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Homework Help: Coulombs Law and Q + 2Q

  1. Sep 21, 2007 #1
    1. The problem statement, all variables and given/known data
    A Positive charge q is between 2 other postive charges Q and 2Q. the distance between Q and 2Q is 1 meter. There is a point on the line between Q and 2Q where the power in the charge q is zero. Where is that point?

    <--------1 meter----->

    2. Relevant equations

    3. The attempt at a solution
    Shouldent it be somthing like Q+2Q = 3/1-d. With out getting to heavy into the math shouldent the answer be somthing near 1 third meter 0.33m

    If anyone could give me a hand solving this you would be doing me a great service.
  2. jcsd
  3. Sep 21, 2007 #2

    Doc Al

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    That should be kqQ/r^2.

    If you let "d" be the distance from Q, then the distance from 2Q must be 1-d. Now set the forces equal and solve for d.
  4. Sep 21, 2007 #3
    OK am really new to all this. when you say set the forces equal you mean make both Q and 2Q into Q or into 1 and 2 somthing like k1*2/1-d or KQ*Q/0.5 ( 0.5 is half a meter)
  5. Sep 21, 2007 #4

    Doc Al

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    I mean:
    (1) Find the force that the left side charge (Q) exerts on q. (Use the formula; the distance is "d".)
    (2) Find the force that the right side charge (2Q) exerts on q. (Use the formula; the distance is "1 - d".)

    The magnitude of the force between any two charges is given by Coulomb's law:
    [tex]F = k \frac{Q_1 Q_2}{r^2}[/tex]

    Just substitute the correct values (for charges and distance) into the formula, set those two force expressions equal, and solve for d. Do it step by step.
  6. Sep 21, 2007 #5
    ok this mite be way off but what if i do like this.

    k= 8.99*10^9( this mite be wong but i found that value in one of my notes )

    and kqQ/r^2 should be 8.99*10^9*(2*10^-6*1.10^-6)/1-d this should be for the force from 2Q to q

    so 8.99*10^9*(2*1e-6*1e-6)/1-d

    so 1-d= (root) 8.99*10^9*(2*10^-6*1.10^-6) i got the answer 0.134 so then i divid 0.134 with 1 = and get d= 0.134

    so the distance from 2Q and q is 0.134m ? or 0.134C

    it feels like am way off, with a little logic i see the distance should atlest be 0.6 or so
    Last edited: Sep 21, 2007
  7. Sep 21, 2007 #6

    Doc Al

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    There's no need to use all those numbers. (Not sure where you're getting the charges--I thought they were just Q & 2Q?) That just distracts from solving the problem and allows more room for mistakes.

    Following what I suggested in my last post:

    (1) Find the force that the left side charge (Q) exerts on q:
    [tex]F = k \frac{Q q}{d^2}[/tex]

    (2) Find the force that the right side charge (2Q) exerts on q:
    [tex]F = k \frac{2Q q}{(1-d)^2}[/tex]

    Set these two force expressions equal to each other. Note that the k, Q, and q all cancel from both sides--so you don't need them. You'll get a (relatively) simple equation which you can solve for d.
  8. Sep 22, 2007 #7
    [tex]F = k \frac{Q q}{d^2}[/tex] = [tex]F = k \frac{2Q q}{(1-d)^2}[/tex]

    In turn are = [tex]\frac{2}{-d^2(1-d)^2}[/tex]

    (not sure about this step or the others really but here it goes )


    so then i make it all = 0

    1+d= 0


    ok d is 1m, that has to be wrong

    ok what if i try

    2=[tex]{2d^2} +1 +2d[/tex]

    then i get 0= [tex]2{d^2}-1 +2d[/tex]

    i divid it all with 2 and get

    d= [tex]-0.5+d[/tex]

    d=0.5(+-)(the root of)0.25+1
    D2=0.5-1.11= -0.61
    That seems wrong aswell, i dont know where am going wrong. seems like am lacking a basic fact some place but i dont know what it could be
    Last edited: Sep 22, 2007
  9. Sep 22, 2007 #8
    i edited my equation a few times to try and work it all out but it seems it got updated wrong someplace. all the same no matter what i tired nothing seemed to work
  10. Sep 22, 2007 #9


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    Check your algebra.
    (Suggestion for problem solving... rather than using "1" or "1 m", use something like "L"... so that you write terms like (L-d) which helps you keep track of units.)

    With the rather nice choice of charges Q and 2Q, you can do this problem (or later re-interpret your algebraic solution) with a simple ratio of the inverse square forces [of course, paying attention to the directions of each force]. That is, you can look at it and give an answer without a calculator.
  11. Sep 22, 2007 #10
    ok ill try this with L ( Am sorry if this is really basic and am missing some basic principle but i have only been studying Physics for 4 weeks so please bare with me :) )

    [tex]F = k \frac{Q q}{d^2}[/tex]
    ok so if F K and Qq all dont matter d^2 is that same as 1^2 so it all adds up to be L? ( if I use L instead of 1 cuz 1^2 is 1 )

    so L = [tex]F = k \frac{2Q q}{(1-d)^2}[/tex]

    [tex]F = k \frac{2Q q}{(L-d)^2}[/tex]

    using this same way of thinking what should be left is [tex]\frac{2}{(L-d)^2}[/tex] and if i break out (L-d)^2 i should get [tex]L^2 -2d +d^2[/tex]

    when i set them both equal

    [tex]L=L^2 -2d +d^2[/tex]

    giving me ( if i exchange the L for 1 )

