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Homework Help: Two charged spheres are connected

  1. Jul 20, 2012 #1
    1. The problem statement, all variables and given/known data

    Two identical conducting spheres, fixed in place, attract each other with an electrostatic force of 0.108N when their center-to-center separation is 50 cm. The spheres are then connected by a thin conducting wire. When the wire is removed, the spheres repel each other with an electrostatic force of 0.0360N. Of the initial charges on the spheres, with a positive net charge, what was the negative charge on one of them and the positive on the other before they were connected?

    2. Relevant equations

    F = kqqr^-2

    3. The attempt at a solution

    First I noted that the charges must be equal after being connected. I called this charge q3.

    Then (.25)0.036 = k(q3)^2
    .009 = k(q3)^2
    1x10^-6 = q^3

    Since some charge n was removed from q1 and given to q2 for this to happen,

    q1 - n = `1x10^-6
    q2 + n = 1x10^-6

    Suggesting that q1 + q2 = 2x10^-6.

    With that in mind, since

    0.108(.25) = k(q1)(q2),

    (q1)(q2) = 3x10^-12

    Substituting results in the quadratic

    0 = -(q2)^2 + (2x10^-6)(q2) - (3x10^-12)

    which has no real solution.

  2. jcsd
  3. Jul 20, 2012 #2


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    If one force is attractive and one force is repulsive, the signs are...
  4. Jul 20, 2012 #3


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    Note that one of the charges q1 or q2 is given to be positive while the other is negative. So, this equation is inconsistent (left side positive, right side negative).
    [oops, I see Ibix essentially already gave you this hint.]
  5. Jul 21, 2012 #4
    well the ans are 10^-6, -10^-6 ,3 x 10^-6,-3x 10^-6 is tht rite?
  6. Jul 21, 2012 #5
    I corrected the sign error and solved the quadratic which has two solutions, 3x10^-6 and -10^-6.

    I don't know how to get those other results you got. The problem specifies a positive net charge but i would still like to know.
  7. Jul 21, 2012 #6
    well after chrge becomes q1-n and q2+n. and force wud become k(q1-n)(q2+n)/r^2. and
    q1-n=q2+n u cn get value of n in terms of q1 and q2. n frm first eqn u have value of q1xq2...frm tht u will get four values of charges...even if ques demands positive chrge it will be both +10^6 and 3x10^6.
  8. Jul 21, 2012 #7


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    You got the possible values of charge on the second capacitor: it can be charged with 3x10^-6 C or -10^-6 C. The charge of the first capacitor is either -10^-6 C or 3x10^-6 C, respectively, as the sum of charges is 2x10^-6 C.
    The equation for q3 had a positive and a negative solution, but the negative has been excluded.

  9. Jul 21, 2012 #8


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    Further to ehild's response, either 10-6 or -10-6 solve your first equation. The second is forbidden by the requirement of a net positive charge, but if you run with it anyway you get the other two answers sirisha gave. I believe that you are correct and sirisha has overlooked the positivity requirement.
  10. Jul 22, 2012 #9
    ya didnt see the positive charge ...bt if only positive charges are to be included it will be both...
  11. Jul 22, 2012 #10


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    The question only requires a net positive charge. There could not be an attractive force initially if the charges were not opposite signs.
  12. Jul 22, 2012 #11
    i did ths que a fewdaysbk n in my que net positive chrge wasnt specified so all the four ans came..
  13. Jul 22, 2012 #12
    sry few days
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