1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Electric field at center of semi-circular bent rod of charge

  1. Jul 19, 2008 #1
    I have the correct answer just the wrong direction

    1. The problem statement, all variables and given/known data
    A uniformly charged insulating rod of length 14.0cm is bent into the shape of a semicircle as shown in figure XXX. The rod has a total charge of 7.5 micro C. Find the magnitude and direction of the electric field at P, the center of the semicircle.

    2. Relevant equations

    dE = k.dq.1/r^2

    3. The attempt at a solution
    I set things up like this
    http://img210.imageshack.us/img210/4690/physprobub2.png [Broken]

    Since the vertical components would cancel I just looked at the x components.

    dE_{x} = \frac{k_{e}\cdot dq\cdot\sin\theta}{r^2}

    = \frac{k_{e}\cdot dx\cdot\lambda\cdot\sin\theta}{r^2} \ \ \mbox{as dq = dx\cdot\lambda}

    = \frac{k_{e}\cdot r\cdot d\theta\cdot\lambda\cdot\sin\theta}{r^2} \ \ \mbox{as dx = ds = r\cdot d\theta}

    = \frac{k_{e}\cdot d\theta\cdot\lambda\cdot\sin\theta}{r} \ \ \mbox{as dx = ds = r\cdot d\theta}

    E = \frac{k_{e}\cdot \lambda}{r} \int d\theta\cdot\sin\theta}
    (integral is from zero to pi)

    E = \frac{k_{e}\cdot \lambda}{r} ( -1 - 1)

    E = \frac{-2\cdot k_{e}\cdot Q}{L\cdot r} \ \ \mbox{as \lambda = \frac{Q}{L}}
    When I substitute the values is I get the correct answer but it's negative. The bent rod is positive so the resultant electric field should be AWAY from the semicircle and therefor positve?!?

    What have I done wrong?
    Last edited by a moderator: May 3, 2017
  2. jcsd
  3. Jul 19, 2008 #2


    User Avatar
    Science Advisor
    Homework Helper

    The integral of sin from 0 to pi is 2. How did you get (-1-1)? BTW the antiderivative of sin(x) is -cos(x).
  4. Feb 10, 2009 #3
    Sorry for reactivating this old thread, but I figured better that than starting a whole new thread.

    I have gotten to the point of [tex]

    E = \frac{2\cdot k_{e}\cdot Q}{L\cdot r} \


    but now my question is what is the r value? Of what my reasoning is telling me, it should be [tex]

    E = \frac{2\cdot k_{e}\cdot Q}{L^{2}} \

    Is this right? Can any one assist me?
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook