1. ## 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

2. Originally Posted by JBswe
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?

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

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

now expand the left hand side and continue

3. Another attempt

Rearrange

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

and extract the respectively square root.

4. ok so then i get $
(1 - d)^2 = 2d^2$

and when i extract it i find

$
1-d= 2d^2$

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

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

I think i did somthing wrong here. that would give me a neg root the way i tired it. where did i go wrong?

5. Originally Posted by JBswe
ok so then i get $
(1 - d)^2 = 2d^2$

and when i extract it i find

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

$(1 - d)^2 \ne 1 - d$

6. 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-->

7. Originally Posted by JBswe
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:
$F_q = \frac{\kappa q(Q)}{d^2} - \frac{\kappa q(2Q)}{(1 - d)^2}$

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

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

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

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

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

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

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

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

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

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

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

(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