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
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?
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