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The missing variable

  1. Nov 25, 2003 #1
    Ok this seems easy enough but I am missing something

    Two objects are identical and small enough that their sizes can be ignored relative to the distance between them, which is 0.200 m. In a vacuum, each object carries a different charge, and they at-tract each other with a force of 1.20 N. The objects are brought into contact, so the net charge is shared equally, and then they are returned to their initial positions. Now it is found that the ob-jects repel one another with a force whose magnitude is equal to that of the initial attractive force. What is the initial charge on each object?

    Coulumbs law states


    f=k * q1*q2/r^2

    f is clearly given
    r is clearly given

    k is a constant 8.99*10^9

    1.20=8.99*10^9 * q1*q2/.200^2

    where do q1 and q2 come into the picture?
  2. jcsd
  3. Nov 25, 2003 #2


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    Well, it says that the charge is shared equally between the two objects, so just call the charge on each object q.


    F = k ((q * q)/r2) = (kq2)/r2

    Now re-arrgange to find q:

    q2 = (Fr2)/k


    q = ±√[(Fr2)/k]

    Since it didn't state whether the charge was positive or negative.

    I'll leave you to compute the answer.
  4. Nov 26, 2003 #3


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    Whoops. It appears that I didn't actually solve the problem. But I think you can get the two charges' sum and product from this and you might be able to find them that way.
  5. Nov 26, 2003 #4

    Doc Al

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    Staff: Mentor

    Right. Call the final charge on each q; use Coulomb's law to find q.

    If the original charges are q1 & q2, you know what the must add up to (2q) and what their product must be (-q2). Do a little algebra and you'll figure them out.
  6. Nov 26, 2003 #5
    So does this look correct

    So does this seem correct


    q = +/-squareroot([(Fr2)/k])


    q=+/- 2.31*10^-6

    it would appear so because

    8.99*10^9*(2.31*10^-6*2.31*10^-6)/.200^2 =1.20N

    so Is this the right way to do it?
  7. Nov 26, 2003 #6
    Not quite.
    This is true, but you're not finished. (Better make it +/- 2.311*10-6).
    Not only does
    q1 ≠ q2
    q1 ≠ -q2

    Do you see why?

    q1 = -q2
    then after the objects touch the final charge q would be 0 and there would be no electrostatic force between them.
  8. Nov 26, 2003 #7
    Kind of lost then

    I am not sure where to go with it then

    I would have thought q1 = -q2

    I am not sure how to solve for q1 and q2

    with q=+/- 2.311*10^-6

    it says that the objects are identical and initialy they repel each other so they have to be opposites
    so I am not sure

    why q1 cannnot = - q2

    it would seem that how it should be since

    they objects are identical except for the charge
  9. Nov 26, 2003 #8

    Doc Al

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    Staff: Mentor

    Re: Kind of lost then

    I thought gnome explained this point well. When the two objects touch, they will distribute the charge between them equally. We know there must be a non-zero charge (q) on each object after they touch since they repel. So q1 cannot equal -q2: since, if they did, there would be no charge on either object after they touch. Do not continue until you understand this point.
    The objects are identical, but not their charges. All you can deduce from the fact that they initially attract is that the two charges (q1, q2) are opposite in sign. You cannot conclude that the magnitudes are equal.
    Read the above carefully.

    To solve for q1 & q2:
    First find q (the final charge on both objects).

    Then consider these two conditions that q1 and q2 must satisfy:
    q1 + q2 = 2q (this comes from the fact that the total charge will distribute equally after touching)
    q1 x q2 = -q2 (this comes from the fact that the force is equal, but opposite, before and after touching; look at Coulomb's law)

    Combine these two equations and solve for q1 & q2. But don't waste time solving until you understand where these two equations come from and what they mean.
  10. Nov 26, 2003 #9
    Yes, the objects are identical except for the charge.
    And since they ATTRACT each other, one must be positive and the other must be negative.
    So let one initial charge be X and the other one Y. One of these numbers is negative, but you'll worry about that later. Meanwhile just call them X and Y.
    Since they are identical except for the charge, if one has charge X and the other has charge Y, after they touch, the final charge q on EACH of them will be (X+Y)/2.
    But if X = -Y, then (X+Y) = 0
    and then each one would have 0 charge and there would be no force between them.

    So, you have two equations and two unknowns. Messy equations, because of the numbers you're dealing with, but its still just algebra at this point. You already solved for q, and you know that
    X + Y = 2q
    so that's one equation.
    The other one is your original equation:
    KXY/r2 = 1.2

    You'll end up with a messy quadratic equation to solve.

    Sorry, I switched to X and Y to avoid typing all those subscripts. Hope that didn't confuse you.
  11. Nov 26, 2003 #10
    Note that there is no contradiction between Doc Al's second equation and mine.

    Remember, way back when you solved for q, you said:
    q2 = (1.2 * r2)/k

    If you rearrange my second equation slightly you get:
    XY = (1.2 * r2)/k

    and I said that either X or Y is negative.

    So XY = -q2
  12. Nov 26, 2003 #11
    I think I understand

    Thanks for all the help I think I got it now

    We know that

    q1 + q2 = 2q


    q1 x q2 = q ^2

    we found q earlier 2.311*10^-6


    q^2 = 5.34*10^-12
    2q = 4.66*10^-6


    5.34*10^-12=q1 * q2

    4.66*10^-6 = q1 +q2


    4.66*10^-6 - q1 =q2


    5.34*10^-12=q1 * (4.66*10^-6 - q1)

    0=-q1^2 + 4.622*10^-6 - 5.34*10^-12

    plug it all into the quadratic equation

    q1 = - 4.622*10^-6 +/- squareroot(((4.622*10^-6)^2 -(4)(-1)(5.34*10^-12))/(2(-1))

    q1= - 4.622*10^-6 + 3.27*10^-6

    q1 = -1.352*10^-6
    4.66*10^-6 - -1.352*10^-6 =q2


    q1= - 4.622*10^-6 - 3.27*10^-6

    q1= -7.89 * 10^-6


    4.66*10^-6 - -7.89 * 10^-6 =q2
    q2 =12.514*10^-6
  13. Nov 26, 2003 #12

    After you solve, you have to check your answer & see if it makes sense. At least, try multiplying q1 x q2 & see what you get.

    Here's the problem:
    No! q1 x q2 = -q^2
    That little minus sign makes a big difference.

    As a result, this:
    should be:
    0=q12 - 4.622*10-6 - 5.34*10-12
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