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Where to place Q so it experiences zero net Force.

  1. Oct 14, 2014 #1
    1. The problem statement, all variables and given/known data

    A charge of +2q (a) is placed at the origin and a second charge of -q (b) is placed at x = 3.0 cm. Where can a third charge +Q (c) be placed so that it experiences a zero net force?


    2. Relevant equations
    F = k(Q1)(Q2)/r^2
    E = kQ/r^2
    Quadratic?


    3. The attempt at a solution

    After many attempts from varying angles I think I have it down to Fac + Fbc = 0

    So: Fc = k(2q)(q)/(0.003 + x)^2 + k(-q)(q)/x^2 = 0

    I can get this down to (k)(2q) + (k)(-q) = 0.003x + x^2, I assume to solve this problem I will need to use the quadratic equation (PLEASE if there is an easier way enlighten me) but for the life of me cannot figure out how to get these figures into the Ax^2 + bx + c format because there are two unknown variables (q and x). I assume because the qs are just factors of each other I probably do not need to actually know the value of q to solve for x. I am so lost... Am I even on the right path?
     
    Last edited by a moderator: Oct 14, 2014
  2. jcsd
  3. Oct 14, 2014 #2

    berkeman

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    You are basically on the right path, but a couple things may help you.

    (First, 3cm is not equal to 0.003m -- be sure to fix that before doing any calculations)

    Second, it helps me to figure out the basic answer intuitively before writing the equations. Where do you think the positive test charge will end up being placed when it feels zero net force? Think about the forces that you get from the + and - stationary charges. What will the forces be on the positive test charge? Where will it feel zero net force?

    So once you know about where it will be, then you can more intuitively write out the sum of the 2 force equations = 0. You are correct that you do not need the absolute value of the charges for this problem when you know the relative amounts...
     
  4. Oct 14, 2014 #3
    Intuitively I can expect the charge would have to be placed to the right of the -q charge. Far enough to the right that the force (due to inverse square of the distance from the 2q charge) will be equal and opposite to the force of the -q charge, with the +2q charge exacting a force in the +x direction and the -q charge exacting an equal but opposite force (-x). Conceptually I think I understand the problem, I feel that I am making it sloppier than it has to be by getting into qudratics though... Is this the only approach to this problem and I just need some remedial math work to get me into the proper form for using the quadratic equation? IE get me from (k)(2q) + (k)(-q) = 0.03x + x^2 to ax^2 + bx + c = 0. Thank you!
     
  5. Oct 14, 2014 #4

    mfb

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    Your expectation for the position is right, and the approach is correct as well. Solving the quadratic equation is easier than you might expect.

    Check the previous steps. k and q should not appear here. You can directly get rid of them in the initial "=0" equation.
     
  6. Oct 14, 2014 #5
    Thank you. So I put k(2q)(Q)/(0.03 + x)^2 = - k(-q)(Q)/x2. This allowed elimination of Q and k. I then worked it to (2q)(X^2)/-q = (0.03 + x)^2 to eliminate q resulting in -2x^2 = -(0.03 + x)^2. After factoring the negative through the brackets and using FOIL I come up with the quadratic 3x^2 + 0.06x + 0.0009. Which is incorrect (provides a negative number to be squared)! What did I do something wrong?
     
  7. Oct 14, 2014 #6

    berkeman

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    Trying with Latex help...

    [tex]\frac{2kqQ}{x^2} - \frac{kqQ}{(x-0.03)^2} = 0[/tex]

    [tex]\frac{2}{x^2} - \frac{1}{(x-0.03)^2} = 0[/tex]

    [tex]2(x-0.03)^2 - x^2 = 0[/tex]

    [tex]2(x^2 -2(0.03x) + 0.0009) - x^2 = 0[/tex]

    [tex]x^2 - 0.12x + 0.0018 = 0[/tex]

    Do you see the small differences between my last equation and yours? Looks to be a simple algebra error (assuming my math is okay). Does my equation give you a reasonable answer?