    [tex]0= -2d+d^2[/tex]


    d=1(+-)(root of ) 1

    D1= 1+1 =2
    D2= 1-1 =0

    Am i even near being on the right track?
  12. Sep 22, 2007 #11
    JBemp, this is just algebra

    you have
    [tex]F_{2Q} = k \frac{2Q q}{(1-d)^2}[/tex]
    and also
    [tex]F_{Q} = k \frac{Q q}{d^2}[/tex]

    now as it was said, set the two forces equal, or:
    [tex] F_{2Q}=F_{Q}[/tex]

    and solve against [tex]d[/tex]

    Correct me if I am wrong.
  13. Sep 23, 2007 #12
    ok here i go agian am trying to set them equal, i know this is algebra but i have always been really bad at algebra

    [tex]F_{2Q} = k \frac{2.00 C*1.00 C}{1-d^2}[/tex]

    [tex]F_{Q} = k \frac{1.00 C*1.00C}{d^2}[/tex]

    [tex]F_{2Q}=\frac {2.00 C*1.00 C}{1-d^2}[/tex]

    [tex]F_{Q}=\frac {1.00 C*1.00 C}{d^2}[/tex]

    [tex]F_{2Q}=\frac {2.00C}{1-d^2}[/tex]


    This is the part am really unsure of if i should set in the F=2Q that d^2 is 1-4pi*8.854*10^12 and in the F=Q that d^2 that i should set it as 4pi*8.854^12 or do i just pull the root out of somthing ( i dont know what ) like d=(root) somthing. man i feel lost.
  14. Sep 23, 2007 #13

    Doc Al

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    This should be:
    [tex]F_{2Q} = k \frac{2Q*1Q}{(1-d)^2} = 2k \frac{Q^2}{(1-d)^2}[/tex]

    OK. But I would use:
    [tex]F_{Q} = k \frac{1Q*1Q}{d^2} = k \frac{Q^2}{d^2}[/tex]

    For some reason you were using actual values for the charges (for example, 2C instead of just 2Q), which is not quite right but it won't matter for this problem since all that matters is their ratio not their actual values. (Maybe you didn't realize that "C" stands for Coulomb and is a unit of charge.)

    Now just set them equal:
    [tex]F_{2Q} = F_Q[/tex]

    Do it. Lots of stuff will cancel out.
  15. Sep 23, 2007 #14
    can i just cancel them out like this

    [tex]2K\frac{Q^2}{(1-d)^2} = {k \frac{Q^2}{d^2}[/tex]


    I guess i need to asign k a value? 8.99*10^9 but that number is huge 8990000000

    [tex]{8990000000=1-2d}[/tex] (this has to be wrong but i dont know what else to do)


    or do i have to find the root of 1-k ?
  16. Sep 23, 2007 #15

    Doc Al

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    This step is correct, but the rest is not.

    Realize that a factor that appears on both sides can be canceled. ("Canceling" just means dividing both sides by the factor.) So the k and the Q^2 can both be canceled.

    What does that leave you with?
  17. Sep 23, 2007 #16
    ok so then somthing like this? q^2/d^2 = q^2/d^2 is = 0
    so turning (1-d)^2 into 1-2d+d^2 was the right thing to do?
    but unsure about 2k/k that should only leave k? or 2?


    i get that -0.5
    Last edited: Sep 23, 2007
  18. Sep 23, 2007 #17

    Doc Al

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    Who says you don't get a fraction anymore?

    Starting with this:
    [tex]2k\frac{Q^2}{(1-d)^2} = {k \frac{Q^2}{d^2}[/tex]

    Canceling the k and Q^2 gives you this:
    [tex]\frac{2}{(1-d)^2} = \frac{1}{d^2}[/tex]
  19. Sep 23, 2007 #18
    [tex]\frac{2}{(1-d)^2} = \frac{1}{d^2}[/tex]

    ok dident know you move the 2 and the 1 up onto the fraction

    [tex]\frac{2}{1-2d+d^2} = \frac{1}{d^2}[/tex]



    am i getting near?
  20. Sep 23, 2007 #19

    Doc Al

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    Instead of doing that (which is correct, just not helpful), take the square root of both sides and go from there.

    And keep the equation! You can't just drop the equal sign.
  21. Sep 23, 2007 #20
    am a little unsure on how to get a root out of a fraction like this when i have (1-d)^2

    [tex]{0} = {\sqrt{\frac{2}{(1-d)^2}[/tex] or [tex]{0} = {\sqrt{\frac{2}{1-2d+d^2}[/tex]

    or somthing like this i know they both say the same thing i just dont know what to do with the d's.

    if i plug [tex]{0} = {\sqrt{\frac{2}{1-2d+d^2}[/tex] in the calc I get 9.055

    and [tex]{0} = {\sqrt{\frac{1}{d^2}[/tex] is 0.1

    these numbers cant be right
  22. Sep 23, 2007 #21

    Doc Al

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    I really don't know where you are getting the "0 =" left side of these equations. In any case, taking the square root of something squared is easy; for example:
    [tex]\sqrt{(1-d)^2} = 1-d[/tex]

    So, go back to this equation:
    [tex]\frac{2}{(1-d)^2} = \frac{1}{d^2}[/tex]

    ...and take the square roots of both sides, setting them equal to each other.
  23. Sep 23, 2007 #22
    ok am even worse at square roots then i am at algebra but here we go.

    [tex]d=\sqrt{1*2} [/tex] =1.41

    [tex]{d^2}=\sqrt{1}[/tex] =1

    setting them equal


    then i try to solve d by setting the problem = 0


    am i on the right track or did i screw it up agian?
  24. Sep 23, 2007 #23

    Doc Al

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    Again, I am somewhat baffled. Start here:
    [tex]\frac{2}{(1-d)^2} = \frac{1}{d^2}[/tex]

    Take the square root of both sides and write the resulting equation.

    Here's a headstart: The square root of the right hand side is just 1/d.
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