Charge on Semicircle: Electric Field Equation

[freezer]

Homework Statement

This is not a homework problem so there is no problem statement. More of a conceptual question.

Consider a charge on a ring:

If this was a semicircle opposed to a ring, how would this change the equation since there is not an opposing dEr force.

<br /> <br /> \boldsymbol{E} = \boldsymbol{\hat{z}} \frac{\rho_l R(-\hat{\boldsymbol{r}}R + \hat{\boldsymbol{z}Z})}{4 \pi \varepsilon_0 (R^2 + Z^2)^{3/2}}\int_{0}^{\pi }d\phi <br /> <br />

 

[mfb]

The vertical component is still easy to calculate (as you just have a smaller part of the ring contributing to it), but the other component (due to the broken symmetry it does not vanish any more) will need its own integral. No idea how easy/messy that gets.

 

[freezer]

So I would have to do it twice.

dE = dEr + dEz

 

[mfb]

The integrals are different, but yes.

 

[freezer]

Does this look reasonable:

<br /> <br /> \frac{Q}{4\pi \varepsilon_0(R^2 + Z^2)^{3/2} }(Z\hat{z}-R\hat{r})<br /> <br />

 

[mfb]

Z and z are the same?

That would mean the electric field points in the same direction as the "r" line in the sketch. No, that cannot work.
It would be easier to find the error if you show your whole work.