Electric Potential, Linear charge with a loop

[Bad-Wolf]

1. A wire of finite length that has a uniform linear charge density \lambda is bent into the shape shown in the figure below. Find the electric potential at point O.
The image has the setup.

2. The answer is k\lambda\pi (\pi + 2ln3) How the hell do I get this? The primary equation I am using is of course \int \frac{dq}{r} where finding a proper dq is the chore.

3. Okay, so this problem is driving me a little bit crazy. I tried integrating the electric potential equation with respect to the loop, integrating from 0 to \pi which gives me k\lambda\pi
It makes sense that this is incorrect as the the linear charge density is going to be spread across the entire wire. However, whenever I try to account for the rest of the wire outside of the loop I get nonsensical and incorrect answers and integrations. I am fairly certain this is what I need to do, I am just clueless as to how to go about it
How do I set this up ? Thanks for any help you can provide.

fixing latex if it something looks funny

 

[Doc Al]

Treat this as three pieces: Two straight segments and one curved. The trick to the curved piece is that it's part of a circle. (No integration needed for that part!)

Looks like you got the curved part, now set up the integration for the straight segments.

 

[Bad-Wolf]

I am having big hangups with the straight pieces and what I am supposed to do with them after I have found them. Already knowing the answer is further messing me up because I know there is no R in there and my functions for the straight pieces spit out natural logs of R and stuff.

\int \frac {k \lambda dx}{x} I don't know how much sense this makes but I was trying to find the potential at 0 in reference to a piece of dx on the right hand segment where it starts at R and ends at 3R.

that is supposed to be marked as Integral from R to 3R but I could not find the proper latex syntax to display it correctly.

 

[Doc Al]

You left out a factor of k, and the limits should be from R to 3R, but otherwise your integral is OK. Hint: \ln (a) - \ln (b) = \ln (a/b).

FYI:
\int_{R}^{3R} \frac{k \lambda}{x}dx

 

[Bad-Wolf]

I didn't leave out the k, my poor latex omitted the k when i had the incorrect syntax for the integral from R to 3R. So then with both sides I have k \lambda 2ln3 and then the loop makes k\lambda\pi Where is that extra PI coming from in the answer ? If I add these it will be missing...

k_{e} \lambda (\pi + 2ln3) If I integrate this from 0 to PI with respect to dtheta I would get the extra PI but why would I do that?

 

[matthenry01]

i just did the same question.
only had one pi in the answer.
i don't know where that first one came from.

 

[Doc Al]

So then with both sides I have k \lambda 2ln3 and then the loop makes k\lambda\pi Where is that extra PI coming from in the answer ? If I add these it will be missing...

k_{e} \lambda (\pi + 2ln3)

This answer is correct. The extra pi in the original post is a typo. (I must not have seen this last post. Oops. )