We have a rod of length L, that holds a total charge of Q. The rod lies on the positive x axis at a distance of d from the origin (see picture below). We want to find the electric field at the origin.
The expression for the electric field of a continuous
charge distribution is given by the following
expression
This particular charge distribution can be taken to
be a linear charge distribution with
, r2 = x2, the limits of integration run from d to L +d and 
points in the
negative x -direction. The field at P will point to the left if Q > 0 and to the right if Q < 0.
Consider a segment of the ring. Let the contribution to the total electric field from this small segment
be 
and will point in the direction shown. Now
consider a segment of charge that is directly opposite from the first segment.
The magnitude of the electric field from this segment will be the same, but
it will point in a direction that will cancel the horizontal components. This
happens to all of the horizontal components, and the net field will be in the
vertical direction. What we need to find is the vertical component of dE.
To find the total electric field, we have to add up the contributions to the
vertical component of the electric field from each little piece of the ring. But
as we move around the ring, everything inside the integral remains constant.
We can treat the disk as a collection of rings. We know the electric field
for a ring from the previous example.
dq is the charge contained in the a ring.
The contribution to the electric field from one ring is
and the total electric field is the sum of the contributions from each of the rings.
Evaluating the integral gives
