Electric Potential, Linear charge with a loop

1. Feb 10, 2008

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

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Last edited: Feb 10, 2008
2. Feb 10, 2008

Staff: Mentor

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.

Last edited: Feb 10, 2008
3. Feb 10, 2008

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.

Last edited: Feb 10, 2008
4. Feb 10, 2008

Staff: Mentor

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

Last edited: Feb 10, 2008
5. Feb 10, 2008

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?

Last edited: Feb 10, 2008
6. Feb 24, 2009

matthenry01

i just did the same question.