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Force between two uniformly charged identical rods

  1. Apr 21, 2012 #1

    j0n

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    1. The problem statement, all variables and given/known data
    Identical thin rods of length 2a carry equal charges, +Q, uniformly distributed along their lengths. The rods lie along the x axis with their centers separated by a distance b > 2a. Find the force, F, exerted by the left rod on the right one.

    Left rod's center @ (0,0) → rod stretches from (-a,0) to (a,0)
    Right rod's center @ (b,0) → rod stretches from (b-a, 0) to (b+a, 0)


    2. Relevant equations
    They give that

    [tex] F = \frac{k Q^2}{4 a^2} \ln \frac{b^2}{b^2 -4a^2} [/tex]

    Also we know that [tex] λ = \frac{Q}{2a} = \frac{dQ}{dx} , \space and \space Q = Q_1 = Q_2 [/tex]

    3. The attempt at a solution

    Electric field produced by rod 1 at a point x > 2a:

    [tex] E_1(x) = \int \frac{k}{χ^2}dQ = \int_{x-a}^{x+a} \frac{k}{χ^2} λ dχ = \int_{x-a}^{x+a} \frac{k}{χ^2} \frac{Q_1}{2a} dχ = \frac{k Q_1}{2a} \int_{x-a}^{x+a} \frac{dχ}{χ^2} [/tex]
    [tex] = \frac{k Q_1}{x^2 - a^2} [/tex]

    Now I understand that FTOT = Ʃ qiE1(xi),
    where qi are the infinitesimal contributions of Q2 (= Q).

    So shouldn't I have [tex] F = \int_{x-a}^{x+a} E_1(χ) λ dχ ? [/tex]

    If so then I have:

    [tex] F = \int_{x-a}^{x+a} \frac{k Q_1}{χ^2 - a^2} \frac{Q_2}{2a} dχ [/tex]
    [tex] = \frac{k Q^2}{2a}\int_{x-a}^{x+a} \frac {dχ}{χ^2 - a^2} [/tex]

    And I'm stuck.
     
    Last edited: Apr 21, 2012
  2. jcsd
  3. Apr 21, 2012 #2

    j0n

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    By the way I used chi (χ) as a dummy variable when integrating since I wanted to obtain E as a function of position along the x-axis.

    Thus, [tex] E_1(x= b) = \frac{k Q_1}{b^2-a^2} [/tex]

    Edit:

    I think I might've solve it now. Instead of cancelling 2a out when I calculated E1(x), I just left it.

    So then [tex] E_1(x) = \frac{k Q_1}{2a} (\frac{1}{x-a} - \frac{1}{x+a}) [/tex]

    Then in finding the Force on rod 2 my range of integration should specifically be from b-a to b+a (not x-a to x+a).

    So, [tex] F = \frac{k Q^2}{4a^2}\int_{b-a}^{b+a} (\frac{1}{χ-a} - \frac{1}{χ+a}) dχ [/tex]
     
    Last edited: Apr 21, 2012
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