Question

the rod below is a uniformly charged semicircle of arc length .14 m. If the total charge on the rod is -7.50 muy c, magnitude and direction of the electric field at point O? Point O is in the center of the semicircle.

Answer 1

Let the semicircle be bent so its symmetrical about the y-axis and its endpoints are at +-a on the x-axis.
The first thing to do is choose a small piece of the rod of arc length "ds" which contains an amount of charge "dq" (we will take care of the sign with the direction of E, so "dq" is a magnitude). It is a distance "r" from the origin and its field "dE" is that due to a point charge. The direction of "dE" is along "r" toward "dq" , since its really negative charge.
dE = kdq/r^2

Now we have to add these "dE's" up for all the "dq's" in the rod. This has to be done by components, so write the component eqs;
dEx = kdqCos()/r^2
dEy = kdqSin()/r^2

The angle is measured between "r" and the x-axis.
You now add up the components due to each "dq" in the rod, by integration;(note that "r" is constant in magnitude)
Ex = (k/r^2)INT[Cos()dq]
Ey = (k/r^2)INT[Sin()dq]

You can't do these integrals without first expressing "dq" in terms of the angle.
This is done by definig a density "D" as
D = total charge/total length = Q/L = Q/pir

Since the rod is uniformly charged the density is constant and it is also equal to;
D = dq/ds (charge on ds divided by ds)

So
dq = Dds = Drd() = (Q/pir)rd() = (Q/pi)d()

And the integrals are now;
Ex = (kQ/pir^2)INT[Cos()d()]
Ey = (kQ/pir^2)INT[Sin()d()]

These are now easily integrated from 0 to pi .
You expect Ex will be zero from the symmetry. So only Ey is non zero, pointing up toward the center of the rod.

Ey = 2kQ/pir^2

in terms of the given length "L" of the rod;
pir = L

Ey = E = 2kpiQ/L^2 (up)

and

First, you know by symmetry that the electric field at the center is directed towards the mid-point of the semicircle.
Then: an elemental arc ds of the semicircle = Rd? has a charge dq = (q/L)Rd? & so will contribute to the E field at the semicircle's center of
dE = kdq/R^2