# Electric potential again.

1. Mar 2, 2008

### wakko101

electric potential....again.

So, I've still got my two infinite lines of charge that run parallel to the x axis running along the lines x = a and x = -a and with charge density + and - lambda respectively.

I've figured out that using the origin as my reference point allows me to integrate the electric field of my lines from a to the point in question (r). However, because the field goes like 1/r, we end up with a ln of something in the potential. For the line that's positive a away from the x axis, there's no problem. But, for the line that runs along x = -a, how can I integrate? should I be integrating from a to r again, or from -a to r. The problem with the latter is that you can't have the natural logarithm of a negative number.

I would REALLY appreciate any advice. I'm a little desperate....

Cheers,
W. =)

2. Mar 2, 2008

### eok20

Remember, the r in 1/r is the distance to the charge so it should always be positive. If your point is (x, y) then the distance to the line x = -a is |x + a| and the distance to the line x=a is |x-a|

3. Mar 2, 2008

### wakko101

that makes sense....I figured it was a that I needed to use, but I just wanted to be sure.

Thanks for the help. =)

Share this great discussion with others via Reddit, Google+, Twitter, or Facebook