Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Electric Potential, Linear charge with a loop

  1. Feb 10, 2008 #1
    1. A wire of finite length that has a uniform linear charge density [tex]\lambda[/tex] 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 [tex]k\lambda\pi (\pi + 2ln3) [/tex] How the hell do I get this? The primary equation I am using is of course [tex] \int \frac{dq}{r} [/tex] 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 [tex]\pi [/tex] which gives me [tex]k\lambda\pi [/tex]
    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
     

    Attached Files:

    • wire.bmp
      wire.bmp
      File size:
      75.8 KB
      Views:
      1,075
    Last edited: Feb 10, 2008
  2. jcsd
  3. Feb 10, 2008 #2

    Doc Al

    User Avatar

    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
  4. Feb 10, 2008 #3
    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.

    [tex] \int \frac {k \lambda dx}{x} [/tex] 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
  5. Feb 10, 2008 #4

    Doc Al

    User Avatar

    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: [itex]\ln (a) - \ln (b) = \ln (a/b)[/itex]. :wink:

    FYI:
    [tex]\int_{R}^{3R} \frac{k \lambda}{x}dx[/tex]
     
    Last edited: Feb 10, 2008
  6. Feb 10, 2008 #5
    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 [tex]k \lambda 2ln3 [/tex] and then the loop makes [tex]k\lambda\pi[/tex] Where is that extra PI coming from in the answer ? If I add these it will be missing...

    [tex]k_{e} \lambda (\pi + 2ln3)[/tex] 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
  7. Feb 24, 2009 #6
    i just did the same question.
    only had one pi in the answer.
    i don't know where that first one came from.
     
  8. Feb 24, 2009 #7

    Doc Al

    User Avatar

    Staff: Mentor

    This answer is correct. The extra pi in the original post is a typo. (I must not have seen this last post. Oops. :uhh:)
     
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook