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Some more trouble with Electric Charge Problems

  1. May 12, 2007 #1
    I've never been good at "Find the maximum whatever" (Area, charge in this case, etc...) type problems, so I could use some coaching on this one.

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

    Of the charge Q initially on a tiny sphere, a portion q is to be transferred to a second, nearby sphere. Both spheres can be treated as particles. For what value of q/Q will the electrostatic force between the two spheres be maximized?

    2. Relevant equations

    q = Q/2, Fmax = kQ^2/4r^2

    3. The attempt at a solution

    Admittedly it's been pretty rough going on this one, and it's only the 3rd homework problem of the semester... but I've resolved to get through it all somehow. Through using the text's website companion I've gotten it down to the final step, which is where I get stumped. I've got it all down to:

    k/r^2(qQ - q^2) = (3/4) (kQ^2/4r^2)

    At this point the interactive help says all I gotta do is solve for q. But I can't seem to get q by itself. How can I manipulate the equation in order to isolate q? I tried a few things, such as dividing both sides by k to eliminate k, then multiplying both sides by r^2 to eliminate r^2. However, at that point I get stumped and have no idea how to further simplify the equation. Any help would be appreciated.
     
    Last edited: May 12, 2007
  2. jcsd
  3. May 13, 2007 #2
    From where you are:
    [tex]\frac{1}{qQ-q^2} = \frac{3Q^2}{4}[/tex]

    Clear all fractions (by multiplying both sides by each denominator), then get your equation in the form ax+bx+c=0 so you can use the quadratic equation.
     
  4. May 13, 2007 #3
    Oh, I'm sorry, I wrote down my stumped point incorrectly. It's supposed to be this:

    k(qQ - q^2)/r^2 = (3/4) (kQ^2/4r^2)

    After performing the two steps I tried I'm left with:

    (qQ - q^2) = 3Q^2/16
     
  5. May 13, 2007 #4
    OK. You can still multiply both sides by 16, then get your equation into the form ax^2+bx+c=0.

    Oops. Sorry, last post I forgot to write squared after the first term.
     
  6. May 13, 2007 #5
    I tried doing that but I'm still confused. Now I've got -16q^2 + 16qQ -3Q^2.

    I'm not sure if that's the correct next step, because when I try to factor it I get something like (-4q - 3)(4q + 1). Plugging in I end up with -(3/4)Q and -(1/4)Q. Which is the answer except the answer isn't negative.
     
  7. May 13, 2007 #6
    All of which =0.

    Actually, you get (-4q-3Q)(4q+Q)=0.

    I get positive 3/4

    Hmmm... which was the answer? 1/4 or 3/4?
     
  8. May 13, 2007 #7
    Ah you are right on the positive (3/4)Q. It's listing both as the answer, but listing both as positive. The (1/4)Q still ends up negative on my end.
     
  9. May 13, 2007 #8
    I also get -1/4.

    Conceptually, that makes no sense at all: if you transfer charge from one object to another, how can the charge change sign?

    Mathematically, I just don't know how to get that -1/4 to be positive, or if you can just ignore the negative, or what.
     
  10. May 13, 2007 #9
    Well, that I can clarify with my professor. Thanks for your help!
     
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