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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
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