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Coulomb's Law question - find initial charge

  1. Feb 9, 2008 #1
    [SOLVED] Coulomb's Law question - find initial charge

    1. The problem statement, all variables and given/known data

    Two identical small metal spheres with q1>0 and |q1| > |q2| attract each other with a force of magnitude 55.4 mN. They are separated by a distance of 3.94 m. The radius of each sphere is 40 micro-Newtons.

    The spheres are then brought together until they are touching. At this point, the spheres are in electrical contact so that the charges can move from one sphere to the other until both spheres have the same final charge, q.

    After the charges on the spheres have come to equilibrium, the spheres are moved so that they are 3.94 m apart again. Now the spheres repel each other with a force of magnitude 6.648 mN.

    The Coulomb constant k is 8.98755x10^9 N*m^2/C^2.

    What is the initial charge q1 on the first sphere?

    2. Relevant equations
    F = k*|q1*q2|/r^2


    3. The attempt at a solution
    I can get as far as figuring out the TOTAL initial charge of q1*q2. That's easy: |q1*q2| = F*r^2/k. What stumps me is how I can figure out what only ONE of them is -- the problem states that they're not equal, but I can't for the life of me figure out how to get an exact number. I'm guessing it has something to do with the fact that first, they are attracted to one another and then they repel after reaching equilibrium.

    ----EDIT 2/10-----
    OK, here's what I've done since I posted, it's still not giving me the right answer, though.

    Before touching:
    F1 = k * |q1*q2| / r^2

    After touching:
    F2 = k * q^2 / r^2 (where q is the charge on each of the spheres, since they both have the same charge)

    The initial charge on q1 is equal to the final charge on q plus the change in charge; that is,
    q1 = q + Δq
    Similarly, q2 = q - Δq

    Now solve for Δq:
    F1 = k * |q1*q2|/r^2
    Rearranging and replacing q1 and q2,
    F1 * r^2 / k = | (q/2 + Δq)*(q/2 - Δq) |
    = | (q/2)^2 - (Δq)^2 |

    This means that F1*r^2/k is equal to either (q/2)^2 - (Δq)^2 or (Δq)^2 - (q/2)^2, due to the absolute value signs.

    This yields two possible values for Δq. The second one I get is greater than the value for q, so I assume this is a 'garbage' value (because this would mean that q2 = q - Δq would yield a negative value, which makes no sense in this situation) so I use the first value I get for Δq.

    Then, I solve for q1: q1 = q + Δq. However, this answer is wrong.

    -----EDIT again, 2/10-----

    OK, got it. Here's what I did:

    EQUATION 1: F1 = k * |q1*q2| / r^2
    EQUATION 2: F2 = k * q^2 / r^2 (where q is the charge in one of the spheres -- they are equal. Absolute value signs are left out since they are redundant, as q is squared)

    EQUATION 3: q1 = q/2 + Δq
    (In words, the initial charge for q1 is equal to the final charge of q1 plus however much charge left q1 when the spheres touched).

    EQUATION 4: q2 = q/2 - Δq
    (In words, the initial charge for q2 is equal to the final charge of q2 minus however much charge it gained when the spheres touched).

    Replace q1 and q2 in equation 1 with their values from equations 3 and 4:

    F1 = k * | (q/2 + Δq)*(q/2 + Δq) | /r^2

    Rearrange this equation to bring k and r to the other side, and expand what's inside the abs. value signs:

    F1 * r^2 / k = | (q/2)^2 - (Δq)^2 |

    Now, the left side of the equation can be equal to two possible things (because of the abs. value signs):

    -[(q/2)^2 - (Δq)^2] or +[(q/2)^2 - (Δq)^2]

    Expanding: (Δq)^2 - (q/2)^2 or (q/2)^2 - (Δq)^2

    Set each of these equal to F1 * r^2 /k and solve for Δq. The first one yields a non-real result (Δq^2 = -8.40899528017e-11...you can't take the square root of a negative number) so the correct one to use is the second:

    F1 * r^2 /k = (q/2)^2 - (Δq)^2

    Plugging in the values for F1, r^2, k, and q, and solving for Δq, I get Δq=1.0357970386e-5.

    Now plug this value into equation 3 to get the value for q1.

    Hope this solution helps anyone else who's stuck on a problem like this!
     
    Last edited: Feb 10, 2008
  2. jcsd
  3. Feb 9, 2008 #2
    Do you have the final answer this by any chance?
     
  4. Feb 10, 2008 #3
    Nope, still working on it...
     
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