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Help with Coulomb's Law

  1. Feb 23, 2008 #1
    !!Coulomb's Law!!

    So I have already solved this one before, but I was redoing it fir practice when I encountered something that is troubling me. Depending on how I choose to solve my two equations, I get different results. Surely I am doing something wrong, but I cannot see it.

    Two identical conducting spheres, fixed in place, attract each other with an electrostatic force of .108 N when their center-to-center separation is 50 cm. The spheres are then connected by a thin conducting wire. When the wire is disconnected, the spheres repel each other with an electrostatic force of .036 N. Of the initial charges on the spheres, with a positive net charge, what was the (a) negative charge of one of them and (b) the positive charge of the other?

    Now I have used conservation of charge for after they connect and I end up with two equations and two unknowns:

    [tex]F_e=\frac{kq_1q_2}{r^2}\Rightarrow q_1q_2=3.00(10^{-12})[/tex] (1)

    [tex]F_e'=k\frac{(\frac{q_1+q_2}{2})^2}{r^2}\Rightarrow q_1+q_2=2.00(10^{-6})[/tex] (2)

    Attempt 1:

    If I solve (2) for q_1 then [itex]q_1=2(10^{-6})-q_2[/itex]

    plugging the above in to (1) [itex] -q_2^2+2(10^{-6})q_2-3(10^{-12})=0[/itex] gets me a nonreal answer.

    Attempt 2:

    BUT if I solve (1) for q_1 then [itex]q_1=\frac{3(10^{-12})}{q_2}[/itex] and plugging that into (2) I get [itex]q_2^2-2(10^{-6})q_2-3(10^{-12})=0[/itex] which solves correctly.

    I am consistently of by a sign in the first attempt. Can anyone see what the problem is?
  2. jcsd
  3. Feb 23, 2008 #2
    Wouldn't that be + (plus) 3e-12? (Resulting in an equivalent expression to than in attempt 1.)
  4. Feb 23, 2008 #3
    No. Attempt 2 is correct. Attempt 1 is not.
  5. Jun 4, 2008 #4
    Awaking this thread. I just went back to try this again and have the same problem... so conceptually, something is off.
  6. Jun 4, 2008 #5


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    Homework Helper

    Hi Saladsamurai,

    In your original post you have an error in equation 1. There is then another mathematical error in method 2 that counteracts the error in equation 1 and gives the right answer.

    For your equation 1, you have:

    [tex]F_e=\frac{kq_1q_2}{r^2}\Rightarrow q_1q_2=3.00(10^{-12})[/tex]

    but that is not right. The formula for Coulomb's law is normally given in terms of magnitudes:

    [tex]F_e=\frac{k |q_1||q_2|}{r^2}[/tex]

    and so you get:

    |q_1 q_2|=3 \times 10^{-12}

    You know that either q1 or q2 is negative, and the other is positive, so your equation 1 should be:

    q_1 q_2 = - 3 \times 10^{-12}

    In your method 2, the quadratic equation you get is the correct equation to get; however, it is not what you get from your original equations 1 and 2 in your post. If you use your original equations, the [itex] 3\times 10^{-12}[/itex] term turns out to be positive, which gives the same (nonreal) answer as method 1.
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