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Homework Help: Electric field on a piece of wire

  1. Jul 18, 2010 #1
    1. The problem statement, all variables and given/known data
    A thin piece of wire, with charge per unit length [itex]\lambda[/itex] lies on the x-axis and extends from x=-L/2 to x=+L/2 so that it has a length L. Show that the electric field (at the point (x,y) = (0,D)) in the +y-direction has magnitude given by [tex]2\frac{\lambda k_{e}}{D}\sin(\theta_{0})[/tex].
    The file attached below is to show where theta is pulled from.

    2. Relevant equations

    [tex]d\mathbf{E}=k_{e}\frac{dq}{r^2}\mathbf{\hat{r}}[/tex]

    3. The attempt at a solution
    Here is what I managed to boil up
    First let
    [tex]\theta+\phi=\frac{\pi}{2}[/tex]
    then
    [tex]r=\frac{D}{\sin(\phi)}[/tex]
    [tex]dq=\lambda dx = -\lambda D d\phi[/tex]
    [tex]\hat{r}=\left(\begin{array}{cc}\cos(\phi)\\\sin(\phi)\end{array}\right)[/tex]
    [tex]\int dE_{y}=-\frac{\lambda k_{e}}{D}\int_{\phi_{0}+\frac{\pi}{2}}^{\theta_{0}-\frac{\pi}{2}}\sin^3(\theta)d\theta[/tex]
    Finally
    [tex]E_{y}=\frac{\lambda k_{e}}{D}(2\sin(\theta_{0})-\frac{2}{3}\sin^3(\theta_{0}))[/tex]
    Did I do something wrong or did they assume that the sin cubed term is small and hence got rid of it?

    Thanks,
     

    Attached Files:

    • pf.png
      pf.png
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  2. jcsd
  3. Jul 18, 2010 #2
    As [tex] sin(\theta_{0}) [/tex] is less than 1, [tex] sin^3(\theta_{0}) [/tex] is alot smaller than [tex] sin(\theta_{0}) [/tex] so can basically be ignored I believe. So yes you were right.
     
  4. Jul 18, 2010 #3
    I think using [tex]\theta[/tex] will avoid some mistakes that using [tex]\phi[/tex] may make.



    It's wrong.
    You may find it simple if using this: [tex]x=Dtan\theta[/tex] :wink:
     
  5. Jul 19, 2010 #4
    Ahh damn, that was careless.

    I used [itex]\phi[/itex] because it helps with getting the right direction for the unit vector, but screwed up the differential unfortunately. (actually I did a similar error the second time aswell)

    Thanks for the help.
     
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