Question

A total charge Q = -4.6x10^-6C is distributed uniformly over a quarter circle arc of radius a = 7.4 cm as shown.What is Ey, the value of the y-component of the electric field at the origin (x,y) = (0,0) ?

Answer 1

Choose an arbitraraly located small piece of the charge located in the arc and call it "dq" . This will be small enough to be treated as a point charge.
Draw a position vector "r" from your "dq" to the origin.
From Coulomb's law the field "dE" ,at the origin, due to the point charge "dq" is
dE = kdq/r^2

The position vector makes some angle "()" with the horizontal, so the components are;
dEx = dECos() = kdqCos()/r^2
dEy = dESin() = kdqSin()/r^2

You then integrate these around the arc.
However first you must write "dq" in metrical terms. You do this by defining a linear charge density "D" as D=dq/ds , where "ds" is the element of arc length that "dq" occupies.
Then dq = Dds

You next want to express "ds" in terms of the angle it subtends as ,ds=rd() .
This is a piece of the same angle used to find the components.

Now your integrands are;
dEx = (kD/r)Cos()d()
dEy = (kD/r)Sin()d()

and the total field components are the integrals;
Ex = INTEGRAL[dEx]
Ey = INTEGRAL[dEy]

Now just integrate from -pi/2 to +pi/2
k & r are constant. So i8s "D" since you are given the charge is uniformly distributed, and in fact D = Q/pir

One of the integrals will be zero, which should be obvious from the symmetry.