- #1
vladimir69
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Homework Statement
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.
Homework Equations
[tex]d\mathbf{E}=k_{e}\frac{dq}{r^2}\mathbf{\hat{r}}[/tex]
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,