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

Electromagnetism linear charge density question

  1. Sep 21, 2010 #1
    1. The problem statement, all variables and given/known data

    Two rods, of lengths l1 and l2, have charges q1 and q2

    a)Find the charges per unit length for each rod individually.
    b)Find the charge per unit length, averaged over both rods.
    c)Check your result for l1 approaching 0
    d)Check your result for l1=l2

    2. Relevant equations

    q = integral dq = integral lamda(ds) = lamda integral ds = lamda(l)

    avg lamda equvilant to q/l equals lamda(l)/(l) = lamda (for uniform charge)

    if you know ^^^ then nonuniformly charged is pretty simple to know (integral (x= 0-l) cxdx)

    3. The attempt at a solution

    i'm not really having trouble with this problem, its that i'm not sure how much work it requires since i'm given bare bones to work with and questions surrounding this one, had integrals and charges on spheres annulus cylinders and what have you.

    If i assume it is as easy as it looks

    a) lamda1 = q1/l1 and lamda2 = q2/l2

    b) lamda3 = (q1+q2)/(l1+l2)

    c) lamda4 = (q1+q2)/(0+l2)

    d) lamda5 = (q1+q2)/(2l)

    so do you think it is that simple?
  2. jcsd
  3. Sep 21, 2010 #2


    User Avatar
    Homework Helper
    Gold Member

    Your attempted solution seems correct to me.

    Perhaps it is a poorly worded problem (solvable yes, but just not very interesting). On the other hand, if it was worded slightly differently such as something like

    ..........Two rods, of lengths l1 and l2, have charge densities λ1 and λ2.

    ..........a) Find the total charge for each rod individually.
    ..........b) Find the total charge, averaged over both rods.
    ..........c) Check your result for l1 approaching 0
    ..........d) Check your result for l1 = l2

    it would have made a more interesting problem.
  4. Sep 22, 2010 #3
    lol, that would make it much more understandable to myself, well i talked to my prof and he said it was an easy mark. So i did the easy assumable answer
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook