep in mind that these fields cannot actually be created by any set of static charges, so just treat the field as given. To aid in visualization of the problem, consider that this problem could also have been titled, "The Spiral Staircase."

    Consider an electric field defined by
    E = 0.4*(-sin([tex]\theta[/tex]) [tex]\hat{i}[/tex]+ cos([tex]\theta[/tex]) [tex]\hat{j}[/tex])/r
    (E in N/C, i is the unit vector along the x-axis, j is the unit vector along the y-axis; the coefficient 0.40 varies from person to person)

    Note that we can write any position in the xy-plane in a similar way:
    r = r(cos([tex]\theta[/tex]) i + sin([tex]\theta[/tex]) j ).
    The Dot product of E and r would equal 0.

    We want to find the potential difference between the two points: (2.7,1.3) and (4.0, 4.9).

    a. First integrate radially outward along a line of constant theta. What's the electric potential difference between the points (2.7,1.3) and (5.70,2.74)?

    b. Now integrate along a circle of constant radius. What's the potential difference between the points (5.70,2.74) and (4.0,4.9)?

    c. What's the total potential difference along the above path between the points (2.7,1.3) and (4.0,4.9)?

    Now reverse the order of integration.

    d. First integrate along a circle of constant radius. What's the potential difference between the points (2.7,1.3) and (1.90,2.32)?

    e. Now integrate along a line of constant theta. What's the potential difference between the points (1.90,2.32) and (4.0,4.9)?

    f. What's the total potential difference along this path between the points (2.7,1.3) and (4.0,4.9)?

    Now consider two points at the same radius, but on opposite sides of the circle: (2.7,1.3) and (-2.7,-1.3)

    g. What's the potential difference between (2.7,1.3) and (-2.7,-1.3) integrating along a circle of constant radius, going in the positive theta direction.

    h. What's the potential difference integrating along a circle in the negative theta direction?

    2. Relevant equations
    I know you have to do a line integral

    [tex]\Delta[/tex]V = -[tex]\int[/tex][tex]\vec{E}[/tex][tex]\vec{ds}[/tex]


    3. The attempt at a solution

    I got part a and e they both are zero because they are gonna be on equipotential lines and therefore there wont a potential difference

    however i cant seem to do a line integral.

    for part b, i have to parametrize the curve with the points given,
    r(t) = (1 -t)<2.7,1.3> + t*(1.9,2.32)
    r(t) = <2.7 - .8*t, 1.3 - 1.02*t)
    r'(t) = (-.8,-1.02)
    but now I don't know how to add it into the integral.

    Please help.

    thanks in advanced
     
    Last edited: Feb 7, 2009
  2. jcsd
  3. Feb 7, 2009 #2

    AEM

    User Avatar

    Here's how I would do your line integral [tex] \Delta V = -\int \vec{E} \cdot d \vec{S} [/tex]

    First, remember that your path is along a circular arc from (5.70, 2.74) to (4.0, 4.9). Express
    [tex] d \vec{S} [/tex] in terms of distance along that arc as

    [tex] d \vec{S} = \hat{\theta} r d \theta [/tex]

    The unit vector [tex] \hat{\theta} [/tex] is given by

    [tex] \hat{\theta} = \frac{-sin(\theta)}{\sqrt{2}} \hat{i} + \frac{cos( \theta)}{\sqrt{2}} \hat{j} [/tex]

    You will then need the angle (in radians) between the x axis and the line from the origin to the point (5.70, 2.74) and the angle (in radians) between the x axis and the line from the origin to the point (4.0, 4.9). Those two angles will be the limits of your integration. Your integral now becomes

    [tex] \Delta V = - \int_1 ^2 \frac{.4}{r}( -sin(\theta) \hat{i} + cos(\theta) \hat{j} ) \cdot (\frac{ -sin(\theta)}{\sqrt{2}} \hat{i} + \frac{cos(\theta)}{\sqrt{2}} \hat{j}) r d \theta [/tex]

    Notice the good things that happen: The r's cancel and the trig functions will disappear when you take the dot product. That will make the integral rather easy.
     
  4. Feb 7, 2009 #3

    ok.. so got r's canceling out, and trigs just turn into [tex] .4*(1/{\sqrt{2}}) [/tex] once you dot them. so will the integral be just [tex] - \int .4*(1/{\sqrt{2}}) d\theta [/tex] & for the bounds for part b would it be from 1 to 2 or arctan(2.74/5.7) to arctan(4.9/4)

    thanx for ur help
     
  5. Feb 8, 2009 #4
    while doing what i described above i got my answer as -.2828
    tat was not the right answer. Did I do something wrong? Please help!
     
  6. Feb 9, 2009 #5

    AEM

    User Avatar

    Well, without seeing the details of your work, I can't do anything more than guess where you might have made a mistake. I just calculated out the answer to be -1.24 . One question: is your calculator in radians mode, or degrees mode? That's a common mistake.
     
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