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Insulating Spheres in Electric Fields

  1. Oct 12, 2008 #1
    Hi, I'd like to ask the good people of this forum for some help.
    Here's a problem I've been working on for a while, and I'm seriously at my wit's end. I guess there's something I'm missing here...


    1. The problem statement, all variables and given/known data
    Two insulating spheres have radii r1 and r2, masses m1 and m2, and uniformly distributed charges -q1 and q2. They are released from rest when their centers are separated by a distance d. How fast is each moving when they collide? Suggestion: Consider conservation of energy and of linear momentum.

    2. Relevant equations
    I thought these were relevant:
    Momentum=mv
    Kinetic energy = 1/2(mv^2)
    [tex]\Delta[/tex]U = -q[tex]\int[/tex]E dr

    3. The attempt at a solution
    First I solved for the potential energy that this system gains when the two spheres are moved apart:
    [tex]\Delta[/tex]U = q1[tex]\int^{d}_{d-r1-r2}[/tex]E dr = k(q1)(q2)([tex]\frac{1}{d-r1-r2}[/tex] - 1/d)

    I figured this is the amount of energy the spheres would have when they collide, so...
    [tex]\Delta[/tex]U = [tex]\frac{1}{2}[/tex](m1)v[tex]^{2}_{1}[/tex] + [tex]\frac{1}{2}[/tex](m2)v[tex]^{2}_{2}[/tex]

    From conservation of momentum, v2 = (m1/m2)v1 so subbing that into the above equation I got:
    [tex]\Delta[/tex]U = [tex]\frac{1}{2}[/tex]m1v[tex]^{2}_{1}[/tex] + [tex]\frac{1}{2}[/tex][tex]\frac{m^{2}_{1}}{m_{2}}[/tex]v[tex]^{2}_{1}[/tex]

    So then I solved for v1 to get:

    v1 = [tex]\sqrt{\frac{2kq_{1}q_{2}((1/(d-r1-r2)-(1/d))}{m_{1}+\frac{m^{2}_{1}}{m_{2}}}}[/tex]

    And v2 can be figured out the same way. However, I know for a fact this isn't the right answer.


    In closing
    I'm don't really know what I did wrong, but I suspect it's because I treated the two spheres as point charges, and I'm not sure if I'm justified in doing that.
     
  2. jcsd
  3. Oct 12, 2008 #2
    They move from a distance d to a distance r1+r2.
    What is d-r1-r2?
    Look at your potential energy again. You don't need any integral. Just take the difference between the final energy and initial energy. I would treat them as point charges.
     
  4. Oct 13, 2008 #3
    Yup lol, that would do it. Thanks a lot for your help Nasu! :biggrin:
     
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