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!

Homework Help: Electric Field of a thin rod

  1. Apr 17, 2010 #1
    1. The problem statement, all variables and given/known data
    We have a thin rod of length 10 cm uniformly charged with a total charge of + 40 nC . A small glass bead charged to + 6 nC is located 4 cm from the center of the rod. What is the force on the bead?

    2. Relevant equations
    E = integral of K dq /r^2 and Q = lambda L and therefore dq = lambda dy since y will be variable of length
    L = length of rod d = distance from rod Lambda = Q/L

    3. The attempt at a solution
    The y component of the elec field is 0 due to symmetry.

    Esub x = KQ / L [integral from - L/2 to + L/2 of d / (y^2 + d^2)^(3/2) dy

    I can do this problem except for evaluating this integral. My book shows that this integral

    d divided by (y^2 + d^2)^(3/2) dy when evaluated is y divided by [ d (y^2 + d^2)^(1/2) ] Can someone go over the steps to this integration PLEASE. I can take it after that and get a force but I can not figure out how this integral is done. THanks Frostking
  2. jcsd
  3. Apr 18, 2010 #2
    You need to make a trig substitution to solve this integral. With [tex]tan(\theta) = y/d [/tex], [tex]y = dtan(\theta)[/tex] and [tex]dy = dsec^{2}(\theta)d(\theta)[/tex]. By substitution, and then using a trig identity on the denominator of the integral ( [tex]tan^{2}(\theta) + 1 = sec^{2}(\theta)[/tex], which is why we chose [tex]tan(\theta)[/tex] for the substitution ), you should get [tex]sec^{2}(\theta)/sec^{3}(\theta)[/tex], which is [tex]cos(\theta)[/tex]. The evaluated integral is therefore [tex]sin(\theta)[/tex], which is equal to [tex]y/\sqrt{y^{2} + d^{2}}[/tex].
    Last edited: Apr 18, 2010
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook