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March 29, 2015

Posted by **Lindsey** on Monday, February 4, 2013 at 9:22pm.

- Physics/Charges -
**Devron**, Tuesday, February 5, 2013 at 7:20amI'm a little rusty on this, but I think you need the charge of the one that you are putting in the middle. Was that provided to you?

- Physics/Charges -
**Devron**, Tuesday, February 5, 2013 at 7:59amI apologize, you do not have to know it, it is not needed.

Let the distance between Q1 and Qo equal x(r1), and the distance between Q2 and Qo equal x-2m (r2).

Using the sum of forces,

F1onO=F2onO

Ko[(Q1Qo)]/r^2=Ko[(Q1Qo)/r^2]

Ko and Qo cancel each other out leaving

Q1/r^2=Q2/r^2

Plugging in my values for Q1 and r, and r2

(50 x10^-6 C/x^2)=( -20 x10^-6 C)/(2-x)^2

Since they are both in µ C, rewrite the equation as

(50/x^2)=( -20)/(2-x)^2)

50(2-x)^2=-20(x^2)

50(x^2-4x+4)=-20x^2

50x^2-200x+200=-20x^2

factoring out 10,

5x^2-20x+20=-2x^2

7x^20x+20=0

Use the quadratic equation to solve for x

I hope this helps, but I am not that positive.

- Physics/Charges -
**Devron**, Tuesday, February 5, 2013 at 8:38amLast set of values should read:

7x^2-20x+20=0

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