    EDIT -- I fixed my 2nd term just now to 0.12x from 0.06x
    EDIT2 -- and fixed my 3rd term to 0.0018 from 0.0009. Duh
     
    Last edited: Oct 14, 2014
  8. Oct 14, 2014 #7
    At this point, I would move the x2 to the other side of the equation, and take the square root of both sides.

    Chet
     
  9. Oct 14, 2014 #8
    I just can't understand how this reflects the =0 equation. Isn't the 2kqQ over (0.03 + x)^2 because the 2q acts on Q from the distance 0.03 + x? If you somehow rearranged 2kqQ/(0.03 + x)^2 - kqQ/x^2 = 0 to 2kqQ/x^2 - kqQ/(0.03 - x)^2 = 0 can you please explain how you did it and why you felt the need? I know this stuff is so basic, and 90% of the time my math skills are sufficient for solving the problems, but every once in awhile I run into one where my math foundation clearly isn't up to par. After physics I will be taking math haha...
     
    Last edited by a moderator: Oct 16, 2014
  10. Oct 14, 2014 #9
    This step will make it easier for me if I ever get there! Thank you.
     
  11. Oct 14, 2014 #10
    You calculations reveal x = 0.1024 m which when using your formula provides F = 190.8 N +/- x direction. Now if only I could understand why... Thanks for your help! It is very appreciated.
     
  12. Oct 16, 2014 #11

    berkeman

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    I think I got two real solutions when I used the quadratic formula to solve for x (but I've tossed the Post-It notes where I did the calc). I do seem to recall one of the solutions being the one you list. Did you not get a 2nd real solution?
     
  13. Oct 17, 2014 #12
    I think one was the answer I gave and one was a negative number so I dismissed it.

    I am not sure you noticed this question?
     
  14. Oct 17, 2014 #13
    Sorry, the other answer was not a negative it was 0.01757. I also realized that I was looking at the problem the wrong way and that the distance from Q1 to Q3 was x while the distance from Q2 to Q3 was x - 0.03.I understand now. Thank you for your help!
     
  15. Oct 17, 2014 #14

    berkeman

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    Great! I think that was the 2nd answer that I got on my Post-It notes. I was surprised that there were 2 real answers, but it seems plausible that there could be 2 places along the x axis where the conditions for zero force were met. Good job! :-)
     
  16. Oct 17, 2014 #15

    mfb

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    How? That looks like a sign error if you get a second solution.

    To the left of the larger charge, the larger charge always dominates -> no stable position.
    In between, both forces always go in the same direction -> no stable position.
    To the right, there is exactly one point where the distance ratio is correct.
     
  17. Oct 17, 2014 #16

    berkeman

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    I'll re-check my calc in a bit. Too bad I threw away those Post-Its... :-)
     
  18. Oct 17, 2014 #17

    berkeman

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    So assuming my algebra was correct, here is the quadratic equation:

    [tex]x^2 - 0.12x + 0.0018 = 0[/tex]

    [tex]x = \frac{-B +/- SQRT(B^2 -4AC)}{2A}[/tex]

    [tex]x = \frac{0.12 +/- SQRT((0.12)^2 - 4(0.0018)}{2}[/tex]

    [tex]x = \frac{0.12 +/- SQRT(0.0144 - 0.0064)}{2}[/tex]

    [tex]x = \frac{0.12 +/- 0.008}{2}[/tex]

    [tex]x = 0.064, 0.056[/tex]

    Not sure why my answers are different from before, but hopefully showing my work helps to find any errors... :)

    EDIT -- fixing some typos -- please wait...
     
    Last edited: Oct 17, 2014
  19. Oct 18, 2014 #18

    mfb

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    Check the result of the sqrt.
    WolframAlpha
    The smaller solution corresponds to the sign problem I mentioned (we would have to reverse the force direction there manually, then the solution vanishes).
     
  20. Oct 18, 2014 #19
    The solution I get is ##x=0.03(2\pm \sqrt{2})##=0.10243, 0.01757

    The 0.01757 is not the physically realistic result, since it is less than the 0.03.

    Chet
     
  21. Oct 18, 2014 #20

    berkeman

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    Thanks a lot, guys :-)
     
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