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Math Help - Really simple problem - rearranging equations, getting it upside down.

  1. #1
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    Really simple problem - rearranging equations, getting it upside down.

    Hey I know this is really simple and silly, I used to find this really easy, I've been confused by a book and now feel rather stupid at not being able to even rearrange equations.. I get it right, right, right and then suddenly I get one written exactly upside down when compared to the book's answer.

    Here's an upside down one:

    F_{el} = \frac{1}{4\pi\epsilon_{0}} \frac{q1q2}{r^2} to make r the subject

    I did

    4\pi\epsilon_{0} F_{el} =  \frac{q1q2}{r^2}

    then

     \frac{4\pi\epsilon_{0}F_{el}}{q1q2} = r^2

    then

     \sqrt\frac{4\pi\epsilon_{0}F_{el}}{q1q2} = r or r=\pm\sqrt\frac{4\pi\epsilon_{0}F_{el}}{q1q2}

    Their answer was
    r=\pm\sqrt\frac{q1q2}{4\pi\epsilon_{0}F_{el}}

    I've spent a few hours rereading my chapters and getting a bit upset...what have I done wrong?
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  2. #2
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    Quote Originally Posted by SillyRabbit View Post
    F_{el} = \frac{1}{4\pi\epsilon_{0}} \frac{q1q2}{r^2} to make r the subject

    4\pi\epsilon_{0} F_{el} =  \frac{q1q2}{r^2}
     \frac{4\pi\epsilon_{0}F_{el}}{q1q2} = r^2 THIS STEP IS WRONG.
     \sqrt\frac{4\pi\epsilon_{0}F_{el}}{q1q2} = r or r=\pm\sqrt\frac{4\pi\epsilon_{0}F_{el}}{q1q2}
    This was the step before your mistake:
    4\pi\epsilon_{0} F_{el} =  \frac{q1q2}{r^2}

    Take the reciprocal of both sides:
    \frac{1}{4\pi\epsilon_{0} F_{el}} =  \frac{r^2}{q1q2}

    Isolate the r squared:
    \frac{q1q2}{4\pi\epsilon_{0} F_{el}} = r^2

    Take the square root of both sides:
    r = \pm\sqrt{\frac{q1q2}{4\pi\epsilon_{0} F_{el}}}


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  3. #3
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    That's what my book says, my problem is...I don't understand... I mean...ARGH I have been royally confused. Firstly I was sticking to BEDMAS as I was taught (which is PEMDAS or whichever is used in the US) but half the time my book doesn't seem to...and now I'm all over the place.

    If I just always always stick to BEDMAS/PEMDAS, will I ALWAYS get a correct rearrangement?? Because sometimes my book does not handle the exponents before multiplication for example. And then I want to stab it. Is there something I'm missing, something to do with isolating the subject or something?

    PS thanks for taking the time to reply!
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  4. #4
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    4\pi\epsilon_{0} F_{el} =  \frac{q1q2}{r^2}
     \frac{4\pi\epsilon_{0}F_{el}}{q1q2} = r^2 THIS STEP IS WRONG.
    It looks like what you tried to do was to divide both sides by q1q2. But note that the r^2 was in the denominator, and suddenly, you put in the numerator, which is a no-no. \left(r^2 = \frac{r^2}{1}\right). If you want to divide both sides by q1q2, then you're going to have to make the numerator on the right side 1, like this:
    4\pi\epsilon_{0} F_{el} =  \frac{q1q2}{r^2}
     \frac{4\pi\epsilon_{0}F_{el}}{q1q2} = \frac{1}{r^2}

    Although we've deviated from the work I showed above, we'll still get the same answer. If we now take the reciprocal of both sides we'll get
     \frac{q1q2}{4\pi\epsilon_{0}F_{el}} = \frac{r^2}{1}
    or
     \frac{q1q2}{4\pi\epsilon_{0}F_{el}} = r^2
    which does match what I have above.


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