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Force On Charges

  1. Jan 9, 2009 #1

    Air

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    1. The problem statement, all variables and given/known data
    A small object has charge [itex]Q[/itex]. Charge [itex]q[/itex] is removed from it and placed on a second small object. The two objects are placed [itex]1[/itex] meter apart. For the force that each object exerts on each other to be a maximum, q should be:


    2. Relevant equations
    [itex]F=\frac{kQ_1Q_2}{r^2}[/itex]


    3. The attempt at a solution
    The correct answer is [itex]\frac{Q}{2}[/itex] although I can't seem to get to that. Is my formula selection correct. Also, if force is maximum, which value do I cosider it as. Thanks for the guidence in advance.
     
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  3. Jan 9, 2009 #2

    gabbagabbahey

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    Good; here [itex]r=1\text{m}[/itex], so you might as well write this as [itex]F=kQ_1Q_2[/itex] and not worry too much about the units.

    Now, what are [itex]Q_1[/itex] and [itex]Q_2[/itex] for this problem (in terms of 'Q' and 'q')?


    Well, you should have an equation for your force that is a function of 'q'; if I gave you some function of x and asked you to find the value of x that gives maximum value of that function, how would you do it? Apply the same idea to the force as a function of q.
     
  4. Jan 9, 2009 #3

    Air

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    It it was function of [itex]x[/itex], I would differentiate to find the maximum and equate it zero. How is that relevant?

    EDIT: [itex]F = KQ^2\implies \frac{\mathrm{d}F}{\mathrm{d}Q} = 2KQ[/itex]. Is this correct?
     
    Last edited: Jan 9, 2009
  5. Jan 9, 2009 #4

    gabbagabbahey

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    Well, if you have a function of x you differentiate with respect to x and set it equal to zero.

    So, if you have a function of q, why not differentiate it with respect to q and set it equal to zero? :wink:
     
  6. Jan 9, 2009 #5

    Air

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    Ok, So if [itex]F=kQ_1Q_2[/itex], we can consider the charge to be [itex]F=kQ^2[/itex] hence as [itex]F = KQ^2\implies \frac{\mathrm{d}F}{\mathrm{d}Q} = 2KQ[/itex].

    [itex]2KQ = 0 \implies Q=0[/itex]. Hmmm... Is my error that I changed [itex]Q_1Q_2[/itex] to [itex]Q^2[/itex]? :redface:
     
  7. Jan 9, 2009 #6

    gabbagabbahey

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    Right idea, but are [itex]Q_1[/itex] and [itex]Q_2[/itex] really both equal to [itex]Q[/itex]?

    If the first object starts out with charge [itex]Q[/itex] and you take away an amount [itex]q[/itex] how much is left? Call that [itex]Q_1[/itex].

    If the second object starts out neutral and you add an amount of charge [itex]q[/itex], what is the charge on that object? Call that [itex]Q_2[/itex].
     
  8. Jan 9, 2009 #7

    Air

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    [itex]F=k(Q-q)(q)[/itex]

    [itex]F=KQq - Kq^2[/itex]

    [itex]\frac{\mathrm{d}F}{\mathrm{d}q} = KQ - 2kq[/itex]

    When derivative equated to zero, max can be found: [itex]0 = KQ - 2kq \implies KQ = 2Kq \therefore q = \frac{Q}{2}[/itex].

    Woo, it works. I got the correct answer. Thanks a lot for the great help. You've been very patient and I appreciate the help. Thanks once again. :smile:
     
  9. Jan 9, 2009 #8

    gabbagabbahey

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    That's much better!:smile:

    You should also check that the second derivative is negative so that you can be sure that you've found the maximum value and not the minimum value.
     
  10. Jan 9, 2009 #9

    Air

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    Yes, [itex]\frac{\mathrm{d}^2F}{\mathrm{d}q^2} = -2q[/itex] Which is negative hence it's a maximum.
     
  11. Jan 9, 2009 #10

    gabbagabbahey

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    Looks good to me :approve:
     
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