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Electric potential integratoin

  1. Jul 2, 2011 #1
    1. The problem statement, all variables and given/known data
    . A wire of finite length that has a uniform linear charge density
    λ=6.22×10-9 C/m is bent into the shape shown below.
    [PLAIN]http://lulzimg.com/i23/7498af.jpg [Broken]
    Answer: 2.98e+02 V


    2. Relevant equations
    V = k integral (dq/r)


    3. The attempt at a solution
    So i think my integrals are right not sure. I have to split into line and circle.
    Left line: K (integral -3r->-R) lambda dx/x
    Circle: K (integral -pi->0) lambda d theta
    Right line k(Integral R->3R) lambda dx/x

    So when i integrate and plug stuff in i get
    K*Lambda (ln[-R]-ln[-3r]+pi+ln[3R]-ln[R])
    im not sure how to simplify the terms with ln R

    side note: How can i properly write stuff out on forums? =S
     
    Last edited by a moderator: May 5, 2017
  2. jcsd
  3. Jul 2, 2011 #2

    ideasrule

    User Avatar
    Homework Helper

    This isn't right. In V = k integral (dq/r), r is the distance between the charge and the origin, and can't be negative.

    These are right.

    You can use the identity ln(a)+ln(b)=ln(a*b), or equivalently, ln(a)-ln(b)=ln(a/b).
    The usual way is to use LaTex. See here for a tutorial: http://www.maths.tcd.ie/~dwilkins/LaTeXPrimer/

    For most purposes, you can just search up and use Unicode symbols along with the exponent and underscore tags. There are text symbols for all the Greek characters, for some fractions, for the integral sign, and probably for a lot more things I don't know about.
     
  4. Jul 2, 2011 #3
    i dont understand what to integrate from the left side. How am i supposed to express it with relation to the origin or do i just take the right integral and multiply by 2?
     
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