Okay cool, at least I know that.
So the overall electric field would point left correct? My only concern is at the origin it's pointing to the right while everywhere else it's pointing to the left. After that I'm just not sure how to calculate the magnitude of it. Gonna read over your stuff some more and compare it with whats in my book to try to make something click. It has a few basic integrals, I'm just not sure what would change when its a semicircle instead of a circle and when half of that semicircle is an opposite charge. I appreciate the help.
I'm looking at it now and the process is correct I just had things a bit backwards. The Y components will cancel out, NOT the X. If you draw an attractive force from a -dQ, you can see how the Y components will cancel as you integrate over the semicircle.
Doing physics in Microsoft Paint is annoying so I'll just upload the similar problem I had to do and tell me if you figure it out. If not, I will go into more detail (forgive my scribble handwriting).
In this case, it's the same kind of problem it's just that the semicircle is rotated.
Okay let me know if this is right or not.
E = kq/r^2
(lambda) = q/rd(theta)
dE = k(lambda)d(theta)/r
dE_x = " "*sin(theta)
Then I integrate from 0 to pi/2 and add that to the integration of pi/2 to pi?
(lambda) is your charge, per unit length.
Your charge is Q, your length is a quarter of a circle.
The circumference of a circle is 2pi*r. Divide that by 4 and you get pi*r/2, so that's your length of a quarter circle.
So lambda = Q/(pi*r/2) = 2Q/(pi*r).
You don't put lambda back into your E or dE by itself. Lambda just represents charge per unit length. You want to replace dQ, which = (charge, per unit length) * (length we're considering for the charge) = (lambda) * (ds) = (2Q/(pi*r)) * (r d(theta))