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Math Help - After 8 years, I have forgotten a thing or two

  1. #1
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    Unhappy After 8 years, I have forgotten a thing or two

    Hey everyone, first post. I found your website when I was looking for a forum regarding mathematics help. I am getting close to graduating with a non science major, but I have to complete my lab hours that I started 8 years ago.

    My background is Physics I in 1998 B, and Cal I and Cal II in 1998 and 1999.. B and a C I think...

    Anyway I am slowly relearning everything but I have come to a factoring issue that I am needing someone to step through so I can understand the right way.

    Since it is early algebra, I think, I am not finding any examples in my Physics or Calculus Texts.

    I believe I have the correct beginning formulas, and the book has provided the answer, I am looking for the meat in between.

    Any help would be much appreciated.

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  2. #2
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    if you put a small (+) test charge, +q, at point P, it will be affected by both charges +Q and -Q using the principle of superposition.

    charge +Q exerts a force ...

    F_1 = \frac{kQq}{(x+a)^2}

    charge -Q exerts a force in the opposite direction ...

    F_2 = -\frac{kQq}{(x-a)^2}

    the net force is the sum ...

    F_1 + F_2 = \frac{kQq}{(x+a)^2} - \frac{kQq}{(x-a)^2}

    F_1 + F_2 = kQq \left(\frac{1}{(x+a)^2} - \frac{1}{(x-a)^2}\right)

    F_1 + F_2 = kQq \left(\frac{(x-a)^2}{(x-a)^2(x+a)^2} - \frac{(x+a)^2}{(x-a)^2(x+a)^2}\right)

    F_1 + F_2 = kQq \left(\frac{(x-a)^2 - (x+a)^2}{(x-a)^2(x+a)^2} \right)

    F_1 + F_2 = kQq \left(\frac{(x^2 -2xa + a^2) - (x^2 + 2xa +a^2)}{(x-a)^2(x+a)^2} \right)

    F_1 + F_2 = kQq \left(\frac{-4xa}{(x-a)^2(x+a)^2} \right)

    F_1 + F_2 = -kQq \left(\frac{4xa}{(x-a)^2(x+a)^2} \right)

    the electric field at point P is E_P = \frac{F_1 + F_2}{q}<br />
...

    E_P = -kQ \left(\frac{4xa}{(x-a)^2(x+a)^2} \right) = \frac{-4kQxa}{(x^2 - a^2)^2}

    the (-) sign indicates the direction of the electric field (left) as you would expect since the charge -Q is nearer point P than the +Q charge.

    hope you can follow the algebra ... I tried hard not to leave out any step.
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  3. #3
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    Smile Perfect....

    That is exactly what I was looking for Skeeter... Thanks SO MUCH!
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  4. #4
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    I worked through this step by step, and the only part that I am missing is how (x-a)^2 (x+a)^2 = (x^2 - a^2)^2.

    I broke it down to (x^2 - 2ax +a^2)(x^2 + 2ax +a^2) but I am unsure of the next step, or I could be completely wrong at that.
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  5. #5
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    (x-a)^2(x+a)^2 =

    (x-a)(x-a)(x+a)(x+a) =

    [(x-a)(x+a)][(x-a)(x+a)] =

    (x^2-a^2)(x^2-a^2) = (x^2-a^2)^2
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  6. #6
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    Guess I owe you a beverage...

    Thanks man.

    Re-learning is fun!
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