# Integration involving a physics problem.

1. May 24, 2013

### lsuspence

1. The problem statement, all variables and given/known data
Hello all! I'm currently trying to work a problem for my Physics 2 class (for engineering and science majors). The example problem deals with "Field of a charged line segment." I conceptually understand the problem, but I am having trouble with the details involving the integration.

The problem: "Positive charge Q is distributed uniformly along the y-axis between y= -a and y= +a. Find the electric field at point P on the x-axis at a distance x from the origin."

I know that $\stackrel{\rightarrow}{E}$= $\frac{kQ}{r^{2}}$

λ=$\frac{Q}{2a}$

dQ=λdy=$\frac{Q}{2a}$dy

r=$\sqrt{x^{2}+y^{2}}$

therfore; dE=(k)($\frac{dQ}{r^{2}}$)= (k)($\frac{Q}{2a}$)($\frac{dy}{x^{2}+y^{2}}$)

E$_{y}=$$_{-a}$∫$^{+a}$dE$_{y}$=0

E$_{x}=$$_{-a}$∫$^{+a}$dE$_{x}$=$_{-a}$∫$^{+a}$(k)($\frac{Q}{2a}$)($\frac{dy}{x^{2}+y^{2}}$)(cosθ)

cosθ=$\frac{x}{\sqrt{x^{2}+y^{2}}}$

E$_{x}=$$_{-a}$∫$^{+a}$dE$_{x}$=$_{-a}$∫$^{+a}$(k)($\frac{Q}{2a}$)($\frac{dy}{x^{2}+y^{2}}$)($\frac{x}{\sqrt{x^{2}+y^{2}}}$)

simplifying and factoring out constants gives:

($\frac{kQ}{2a}$)$_{-a}$∫$^{+a}$$\frac{xdy}{(x^{2}+y^{2})^{3/2}}$

Here is where my problem comes in.... I don't know how to integrate this. The book says "a table of integrals will help."

The solution is given to be:

E$_{x}$=$\frac{kQ}{x\sqrt{x^{2}+y^{2}}}$

2. Relevant equations

3. The attempt at a solution

I do have the latest CRC book which has integral tables in it. I looked at the general forms containing: c2+x2. The one it looked the closest to was: $\frac{dx}{(c^{2}+x^{2})^{n}}$. But I'm not sure... I believe the x's may be treated as constants since I'm integrating with respect to y, but I'm not exactly sure how to go about working it. I looked a U-substitution but I get bogged down and confused by the fact that I am integrating a function that includes two variables. Any help would greatly be appreciated. Again, I understand the concept, but I'm getting confused on the calculus part of it (the integration/last step). Thank you.

2. May 24, 2013

### LCKurtz

It looks like a trig substitution might work. Try $y = x\tan\theta,\, dy=x\sec^2\theta\,d\theta,\, \sqrt{x^2+y^2}=x\sec\theta$ and see if that does anything for you.

3. May 24, 2013

### Staff: Mentor

Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted