Electric field on a piece of wire

[vladimir69]

Homework Statement

A thin piece of wire, with charge per unit length $\lambda$ 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 $$2\frac{\lambda k_{e}}{D}\sin(\theta_{0})$$.
The file attached below is to show where theta is pulled from.

Homework Equations

$$d\mathbf{E}=k_{e}\frac{dq}{r^2}\mathbf{\hat{r}}$$

The Attempt at a Solution

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

Thanks,

 

[Joe_UK]

As $$sin(\theta_{0})$$ is less than 1, $$sin^3(\theta_{0})$$ is a lot smaller than $$sin(\theta_{0})$$ so can basically be ignored I believe. So yes you were right.

 

[hikaru1221]

I think using $$\theta$$ will avoid some mistakes that using $$\phi$$ may make.

$$dq=\lambda dx = -\lambda D d\phi$$

It's wrong.
You may find it simple if using this: $$x=Dtan\theta$$

 

[vladimir69]

Ahh damn, that was careless.

I used $\phi$ 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.