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Thread: algebra and roots

  1. #1
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    algebra and roots

    Hi, I need a little help. I got this far on a problem but now am stuck, i know i need to find the root but i just cant get it right

    2/(1-d)^2=1/d^2
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by JBswe View Post
    Hi, I need a little help. I got this far on a problem but now am stuck, i know i need to find the root but i just cant get it right

    2/(1-d)^2=1/d^2
    do you know what "roots" are?


    $\displaystyle \frac 2{(1 - d)^2} = \frac 1{d^2}$ ........cross multiply

    $\displaystyle \Rightarrow (1 - d)^2 = 2d^2$

    now expand the left hand side and continue
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  3. #3
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    Another attempt

    Rearrange

    $\displaystyle \frac{(1-d)^2}{d^2}=2$

    and extract the respectively square root.
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  4. #4
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    ok so then i get $\displaystyle
    (1 - d)^2 = 2d^2$

    and when i extract it i find

    $\displaystyle
    1-d= 2d^2$

    $\displaystyle 2d^2+d-1=0$

    $\displaystyle d^2+0.5d-0.5=0$

    I think i did somthing wrong here. that would give me a neg root the way i tired it. where did i go wrong?
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by JBswe View Post
    ok so then i get $\displaystyle
    (1 - d)^2 = 2d^2$

    and when i extract it i find

    $\displaystyle
    1-d= 2d^2$
    huh?!

    $\displaystyle (1 - d)^2 \ne 1 - d$
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  6. #6
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    the whole problem looks like this.

    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----->
    +Q---+q-----------+2Q
    <--d-->
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  7. #7
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    Quote Originally Posted by JBswe View Post
    the whole problem looks like this.

    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----->
    +Q---+q-----------+2Q
    <--d-->
    The "power" in the charge is zero. What does this mean? Are you looking for the point in space where the force on q is zero?

    If so then the net force on charge q at a position d (measuring a positive direction to the right) is:
    $\displaystyle F_q = \frac{\kappa q(Q)}{d^2} - \frac{\kappa q(2Q)}{(1 - d)^2}$

    So if the net force is 0:
    $\displaystyle \frac{\kappa q(Q)}{d^2} - \frac{\kappa q(2Q)}{(1 - d)^2} = 0$

    $\displaystyle \frac{1}{d^2} - \frac{2}{(1 - d)^2} = 0$

    $\displaystyle \frac{1}{d^2} = \frac{2}{(1 - d)^2}$<-- Multiply both sides by $\displaystyle d^2(1 - d)^2$

    $\displaystyle (1 - d)^2 = 2d^2$

    $\displaystyle 1 - 2d + d^2 = 2d^2$

    $\displaystyle d^2 + 2d - 1 = 0$<-- Use the quadratic formula:

    $\displaystyle d = \frac{-2 \pm \sqrt{2^2 - 4(1)(-1)}}{2}$

    $\displaystyle d = \frac{-2 \pm \sqrt{8}}{2}$

    $\displaystyle d = \frac{-2 \pm 2 \sqrt{2}}{2}$

    $\displaystyle d = -1 \pm \sqrt{2}$

    Now, since $\displaystyle \sqrt{2} > 1$ we must take the "+" solution (since the "-" solution gives a negative d.) Thus
    $\displaystyle d = \sqrt{2} - 1$

    (Note also that d is less than 1, which makes sense as that puts it between the charges.)

    For the record, the negative value of d also makes sense: There is a point on the far side of Q where the repulsion of q from Q is equal to the attraction of q to the 2Q charge. But, of course, that isn't what we are specifically looking for.

    -Dan
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