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Math Help - is my solution correct ..

  1. #1
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    is my solution correct ..

    here is a question:
    http://i25.tinypic.com/8xlcfl.gif
    there is endless wire uniformly charged \lambda=2*10^{-4}C/m
    and close to it located a wire with l=0.12_m
    which has 30 degree angle between him and the endless wire,and it charged in uniformed way so that the total charge on it is q=3*10^{-9}
    the distance of the finite wire and the infinite wire is 0.08 meters

    A.find the total force acted on the finite wire.
    B.find it when \alpha=0 degrees and when \alpha=90 degrees.

    gaus law for infinite wire i presume its length is l and distance r.
    E2\pi r h=\frac{q}{\epsilon_0}\\
    E=\frac{q}{\epsilon_0 2\pi r h}=\frac{\lambda h}{\epsilon_0 2\pi r h}=\frac{\lambda }{\epsilon_0 2\pi r }\\
    \frac{q_2}{l_2}=\lambda_2\\
    l_0=l_2\cos \alpha\\
    \int_{0}^{l_0}Edl=\int_{0}^{l_0}\frac{\lambda }{\epsilon_0 2\pi r }dl

    i sum along the finite wire
    but the ditance from the infinite changes too.
    so it should be a double integral
    but its not a 2d body

    ??
    Last edited by transgalactic; August 22nd 2009 at 03:57 AM.
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  2. #2
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    you are right, if its a 1 dimensional body you will need only 1 integral

    but you also have to find out the relation of r with l.


    hint:
    if you draw a xy coordinate axis (parallel and perpendicular to the infinite wire), you can do it

    then
    dl=\sqrt{dx^2+dy^2}
    dl=\sqrt{(\frac{dx}{dx})^2+(\frac{dy}{dx})^2}dx
    dl=\sqrt{1+(\frac{dy}{dx})^2}dx
    and you got a 1d integral
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  3. #3
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    i cant use expression using dx and dy
    because thus i wil have a double integral

    i need to link l with r
    r=0.08+(temporary hipotenuse)*cos alpha

    i dont know how to express this temporary hipotenuse
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  4. #4
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    Quote Originally Posted by transgalactic View Post
    i cant use expression using dx and dy
    because thus i wil have a double integral
    no.
    its not dx and dy. its dx and \frac{dy}{dx}.


    Imagine these axis. y-axis parallel the infinite wire. x-axis perpendicular.
    that way r will simply be x.

    now you need to integrate dl, as you wrote in the end of your 1st post. but you need to write dl into a function of dx.




    notice the 90 angle triangle, you use





    now you just need to find the wire equation
    y=mx+b
    calculate \frac{dy}{dx} and you get dl in terms of dx.

    Then substitute dl in your equation.
    Actually your equation isnt entirely correct (Ithink)

    F=E q
    so you need to use
    dF=E \times dq
    dF=E \times \lambda_2 \times dl
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  5. #5
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    i cant find
    y=mx+b

    m=dy/dx
    but b cannot be found
    i dont have enough data to know where the line crosses x axes

    ??
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  6. #6
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    m is all you need
    you only need dy/dx so find out m
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  7. #7
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    my expestion is correct because its F=dq*E

    and we substitute dl by an expression of dx
    but i cant express dy only with dx
    <br />
\int_{0}^{l_0}Edl=\int_{0}^{xpretion}\frac{\lambda }{\epsilon_0 2\pi r }\sqrt{1+(\frac{dy}{dx})^2}dx<br /> <br />
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  8. #8
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    i dont know how to express dy/dx using only dx
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  9. #9
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    y=mx+b

    if you derive that in relation to x

    \frac{dy}{dx}=m

    you look to the line and you see its 0.12m long and makes 30 with vertical, so
    the x component of the line, \Delta x, is
    \Delta x=0.12 \times sin(30)=0.06 m

    the y component of the line, \Delta y, is
    \Delta y=0.12 \times cos(30)=0.06 \sqrt 3 m

    m=\frac{\Delta y}{\Delta x}=\sqrt 3

    \frac{dy}{dx}=\sqrt 3



    my expestion is correct because its F=dq*E

    and we substitute dl by an expression of dx
    but i cant express dy only with dx
    I don't understand.
    if you say F=dq*E, why do you use in the integral E dl.
    If you are calculating the Force you need to use E dq.

    Note that nowhere in your expression the electric charge of the finite line is included. And the force clearly depends on that charge.
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  10. #10
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    <br />
E=\frac{\lambda }{\epsilon_0 2\pi r }\\
    <br />
\int_{0}^{0.12}Edq=\int_{0}^{0.12}\frac{\lambda^2\  sqrt{1+3}dx }{\epsilon_0 2\pi r }=\int_{0.08}^{0.08+0.06 \sqrt 3}\frac{\lambda^22dx }{\epsilon_0 2\pi r }=\int_{0.08}^{0.08+0.06 \sqrt 3}\frac{\lambda^22dx }{\epsilon_0 2\pi x }

    correct?
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  11. #11
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    it looks good

    just use \lambda_1 \lambda_2 instead of \lambda^2, because the wires have different values.

    and the integration limit should be
    \int_{0.08}^{0.08+0.06}
    because its l \times sin(30)
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  12. #12
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    According to the figure, the distance from the center of the finite wire to the infinite wire is  x_0 . Integrate along the finite wire,

    <br />
F=\int_{-\tfrac{l}{2}}^{+\tfrac{l}{2}} \frac{\lambda}{2\pi \epsilon_0 (x_0+x\sin\alpha)} \ \frac{q}{l} \ dx = \frac{\lambda \ q}{2\pi \epsilon_0 l} \left(\frac{1}{\sin\alpha} \ln \frac{x_0+\frac{l}{2}\sin\alpha}{x_0-\frac{l}{2}\sin\alpha}\right)<br />
    Last edited by mr fantastic; September 18th 2009 at 08:51 AM. Reason: Restored original reply
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  13. #13
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    thanks
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