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Electrostatics, Even Distrobution on wire dumb question

  1. Jan 31, 2010 #1
    1. The problem statement, all variables and given/known data
    There is a charged wire aligned/centered along the Y 2L in length. A test charge is placed D distance from the wire on the X axis. The wire has a total charge of Q. calculate the integral to figure out how much force the test charge experiences.

    svnziu.jpg
    a semi decent pic of the layout



    2. Relevant equations
    Columbs law + some calc


    3. The attempt at a solution
    My question isn't about getting the answer, I have it (too bad I didn't have the problem right when I looked at it), but I don't understand some of it and am hoping for some clarification.

    [tex]F_x = \int_{-L}^{L} \frac{kQq}{2Lr^2}cos(\theta)dy[/tex]

    This is all well and good, the integral is over the length of the wire, the rest of it is basically Columbs law with Q/2L * dy as the charge density * tiny section of wire for integration. It took me a while (and this is the part i am hung up on) but the cos theta is in there because we need to deal with the fact that all the little tiny chunks of wire are each at a different angle to the test particle so different ratios of force are felt in the X and Y directions.

    now for what I don't understand when this integral is worked you get this
    [tex]F_x = \frac{kQq}{2L}\int_{-L}^{L} \frac{D}{\sqrt{D^2 + y^2}^{3/2}}dy[/tex]

    Inside the integral the denomanator is obviously pythagorians theorum to find the distance between wire chunk and particle. How does [itex]cos{\theta}[/itex] convert to just plain D (the distance between the wire and particle at the x axis)?
     
    Last edited: Jan 31, 2010
  2. jcsd
  3. Jan 31, 2010 #2

    gabbagabbahey

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    Homework Helper
    Gold Member

    Hi mrtubby, welcome to PF!:smile:

    It doesn't!:wink:

    Remember your unit circle...[itex]\cos\theta[/itex] equals adjacent/hypoteneuse. And take note of the fact that the denominator in your integrand is actually [tex](D^2+y^2)^{3/2}[/itex] and Pythagoras Law tells you [itex]r^2=D^2+y^2[/itex].
     
  4. Jan 31, 2010 #3
    Fantastic, I had come up with some kind of flawed reasoning as to why i was looking at [itex](D^2+y^2)^{3/2} [/itex] that basically boils down to some weak algebra skills. Now the math makes sense and I can continue on with my day.

    Thanks Bunches!!!
     
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