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

1. ## 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 $r=\pm\sqrt\frac{4\pi\epsilon_{0}F_{el}}{q1q2}$

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

I've spent a few hours rereading my chapters and getting a bit upset...what have I done wrong?

2. Originally Posted by SillyRabbit
$F_{el} = \frac{1}{4\pi\epsilon_{0}} \frac{q1q2}{r^2}$ 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. 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!

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