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Charges are connected and then kicked by chuck norris

  1. Aug 21, 2012 #1
    1. The problem statement, all variables and given/known data

    Two identical conducting spheres, fixed in place, attract each other with a force of 0.111 N when their center to center separation is 45.00 cm. The spheres are then connected by a thin conducting wire. When the wire is removed, the spheres have a net positive charge and repel each other with an electrostatic force of 0.043 N. What was the initial negative charge on one of the spheres, and what was the initial positive charge on the other?

    2. Relevant equations

    Coulomb's Law

    3. The attempt at a solution

    I posted a problem nearly EXACTLY like this and I still don't understand.

    First, since they are attracting initially I know they have opposite signs. To be mathematically consistent, I will then call this force in the negative direction.

    [itex]-.111 = kq_{1}q_{2}r^{-2}[/itex]

    I also know that after being connected, both spheres have a charge of

    [itex]q_{f} = 0.5(q_{1} + q_{2})[/itex]

    So then I can use Coulomb's Law with the final force,

    [itex]0.43 = kq^{2}_{f}r^{-2}[/itex]

    Which allows me to solve for that final charge.

    [itex]9.84x10^{-8} = q_{f}[/itex]

    Then I can use this numerical value to solve for one of the initial forces.

    [itex]1.96x10^{-7} - q_{1} = q_{2}[/itex]

    Then I can substitute this into the initial force equation for a quadratic.

    [itex]0 = 1.96x10^{-7}q_{1} - q^{2}_{1} + 2.5x10^{-14}[/itex]

    Then I can use the quadratic formula to get the two solutions,

    -8.8x10^-8, and 2.84x10^-7.

    Then I can hit submit to waste one of my attempts because I am doing it wrong.
     
  2. jcsd
  3. Aug 21, 2012 #2
    Because of this?
     
  4. Aug 21, 2012 #3
    No, typo.
     
  5. Aug 21, 2012 #4
    Well, here's how I would have done it.

    Fi = kq1q2r-2

    Ff = kqf2r-2
    ↓ qf = 0.5(q1+q2)
    = 0.25k(q1+q2)2r-2

    q2 = (4Ff k-1r2)1/2 -q1

    Ff = 0.25k(q12+2q1q2+q22)r-2
    = 0.25k(q12+q22)r-2+0.5Fi

    4(Ff -0.5Fi)k-1r2 = q12+q22
    4(Ff -0.5Fi)k-1r2 = q12+((4Ff k-1r2)1/2 -q1)2
    4(Ff -0.5Fi)k-1r2 = q12+(q12-2(4Ff k-1r2)1/2q1+4Ff k-1r2)
    2q12-2(4Ff k-1r2)1/2q1+4Ff k-1r2-4(Ff -0.5Fi)k-1r2 = 0
    q12-(4Ff k-1r2)1/2q1+Fi k-1r2 = 0

    q1 = (4Ff k-1r2)1/2 ± √(4Ff k-1r2 -4Fi k-1r2)/2
    q1 = (4Ff k-1r2)1/2 ± ((Ff -Fi)k-1r2)1/2
    q1 = ((4Ff ±(Ff -Fi))k-1r2)1/2

    Then plug in numbers. The reason it is better this way is that now you have solved this problem for all possible given numbers, rather than only the one specifically given. It's also easy to check if the equation makes dimensional sense (if it fits the form of the Coulomb's law).

    For this problem:
    Ff = 0.043N and Fi = -0.111N. Since k is normally given in Nm2C-2, if you convert the r = 45.00 cm to meters, then the units of q is in Coulombs.

    You should go back and check your previous "problem nearly EXACTLY like this" using this solution.
     
    Last edited: Aug 21, 2012
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