Calculate the electric field above a ring of charge of radius R and charge density lambda at a point P centered on the axis of the ring.

In general, the contribution to the total electric field E from an infinitesimal piece of charge dE is
dE=(k*dq)/r^2
The total field is then calculated by integrating over all the charge:
E=int(k*dq)/r^2
Where k=1/(4pi epsilon_0) and r is the distance from dq to the point P . 
For this problem, if the point P is a distance z above the ring then we can set up a triangle to get a relationship between z and the distance r.
r^2=z^2+R^2
The vector dE pointing at point P from a chunk of charge, can be broken up into two components, one in the radial direction and one in the vertical direction.
dE=dE_r+dE_z
When considering the total electric field contribution from the ring, the radial contribution of every piece of charge will sum to zero and only the z contribution will be left due to the symmetry of the ring.
dE_z=dE*cos(phi)=dE*(z/r)=(k*dq)/r^2*(z/r)
E_z=int(k*dq)/r^2 *(z/r)
E_z=int (k*z*dq)/r^3
lambda is the piece of charge divided by a piece of length along the ring.
lambda=(dq)/(dl)
From the fact that r*theta is equal to the arc length around a circle we can say that
dl=R*d(theta)
Therefore,
E_z=int (k*z*lambda dl)/r^3
E_z=int (k*z*lambda R d(theta))/r^3
Now substitute for r and integrate theta from [0,2pi] .
E_z=int_0^(2pi) (k*z*lambda R d(theta))/(sqrt(z^2+R^2))^3
E_z=(k*z*lambda R)/(sqrt(z^2+R^2))^3 int_0^(2pi) d(theta)
E_z=(2pi k*z*lambda*R )/(sqrt(z^2+R^2))^3
Therefore the electric field at point P is entirely in the z -direction with magnitude:
E_z=(z*lambda*R)/(2epsilon_0(sqrt(z^2+R^2))^3)
http://hyperphysics.phy-astr.gsu.edu/hbase/electric/elelin.html