Solve Coulomb's Law 1D: Find q3 for Zero Force

[Mitchtwitchita]

Homework Statement

Three charges lie along the x-axis. The positive charge q1 = 10.0 x 10^-6 C is at x = 1.00 m, and the negative charge q2 = -2.00 x 10^-6 C is at the origin. Where must a positive charge q3 be placed on the x-axis so that the resultant force on it is zero?

Homework Equations

F=kq1q2/r^2

The Attempt at a Solution

F23 = -F13

F23 = k(-2.00 x 10^-6 C)q3/-x^2
F13 = -k(10.0 x 10^-6 C)q3/(1.0 m - x)^2

[k(-2.00 x 10^-6 C)q3/-x^2] - [k(10.0 x 10^-6 C)q3/(1.0 - x)^2] = 0
(-2.00 x 10^-6 C)/-x^2 = (10.0 x 10^-6 C)/(1.0 -x)^2
2/x^2 = 10/(1- x)^2
2 - 4x + 2x^2 = 10x^2
-8x^2 - 4x +2 = 0

My quadratic equation doesn't work out so I'm assuming there was a terrible malfunction earlier on. Can somebody please help me out with this one?

 

[Doc Al]

Hint: First figure out what region the positive charge q3 must be in. ( x < 0; 0 < x < 1; x > 1) The field from each charge may have a different sign in different regions, which affects how you'd write the equation.

 

[Mitchtwitchita]

It's going to be left of the origin (-x). How would this affect my initial signs?

 

[Doc Al]

It's going to be left of the origin (-x). How would this affect my initial signs?

Actually your equations are fine. (I just saw the double negative.) Why do you think your quadratic doesn't work out?

 

[Mitchtwitchita]

Because I end up with 4 +/- sqr.[(-4_^2) - (4)(-8)(2)]/2 x 2
= (4 - 80)/4
= -19 which is much too high.

 

[Mitchtwitchita]

Er, I meant low in the last reply.

 

[Doc Al]

Because I end up with 4 +/- sqr.[(-4_^2) - (4)(-8)(2)]/2 x 2
= (4 - 80)/4
= -19 which is much too high.

I assume you are trying to use the quadratic formula. Try again; you're making an error somewhere.

Edit: You forgot to take the square root.

 

[Mitchtwitchita]

Is my error in the use of the quadratic formula or in the work leading up to it?

 

[Doc Al]

Is my error in the use of the quadratic formula or in the work leading up to it?

Your work is fine. You just forgot to take the square root. (See my last post.)

 

[Mitchtwitchita]

Oooooooooh...duh. Thanks!