Finding the Equilibrium Position for Three Charges

[Fireupchip]

Homework Statement

A charge of 2.21E-9 C is placed at the origin, and a charge of 3.78E-9 C is placed at x = 1.60 m. Find the position at which a third charge of 2.94E-9 C can be placed so that the net electrostatic force on it is zero.

Homework Equations

Coloumb's law

The Attempt at a Solution

All 3 charges are positive, so for the net force to be zero, I believe that the 3rd charge will go in the middle of the other 2 charges.

+q1 |----1.60m - r -----| +q3 |------ r------| +q2

Knowns:
q1 = 2.21 x 10^-9 C
q2 = 3.78 x 10^-9 C
q3 = 2.94 x 10^-9 C
r (between q1 and q2) = 1.60 m

F1 = F2

the k and q3 both cancel, leaving me with:

q1 / (1.60m - r)^2 = q2 / r^2

I cross multiplied, giving me:

q1 / q2 = (1.60m - r)^2 / r^2

This is where I think I'm having a problem. From this point, I found 2 different answers, and both turned out to be wrong: 0.91m and 1.05m (both from q1). Can anyone please help?

 

[HallsofIvy]

You didn't completely "cross multiply" (and I dislike that term). You did multiply on both sides by the denominator on the left, (160- r)^2. Now multiply on both sides by the denominator on the right to get (q1/q2)r^2= (160- r)^2. Multiply out the square on the right and you a quadratic equation to solve for r.

 

[Fireupchip]

So if I multiply the square out on the right, there'll still be an r on the left side.

 

[Fireupchip]

My physics professor said to avoid using the quadratic equation in this situation. Where am I going wrong?

 

[Fireupchip]

Anyone?