10-13-14 potential of continuous charge distribution

This shows that V=k int(dq/r)=k int(dq/(x^2+a^2)^1/2)
replacing the r with an angular radius. as well the integration is for q and so the denominator can be pulled out of the integral and it become kq/(x^2+a^2)^1/2
BUT
dq=lambdadx
this way you can integrate along a rod instead of a disk
YAY!!!!1!

The potential from a continuous charge distribution can be calculated several ways.  Each method should yield approximately the same result.  First, we can use an integral method in which the potential dV from each element of charge dq is integrated mathematically to give a total potential at the location of interest.  Second, we can approximate the value of the potential V by summing up several finite elements of charge Dq by using a computer spreadsheet or hand calculations.  Finally, we can use Gauss’ law to find the electric field along with the defining equation for potential difference to set up the appropriate line integral.

            Again, let’s consider a relatively simple charge distribution.  In this case we will look at a ring with charge uniformly distributed on it.  We will calculate the potential on the axis passing through the center of the ring as shown in the diagram below.  (Later on you could find the potential difference from a disk or a sheet of charge by considering a collection of nested rings.)

There are several ways that you could evaluate this integral. If you have a program like Wolfram Alpha or Maple, you could get the symbolic solution directly. For example using wolfram Alpha you get the following
One way to see if our answer is in the correct ball park is to imagine that we replace the line charge with a point charge at the center of the rod. What is the equivalent magnitude of the point charge, given that it has to be equal to the entire charge of the rod

This is a huge mess.