# 1251 HW 1 #11

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I was working on old homework problems in order to study for the exam, and I can't figure out how to solve number 11 on the first homework. The problem asks where the third charge should be placed in order to minimize the force on the third charge. I set up the problem normally, using the F  = kqq/r^2 equation and set that equal to zero. However, when I get to the point where I'm ready to use the quadratic formula, I am unable to proceed because my value in the square root is always negative. This is leading me to believe that I should not be solving for zero but I'm confused as to how to set up my equation to minimize the force without it?

Be sure to add the force due to both charges (the force from $q_1$ points to the right and the force from $q_2$ points to the left):

$$F = \frac{k q_1 q_3}{x^2} - \frac{k q_2 q_3}{(d-x)^2} = 0$$

Then we are able to set the force equal to zero and algebraically solve for the position:

$$q_1 (d-x)^2 = q_2 x^2$$

$$(q_1 - q_2) x^2 - (2d q_1) x + q_1 d^2 = 0$$

From here we must use the quadratic equation:

$$\frac{-b \pm \sqrt{ b^2 - 4 a c}}{ 2 a}$$

Which leaves us with

$$\frac{2d q_1 \pm \sqrt{(2d q_1)^2 - 4 (q_1 - q_2) ( q_1 d^2)}}{2 (q_1 - q_2)}$$

Here the quantity in the root is always positive since:

$$4 d^2 q_1^2 - 4 q_1^2 d^2 + 4 q_1 q_2 d^2 = 4 q_1 q_2 d^2 >0$$
answered Oct 7, 2014 by (2,140 points)
edited Oct 7, 2014 by jcf1559