What is the Electric Field at the Origin in a Semicircle?

[amb0027]

Homework Statement

You have a semicircle of radius R and charge Q. We also have available a test charge +qo.

Find the Magnitude and direction of the electric field at the origin.

Homework Equations

Electric Field:
F = k (Q * qo)/(R^2)
E = F/qo = kQ/R^2

The Attempt at a Solution

I'm not sure how to solve this, I assumed it would be kQ/R^2 but I am mistaken. Can someone please explain? I'm looking for an explanation on line but not having any luck

 

[Amok]

Wait? The charge Q is distributed over a semicircle? I'm guessing you're going to have to use Gauss' law. But could you make the problem statement clearer? Where's the origin?

 

[amb0027]

Sorry, I've figured it out. I just thought it was a more simple problem than this... For future reference, this is how you solve it.

It's a continuous Distribution of charge so
1) Divide the charge into segments dQ for which you already know the field
2) Find the field of each dQ
3) Find E by summing all dQ

So, the charge per unit length would be : $$\lambda$$ = Q/$$\pi$$R
The charge on the slice dq = $$\lambda$$Rd$$\theta$$

The field generated by the slice would be dE = k dq/R^2 = k $$\lambda$$/R d$$\theta$$

Components of dE would be: dEx = dEcos$$\theta$$, dEy = -dEsin$$\theta$$

Add them all up you get:
Ex = k$$\lambda$$/R $$\int$$ from 0 to pi of cos$$\theta$$ d$$\theta$$ = k$$\lambda$$/R sin$$\theta$$ from 0 to pi which = 0

Ey = -k$$\lambda$$/R $$\int$$ from 0 to pi of sin$$\theta$$ d$$\theta$$ = k$$\lambda$$/R cos$$\theta$$ from 0 to pi which = -2k$$\lambda$$/R